PHYSICS

PART – I

TEXTBOOK FOR CLASS XII

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12089 – PHYSICS PART I ISBN 81-7450-631-4
Textbook for Class XII

First Edition
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December 2006 Pausa 1928
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 FOREWORD

The National Curriculum Framework (NCF), 2005 recommends that children’s life at school must
be linked to their life outside the school. This principle marks a departure from the legacy of bookish
learning which continues to shape our system and causes a gap between the school, home and
community. The syllabi and textbooks developed on the basis of NCF signify an attempt to implement
this basic idea. They also attempt to discourage rote learning and the maintenance of sharp
boundaries between different subject areas. We hope these measures will take us significantly
further in the direction of a child-centred system of education outlined in the National Policy on
Education (NPE), 1986.
The success of this effort depends on the steps that school principals and teachers will take to
encourage children to reflect on their own learning and to pursue imaginative activities and questions.
We must recognise that, given space, time and freedom, children generate new knowledge by engaging
with the information passed on to them by adults. Treating the prescribed textbook as the sole basis
of examination is one of the key reasons why other resources and sites of learning are ignored.
Inculcating creativity and initiative is possible if we perceive and treat children as participants in
learning, not as receivers of a fixed body of knowledge.
These aims imply considerable change in school routines and mode of functioning. Flexibility in
the daily time-table is as necessary as rigour in implementing the annual calendar so that the
required number of teaching days are actually devoted to teaching. The methods used for teaching
and evaluation will also determine how effective this textbook proves for making children’s life at
school a happy experience, rather than a source of stress or boredom. Syllabus designers have tried
to address the problem of curricular burden by restructuring and reorienting knowledge at different
stages with greater consideration for child psychology and the time available for teaching. The textbook
attempts to enhance this endeavour by giving higher priority and space to opportunities for
contemplation and wondering, discussion in small groups, and activities requiring hands-on
experience.
The National Council of Educational Research and Training (NCERT) appreciates the hard
work done by the textbook development committee responsible for this book. We wish to thank the
Chairperson of the advisory group in science and mathematics, Professor J.V. Narlikar and the
Chief Advisor for this book, Professor A.W. Joshi for guiding the work of this committee. Several
teachers contributed to the development of this textbook; we are grateful to their principals for
making this possible. We are indebted to the institutions and organisations which have generously
permitted us to draw upon their resources, material and personnel. We are especially grateful to
the members of the National Monitoring Committee, appointed by the Department of Secondary
and Higher Education, Ministry of Human Resource Development under the Chairpersonship of
Professor Mrinal Miri and Professor G.P. Deshpande, for their valuable time and contribution. As
an organisation committed to systemic reform and continuous improvement in the quality of its
products, NCERT welcomes comments and suggestions which will enable us to undertake further
revision and refinement.

Director
New Delhi National Council of Educational
20 December 2006 Research and Training

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Rationalised 2023-24
 RATIONALISATION OF CONTENT IN THE TEXTBOOKS

In view of the COVID-19 pandemic, it is imperative to reduce content load on

students. The National Education Policy 2020, also emphasises reducing the
content load and providing opportunities for experiential learning with creative
mindset. In this background, the NCERT has undertaken the exercise to rationalise
the textbooks across all classes. Learning Outcomes already developed by the NCERT
across classes have been taken into consideration in this exercise.
Contents of the textbooks have been rationalised in view of the following:
• Overlapping with similar content included in other subject areas in the same
class
• Similar content included in the lower or higher class in the same subject
• Difficulty level
• Content, which is easily accessible to students without much interventions
from teachers and can be learned by children through self-learning or peer-
learning
• Content, which is irrelevant in the present context

This present edition, is a reformatted version after carrying out the changes
given above.

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Rationalised 2023-24
 TEXTBOOK DEVELOPMENT COMMITTEE

CHAIRPERSON, ADVISORY GROUP FOR TEXTBOOKS IN SCIENCE AND MATHEMATICS

J.V. Narlikar, Emeritus Professor, Inter-University Centre for Astronomy and Astrophysics
(IUCAA), Ganeshkhind, Pune University Campus, Pune

CHIEF ADVISOR
A.W. Joshi, Honorary Visiting Scientist, National Centre for Radio Astrophysics (NCRA), Pune
University Campus, Pune (Formerly Professor at Department of Physics, University of Pune)

MEMBERS
A.K. Ghatak, Emeritus Professor, Department of Physics, Indian Institute of Technology,
New Delhi
Alika Khare, Professor, Department of Physics, Indian Institute of Technology, Guwahati
Anjali Kshirsagar, Reader, Department of Physics, University of Pune, Pune
Anuradha Mathur, PGT , Modern School, Vasant Vihar, New Delhi
Atul Mody, Lecturer (S.G.), VES College of Arts, Science and Commerce, Mumbai
B.K. Sharma, Professor, DESM, NCERT, New Delhi
Chitra Goel, PGT, Rajkiya Pratibha Vikas Vidyalaya, Tyagraj Nagar, New Delhi
Gagan Gupta, Reader, DESM, NCERT, New Delhi
H.C. Pradhan, Professor, Homi Bhabha Centre of Science Education (TIFR), Mumbai
N. Panchapakesan, Professor (Retd.), Department of Physics and Astrophysics, University of
Delhi, Delhi
R. Joshi, Lecturer (S.G.), DESM, NCERT, New Delhi
S.K. Dash, Reader, DESM, NCERT, New Delhi
S. Rai Choudhary, Professor, Department of Physics and Astrophysics, University of Delhi, Delhi
S.K. Upadhyay, PGT, Jawahar Navodaya Vidyalaya, Muzaffar Nagar
S.N. Prabhakara, PGT, DM School, Regional Institute of Education (NCERT), Mysore
V.H. Raybagkar, Reader, Nowrosjee Wadia College, Pune
Vishwajeet Kulkarni, Teacher (Grade I ), Higher Secondary Section, Smt. Parvatibai Chowgule
College, Margao, Goa

MEMBER-COORDINATOR
V.P. Srivastava, Reader, DESM, NCERT, New Delhi

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 Constitution of India
Part IV A (Article 51 A)

Fundamental Duties
It shall be the duty of every citizen of India —
(a) to abide by the Constitution and respect its ideals and institutions, the
National Flag and the National Anthem;
(b) to cherish and follow the noble ideals which inspired our national struggle
for freedom;
(c) to uphold and protect the sovereignty, unity and integrity of India;
(d) to defend the country and render national service when called upon to
do so;
(e) to promote harmony and the spirit of common brotherhood amongst all
the people of India transcending religious, linguistic and regional or
sectional diversities; to renounce practices derogatory to the dignity of
women;
(f) to value and preserve the rich heritage of our composite culture;
(g) to protect and improve the natural environment including forests, lakes,
rivers, wildlife and to have compassion for living creatures;
(h) to develop the scientific temper, humanism and the spirit of inquiry and
reform;
(i) to safeguard public property and to abjure violence;
(j) to strive towards excellence in all spheres of individual and collective
activity so that the nation constantly rises to higher levels of endeavour
and achievement;
*(k) who is a parent or guardian, to provide opportunities for education to
his child or, as the case may be, ward between the age of six and
fourteen years.

Note: The Article 51A containing Fundamental Duties was inserted by the Constitution
(42nd Amendment) Act, 1976 (with effect from 3 January 1977).
*(k) was inserted by the Constitution (86th Amendment) Act, 2002 (with effect from
1 April 2010).

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 ACKNOWLEDGEMENTS

The National Council of Educational Research and Training acknowledges the valuable
contribution of the individuals and organisations involved in the development of Physics Textbook
for Class XII. The Council also acknowledges the valuable contribution of the following academics
for reviewing and refining the manuscripts of this book:
Anu Venugopalan, Lecturer, School of Basic and Applied Sciences, GGSIP University, Delhi;
A.K. Das, PGT, St. Xavier’s Senior Secondary School, Delhi; Bharati Kukkal, PGT, Kendriya
Vidyalaya, Pushp Vihar, New Delhi; D.A. Desai, Lecturer (Retd.), Ruparel College, Mumbai;
Devendra Kumar, PGT, Rajkiya Pratibha Vikas Vidyalaya, Yamuna Vihar, Delhi; I.K. Gogia, PGT,
Kendriya Vidyalaya, Gole Market, New Delhi; K.C. Sharma, Reader, Regional Institute of Education
(NCERT), Ajmer; M.K. Nandy, Associate Professor, Department of Physics, Indian Institute of
Technology, Guwahati; M.N. Bapat, Reader, Regional Institute of Education (NCERT), Mysuru;
R. Bhattacharjee, Assistant Professor, Department of Electronics and Communication
Engineering, Indian Institute of Technology, Guwahati; R.S. Das, Vice-Principal (Retd.), Balwant
Ray Mehta Senior Secondary School, Lajpat Nagar, New Delhi; Sangeeta D. Gadre, Reader, Kirori
Mal College, Delhi; Suresh Kumar, PGT, Delhi Public School, Dwarka, New Delhi; Sushma Jaireth,
Reader, Department of Women’s Studies, NCERT, New Delhi; Shyama Rath, Reader, Department
of Physics and Astrophysics, University of Delhi, Delhi; Yashu Kumar, PGT, Kulachi Hans Raj
Model School, Ashok Vihar, Delhi.
The Council also gratefully acknowledges the valuable contribution of the following academics
for the editing and finalisation of this book: B.B. Tripathi, Professor (Retd.), Department of Physics,
Indian Institute of Technology, New Delhi; Dipan K. Ghosh, Professor, Department of Physics,
Indian Institute of Technology, Mumbai; Dipanjan Mitra, Scientist, National Centre for Radio
Astrophysics (TIFR), Pune; G.K. Mehta, Raja Ramanna Fellow, Inter-University Accelerator
Centre, New Delhi; G.S. Visweswaran, Professor, Department of Electrical Engineering, Indian
Institute of Technology, New Delhi; H.C. Kandpal, Head, Optical Radiation Standards, National
Physical Laboratory, New Delhi; H.S. Mani, Raja Ramanna Fellow, Institute of Mathematical
Sciences, Chennai; K. Thyagarajan, Professor, Department of Physics, Indian Institute of
Technology, New Delhi; P.C. Vinod Kumar, Professor, Department of Physics, Sardar Patel
University, Vallabh Vidyanagar, Gujarat; S. Annapoorni, Professor, Department of Physics and
Astrophysics, University of Delhi, Delhi; S.C. Dutta Roy, Emeritus Professor, Department of
Electrical Engineering, Indian Institute of Technology, New Delhi; S.D. Joglekar, Professor,
Department of Physics, Indian Institute of Technology, Kanpur; and V. Sundara Raja, Professor,
Sri Venkateswara University, Tirupati.
The Council also acknowledges the valuable contributions of the following academics for
refining the text in 2017: A.K. Srivastava, Assistant Professor, DESM, NCERT, New Delhi; Arnab
Sen, Assistant Professor, NERIE, Shillong; L.S. Chauhan, Assistant Professor, RIE, Bhopal;
O.N. Awasthi, Professor (Retd.), RIE, Bhopal; Rachna Garg, Professor, DESM, NCERT, New
Delhi; Raman Namboodiri, Assistant Professor, RIE, Mysuru; R.R. Koireng, Assistant Professor,
DCS, NCERT, New Delhi; Shashi Prabha, Professor, DESM, NCERT, New Delhi; and S.V. Sharma,
Professor, RIE, Ajmer.
Special thanks are due to Hukum Singh, Professor and Head, DESM, NCERT for his support.
The Council also acknowledges the support provided by the APC office and the administrative
staff of the DESM; Deepak Kapoor, Incharge, Computer Station; Inder Kumar, DTP Operator;
Mohd. Qamar Tabrez, Copy Editor; Ashima Srivastava, Proof Reader in shaping this book.
The contributions of the Publication Department in bringing out this book are also duly
acknowledged.

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 CONSTITUTION OF INDIA
Part III (Articles 12 – 35)
(Subject to certain conditions, some exceptions
and reasonable restrictions)
guarantees these
Fundamental Rights
Right to Equality
· before law and equal protection of laws;
· irrespective of religion, race, caste, sex or place of birth;
· of opportunity in public employment;
· by abolition of untouchability and titles.
Right to Freedom
· of expression, assembly, association, movement, residence and profession;
· of certain protections in respect of conviction for offences;
· of protection of life and personal liberty;
· of free and compulsory education for children between the age of six and fourteen years;
· of protection against arrest and detention in certain cases.
Right against Exploitation
· for prohibition of traffic in human beings and forced labour;
· for prohibition of employment of children in hazardous jobs.
Right to Freedom of Religion
· freedom of conscience and free profession, practice and propagation of religion;
· freedom to manage religious affairs;
· freedom as to payment of taxes for promotion of any particular religion;
· freedom as to attendance at religious instruction or religious worship in educational
institutions wholly maintained by the State.
Cultural and Educational Rights
· for protection of interests of minorities to conserve their language, script and culture;
· for minorities to establish and administer educational institutions of their choice.
Right to Constitutional Remedies
· by issuance of directions or orders or writs by the Supreme Court and High
Courts for enforcement of these Fundamental Rights.

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 PREFACE

It gives me pleasure to place this book in the hands of the students, teachers and the public at
large (whose role cannot be overlooked). It is a natural sequel to the Class XI textbook which
was brought out in 2006. This book is also a trimmed version of the textbooks which existed so
far. The chapter on thermal and chemical effects of current has been cut out. This topic has also
been dropped from the CBSE syllabus. Similarly, the chapter on communications has been
substantially curtailed. It has been rewritten in an easily comprehensible form.
Although most other chapters have been based on the earlier versions, several parts and
sections in them have been rewritten. The Development Team has been guided by the feedback
received from innumerable teachers across the country.
In producing these books, Class XI as well as Class XII, there has been a basic change of
emphasis. Both the books present physics to students without assuming that they would pursue
this subject beyond the higher secondary level. This new view has been prompted by the various
observations and suggestions made in the National Curriculum Framework (NCF), 2005.
Similarly, in today’s educational scenario where students can opt for various combinations of
subjects, we cannot assume that a physics student is also studying mathematics. Therefore,
physics has to be presented, so to say, in a standalone form.
As in Class XI textbook, some interesting box items have been inserted in many chapters.
They are not meant for teaching or examinations. Their purpose is to catch the attention of the
reader, to show some applications in daily life or in other areas of science and technology, to
suggest a simple experiment, to show connection of concepts in different areas of physics, and
in general, to break the monotony and enliven the book.
Features like Summary, Points to Ponder, Exercises and Additional Exercises at the end of
each chapter, and Examples have been retained. Several concept-based Exercises have been
transferred from end-of-chapter Exercises to Examples with Solutions in the text. It is hoped
that this will make the concepts discussed in the chapter more comprehensible. Several new
examples and exercises have been added. Students wishing to pursue physics further would
find Points to Ponder and Additional Exercises very useful and thoughtful. To provide resources
beyond the textbook and to encourage eLearning, each chapter has been provided with
some relevant website addresses under the title ePhysics. These sites provide additional
material on specific topics and also provide learners with opportunites for interactive
demonstrations/experiments.
The intricate concepts of physics must be understood, comprehended and appreciated.
Students must learn to ask questions like ‘why’, ‘how’, ‘how do we know it’. They will find
almost always that the question ‘why’ has no answer within the domain of physics and science
in general. But that itself is a learning experience, is it not? On the other hand, the question
‘how’ has been reasonably well answered by physicists in the case of most natural phenomena.
In fact, with the understanding of how things happen, it has been possible to make use of many
phenomena to create technological applications for the use of humans.
For example, consider statements in a book, like ‘A negatively charged electron is attracted
by the positively charged plate’, or ‘In this experiment, light (or electron) behaves like a wave’.
You will realise that it is not possible to answer ‘why’. This question belongs to the domain of
philosophy or metaphysics. But we can answer ‘how’, we can find the force acting, we can find

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the wavelength of the photon (or electron), we can determine how things behave under different
conditions, and we can develop instruments which will use these phenomena to our advantage.
It has been a pleasure to work for these books at the higher secondary level, along with a
team of members. The Textbook Development Team, Review Team and Editing Teams involved
college and university teachers, teachers from Indian Institutes of Technology, scientists from
national institutes and laboratories, as well as, higher secondary teachers. The feedback and
critical look provided by higher secondary teachers in the various teams are highly laudable.
Most box items were generated by members of one or the other team, but three of them were
generated by friends and well-wishers not part of any team. We are thankful to Dr P.N. Sen of
Pune, Professor Roopmanjari Ghosh of Delhi and Dr Rajesh B Khaparde of Mumbai for allowing
us to use their box items, respectively, in Chapters 3, 4 (Part I) and 9 (Part II). We are thankful
to the members of the review and editing workshops to discuss and refine the first draft of the
textbook. We also express our gratitude to Prof. Krishna Kumar, Director, NCERT, for entrusting
us with the task of presenting this textbook as a part of the national effort for improving science
education. I also thank Prof. G. Ravindra, Joint Director, NCERT, for his help from time-to-
time. Prof. Hukum Singh, Head, Department of Education in Science and Mathematics, NCERT,
was always willing to help us in our endeavour in every possible way.
We welcome suggestions and comments from our valued users, especially students and
teachers. We wish our young readers a happy journey into the exciting realm of physics.

A. W. JOSHI
Chief Advisor
Textbook Development Committee

xii

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 CONTENTS

FOREWORD iii
RATIONALISATION OF CONTENT IN THE TEXTBOOKS v

CHAPTER ONE
ELECTRIC CHARGES AND FIELDS
1.1 Introduction 1
1.2 Electric Charge 1
1.3 Conductors and Insulators 3
1.4 Basic Properties of Electric Charge 4
1.5 Coulomb’s Law 6
1.6 Forces between Multiple Charges 11
1.7 Electric Field 14
1.8 Electric Field Lines 19
1.9 Electric Flux 21
1.10 Electric Dipole 23
1.11 Dipole in a Uniform External Field 27
1.12 Continuous Charge Distribution 28
1.13 Gauss’s Law 29
1.14 Applications of Gauss’s Law 33

CHAPTER TWO
ELECTROSTATIC POTENTIAL AND CAPACITANCE
2.1 Introduction 45
2.2 Electrostatic Potential 47
2.3 Potential due to a Point Charge 48
2.4 Potential due to an Electric Dipole 49
2.5 Potential due to a System of Charges 51
2.6 Equipotential Surfaces 54
2.7 Potential Energy of a System of Charges 55
2.8 Potential Energy in an External Field 58
2.9 Electrostatics of Conductors 61
2.10 Dielectrics and Polarisation 65
2.11 Capacitors and Capacitance 67
2.12 The Parallel Plate Capacitor 68
2.13 Effect of Dielectric on Capacitance 69
2.14 Combination of Capacitors 71
2.15 Energy Stored in a Capacitor 73

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CHAPTER THREE
CURRENT ELECTRICITY
3.1 Introduction 81
3.2 Electric Current 81
3.3 Electric Currents in Conductors 82
3.4 Ohm’s law 83
3.5 Drift of Electrons and the Origin of Resistivity 85
3.6 Limitations of Ohm’s Law 89
3.7 Resistivity of Various Materials 89
3.8 Temperature Dependence of Resistivity 90
3.9 Electrical Energy, Power 92
3.10 Cells, emf, Internal Resistance 93
3.11 Cells in Series and in Parallel 95
3.12 Kirchhoff’s Rules 97
3.13 Wheatstone Bridge 100

CHAPTER FOUR
MOVING CHARGES AND MAGNETISM
4.1 Introduction 107
4.2 Magnetic Force 108
4.3 Motion in a Magnetic Field 112
4.4 Magnetic Field due to a Current Element, Biot-Savart Law 113
4.5 Magnetic Field on the Axis of a Circular Current Loop 115
4.6 Ampere’s Circuital Law 117
4.7 The Solenoid 121
4.8 Force between Two Parallel Currents, the Ampere 122
4.9 Torque on Current Loop, Magnetic Dipole 124
4.10 The Moving Coil Galvanometer 129

CHAPTER FIVE
MAGNETISM AND MATTER
5.1 Introduction 136
5.2 The Bar Magnet 137
5.3 Magnetism and Gauss’s Law 142
5.4 Magnetisation and Magnetic Intensity 145
5.5 Magnetic Properties of Materials 147

xiv

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CHAPTER SIX
ELECTROMAGNETIC INDUCTION
6.1 Introduction 154
6.2 The Experiments of Faraday and Henry 155
6.3 Magnetic Flux 156
6.4 Faraday’s Law of Induction 157
6.5 Lenz’s Law and Conservation of Energy 160
6.6 Motional Electromotive Force 162
6.7 Inductance 165
6.8 AC Generator 170

CHAPTER SEVEN
ALTERNATING CURRENT
7.1 Introduction 177
7.2 AC Voltage Applied to a Resistor 178
7.3 Representation of AC Current and Voltage by
Rotating Vectors — Phasors 181
7.4 AC Voltage Applied to an Inductor 181
7.5 AC Voltage Applied to a Capacitor 184
7.6 AC Voltage Applied to a Series LCR Circuit 186
7.7 Power in AC Circuit: The Power Factor 190
7.8 Transformers 194

CHAPTER EIGHT
ELECTROMAGNETIC WAVES
8.1 Introduction 201
8.2 Displacement Current 202
8.3 Electromagnetic Waves 205
8.4 Electromagnetic Spectrum 208

ANSWERS 215

xv

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 COVER DESIGN
(Adapted from http://nobelprize.org and
the Nobel Prize in Physics 2006)

Different stages in the evolution of

the universe.

BACK COVER
(Adapted from http://www.iter.org and
http://www.dae.gov.in)

Cut away view of International Thermonuclear Experimental Reactor (ITER)

device. The man in the bottom shows the scale.
ITER is a joint international research and development project that
aims to demonstrate the scientific and technical feasibility of fusion power.
India is one of the seven full partners in the project, the others being
the European Union (represented by EURATOM), Japan, the People’s
Republic of China, the Republic of Korea, the Russian Federation and the
USA. ITER will be constructed in Europe, at Cadarache in the South of
France and will provide 500 MW of fusion power.
Fusion is the energy source of the sun and the stars. On earth, fusion
research is aimed at demonstrating that this energy source can be used to
produce electricity in a safe and environmentally benign way, with
abundant fuel resources, to meet the needs of a growing world population.
For details of India’s role, see Nuclear India, Vol. 39, Nov. 11-12/
May-June 2006, issue available at Department of Atomic Energy (DAE)
website mentioned above.

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Chapter One

ELECTRIC CHARGES
AND FIELDS

1.1 INTRODUCTION
All of us have the experience of seeing a spark or hearing a crackle when
we take off our synthetic clothes or sweater, particularly in dry weather.
Have you ever tried to find any explanation for this phenomenon? Another
common example of electric discharge is the lightning that we see in the
sky during thunderstorms. We also experience a sensation of an electric
shock either while opening the door of a car or holding the iron bar of a
bus after sliding from our seat. The reason for these experiences is
discharge of electric charges through our body, which were accumulated
due to rubbing of insulating surfaces. You might have also heard that
this is due to generation of static electricity. This is precisely the topic we
are going to discuss in this and the next chapter. Static means anything
that does not move or change with time. Electrostatics deals with
the study of forces, fields and potentials arising from
static charges.

1.2 ELECTRIC CHARGE

Historically the credit of discovery of the fact that amber rubbed with
wool or silk cloth attracts light objects goes to Thales of Miletus, Greece,
around 600 BC. The name electricity is coined from the Greek word

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 Physics

FIGURE 1.1 Rods: like charges repel and unlike charges attract each other.

elektron meaning amber. Many such pairs of materials were known which
on rubbing could attract light objects like straw, pith balls and bits of
papers.
It was observed that if two glass rods rubbed with wool or silk cloth
are brought close to each other, they repel each other [Fig. 1.1(a)]. The
two strands of wool or two pieces of silk cloth, with which the rods were
rubbed, also repel each other. However, the glass rod and wool attracted
each other. Similarly, two plastic rods rubbed with cat’s fur repelled each
other [Fig. 1.1(b)] but attracted the fur. On the other hand, the plastic
rod attracts the glass rod [Fig. 1.1(c)] and repel the silk or wool with
which the glass rod is rubbed. The glass rod repels the fur.
These seemingly simple facts were established from years of efforts
and careful experiments and their analyses. It was concluded, after many
careful studies by different scientists, that there were only two kinds of
an entry which is called the electric charge. We say that the bodies like
glass or plastic rods, silk, fur and pith balls are electrified. They acquire
an electric charge on rubbing. There are two kinds of electrification and
we find that (i) like charges repel and (ii) unlike charges attract each
other. The property which differentiates the two kinds of charges is called
the polarity of charge.
When a glass rod is rubbed with silk, the rod acquires one kind of
charge and the silk acquires the second kind of charge. This is true for
any pair of objects that are rubbed to be electrified. Now if the electrified
glass rod is brought in contact with silk, with which it was rubbed, they
no longer attract each other. They also do not attract or repel other light
objects as they did on being electrified.
Thus, the charges acquired after rubbing are lost when the charged
bodies are brought in contact. What can you conclude from these
observations? It just tells us that unlike charges acquired by the objects
neutralise or nullify each other’s effect. Therefore, the charges were named
as positive and negative by the American scientist Benjamin Franklin.
By convention, the charge on glass rod or cat’s fur is called positive and
that on plastic rod or silk is termed negative. If an object possesses an
electric charge, it is said to be electrified or charged. When it has no charge
it is said to be electrically neutral.
2

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 Electric Charges
and Fields
A simple apparatus to detect charge on a body is the gold-leaf
electroscope [Fig. 1.2(a)]. It consists of a vertical metal rod housed in a
box, with two thin gold leaves attached to its bottom end. When a charged
object touches the metal knob at the top of the rod, charge flows on to
the leaves and they diverge. The degree of divergance is an indicator of
the amount of charge.
Try to understand why material bodies acquire charge. You know that
all matter is made up of atoms and/or molecules. Although normally the
materials are electrically neutral, they do contain charges; but their charges
are exactly balanced. Forces that hold the molecules together, forces that
hold atoms together in a solid, the adhesive force of glue, forces associated
with surface tension, all are basically electrical in nature, arising from the
forces between charged particles. Thus the electric force is all pervasive and
it encompasses almost each and every field associated with our life. It is
therefore essential that we learn more about such a force.
To electrify a neutral body, we need to add or remove one kind of
charge. When we say that a body is charged, we always refer to this
excess charge or deficit of charge. In solids, some of the electrons, being
less tightly bound in the atom, are the charges which are transferred
from one body to the other. A body can thus be charged positively by
losing some of its electrons. Similarly, a body can be charged negatively
by gaining electrons. When we rub a glass rod with silk, some of the
electrons from the rod are transferred to the silk cloth. Thus the rod gets
positively charged and the silk gets negatively charged. No new charge is
created in the process of rubbing. Also the number of electrons, that are
transferred, is a very small fraction of the total number of electrons in the
material body.

1.3 CONDUCTORS AND INSULATORS

Some substances readily allow passage of electricity through them, others
do not. Those which allow electricity to pass through them easily are
called conductors. They have electric charges (electrons) that are
comparatively free to move inside the material. Metals, human and animal
bodies and earth are conductors. Most of the non-metals like glass,
porcelain, plastic, nylon, wood offer high resistance to the passage of
electricity through them. They are called insulators. Most substances
fall into one of the two classes stated above*.
When some charge is transferred to a conductor, it readily gets
distributed over the entire surface of the conductor. In contrast, if some
charge is put on an insulator, it stays at the same place. You will learn
why this happens in the next chapter.
This property of the materials tells you why a nylon or plastic comb
gets electrified on combing dry hair or on rubbing, but a metal article

* There is a third category called semiconductors, which offer resistance to the

movement of charges which is intermediate between the conductors and
insulators.
3

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 Physics
like spoon does not. The charges on metal leak through
our body to the ground as both are conductors of
electricity. However, if a metal rod with a wooden or plastic
handle is rubbed without touching its metal part, it shows
signs of charging.

1.4 BASIC PROPERTIES OF ELECTRIC

CHARGE
We have seen that there are two types of charges, namely
positive and negative and their effects tend to cancel each
other. Here, we shall now describe some other properties
of the electric charge.
If the sizes of charged bodies are very small as
compared to the distances between them, we treat them
as point charges. All the charge content of the body is
assumed to be concentrated at one point in space.

1.4.1 Additivity of charges

FIGURE 1.2 Electroscopes: (a) We have not as yet given a quantitative definition of a
The gold leaf electroscope, (b) charge; we shall follow it up in the next section. We shall
Schematics of a simple tentatively assume that this can be done and proceed. If
electroscope.
a system contains two point charges q1 and q2, the total
charge of the system is obtained simply by adding
algebraically q 1 and q2 , i.e., charges add up like real numbers or they
are scalars like the mass of a body. If a system contains n charges q1,
q2, q3, …, qn, then the total charge of the system is q1 + q2 + q3 + … + qn
. Charge has magnitude but no direction, similar to mass. However,
there is one difference between mass and charge. Mass of a body is
always positive whereas a charge can be either positive or negative.
Proper signs have to be used while adding the charges in a system. For
example, the total charge of a system containing five charges +1, +2, –3,
+4 and –5, in some arbitrary unit, is (+1) + (+2) + (–3) + (+4) + (–5) = –1 in
the same unit.

1.4.2 Charge is conserved

We have already hinted to the fact that when bodies are charged by
rubbing, there is transfer of electrons from one body to the other; no new
charges are either created or destroyed. A picture of particles of electric
charge enables us to understand the idea of conservation of charge. When
we rub two bodies, what one body gains in charge the other body loses.
Within an isolated system consisting of many charged bodies, due to
interactions among the bodies, charges may get redistributed but it is
found that the total charge of the isolated system is always conserved.
Conservation of charge has been established experimentally.
It is not possible to create or destroy net charge carried by any isolated
system although the charge carrying particles may be created or destroyed
4

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 Electric Charges
and Fields
in a process. Sometimes nature creates charged particles: a neutron turns
into a proton and an electron. The proton and electron thus created have
equal and opposite charges and the total charge is zero before and after
the creation.

1.4.3 Quantisation of charge

Experimentally it is established that all free charges are integral multiples
of a basic unit of charge denoted by e. Thus charge q on a body is always
given by
q = ne
where n is any integer, positive or negative. This basic unit of charge is
the charge that an electron or proton carries. By convention, the charge
on an electron is taken to be negative; therefore charge on an electron is
written as –e and that on a proton as +e.
The fact that electric charge is always an integral multiple of e is termed
as quantisation of charge. There are a large number of situations in physics
where certain physical quantities are quantised. The quantisation of charge
was first suggested by the experimental laws of electrolysis discovered by
English experimentalist Faraday. It was experimentally demonstrated by
Millikan in 1912.
In the International System (SI) of Units, a unit of charge is called a
coulomb and is denoted by the symbol C. A coulomb is defined in terms
the unit of the electric current which you are going to learn in a
subsequent chapter. In terms of this definition, one coulomb is the charge
flowing through a wire in 1 s if the current is 1 A (ampere), (see Chapter 1
of Class XI, Physics Textbook , Part I). In this system, the value of the
basic unit of charge is
e = 1.602192 × 10–19 C
Thus, there are about 6 × 1018 electrons in a charge of –1C. In
electrostatics, charges of this large magnitude are seldom encountered
and hence we use smaller units 1 mC (micro coulomb) = 10–6 C or 1 mC
(milli coulomb) = 10–3 C.
If the protons and electrons are the only basic charges in the
universe, all the observable charges have to be integral multiples of e.
Thus, if a body contains n1 electrons and n2 protons, the total amount
of charge on the body is n2 × e + n1 × (–e) = (n2 – n1) e. Since n1 and n2
are integers, their difference is also an integer. Thus the charge on any
body is always an integral multiple of e and can be increased or
decreased also in steps of e.
The step size e is, however, very small because at the macroscopic
level, we deal with charges of a few mC. At this scale the fact that charge of
a body can increase or decrease in units of e is not visible. In this respect,
the grainy nature of the charge is lost and it appears to be continuous.
This situation can be compared with the geometrical concepts of points
and lines. A dotted line viewed from a distance appears continuous to
us but is not continuous in reality. As many points very close to
5

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 Physics
each other normally give an impression of a continuous line, many
small charges taken together appear as a continuous charge distribution.
At the macroscopic level, one deals with charges that are enormous
compared to the magnitude of charge e. Since e = 1.6 × 10–19 C, a charge
of magnituOde, say 1 mC, contains something like 1013 times the electronic
charge. At this scale, the fact that charge can increase or decrease only in
units of e is not very different from saying that charge can take continuous
values. Thus, at the macroscopic level, the quantisation of charge has no
practical consequence and can be ignored. However, at the microscopic
level, where the charges involved are of the order of a few tens or hundreds
of e, i.e., they can be counted, they appear in discrete lumps and
quantisation of charge cannot be ignored. It is the magnitude of scale
involved that is very important.

Example 1.1 If 109 electrons move out of a body to another body

every second, how much time is required to get a total charge of 1 C
on the other body?
Solution In one second 109 electrons move out of the body. Therefore
the charge given out in one second is 1.6 × 10–19 × 109 C = 1.6 × 10–10 C.
The time required to accumulate a charge of 1 C can then be estimated
to be 1 C ÷ (1.6 × 10–10 C/s) = 6.25 × 109 s = 6.25 × 109 ÷ (365 × 24 ×
3600) years = 198 years. Thus to collect a charge of one coulomb,
from a body from which 109 electrons move out every second, we will
need approximately 200 years. One coulomb is, therefore, a very large
EXAMPLE 1.1

unit for many practical purposes.

It is, however, also important to know what is roughly the number of
electrons contained in a piece of one cubic centimetre of a material.
A cubic piece of copper of side 1 cm contains about 2.5 × 10 24
electrons.

Example 1.2 How much positive and negative charge is there in a

cup of water?
Solution Let us assume that the mass of one cup of water is
250 g. The molecular mass of water is 18g. Thus, one mole
(= 6.02 × 1023 molecules) of water is 18 g. Therefore the number of
EXAMPLE 1.2

molecules in one cup of water is (250/18) × 6.02 × 1023.

Each molecule of water contains two hydrogen atoms and one oxygen
atom, i.e., 10 electrons and 10 protons. Hence the total positive and
total negative charge has the same magnitude. It is equal to
(250/18) × 6.02 × 1023 × 10 × 1.6 × 10–19 C = 1.34 × 107 C.

1.5 COULOMB’S LAW

Coulomb’s law is a quantitative statement about the force between two
point charges. When the linear size of charged bodies are much smaller
than the distance separating them, the size may be ignored and the
charged bodies are treated as point charges. Coulomb measured the
force between two point charges and found that it varied inversely as
the square of the distance between the charges and was directly
6 proportional to the product of the magnitude of the two charges and

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 Electric Charges
and Fields

acted along the line joining the two charges. Thus, if two
point charges q1, q2 are separated by a distance r in vacuum,
the magnitude of the force (F) between them is given by
q1 q 2
F =k (1.1)
r2
How did Coulomb arrive at this law from his experiments?
Coulomb used a torsion balance* for measuring the force
between two charged metallic spheres. When the separation
between two spheres is much larger than the radius of each
sphere, the charged spheres may be regarded as point charges.
However, the charges on the spheres were unknown, to begin
with. How then could he discover a relation like Eq. (1.1)?
Coulomb thought of the following simple way: Suppose the Charles Augustin de
Coulomb (1736 – 1806)
charge on a metallic sphere is q. If the sphere is put in contact
Coulomb, a French
with an identical uncharged sphere, the charge will spread over
physicist, began his
the two spheres. By symmetry, the charge on each sphere will

CHARLES AUGUSTIN DE COULOMB (1736 –1806)

career as a military
be q/2*. Repeating this process, we can get charges q/2, q/4, engineer in the West
etc. Coulomb varied the distance for a fixed pair of charges and Indies. In 1776, he
measured the force for different separations. He then varied the returned to Paris and
charges in pairs, keeping the distance fixed for each pair. retired to a small estate
Comparing forces for different pairs of charges at different to do his scientific
distances, Coulomb arrived at the relation, Eq. (1.1). research. He invented a
Coulomb’s law, a simple mathematical statement, was torsion balance to
initially experimentally arrived at in the manner described measure the quantity of
above. While the original experiments established it at a a force and used it for
determination of forces
macroscopic scale, it has also been established down to
of electric attraction or
subatomic level (r ~ 10–10 m). repulsion between small
Coulomb discovered his law without knowing the explicit charged spheres. He
magnitude of the charge. In fact, it is the other way round: thus arrived in 1785 at
Coulomb’s law can now be employed to furnish a definition the inverse square law
for a unit of charge. In the relation, Eq. (1.1), k is so far relation, now known as
arbitrary. We can choose any positive value of k. The choice Coulomb’s law. The law
of k determines the size of the unit of charge. In SI units, the had been anticipated by
Nm 2 Priestley and also by
value of k is about 9 × 109 . The unit of charge that Cavendish earlier,
C2 though Cavendish
results from this choice is called a coulomb which we defined
never published his
earlier in Section 1.4. Putting this value of k in Eq. (1.1), we results. Coulomb also
see that for q1 = q2 = 1 C, r = 1 m found the inverse
F = 9 × 109 N square law of force
That is, 1 C is the charge that when placed at a distance between unlike and like
of 1 m from another charge of the same magnitude in vacuum magnetic poles.
experiences an electrical force of repulsion of magnitude

* A torsion balance is a sensitive device to measure force. It was also used later
by Cavendish to measure the very feeble gravitational force between two objects,
to verify Newton’s Law of Gravitation.
* Implicit in this is the assumption of additivity of charges and conservation:
two charges (q/2 each) add up to make a total charge q. 7

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 Physics
9 × 109 N. One coulomb is evidently too big a unit to
be used. In practice, in electrostatics, one uses
smaller units like 1 mC or 1 mC.
The constant k in Eq. (1.1) is usually put as
k = 1/4pe0 for later convenience, so that Coulomb’s
law is written as

1 q1 q2
F = (1.2)
4 π ε0 r2

e0 is called the permittivity of free space . The value

of e0 in SI units is
 0 = 8.854 × 10–12 C2 N–1m–2
Since force is a vector, it is better to write
Coulomb’s law in the vector notation. Let the position
vectors of charges q1 and q2 be r1 and r2 respectively
[see Fig.1.3(a)]. We denote force on q1 due to q2 by
FIGURE 1.3 (a) Geometry and F12 and force on q2 due to q1 by F21. The two point
(b) Forces between charges. charges q1 and q2 have been numbered 1 and 2 for
convenience and the vector leading from 1 to 2 is
denoted by r21:
r21 = r2 – r1
In the same way, the vector leading from 2 to 1 is denoted by r12:
r12 = r1 – r2 = – r21
The magnitude of the vectors r21 and r12 is denoted by r21 and r12 ,
respectively (r12 = r21). The direction of a vector is specified by a unit
vector along the vector. To denote the direction from 1 to 2 (or from 2 to
1), we define the unit vectors:
r r
rɵ 21 = 21 , rɵ 12 = 12 , rɵ 21 − rɵ 12
r21 r12
Coulomb’s force law between two point charges q1 and q2 located at
r1 and r2, respectively is then expressed as

1 q1 q 2 ɵ
F21 = r 21 (1.3)
4 π εo 2
r21
Some remarks on Eq. (1.3) are relevant:
· Equation (1.3) is valid for any sign of q1 and q2 whether positive or
negative. If q1 and q2 are of the same sign (either both positive or both
negative), F21 is along r̂ 21, which denotes repulsion, as it should be for
like charges. If q1 and q2 are of opposite signs, F21 is along – rɵ 21(= rɵ 12),
which denotes attraction, as expected for unlike charges. Thus, we do
not have to write separate equations for the cases of like and unlike
charges. Equation (1.3) takes care of both cases correctly [Fig. 1.3(b)].
8

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 Electric Charges
and Fields
· The force F12 on charge q1 due to charge q2, is obtained from Eq. (1.3),
by simply interchanging 1 and 2, i.e.,
1 q1 q 2
F12 = rˆ12 = −F21
4 π ε0 2
r12

Thus, Coulomb’s law agrees with the Newton’s third law.

· Coulomb’s law [Eq. (1.3)] gives the force between two charges q1 and
q2 in vacuum. If the charges are placed in matter or the intervening
space has matter, the situation gets complicated due to the presence

http://webphysics.davidson.edu/physlet_resources/bu_semester2/menu_semester2.html
Interactive animation on Coulomb’s law:
of charged constituents of matter. We shall consider electrostatics in
matter in the next chapter.

Example 1.3 Coulomb’s law for electrostatic force between two point
charges and Newton’s law for gravitational force between two stationary
point masses, both have inverse-square dependence on the distance
between the charges and masses respectively. (a) Compare the strength
of these forces by determining the ratio of their magnitudes (i) for an
electron and a proton and (ii) for two protons. (b) Estimate the
accelerations of electron and proton due to the electrical force of their
mutual attraction when they are 1 Å (= 10-10 m) apart? (mp = 1.67 ×
10–27 kg, me = 9.11 × 10–31 kg)
Solution
(a) (i) The electric force between an electron and a proton at a distance
r apart is:
1 e2
Fe = −
4 πε 0 r 2
where the negative sign indicates that the force is attractive. The
corresponding gravitational force (always attractive) is:
m p me
FG = −G
r2
where mp and me are the masses of a proton and an electron
respectively.
Fe e2
= = 2.4 × 1039
FG 4 πε 0Gm pm e
(ii) On similar lines, the ratio of the magnitudes of electric force
to the gravitational force between two protons at a distance r
apart is:
Fe e2
= = 1.3 × 1036
FG 4πε 0Gm p m p
However, it may be mentioned here that the signs of the two forces
are different. For two protons, the gravitational force is attractive
in nature and the Coulomb force is repulsive. The actual values
EXAMPLE 1.3

of these forces between two protons inside a nucleus (distance

between two protons is ~ 10-15 m inside a nucleus) are Fe ~ 230 N,
whereas, FG ~ 1.9 × 10–34 N.
The (dimensionless) ratio of the two forces shows that electrical
forces are enormously stronger than the gravitational forces.

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 Physics

(b) The electric force F exerted by a proton on an electron is same in

magnitude to the force exerted by an electron on a proton; however,
the masses of an electron and a proton are different. Thus, the
magnitude of force is
1 e2
|F| = = 8.987 × 109 Nm2/C2 × (1.6 ×10–19C)2 / (10–10m)2
4 πε 0 r 2
= 2.3 × 10–8 N
Using Newton’s second law of motion, F = ma, the acceleration
that an electron will undergo is
a = 2.3×10–8 N / 9.11 ×10–31 kg = 2.5 × 1022 m/s2
Comparing this with the value of acceleration due to gravity, we
can conclude that the effect of gravitational field is negligible on
EXAMPLE 1.3

the motion of electron and it undergoes very large accelerations

under the action of Coulomb force due to a proton.
The value for acceleration of the proton is
2.3 × 10–8 N / 1.67 × 10–27 kg = 1.4 × 1019 m/s2

Example 1.4 A charged metallic sphere A is suspended by a nylon

thread. Another charged metallic sphere B held by an insulating
EXAMPLE 1.4

10 FIGURE 1.4

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 Electric Charges
and Fields

handle is brought close to A such that the distance between their

centres is 10 cm, as shown in Fig. 1.4(a). The resulting repulsion of A
is noted (for example, by shining a beam of light and measuring the
deflection of its shadow on a screen). Spheres A and B are touched
by uncharged spheres C and D respectively, as shown in Fig. 1.4(b).
C and D are then removed and B is brought closer to A to a
distance of 5.0 cm between their centres, as shown in Fig. 1.4(c).
What is the expected repulsion of A on the basis of Coulomb’s law?
Spheres A and C and spheres B and D have identical sizes. Ignore
the sizes of A and B in comparison to the separation between their
centres.
Solution Let the original charge on sphere A be q and that on B be
q¢. At a distance r between their centres, the magnitude of the
electrostatic force on each is given by
1 qq ′
F =
4 πε 0 r 2
neglecting the sizes of spheres A and B in comparison to r. When an
identical but uncharged sphere C touches A, the charges redistribute
on A and C and, by symmetry, each sphere carries a charge q/2.
Similarly, after D touches B, the redistributed charge on each is
q¢/2. Now, if the separation between A and B is halved, the magnitude

EXAMPLE 1.4
of the electrostatic force on each is
1 (q / 2 )(q ′ / 2) 1 (qq ′ )
F′ = = =F
4 πε 0 (r / 2)2 4 πε 0 r 2
Thus the electrostatic force on A, due to B, remains unaltered.

1.6 FORCES BETWEEN MULTIPLE CHARGES

The mutual electric force between two charges is given by
Coulomb’s law. How to calculate the force on a charge where
there are not one but several charges around? Consider a
system of n stationary charges q1, q2, q3, ..., qn in vacuum.
What is the force on q1 due to q2, q3, ..., qn? Coulomb’s law is
not enough to answer this question. Recall that forces of
mechanical origin add according to the parallelogram law of
addition. Is the same true for forces of electrostatic origin?
Experimentally, it is verified that force on any charge due
to a number of other charges is the vector sum of all the forces
on that charge due to the other charges, taken one at a time.
The individual forces are unaffected due to the presence of
other charges. This is termed as the principle of superposition.
To better understand the concept, consider a system of
three charges q1, q2 and q3, as shown in Fig. 1.5(a). The force
on one charge, say q1, due to two other charges q2, q3 can
therefore be obtained by performing a vector addition of the
forces due to each one of these charges. Thus, if the force on q1
due to q2 is denoted by F12, F12 is given by Eq. (1.3) even
though other charges are present.
FIGURE 1.5 A system of
1 q1q 2 (a) three charges
Thus, F12 = r̂12 (b) multiple charges. 11
4 πε 0 r12
2

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 Physics
In the same way, the force on q1 due to q3, denoted by F13, is given by
1 q1q3
F13 = rˆ13
4 πε 0 r13
2

which again is the Coulomb force on q1 due to q3, even though other
charge q2 is present.
Thus the total force F1 on q1 due to the two charges q2 and q3 is
given as
1 q1q 2 1 q1q 3
F1 = F12 + F13 = rˆ12 + rˆ13 (1.4)
4 πε 0 r12
2
4 πε 0 r13
2

The above calculation of force can be generalised to a system of

charges more than three, as shown in Fig. 1.5(b).
The principle of superposition says that in a system of charges q1,
q2, ..., qn, the force on q1 due to q2 is the same as given by Coulomb’s law,
i.e., it is unaffected by the presence of the other charges q3, q4, ..., qn. The
total force F1 on the charge q1, due to all other charges, is then given by
the vector sum of the forces F12, F13, ..., F1n:
i.e.,

1  q1q 2 q1q 3 q1qn 

F1 = F12 + F13 + ...+ F1n =  2 rˆ12 + 2 rˆ13 + ... + 2 rˆ1n 
4 πε 0  r12 r13 r1n 
n
q1 q
=
4πε 0
∑ r 2i r̂1i (1.5)
i = 2 1i

The vector sum is obtained as usual by the parallelogram law of

addition of vectors. All of electrostatics is basically a consequence of
Coulomb’s law and the superposition principle.

Example 1.5 Consider three charges q1, q2, q3 each equal to q at the
vertices of an equilateral triangle of side l. What is the force on a
charge Q (with the same sign as q) placed at the centroid of the
triangle, as shown in Fig. 1.6?
EXAMPLE 1.5

FIGURE 1.6

Solution In the given equilateral triangle ABC of sides of length l, if

we draw a perpendicular AD to the side BC,
AD = AC cos 30º = ( 3 /2 ) l and the distance AO of the centroid O
12 from A is (2/3) AD = ( 1/ 3 ) l. By symmatry AO = BO = CO.

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 Electric Charges
and Fields

Thus,
3 Qq
Force F1 on Q due to charge q at A = along AO
4 πε 0 l2
3 Qq
Force F2 on Q due to charge q at B = 4 πε l 2 along BO
0

3 Qq
Force F3 on Q due to charge q at C = 4 πε l 2 along CO
0

3 Qq
The resultant of forces F 2 and F 3 is 4 πε l 2 along OA, by the
0

3 Qq
parallelogram law. Therefore, the total force on Q = 4 πε l 2 ( rˆ − rˆ )
0

EXAMPLE 1.5
= 0, where r̂ is the unit vector along OA.
It is clear also by symmetry that the three forces will sum to zero.
Suppose that the resultant force was non-zero but in some direction.
Consider what would happen if the system was rotated through 60°
about O.

Example 1.6 Consider the charges q, q, and –q placed at the vertices

of an equilateral triangle, as shown in Fig. 1.7. What is the force on
each charge?

FIGURE 1.7

Solution The forces acting on charge q at A due to charges q at B

and –q at C are F12 along BA and F13 along AC respectively, as shown
in Fig. 1.7. By the parallelogram law, the total force F1 on the charge
q at A is given by
F1 = F r̂1 where r̂1 is a unit vector along BC.
The force of attraction or repulsion for each pair of charges has the
EXAMPLE 1.6

q2
same magnitude F =
4 π ε0 l 2

The total force F2 on charge q at B is thus F2 = F r̂ 2, where r̂ 2 is a

unit vector along AC. 13

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 Physics
Similarly the total force on charge –q at C is F3 = 3 F n̂ , where n̂ is
the unit vector along the direction bisecting the ÐBCA.
It is interesting to see that the sum of the forces on the three charges
EXAMPLE 1.6 is zero, i.e.,
F1 + F2 + F3 = 0
The result is not at all surprising. It follows straight from the fact
that Coulomb’s law is consistent with Newton’s third law. The proof
is left to you as an exercise.

1.7 ELECTRIC FIELD

Let us consider a point charge Q placed in vacuum, at the origin O. If we
place another point charge q at a point P, where OP = r, then the charge Q
will exert a force on q as per Coulomb’s law. We may ask the question: If
charge q is removed, then what is left in the surrounding? Is there
nothing? If there is nothing at the point P, then how does a force act
when we place the charge q at P. In order to answer such questions, the
early scientists introduced the concept of field. According to this, we say
that the charge Q produces an electric field everywhere in the surrounding.
When another charge q is brought at some point P, the field there acts on
it and produces a force. The electric field produced by the charge Q at a
point r is given as
1 Q 1 Q
E ( r) = rˆ = rˆ (1.6)
4πε 0 r 2 4πε 0 r 2
where rˆ = r/r, is a unit vector from the origin to the point r. Thus, Eq.(1.6)
specifies the value of the electric field for each value of the position
vector r. The word “field” signifies how some distributed quantity (which
could be a scalar or a vector) varies with position. The effect of the charge
has been incorporated in the existence of the electric field. We obtain the
force F exerted by a charge Q on a charge q, as
1 Qq
F= rˆ (1.7)
4 πε 0 r 2
Note that the charge q also exerts an equal and opposite force on the
charge Q. The electrostatic force between the charges Q and q can be
looked upon as an interaction between charge q and the electric field of
Q and vice versa. If we denote the position of charge q by the vector r, it
experiences a force F equal to the charge q multiplied by the electric
field E at the location of q. Thus,
F(r) = q E(r) (1.8)
Equation (1.8) defines the SI unit of electric field as N/C*.
Some important remarks may be made here:
(i) From Eq. (1.8), we can infer that if q is unity, the electric field due to
FIGURE 1.8 Electric a charge Q is numerically equal to the force exerted by it. Thus, the
field (a) due to a electric field due to a charge Q at a point in space may be defined
charge Q, (b) due to a as the force that a unit positive charge would experience if placed
charge –Q.
14 * An alternate unit V/m will be introduced in the next chapter.

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 Electric Charges
and Fields
at that point. The charge Q, which is producing the electric field, is
called a source charge and the charge q, which tests the effect of a
source charge, is called a test charge. Note that the source charge Q
must remain at its original location. However, if a charge q is brought
at any point around Q, Q itself is bound to experience an electrical
force due to q and will tend to move. A way out of this difficulty is to
make q negligibly small. The force F is then negligibly small but the
ratio F/q is finite and defines the electric field:
 F
E = lim   (1.9)
q →0  q 

A practical way to get around the problem (of keeping Q undisturbed

in the presence of q) is to hold Q to its location by unspecified forces!
This may look strange but actually this is what happens in practice.
When we are considering the electric force on a test charge q due to a
charged planar sheet (Section 1.14), the charges on the sheet are held to
their locations by the forces due to the unspecified charged constituents
inside the sheet.
(ii) Note that the electric field E due to Q, though defined operationally in
terms of some test charge q, is independent of q. This is because
F is proportional to q, so the ratio F/q does not depend on q. The
force F on the charge q due to the charge Q depends on the particular
location of charge q which may take any value in the space around
the charge Q. Thus, the electric field E due to Q is also dependent on
the space coordinate r. For different positions of the charge q all over
the space, we get different values of electric field E. The field exists at
every point in three-dimensional space.
(iii) For a positive charge, the electric field will be directed radially
outwards from the charge. On the other hand, if the source charge is
negative, the electric field vector, at each point, points radially inwards.
(iv) Since the magnitude of the force F on charge q due to charge Q
depends only on the distance r of the charge q from charge Q,
the magnitude of the electric field E will also depend only on the
distance r. Thus at equal distances from the charge Q, the magnitude
of its electric field E is same. The magnitude of electric field E due to
a point charge is thus same on a sphere with the point charge at its
centre; in other words, it has a spherical symmetry.

1.7.1 Electric field due to a system of charges

Consider a system of charges q1, q2, ..., qn with position vectors r1,
r2, ..., rn relative to some origin O. Like the electric field at a point in
space due to a single charge, electric field at a point in space due to the
system of charges is defined to be the force experienced by a unit
test charge placed at that point, without disturbing the original
positions of charges q1, q2, ..., qn. We can use Coulomb’s law and the
superposition principle to determine this field at a point P denoted by
position vector r. 15

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 Physics
Electric field E1 at r due to q1 at r1 is given by
1 q1
E1 = r̂1P
4 πε 0 r12P
where r̂1P is a unit vector in the direction from q1 to P,
and r1P is the distance between q1 and P.
In the same manner, electric field E2 at r due to q2 at
r2 is
1 q2
E2 = r̂2P
4 πε 0 r22P
where r̂2P is a unit vector in the direction from q2 to P
FIGURE 1.9 Electric field at a point and r 2P is the distance between q 2 and P. Similar
due to a system of charges is the expressions hold good for fields E3, E4, ..., En due to
vector sum of the electric fields at charges q3, q4, ..., qn.
the point due to individual charges. By the superposition principle, the electric field E at r
due to the system of charges is (as shown in Fig. 1.9)
E(r) = E1 (r) + E2 (r) + … + En(r)
1 q1 1 q2 1 qn
= rˆ +
2 1P
rˆ2 P + ... + rˆnP
4 πε 0 r1P 4 πε 0 r2 P
2
4 πε 0 rn2P

1 n
q
E(r) =
4π ε 0
∑ r 2i r̂i P (1.10)
i =1 i P

E is a vector quantity that varies from one point to another point in space
and is determined from the positions of the source charges.

1.7.2 Physical significance of electric field

You may wonder why the notion of electric field has been introduced
here at all. After all, for any system of charges, the measurable quantity
is the force on a charge which can be directly determined using Coulomb’s
law and the superposition principle [Eq. (1.5)]. Why then introduce this
intermediate quantity called the electric field?
For electrostatics, the concept of electric field is convenient, but not
really necessary. Electric field is an elegant way of characterising the
electrical environment of a system of charges. Electric field at a point in
the space around a system of charges tells you the force a unit positive
test charge would experience if placed at that point (without disturbing
the system). Electric field is a characteristic of the system of charges and
is independent of the test charge that you place at a point to determine
the field. The term field in physics generally refers to a quantity that is
defined at every point in space and may vary from point to point. Electric
field is a vector field, since force is a vector quantity.
The true physical significance of the concept of electric field, however,
emerges only when we go beyond electrostatics and deal with time-
dependent electromagnetic phenomena. Suppose we consider the force
between two distant charges q1, q2 in accelerated motion. Now the greatest
speed with which a signal or information can go from one point to another
16 is c, the speed of light. Thus, the effect of any motion of q1 on q2 cannot

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 Electric Charges
and Fields
arise instantaneously. There will be some time delay between the effect
(force on q2) and the cause (motion of q1). It is precisely here that the
notion of electric field (strictly, electromagnetic field) is natural and very
useful. The field picture is this: the accelerated motion of charge q1
produces electromagnetic waves, which then propagate with the speed
c, reach q2 and cause a force on q2. The notion of field elegantly accounts
for the time delay. Thus, even though electric and magnetic fields can be
detected only by their effects (forces) on charges, they are regarded as
physical entities, not merely mathematical constructs. They have an
independent dynamics of their own, i.e., they evolve according to laws
of their own. They can also transport energy. Thus, a source of time-
dependent electromagnetic fields, turned on for a short interval of time
and then switched off, leaves behind propagating electromagnetic fields
transporting energy. The concept of field was first introduced by Faraday
and is now among the central concepts in physics.

Example 1.7 An electron falls through a distance of 1.5 cm in a

uniform electric field of magnitude 2.0 × 104 N C–1 [Fig. 1.10(a)]. The
direction of the field is reversed keeping its magnitude unchanged
and a proton falls through the same distance [Fig. 1.10(b)]. Compute
the time of fall in each case. Contrast the situation with that of ‘free
fall under gravity’.

FIGURE 1.10
Solution In Fig. 1.10(a) the field is upward, so the negatively charged
electron experiences a downward force of magnitude eE where E is
the magnitude of the electric field. The acceleration of the electron is
ae = eE/me
where me is the mass of the electron.

Starting from rest, the time required by the electron to fall through a
2h 2h m e
distance h is given by t e = =
ae eE
For e = 1.6 × 10–19C, me = 9.11 × 10–31 kg,
E = 2.0 × 104 N C–1, h = 1.5 × 10–2 m,
te = 2.9 × 10–9s
In Fig. 1.10 (b), the field is downward, and the positively charged
proton experiences a downward force of magnitude eE . The
EXAMPLE 1.7

acceleration of the proton is

ap = eE/mp
where mp is the mass of the proton; mp = 1.67 × 10–27 kg. The time of
fall for the proton is
17

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 Physics
2h 2h m p
tp = = = 1.3 × 10 –7 s
ap eE
Thus, the heavier particle (proton) takes a greater time to fall through
the same distance. This is in basic contrast to the situation of ‘free
fall under gravity’ where the time of fall is independent of the mass of
the body. Note that in this example we have ignored the acceleration
due to gravity in calculating the time of fall. To see if this is justified,
let us calculate the acceleration of the proton in the given electric
field:
eE
ap =
mp

(1.6 × 10−19 C) × (2.0 × 10 4 N C −1 )

=
1.67 × 10 −27 kg
EXAMPLE 1.7

= 1.9 × 1012 m s –2
which is enormous compared to the value of g (9.8 m s –2), the
acceleration due to gravity. The acceleration of the electron is even
greater. Thus, the effect of acceleration due to gravity can be ignored
in this example.

Example 1.8 Two point charges q1 and q2, of magnitude +10–8 C and
–10–8 C, respectively, are placed 0.1 m apart. Calculate the electric
fields at points A, B and C shown in Fig. 1.11.

FIGURE 1.11
Solution The electric field vector E1A at A due to the positive charge
q1 points towards the right and has a magnitude
(9 × 109 Nm 2C-2 ) × (10 −8 C)
E1A = = 3.6 × 104 N C–1
(0.05 m)2
The electric field vector E2A at A due to the negative charge q2 points
EXAMPLE 1.8

towards the right and has the same magnitude. Hence the magnitude
of the total electric field EA at A is
EA = E1A + E2A = 7.2 × 104 N C–1
EA is directed toward the right.
18

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 Electric Charges
and Fields

The electric field vector E1B at B due to the positive charge q1 points
towards the left and has a magnitude
(9 × 109 Nm2 C –2 ) × (10 −8 C)
E1B = = 3.6 × 104 N C–1
(0.05 m)2
The electric field vector E2B at B due to the negative charge q2 points
towards the right and has a magnitude
(9 × 109 Nm 2 C –2 ) × (10 −8 C)
E 2B = = 4 × 103 N C–1
(0.15 m)2
The magnitude of the total electric field at B is
EB = E1B – E2B = 3.2 × 104 N C–1
EB is directed towards the left.
The magnitude of each electric field vector at point C, due to charge
q1 and q2 is
(9 × 109 Nm 2C –2 ) × (10−8 C)
E1C = E2C = = 9 × 103 N C–1
(0.10 m)2
The directions in which these two vectors point are indicated in

EXAMPLE 1.8
Fig. 1.11. The resultant of these two vectors is
π π
EC = E1c cos + E 2c cos = 9 × 103 N C–1
3 3
EC points towards the right.

1.8 ELECTRIC FIELD LINES

We have studied electric field in the last section. It is a vector quantity
and can be represented as we represent vectors. Let us try to represent E
due to a point charge pictorially. Let the point charge be placed at the
origin. Draw vectors pointing along the direction of the
electric field with their lengths proportional to the strength
of the field at each point. Since the magnitude of electric
field at a point decreases inversely as the square of the
distance of that point from the charge, the vector gets
shorter as one goes away from the origin, always pointing
radially outward. Figure 1.12 shows such a picture. In
this figure, each arrow indicates the electric field, i.e., the
force acting on a unit positive charge, placed at the tail of
that arrow. Connect the arrows pointing in one direction
and the resulting figure represents a field line. We thus
get many field lines, all pointing outwards from the point
charge. Have we lost the information about the strength
or magnitude of the field now, because it was contained
in the length of the arrow? No. Now the magnitude of the
field is represented by the density of field lines. E is strong
near the charge, so the density of field lines is more near
the charge and the lines are closer. Away from the charge, FIGURE 1.12 Field of a point charge.
the field gets weaker and the density of field lines is less,
resulting in well-separated lines.
Another person may draw more lines. But the number of lines is not
important. In fact, an infinite number of lines can be drawn in any region.
19

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 Physics
It is the relative density of lines in different regions which is
important.
We draw the figure on the plane of paper, i.e., in two-
dimensions but we live in three-dimensions. So if one wishes
to estimate the density of field lines, one has to consider the
number of lines per unit cross-sectional area, perpendicular
to the lines. Since the electric field decreases as the square of
the distance from a point charge and the area enclosing the
charge increases as the square of the distance, the number
of field lines crossing the enclosing area remains constant,
whatever may be the distance of the area from the charge.
We started by saying that the field lines carry information
about the direction of electric field at different points in space.
FIGURE 1.13 Dependence of
Having drawn a certain set of field lines, the relative density
electric field strength on the
distance and its relation to the (i.e., closeness) of the field lines at different points indicates
number of field lines. the relative strength of electric field at those points. The field
lines crowd where the field is strong and are spaced apart
where it is weak. Figure 1.13 shows a set of field lines. We
can imagine two equal and small elements of area placed at points R and
S normal to the field lines there. The number of field lines in our picture
cutting the area elements is proportional to the magnitude of field at
these points. The picture shows that the field at R is stronger than at S.
To understand the dependence of the field lines on the area, or rather
the solid angle subtended by an area element, let us try to relate the
area with the solid angle, a generalisation of angle to three dimensions.
Recall how a (plane) angle is defined in two-dimensions. Let a small
transverse line element Dl be placed at a distance r from a point O. Then
the angle subtended by Dl at O can be approximated as Dq = Dl/r.
Likewise, in three-dimensions the solid angle* subtended by a small
perpendicular plane area DS, at a distance r, can be written as
DW = DS/r2. We know that in a given solid angle the number of radial
field lines is the same. In Fig. 1.13, for two points P1 and P2 at distances
r1 and r2 from the charge, the element of area subtending the solid angle
DW is r12 DW at P1 and an element of area r22 DW at P2, respectively. The
number of lines (say n) cutting these area elements are the same. The
number of field lines, cutting unit area element is therefore n/( r12 DW) at
P1 and n/( r22 DW) at P2 , respectively. Since n and DW are common, the
strength of the field clearly has a 1/r 2 dependence.
The picture of field lines was invented by Faraday to develop an
intuitive non-mathematical way of visualising electric fields around
charged configurations. Faraday called them lines of force. This term is
somewhat misleading, especially in case of magnetic fields. The more
appropriate term is field lines (electric or magnetic) that we have
adopted in this book.
Electric field lines are thus a way of pictorially mapping the electric
field around a configuration of charges. An electric field line is, in general,

* Solid angle is a measure of a cone. Consider the intersection of the given cone
with a sphere of radius R. The solid angle DW of the cone is defined to be equal
20 2
to DS/R , where DS is the area on the sphere cut out by the cone.

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 Electric Charges
and Fields
a curve drawn in such a way that the tangent to it at each
point is in the direction of the net field at that point. An
arrow on the curve is obviously necessary to specify the
direction of electric field from the two possible directions
indicated by a tangent to the curve. A field line is a space
curve, i.e., a curve in three dimensions.
Figure 1.14 shows the field lines around some simple
charge configurations. As mentioned earlier, the field lines
are in 3-dimensional space, though the figure shows them
only in a plane. The field lines of a single positive charge
are radially outward while those of a single negative
charge are radially inward. The field lines around a system
of two positive charges (q, q) give a vivid pictorial
description of their mutual repulsion, while those around
the configuration of two equal and opposite charges
(q, –q), a dipole, show clearly the mutual attraction
between the charges. The field lines follow some important
general properties:
(i) Field lines start from positive charges and end at
negative charges. If there is a single charge, they may
start or end at infinity.
(ii) In a charge-free region, electric field lines can be taken
to be continuous curves without any breaks.
(iii) Two field lines can never cross each other. (If they did,
the field at the point of intersection will not have a
unique direction, which is absurd.)
(iv) Electrostatic field lines do not form any closed loops.
This follows from the conservative nature of electric
field (Chapter 2).

1.9 ELECTRIC FLUX

Consider flow of a liquid with velocity v, through a small
flat surface dS, in a direction normal to the surface. The
rate of flow of liquid is given by the volume crossing the
area per unit time v dS and represents the flux of liquid
flowing across the plane. If the normal to the surface is
not parallel to the direction of flow of liquid, i.e., to v, but
makes an angle q with it, the projected area in a plane
perpendicular to v is δ dS cos q. Therefore, the flux going
out of the surface dS is v. n̂ dS. For the case of the electric
field, we define an analogous quantity and call it electric
flux. We should, however, note that there is no flow of a
physically observable quantity unlike the case of
liquid flow.
In the picture of electric field lines described above, FIGURE 1.14 Field lines due to
we saw that the number of field lines crossing a unit area, some simple charge configurations.
placed normal to the field at a point is a measure of the
strength of electric field at that point. This means that if 21

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 Physics
we place a small planar element of area DS
normal to E at a point, the number of field lines
crossing it is proportional* to E DS. Now
suppose we tilt the area element by angle q.
Clearly, the number of field lines crossing the
area element will be smaller. The projection of
the area element normal to E is DS cosq. Thus,
the number of field lines crossing DS is
proportional to E DS cosq. When q = 90°, field
lines will be parallel to DS and will not cross it
at all (Fig. 1.15).
The orientation of area element and not
merely its magnitude is important in many
contexts. For example, in a stream, the amount
of water flowing through a ring will naturally
depend on how you hold the ring. If you hold
it normal to the flow, maximum water will flow
FIGURE 1.15 Dependence of flux on the
inclination q between E and n̂ . through it than if you hold it with some other
orientation. This shows that an area element
should be treated as a vector. It has a
magnitude and also a direction. How to specify the direction of a planar
area? Clearly, the normal to the plane specifies the orientation of the
plane. Thus the direction of a planar area vector is along its normal.
How to associate a vector to the area of a curved surface? We imagine
dividing the surface into a large number of very small area elements.
Each small area element may be treated as planar and a vector associated
with it, as explained before.
Notice one ambiguity here. The direction of an area element is along
its normal. But a normal can point in two directions. Which direction do
we choose as the direction of the vector associated with the area element?
This problem is resolved by some convention appropriate to the given
context. For the case of a closed surface, this convention is very simple.
The vector associated with every area element of a closed surface is taken
to be in the direction of the outward normal. This is the convention used
in Fig. 1.16. Thus, the area element vector DS at a point on a closed
surface equals DS n̂ where DS is the magnitude of the area element and
n̂ is a unit vector in the direction of outward normal at that point.
We now come to the definition of electric flux. Electric flux Df through
an area element DS is defined by
Df = E.DS = E DS cosq (1.11)
which, as seen before, is proportional to the number of field lines cutting
the area element. The angle q here is the angle between E and DS. For a
closed surface, with the convention stated already, q is the angle between
FIGURE 1.16 E and the outward normal to the area element. Notice we could look at
Convention for the expression E DS cosq in two ways: E (DS cosq ) i.e., E times the
defining normal
n̂ and DS. * It will not be proper to say that the number of field lines is equal to EDS. The
number of field lines is after all, a matter of how many field lines we choose to
draw. What is physically significant is the relative number of field lines crossing
22 a given area at different points.

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 Electric Charges
and Fields
projection of area normal to E, or E^ DS, i.e., component of E along the
normal to the area element times the magnitude of the area element. The
unit of electric flux is N C–1 m2.
The basic definition of electric flux given by Eq. (1.11) can be used, in
principle, to calculate the total flux through any given surface. All we
have to do is to divide the surface into small area elements, calculate the
flux at each element and add them up. Thus, the total flux f through a
surface S is
f ~ S E.DS (1.12)
The approximation sign is put because the electric field E is taken to
be constant over the small area element. This is mathematically exact
only when you take the limit DS ® 0 and the sum in Eq. (1.12) is written
as an integral.

1.10 ELECTRIC DIPOLE

An electric dipole is a pair of equal and opposite point charges q and –q,
separated by a distance 2a. The line connecting the two charges defines
a direction in space. By convention, the direction from –q to q is said to
be the direction of the dipole. The mid-point of locations of –q and q is
called the centre of the dipole.
The total charge of the electric dipole is obviously zero. This does not
mean that the field of the electric dipole is zero. Since the charge q and
–q are separated by some distance, the electric fields due to them, when
added, do not exactly cancel out. However, at distances much larger than
the separation of the two charges forming a dipole (r >> 2a), the fields
due to q and –q nearly cancel out. The electric field due to a dipole
therefore falls off, at large distance, faster than like 1/r 2 (the dependence
on r of the field due to a single charge q). These qualitative ideas are
borne out by the explicit calculation as follows:

1.10.1 The field of an electric dipole

The electric field of the pair of charges (–q and q) at any point in space
can be found out from Coulomb’s law and the superposition principle.
The results are simple for the following two cases: (i) when the point is on
the dipole axis, and (ii) when it is in the equatorial plane of the dipole,
i.e., on a plane perpendicular to the dipole axis through its centre. The
electric field at any general point P is obtained by adding the electric
fields E–q due to the charge –q and E+q due to the charge q, by the
parallelogram law of vectors.
(i) For points on the axis
Let the point P be at distance r from the centre of the dipole on the side of
the charge q, as shown in Fig. 1.17(a). Then
q
E −q = − p [1.13(a)]
4πε0 (r + a )2
where p̂ is the unit vector along the dipole axis (from –q to q). Also
q
E +q = p [1.13(b)] 23
4 π ε 0 (r − a )2

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 Physics
The total field at P is
q  1 1 
E = E +q + E − q =  − p
4 π ε0  (r − a )
2
(r + a )2 

q 4a r
= p (1.14)
4 π εo ( r 2 − a 2 )2
For r >> a
4qa
E= ˆ
p (r >> a) (1.15)
4 π ε 0r 3

(ii) For points on the equatorial plane

The magnitudes of the electric fields due to the two
charges +q and –q are given by
q 1
E +q = [1.16(a)]
4 πε 0 r + a 2
2

q 1
E –q = [1.16(b)]
4 πε 0 r + a 2
2

FIGURE 1.17 Electric field of a dipole and are equal.

at (a) a point on the axis, (b) a point The directions of E +q and E –q are as shown in
on the equatorial plane of the dipole. Fig. 1.17(b). Clearly, the components normal to the dipole
p is the dipole moment vector of
axis cancel away. The components along the dipole axis
magnitude p = q × 2a and
directed from –q to q.
add up. The total electric field is opposite to p̂ . We have
E = – (E +q + E –q ) cosq p̂
2q a
=− p (1.17)
4 π ε o (r 2 + a 2 )3 / 2
At large distances (r >> a), this reduces to
2qa
E=− ˆ
p (r >> a ) (1.18)
4 π εo r 3
From Eqs. (1.15) and (1.18), it is clear that the dipole field at large
distances does not involve q and a separately; it depends on the product
qa. This suggests the definition of dipole moment. The dipole moment
vector p of an electric dipole is defined by
p = q × 2a p̂ (1.19)
that is, it is a vector whose magnitude is charge q times the separation
2a (between the pair of charges q, –q) and the direction is along the line
from –q to q. In terms of p, the electric field of a dipole at large distances
takes simple forms:
At a point on the dipole axis
2p
E= (r >> a) (1.20)
4 πε o r 3
At a point on the equatorial plane
p
E=− (r >> a) (1.21)
24 4πε or 3

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 Electric Charges
and Fields
Notice the important point that the dipole field at large distances
falls off not as 1/r 2 but as1/r 3. Further, the magnitude and the direction
of the dipole field depends not only on the distance r but also on the
angle between the position vector r and the dipole moment p.
We can think of the limit when the dipole size 2a approaches zero,
the charge q approaches infinity in such a way that the product
p = q × 2a is finite. Such a dipole is referred to as a point dipole. For a
point dipole, Eqs. (1.20) and (1.21) are exact, true for any r.

1.10.2 Physical significance of dipoles

In most molecules, the centres of positive charges and of negative charges*
lie at the same place. Therefore, their dipole moment is zero. CO2 and
CH4 are of this type of molecules. However, they develop a dipole moment
when an electric field is applied. But in some molecules, the centres of
negative charges and of positive charges do not coincide. Therefore they
have a permanent electric dipole moment, even in the absence of an electric
field. Such molecules are called polar molecules. Water molecules, H2O,
is an example of this type. Various materials give rise to interesting
properties and important applications in the presence or absence of
electric field.

Example 1.9 Two charges ±10 mC are placed 5.0 mm apart. Determine
the electric field at (a) a point P on the axis of the dipole 15 cm away
from its centre O on the side of the positive charge, as shown in Fig.
1.18(a), and (b) a point Q, 15 cm away from O on a line passing through
O and normal to the axis of the dipole, as shown in Fig. 1.18(b).
EXAMPLE 1.9

FIGURE 1.18

* Centre of a collection of positive point charges is defined much the same way
∑ qi ri
as the centre of mass: rcm = i .
∑ qi 25
i

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 Physics
Solution (a) Field at P due to charge +10 mC
10 −5 C 1
= ×
4 π (8.854 × 10 −12 2
C N −1
m ) −2
(15 − 0.25)2 × 10 −4 m 2
6 –1
= 4.13 × 10 N C along BP
Field at P due to charge –10 mC
10 –5 C 1
= ×
−12 −1
4 π (8.854 × 10 C N m )
2 −2
(15 + 0.25)2
× 10 −4 m 2
= 3.86 × 106 N C–1 along PA
The resultant electric field at P due to the two charges at A and B is
= 2.7 × 105 N C–1 along BP.
In this example, the ratio OP/OB is quite large (= 60). Thus, we can
expect to get approximately the same result as above by directly using
the formula for electric field at a far-away point on the axis of a dipole.
For a dipole consisting of charges ± q, 2a distance apart, the electric
field at a distance r from the centre on the axis of the dipole has a
magnitude
2p
E = (r/a >> 1)
4 πε 0r 3
where p = 2a q is the magnitude of the dipole moment.
The direction of electric field on the dipole axis is always along the
direction of the dipole moment vector (i.e., from –q to q). Here,
p =10–5 C × 5 × 10–3 m = 5 × 10–8 C m
Therefore,
2 × 5 × 10−8 C m 1
E = × = 2.6 × 105 N C–1
4 π (8.854 × 10 −12 2
C N m ) −1 −2
(15) × 10 −6 m 3
3

along the dipole moment direction AB, which is close to the result
obtained earlier.
(b) Field at Q due to charge + 10 mC at B
10−5 C 1
= 4 π (8.854 × 10 −12 C 2 N −1 m −2 ) × [152 + (0.25)2 ] × 10 −4 m 2

= 3.99 × 106 N C–1 along BQ

Field at Q due to charge –10 mC at A

10 −5 C 1
= ×
−12
4 π (8.854 × 10 C N m ) 2
[15 −1 −2 2 2
+ (0.25) ] × 10 −4 m 2
= 3.99 × 106 N C–1 along QA.

Clearly, the components of these two forces with equal magnitudes

cancel along the direction OQ but add up along the direction parallel
to BA. Therefore, the resultant electric field at Q due to the two
charges at A and B is
0.25
=2× × 3.99 × 106 N C –1 along BA
EXAMPLE 1.9

15 2
+ (0.25) 2

= 1.33 × 10 N C–1 along BA.

5

As in (a), we can expect to get approximately the same result by

directly using the formula for dipole field at a point on the normal to
26 the axis of the dipole:

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 Electric Charges
and Fields
p
E = (r/a >> 1)
4 πε 0 r 3

5 × 10−8 C m 1
= ×
4 π (8.854 ×10−12 C2 N –1 m –2 ) (15)3 × 10 −6 m 3

EXAMPLE 1.9
= 1.33 × 105 N C–1.
The direction of electric field in this case is opposite to the direction
of the dipole moment vector. Again, the result agrees with that obtained
before.

1.11 DIPOLE IN A UNIFORM EXTERNAL FIELD

Consider a permanent dipole of dipole moment p in a uniform
external field E, as shown in Fig. 1.19. (By permanent dipole, we
mean that p exists irrespective of E; it has not been induced by E.)
There is a force qE on q and a force –qE on –q. The net force on
the dipole is zero, since E is uniform. However, the charges are
separated, so the forces act at different points, resulting in a torque
on the dipole. When the net force is zero, the torque (couple) is
independent of the origin. Its magnitude equals the magnitude of FIGURE 1.19 Dipole in a
each force multiplied by the arm of the couple (perpendicular uniform electric field.
distance between the two antiparallel forces).
Magnitude of torque = q E × 2 a sinq
= 2 q a E sinq
Its direction is normal to the plane of the paper, coming out of it.
The magnitude of p × E is also p E sinq and its direction
is normal to the paper, coming out of it. Thus,
t =p×E (1.22)
This torque will tend to align the dipole with the field
E. When p is aligned with E, the torque is zero.
What happens if the field is not uniform? In that case,
the net force will evidently be non-zero. In addition there
will, in general, be a torque on the system as before. The
general case is involved, so let us consider the simpler
situations when p is parallel to E or antiparallel to E. In
either case, the net torque is zero, but there is a net force
on the dipole if E is not uniform.
Figure 1.20 is self-explanatory. It is easily seen that
when p is parallel to E, the dipole has a net force in the
direction of increasing field. When p is antiparallel to E,
the net force on the dipole is in the direction of decreasing
field. In general, the force depends on the orientation of p
with respect to E.
This brings us to a common observation in frictional
electricity. A comb run through dry hair attracts pieces of FIGURE 1.20 Electric force on a
paper. The comb, as we know, acquires charge through dipole: (a) E parallel to p, (b) E
friction. But the paper is not charged. What then explains antiparallel to p.
the attractive force? Taking the clue from the preceding 27

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 Physics
discussion, the charged comb ‘polarises’ the piece of paper, i.e., induces
a net dipole moment in the direction of field. Further, the electric field
due to the comb is not uniform. This non-uniformity of the field makes a
dipole to experience a net force on it. In this situation, it is easily seen
that the paper should move in the direction of the comb!

1.12 CONTINUOUS CHARGE DISTRIBUTION

We have so far dealt with charge configurations involving discrete charges
q1, q2, ..., qn. One reason why we restricted to discrete charges is that the
mathematical treatment is simpler and does not involve calculus. For
many purposes, however, it is impractical to work in terms of discrete
charges and we need to work with continuous charge distributions. For
example, on the surface of a charged conductor, it is impractical to specify
the charge distribution in terms of the locations of the microscopic charged
constituents. It is more feasible to consider an area element DS (Fig. 1.21)
on the surface of the conductor (which is very small on the macroscopic
scale but big enough to include a very large number of electrons) and
specify the charge DQ on that element. We then define a surface charge
density s at the area element by
∆Q
σ= (1.23)
∆S
We can do this at different points on the conductor and thus arrive at
a continuous function s, called the surface charge density. The surface
charge density s so defined ignores the quantisation of charge and the
discontinuity in charge distribution at the microscopic level*. s represents
macroscopic surface charge density, which in a sense, is a smoothed out
average of the microscopic charge density over an area element DS which,
as said before, is large microscopically but small macroscopically. The
units for s are C/m2.
FIGURE 1.21 Similar considerations apply for a line charge distribution and a volume
Definition of linear, charge distribution. The linear charge density l of a wire is defined by
surface and volume
∆Q
charge densities. λ = (1.24)
In each case, the ∆l
element (Dl, DS, DV ) where Dl is a small line element of wire on the macroscopic scale that,
chosen is small on however, includes a large number of microscopic charged constituents,
the macroscopic and DQ is the charge contained in that line element. The units for l are
scale but contains C/m. The volume charge density (sometimes simply called charge density)
a very large number
is defined in a similar manner:
of microscopic
constituents. ∆Q
ρ= (1.25)
∆V
where DQ is the charge included in the macroscopically small volume
element DV that includes a large number of microscopic charged
constituents. The units for r are C/m3.
The notion of continuous charge distribution is similar to that we
adopt for continuous mass distribution in mechanics. When we refer to

28 * At the microscopic level, charge distribution is discontinuous, because they are

discrete charges separated by intervening space where there is no charge.

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 Electric Charges
and Fields
the density of a liquid, we are referring to its macroscopic density. We
regard it as a continuous fluid and ignore its discrete molecular
constitution.
The field due to a continuous charge distribution can be obtained in
much the same way as for a system of discrete charges, Eq. (1.10). Suppose
a continuous charge distribution in space has a charge density r. Choose
any convenient origin O and let the position vector of any point in the
charge distribution be r. The charge density r may vary from point to
point, i.e., it is a function of r. Divide the charge distribution into small
volume elements of size DV. The charge in a volume element DV is rDV.
Now, consider any general point P (inside or outside the distribution)
with position vector R (Fig. 1.21). Electric field due to the charge rDV is
given by Coulomb’s law:
1 ρ ∆V
∆E = rˆ' (1.26)
4πε 0 r' 2
where r¢ is the distance between the charge element and P, and r̂ ¢ is a
unit vector in the direction from the charge element to P. By the
superposition principle, the total electric field due to the charge
distribution is obtained by summing over electric fields due to different
volume elements:
1 ρ ∆V
E≅ Σ rˆ' (1.27)
4 πε 0 all ∆V r' 2
Note that r, r¢, rˆ ′ all can vary from point to point. In a strict
mathematical method, we should let DV®0 and the sum then becomes
an integral; but we omit that discussion here, for simplicity. In short,
using Coulomb’s law and the superposition principle, electric field can
be determined for any charge distribution, discrete or continuous or part
discrete and part continuous.

1.13 GAUSS’S LAW

As a simple application of the notion of electric flux, let us consider the
total flux through a sphere of radius r, which encloses a point charge q
at its centre. Divide the sphere into small area elements, as shown in
Fig. 1.22.
The flux through an area element DS is
q
∆φ = E i ∆ S = rˆ i ∆S (1.28)
4 πε 0 r 2
where we have used Coulomb’s law for the electric field due to a single
charge q. The unit vector r̂ is along the radius vector from the centre to
the area element. Now, since the normal to a sphere at every point is
along the radius vector at that point, the area element DS and r̂ have
the same direction. Therefore,
q FIGURE 1.22 Flux
∆φ = ∆S (1.29) through a sphere
4 πε 0 r 2 enclosing a point
since the magnitude of a unit vector is 1. charge q at its centre.
The total flux through the sphere is obtained by adding up flux
through all the different area elements: 29

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 Physics
q
φ= Σ ∆S
all ∆S 4 π ε0 r 2
Since each area element of the sphere is at the same
distance r from the charge,
FIGURE 1.23 Calculation of the q q
flux of uniform electric field
φ= Σ ∆S = S
4 πεo r 2 all ∆S
4 πε 0 r 2
through the surface of a cylinder.
Now S, the total area of the sphere, equals 4pr 2. Thus,
q q
φ= × 4 πr 2 = (1.30)
4 πε 0 r 2
ε0
Equation (1.30) is a simple illustration of a general result of
electrostatics called Gauss’s law.
We state Gauss’s law without proof:
Electric flux through a closed surface S
= q/e0 (1.31)
q = total charge enclosed by S.
The law implies that the total electric flux through a closed surface is
zero if no charge is enclosed by the surface. We can see that explicitly in
the simple situation of Fig. 1.23.
Here the electric field is uniform and we are considering a closed
cylindrical surface, with its axis parallel to the uniform field E. The total
flux f through the surface is f = f1 + f2 + f3, where f1 and f2 represent
the flux through the surfaces 1 and 2 (of circular cross-section) of the
cylinder and f3 is the flux through the curved cylindrical part of the
closed surface. Now the normal to the surface 3 at every point is
perpendicular to E, so by definition of flux, f3 = 0. Further, the outward
normal to 2 is along E while the outward normal to 1 is opposite to E.
Therefore,
f1 = –E S1, f2 = +E S2
S1 = S2 = S
where S is the area of circular cross-section. Thus, the total flux is zero,
as expected by Gauss’s law. Thus, whenever you find that the net electric
flux through a closed surface is zero, we conclude that the total charge
contained in the closed surface is zero.
The great significance of Gauss’s law Eq. (1.31), is that it is true in
general, and not only for the simple cases we have considered above. Let
us note some important points regarding this law:
(i) Gauss’s law is true for any closed surface, no matter what its shape
or size.
(ii) The term q on the right side of Gauss’s law, Eq. (1.31), includes the
sum of all charges enclosed by the surface. The charges may be located
anywhere inside the surface.
(iii) In the situation when the surface is so chosen that there are some
charges inside and some outside, the electric field [whose flux appears
on the left side of Eq. (1.31)] is due to all the charges, both inside and
outside S. The term q on the right side of Gauss’s law, however,
30 represents only the total charge inside S.

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 Electric Charges
and Fields
(iv) The surface that we choose for the application of Gauss’s law is called
the Gaussian surface. You may choose any Gaussian surface and
apply Gauss’s law. However, take care not to let the Gaussian surface
pass through any discrete charge. This is because electric field due
to a system of discrete charges is not well defined at the location of
any charge. (As you go close to the charge, the field grows without
any bound.) However, the Gaussian surface can pass through a
continuous charge distribution.
(v) Gauss’s law is often useful towards a much easier calculation of the
electrostatic field when the system has some symmetry. This is
facilitated by the choice of a suitable Gaussian surface.
(vi) Finally, Gauss’s law is based on the inverse square dependence on
distance contained in the Coulomb’s law. Any violation of Gauss’s
law will indicate departure from the inverse square law.

Example 1.10 The electric field components in Fig. 1.24 are

Ex = ax1/2, Ey = Ez = 0, in which a = 800 N/C m1/2. Calculate (a) the
flux through the cube, and (b) the charge within the cube. Assume
that a = 0.1 m.

FIGURE 1.24
Solution
(a) Since the electric field has only an x component, for faces
perpendicular to x direction, the angle between E and DS is
± p/2. Therefore, the flux f = E.DS is separately zero for each face
of the cube except the two shaded ones. Now the magnitude of
the electric field at the left face is
EL = ax1/2 = aa1/2
(x = a at the left face).
The magnitude of electric field at the right face is
ER = a x1/2 = a (2a)1/2
(x = 2a at the right face).
The corresponding fluxes are
EXAMPLE 1.10

f = E .DS = ∆S E L ⋅ n
L L
ˆ L =E DS cosq = –E DS, since q = 180°
L L

= –ELa2
fR= ER.DS = ER DS cosq = ER DS, since q = 0°
= ERa2
Net flux through the cube 31

Rationalised 2023-24
 Physics
= fR + fL = ERa2 – ELa2 = a2 (ER – EL) = aa2 [(2a)1/2 – a1/2]
= aa5/2 ( 2 –1)
EXAMPLE 1.10 = 800 (0.1)5/2 ( 2 –1 )
2 –1
= 1.05 N m C
(b) We can use Gauss’s law to find the total charge q inside the cube.
We have f = q/e0 or q = fe0. Therefore,
q = 1.05 × 8.854 × 10–12 C = 9.27 × 10–12 C.

Example 1.11 An electric field is uniform, and in the positive x

direction for positive x, and uniform with the same magnitude but in
the negative x direction for negative x. It is given that E = 200 î N/C
for x > 0 and E = –200 î N/C for x < 0. A right circular cylinder of
length 20 cm and radius 5 cm has its centre at the origin and its axis
along the x-axis so that one face is at x = +10 cm and the other is at
x = –10 cm (Fig. 1.25). (a) What is the net outward flux through each
flat face? (b) What is the flux through the side of the cylinder?
(c) What is the net outward flux through the cylinder? (d) What is the
net charge inside the cylinder?
Solution
(a) We can see from the figure that on the left face E and DS are
parallel. Therefore, the outward flux is
f = E.DS = – 200 ˆii ∆S
L

= + 200 DS, since ˆii ∆S = – DS

= + 200 × p (0.05)2 = + 1.57 N m2 C–1
On the right face, E and DS are parallel and therefore
fR = E.DS = + 1.57 N m2 C–1.
(b) For any point on the side of the cylinder E is perpendicular to
DS and hence E.DS = 0. Therefore, the flux out of the side of the
cylinder is zero.
(c) Net outward flux through the cylinder
f = 1.57 + 1.57 + 0 = 3.14 N m2 C–1

FIGURE 1.25
EXAMPLE 1.11

(d) The net charge within the cylinder can be found by using Gauss’s
law which gives
q = e0f
= 3.14 × 8.854 × 10–12 C
= 2.78 × 10–11 C
32

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 Electric Charges
and Fields

1.14 APPLICATIONS OF GAUSS’S LAW

The electric field due to a general charge distribution is, as seen above,
given by Eq. (1.27). In practice, except for some special cases, the
summation (or integration) involved in this equation cannot be carried
out to give electric field at every point in
space. For some symmetric charge
configurations, however, it is possible to
obtain the electric field in a simple way using
the Gauss’s law. This is best understood by
some examples.

1.14.1 Field due to an infinitely

long straight uniformly
charged wire
Consider an infinitely long thin straight wire
with uniform linear charge density l. The wire
is obviously an axis of symmetry. Suppose we
take the radial vector from O to P and rotate it
around the wire. The points P, P¢, P¢¢ so
obtained are completely equivalent with
respect to the charged wire. This implies that
the electric field must have the same magnitude
at these points. The direction of electric field at
every point must be radial (outward if l > 0,
inward if l < 0). This is clear from Fig. 1.26.
Consider a pair of line elements P1 and P2
of the wire, as shown. The electric fields
produced by the two elements of the pair when
summed give a resultant electric field which
is radial (the components normal to the radial
vector cancel). This is true for any such pair
and hence the total field at any point P is
radial. Finally, since the wire is infinite,
electric field does not depend on the position
of P along the length of the wire. In short, the
electric field is everywhere radial in the plane
cutting the wire normally, and its magnitude
depends only on the radial distance r.
To calculate the field, imagine a cylindrical
Gaussian surface, as shown in the Fig. 1.26(b).
Since the field is everywhere radial, flux
through the two ends of the cylindrical
Gaussian surface is zero. At the cylindrical
FIGURE 1.26 (a) Electric field due to an
part of the surface, E is normal to the surface infinitely long thin straight wire is radial,
at every point, and its magnitude is constant, (b) The Gaussian surface for a long thin
since it depends only on r. The surface area wire of uniform linear charge density.
of the curved part is 2prl, where l is the length
of the cylinder. 33

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 Physics
Flux through the Gaussian surface
= flux through the curved cylindrical part of the surface
= E × 2prl
The surface includes charge equal to l l. Gauss’s law then gives
E × 2prl = ll/e0
λ
i.e.,E =
2πε 0r
Vectorially, E at any point is given by
λ
E= ˆ
n (1.32)
2πε0r
where n̂ is the radial unit vector in the plane normal to the wire passing
through the point. E is directed outward if l is positive and inward if l is
negative.
Note that when we write a vector A as a scalar multiplied by a unit
vector, i.e., as A = A â , the scalar A is an algebraic number. It can be
negative or positive. The direction of A will be the same as that of the unit
vector â if A > 0 and opposite to â if A < 0. When we want to restrict to
non-negative values, we use the symbol A and call it the modulus of A .
Thus, A ≥ 0 .
Also note that though only the charge enclosed by the surface (ll )
was included above, the electric field E is due to the charge on the entire
wire. Further, the assumption that the wire is infinitely long is crucial.
Without this assumption, we cannot take E to be normal to the curved
part of the cylindrical Gaussian surface. However, Eq. (1.32) is
approximately true for electric field around the central portions of a long
wire, where the end effects may be ignored.

1.14.2 Field due to a uniformly charged infinite plane sheet

Let s be the uniform surface charge density of an infinite plane sheet
(Fig. 1.27). We take the x-axis normal to the given plane. By symmetry,
the electric field will not depend on y and z coordinates and its direction
at every point must be parallel to the x-direction.
We can take the Gaussian surface to be a
rectangular parallelepiped of cross-sectional area
A, as shown. (A cylindrical surface will also do.) As
seen from the figure, only the two faces 1 and 2 will
contribute to the flux; electric field lines are parallel
to the other faces and they, therefore, do not
contribute to the total flux.
The unit vector normal to surface 1 is in –x
direction while the unit vector normal to surface 2
is in the +x direction. Therefore, flux E.DS through
both the surfaces are equal and add up. Therefore
FIGURE 1.27 Gaussian surface for a the net flux through the Gaussian surface is 2 EA.
uniformly charged infinite plane sheet.
The charge enclosed by the closed surface is sA.
34 Therefore by Gauss’s law,

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 Electric Charges
and Fields
2 EA = sA/e0
or, E = s/2e0
Vectorically,
σ
E= ˆ
n (1.33)
2ε 0
where n̂ is a unit vector normal to the plane and going away from it.
E is directed away from the plate if s is positive and toward the plate
if s is negative. Note that the above application of the Gauss’ law has
brought out an additional fact: E is independent of x also.
For a finite large planar sheet, Eq. (1.33) is approximately true in the
middle regions of the planar sheet, away from the ends.

1.14.3 Field due to a uniformly charged thin spherical shell

Let s be the uniform surface charge density of a thin spherical shell of
radius R (Fig. 1.28). The situation has obvious spherical symmetry. The
field at any point P, outside or inside, can depend only on r (the radial
distance from the centre of the shell to the point) and must be radial (i.e.,
along the radius vector).
(i) Field outside the shell: Consider a point P outside the
shell with radius vector r. To calculate E at P, we take the
Gaussian surface to be a sphere of radius r and with centre
O, passing through P. All points on this sphere are equivalent
relative to the given charged configuration. (That is what we
mean by spherical symmetry.) The electric field at each point
of the Gaussian surface, therefore, has the same magnitude
E and is along the radius vector at each point. Thus, E and
DS at every point are parallel and the flux through each
element is E DS. Summing over all DS, the flux through the
Gaussian surface is E × 4 p r 2. The charge enclosed is
s × 4 p R 2. By Gauss’s law
σ
E × 4 p r2 = 4 π R2
ε0

σ R2 q
Or, E = =
ε0 r 2
4 π ε0 r 2
where q = 4 p R2 s is the total charge on the spherical shell.
Vectorially,
q FIGURE 1.28 Gaussian
E= rˆ (1.34)
4 πε 0 r 2 surfaces for a point with
(a) r > R, (b) r < R.
The electric field is directed outward if q > 0 and inward if
q < 0. This, however, is exactly the field produced by a charge
q placed at the centre O. Thus for points outside the shell, the field due
to a uniformly charged shell is as if the entire charge of the shell is
concentrated at its centre.
(ii) Field inside the shell: In Fig. 1.28(b), the point P is inside the
shell. The Gaussian surface is again a sphere through P centred at O. 35

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 Physics
The flux through the Gaussian surface, calculated as before, is
E × 4 p r 2. However, in this case, the Gaussian surface encloses no
charge. Gauss’s law then gives
E × 4 p r2 = 0
i.e., E = 0 (r < R ) (1.35)
that is, the field due to a uniformly charged thin shell is zero at all points
inside the shell*. This important result is a direct consequence of Gauss’s
law which follows from Coulomb’s law. The experimental verification of
this result confirms the 1/r2 dependence in Coulomb’s law.

Example 1.12 An early model for an atom considered it to have a

positively charged point nucleus of charge Ze, surrounded by a
uniform density of negative charge up to a radius R. The atom as a
whole is neutral. For this model, what is the electric field at a distance
r from the nucleus?

FIGURE 1.29

Solution The charge distribution for this model of the atom is as

shown in Fig. 1.29. The total negative charge in the uniform spherical
charge distribution of radius R must be –Z e, since the atom (nucleus
of charge Z e + negative charge) is neutral. This immediately gives us
the negative charge density r, since we must have
4 πR3
ρ = 0 – Ze
3
3 Ze
or ρ= −
4 π R3
To find the electric field E(r) at a point P which is a distance r away
from the nucleus, we use Gauss’s law. Because of the spherical
symmetry of the charge distribution, the magnitude of the electric
field E(r) depends only on the radial distance, no matter what the
direction of r. Its direction is along (or opposite to) the radius vector r
from the origin to the point P. The obvious Gaussian surface is a
EXAMPLE 1.12

spherical surface centred at the nucleus. We consider two situations,

namely, r < R and r > R.
(i) r < R : The electric flux f enclosed by the spherical surface is
f = E (r ) × 4 p r 2

* Compare this with a uniform mass shell discussed in Section 7.5 of Class XI
36 Textbook of Physics.

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 Electric Charges
and Fields

where E (r ) is the magnitude of the electric field at r. This is because

the field at any point on the spherical Gaussian surface has the
same direction as the normal to the surface there, and has the same
magnitude at all points on the surface.
The charge q enclosed by the Gaussian surface is the positive nuclear
charge and the negative charge within the sphere of radius r,
4 πr3
i.e., q = Z e + ρ
3
Substituting for the charge density r obtained earlier, we have
r3
q = Ze−Ze
R3
Gauss’s law then gives,
Z e 1 r
E (r ) = − 3 ; r < R
4 π ε0 r 2
R
The electric field is directed radially outward.

EXAMPLE 1.12
(ii) r > R: In this case, the total charge enclosed by the Gaussian
spherical surface is zero since the atom is neutral. Thus, from Gauss’s
law,
E (r ) × 4 p r 2 = 0 or E (r ) = 0; r > R
At r = R, both cases give the same result: E = 0.

SUMMARY

1. Electric and magnetic forces determine the properties of atoms,

molecules and bulk matter.
2. From simple experiments on frictional electricity, one can infer that
there are two types of charges in nature; and that like charges repel
and unlike charges attract. By convention, the charge on a glass rod
rubbed with silk is positive; that on a plastic rod rubbed with fur is
then negative.
3. Conductors allow movement of electric charge through them,
insulators do not. In metals, the mobile charges are electrons; in
electrolytes both positive and negative ions are mobile.
4. Electric charge has three basic properties: quantisation, additivity
and conservation.
Quantisation of electric charge means that total charge (q) of a body
is always an integral multiple of a basic quantum of charge (e) i.e.,
q = n e, where n = 0, ±1, ±2, ±3, .... Proton and electron have charges
+e, –e, respectively. For macroscopic charges for which n is a very large
number, quantisation of charge can be ignored.
Additivity of electric charges means that the total charge of a system
is the algebraic sum (i.e., the sum taking into account proper signs)
of all individual charges in the system.
Conservation of electric charges means that the total charge of an
isolated system remains unchanged with time. This means that when
37

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 Physics
bodies are charged through friction, there is a transfer of electric charge
from one body to another, but no cr eation or destruction
of charge.
5. Coulomb’s Law: The mutual electrostatic force between two point
charges q1 and q2 is proportional to the product q1q2 and inversely
proportional to the square of the distance r 21 separating them.
Mathematically,
k (q1q2 )
F21 = force on q2 due to q1 = 2
rˆ21
r21
1
where r̂21 is a unit vector in the direction from q1 to q2 and k =
4 πε 0
is the constant of proportionality.
In SI units, the unit of charge is coulomb. The experimental value of
the constant e0 is
e0 = 8.854 × 10–12 C2 N–1 m–2
The approximate value of k is
k = 9 × 109 N m2 C–2
6. The ratio of electric force and gravitational force between a proton
and an electron is

k e2
≅ 2.4 × 1039
G m em p
7. Superposition Principle: The principle is based on the property that the
forces with which two charges attract or repel each other are not
affected by the presence of a third (or more) additional charge(s). For
an assembly of charges q1, q2, q3, ..., the force on any charge, say q1, is
the vector sum of the force on q1 due to q2, the force on q1 due to q3,
and so on. For each pair, the force is given by the Coulomb’s law for
two charges stated earlier.
8. The electric field E at a point due to a charge configuration is the
force on a small positive test charge q placed at the point divided by
the magnitude of the charge. Electric field due to a point charge q has
a magnitude |q|/4pe0r 2; it is radially outwards from q, if q is positive,
and radially inwards if q is negative. Like Coulomb force, electric field
also satisfies superposition principle.
9. An electric field line is a curve drawn in such a way that the tangent
at each point on the curve gives the direction of electric field at that
point. The relative closeness of field lines indicates the relative strength
of electric field at different points; they crowd near each other in regions
of strong electric field and are far apart where the electric field is
weak. In regions of constant electric field, the field lines are uniformly
spaced parallel straight lines.
10. Some of the important properties of field lines are: (i) Field lines are
continuous curves without any breaks. (ii) Two field lines cannot cross
each other. (iii) Electrostatic field lines start at positive charges and
end at negative charges —they cannot form closed loops.
11. An electric dipole is a pair of equal and opposite charges q and –q
separated by some distance 2a. Its dipole moment vector p has
magnitude 2qa and is in the direction of the dipole axis from –q to q.

38

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 Electric Charges
and Fields

12. Field of an electric dipole in its equatorial plane (i.e., the plane
perpendicular to its axis and passing through its centre) at a distance
r from the centre:

−p 1
E=
4 πε o (a + r 2 )3 / 2
2

−p
≅ , for r >> a
4 πε o r 3
Dipole electric field on the axis at a distance r from the centre:

2 pr
E =
4 πε 0 (r 2 − a 2 )2

2p
≅ for r >> a
4 π ε 0r 3
The 1/r 3 dependence of dipole electric fields should be noted in contrast
to the 1/r 2 dependence of electric field due to a point charge.
13. In a uniform electric field E, a dipole experiences a torque τ given by
τ =p×E
but experiences no net force.
14. The flux Df of electric field E through a small area element DS is
given by
Df = E.DS
The vector area element DS is

DS = DS n̂
where DS is the magnitude of the area element and n̂ is normal to the
area element, which can be considered planar for sufficiently small DS.
For an area element of a closed surface, n̂ is taken to be the direction
of outward normal, by convention.
15. Gauss’s law: The flux of electric field through any closed surface S is
1/e0 times the total charge enclosed by S. The law is especially useful
in determining electric field E, when the source distribution has simple
symmetry:
(i) Thin infinitely long straight wire of uniform linear charge density l
λ
E= ˆ
n
2 πε 0 r
where r is the perpendicular distance of the point from the wire and
n̂ is the radial unit vector in the plane normal to the wire passing
through the point.
(ii) Infinite thin plane sheet of uniform surface charge density s
σ
E= ˆ
n
2 ε0

where n̂ is a unit vector normal to the plane, outward on either side.

39

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 Physics
(iii) Thin spherical shell of uniform surface charge density s
q
E= rˆ (r ≥ R )
4 πε 0 r 2
E=0 (r < R )
where r is the distance of the point from the centre of the shell and R
the radius of the shell. q is the total charge of the shell: q = 4pR2s.
The electric field outside the shell is as though the total charge is
concentrated at the centre. The same result is true for a solid sphere
of uniform volume charge density. The field is zero at all points inside
the shell.

Physical quantity Symbol Dimensions Unit Remarks

Vector area element DS [L2] m2 DS = DS n̂

Electric field E [MLT–3A–1] V m–1

Electric flux f [ML3 T–3A–1] Vm Df = E.DS

Dipole moment p [LTA] Cm Vector directed

from negative to
positive charge

Charge density:

linear l [L–1 TA] C m–1 Charge/length

surface s [L–2 TA] C m–2 Charge/area

volume r [L–3 TA] C m–3 Charge/volume

POINTS TO PONDER

1. You might wonder why the protons, all carrying positive charges,
are compactly residing inside the nucleus. Why do they not fly away?
You will learn that there is a third kind of a fundamental force,
called the strong force which holds them together. The range of
distance where this force is effective is, however, very small ~10-14
m. This is precisely the size of the nucleus. Also the electrons are
not allowed to sit on top of the protons, i.e. inside the nucleus,
due to the laws of quantum mechanics. This gives the atoms their
structure as they exist in nature.
2. Coulomb force and gravitational force follow the same inverse-square
law. But gravitational force has only one sign (always attractive), while

40

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 Electric Charges
and Fields

Coulomb force can be of both signs (attractive and repulsive), allowing

possibility of cancellation of electric forces. This is how gravity, despite
being a much weaker force, can be a dominating and more pervasive
force in nature.
3. The constant of proportionality k in Coulomb’s law is a matter of
choice if the unit of charge is to be defined using Coulomb’s law. In SI
units, however, what is defined is the unit of current (A) via its magnetic
effect (Ampere’s law) and the unit of charge (coulomb) is simply defined
by (1C = 1 A s). In this case, the value of k is no longer arbitrary; it is
approximately 9 × 109 N m2 C–2.
4. The rather large value of k, i.e., the large size of the unit of charge
(1C) from the point of view of electric effects arises because (as
mentioned in point 3 already) the unit of charge is defined in terms of
magnetic forces (forces on current–carrying wires) which are generally
much weaker than the electric forces. Thus while 1 ampere is a unit
of reasonable size for magnetic effects, 1 C = 1 A s, is too big a unit for
electric effects.
5. The additive property of charge is not an ‘obvious’ property. It is related
to the fact that electric charge has no direction associated with it;
charge is a scalar.
6. Charge is not only a scalar (or invariant) under rotation; it is also
invariant for frames of reference in relative motion. This is not always
true for every scalar. For example, kinetic energy is a scalar under
rotation, but is not invariant for frames of reference in relative
motion.
7. Conservation of total charge of an isolated system is a property
independent of the scalar nature of charge noted in point 6.
Conservation refers to invariance in time in a given frame of reference.
A quantity may be scalar but not conserved (like kinetic energy in an
inelastic collision). On the other hand, one can have conserved vector
quantity (e.g., angular momentum of an isolated system).
8. Quantisation of electric charge is a basic (unexplained) law of nature;
interestingly, there is no analogous law on quantisation of mass.
9. Superposition principle should not be regarded as ‘obvious’, or
equated with the law of addition of vectors. It says two things:
force on one charge due to another charge is unaffected by the
presence of other charges, and there are no additional three-body,
four -body, etc., forces which arise only when there are more than
two charges.
10. The electric field due to a discrete charge configuration is not defined
at the locations of the discrete charges. For continuous volume
charge distribution, it is defined at any point in the distribution.
For a surface charge distribution, electric field is discontinuous
across the surface.
11. The electric field due to a charge configuration with total charge zero
is not zero; but for distances large compared to the size of
the configuration, its field falls off faster than 1/r 2, typical of field
due to a single charge. An electric dipole is the simplest example of
this fact.

41

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 Physics
EXERCISES
1.1 What is the force between two small charged spheres having
charges of 2 × 10–7C and 3 × 10–7C placed 30 cm apart in air?
1.2 The electrostatic force on a small sphere of charge 0.4 mC due to
another small sphere of charge – 0.8 mC in air is 0.2 N. (a) What is
the distance between the two spheres? (b) What is the force on the
second sphere due to the first?
1.3 Check that the ratio ke2/G memp is dimensionless. Look up a Table
of Physical Constants and determine the value of this ratio. What
does the ratio signify?
1.4 (a) Explain the meaning of the statement ‘electric charge of a body
is quantised’.
(b) Why can one ignore quantisation of electric charge when dealing
with macroscopic i.e., large scale charges?
1.5 When a glass rod is rubbed with a silk cloth, charges appear on
both. A similar phenomenon is observed with many other pairs of
bodies. Explain how this observation is consistent with the law of
conservation of charge.
1.6 Four point charges qA = 2 mC, qB = –5 mC, qC = 2 mC, and qD = –5 mC are
located at the corners of a square ABCD of side 10 cm. What is the
force on a charge of 1 mC placed at the centre of the square?
1.7 (a) An electrostatic field line is a continuous curve. That is, a field
line cannot have sudden breaks. Why not?
(b) Explain why two field lines never cross each other at any point?
1.8 Two point charges qA = 3 mC and qB = –3 mC are located 20 cm apart
in vacuum.
(a) What is the electric field at the midpoint O of the line AB joining
the two charges?
(b) If a negative test charge of magnitude 1.5 × 10–9 C is placed at
this point, what is the force experienced by the test charge?
1.9 A system has two charges qA = 2.5 × 10–7 C and qB = –2.5 × 10–7 C
located at points A: (0, 0, –15 cm) and B: (0,0, +15 cm), respectively.
What are the total charge and electric dipole moment of the system?
1.10 An electric dipole with dipole moment 4 × 10–9 C m is aligned at 30°
with the direction of a uniform electric field of magnitude 5 × 104 NC–1.
Calculate the magnitude of the torque acting on the dipole.
1.11 A polythene piece rubbed with wool is found to have a negative
charge of 3 × 10–7 C.
(a) Estimate the number of electrons transferred (from which to
which?)
(b) Is there a transfer of mass from wool to polythene?
1.12 (a) Two insulated charged copper spheres A and B have their centres
separated by a distance of 50 cm. What is the mutual force of
electrostatic repulsion if the charge on each is 6.5 × 10–7 C? The
radii of A and B are negligible compared to the distance of
separation.
(b) What is the force of repulsion if each sphere is charged double
the above amount, and the distance between them is halved?
1.13 Figure 1.30 shows tracks of three charged particles in a uniform
electrostatic field. Give the signs of the three charges. Which particle
42 has the highest charge to mass ratio?

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 Electric Charges
and Fields

FIGURE 1.30

1.14 Consider a uniform electric field E = 3 × 103 î N/C. (a) What is the
flux of this field through a square of 10 cm on a side whose plane is
parallel to the yz plane? (b) What is the flux through the same
square if the normal to its plane makes a 60° angle with the x-axis?
1.15 What is the net flux of the uniform electric field of Exercise 1.14
through a cube of side 20 cm oriented so that its faces are parallel
to the coordinate planes?
1.16 Careful measurement of the electric field at the surface of a black
box indicates that the net outward flux through the surface of the
box is 8.0 × 103 Nm2/C. (a) What is the net charge inside the box?
(b) If the net outward flux through the surface of the box were zero,
could you conclude that there were no charges inside the box? Why
or Why not?
1.17 A point charge +10 mC is a distance 5 cm directly above the centre
of a square of side 10 cm, as shown in Fig. 1.31. What is the
magnitude of the electric flux through the square? (Hint: Think of
the square as one face of a cube with edge 10 cm.)

FIGURE 1.31

1.18 A point charge of 2.0 mC is at the centre of a cubic Gaussian

surface 9.0 cm on edge. What is the net electric flux through the
surface?
1.19 A point charge causes an electric flux of –1.0 × 103 Nm2/C to pass
through a spherical Gaussian surface of 10.0 cm radius centred on
the charge. (a) If the radius of the Gaussian surface were doubled,
how much flux would pass through the surface? (b) What is the
value of the point charge?
1.20 A conducting sphere of radius 10 cm has an unknown charge. If
the electric field 20 cm from the centre of the sphere is 1.5 × 103 N/C
and points radially inward, what is the net charge on the sphere? 43

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 Physics
1.21 A uniformly charged conducting sphere of 2.4 m diameter has a
surface charge density of 80.0 mC/m2. (a) Find the charge on the
sphere. (b) What is the total electric flux leaving the surface of the
sphere?
1.22 An infinite line charge produces a field of 9 × 104 N/C at a distance
of 2 cm. Calculate the linear charge density.
1.23 Two large, thin metal plates are parallel and close to each other. On
their inner faces, the plates have surface charge densities of opposite
signs and of magnitude 17.0 × 10–22 C/m2. What is E: (a) in the outer
region of the first plate, (b) in the outer region of the second plate,
and (c) between the plates?

44

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Chapter Two

ELECTROSTATIC
POTENTIAL AND
CAPACITANCE

2.1 INTRODUCTION
In Chapters 6 and 8 (Class XI), the notion of potential energy was
introduced. When an external force does work in taking a body from a
point to another against a force like spring force or gravitational force,
that work gets stored as potential energy of the body. When the external
force is removed, the body moves, gaining kinetic energy and losing
an equal amount of potential energy. The sum of kinetic and
potential energies is thus conserved. Forces of this kind are called
conservative forces. Spring force and gravitational force are examples of
conservative forces.
Coulomb force between two (stationary) charges is also a conservative
force. This is not surprising, since both have inverse-square dependence
on distance and differ mainly in the proportionality constants – the
masses in the gravitational law are replaced by charges in Coulomb’s
law. Thus, like the potential energy of a mass in a gravitational
field, we can define electrostatic potential energy of a charge in an
electrostatic field.
Consider an electrostatic field E due to some charge configuration.
First, for simplicity, consider the field E due to a charge Q placed at the
origin. Now, imagine that we bring a test charge q from a point R to a
point P against the repulsive force on it due to the charge Q. With reference

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 Physics
to Fig. 2.1, this will happen if Q and q are both positive
or both negative. For definiteness, let us take Q, q > 0.
Two remarks may be made here. First, we assume
that the test charge q is so small that it does not disturb
the original configuration, namely the charge Q at the
origin (or else, we keep Q fixed at the origin by some
FIGURE 2.1 A test charge q (> 0) is unspecified force). Second, in bringing the charge q from
moved from the point R to the R to P, we apply an external force Fext just enough to
point P against the repulsive counter the repulsive electric force FE (i.e, Fext= –FE).
force on it by the charge Q (> 0) This means there is no net force on or acceleration of
placed at the origin. the charge q when it is brought from R to P, i.e., it is
brought with infinitesimally slow constant speed. In
this situation, work done by the external force is the negative of the work
done by the electric force, and gets fully stored in the form of potential
energy of the charge q. If the external force is removed on reaching P, the
electric force will take the charge away from Q – the stored energy (potential
energy) at P is used to provide kinetic energy to the charge q in such a
way that the sum of the kinetic and potential energies is conserved.
Thus, work done by external forces in moving a charge q from R to P is

WRP =

= – (2.1)

This work done is against electrostatic repulsive force and gets stored
as potential energy.
At every point in electric field, a particle with charge q possesses a
certain electrostatic potential energy, this work done increases its potential
energy by an amount equal to potential energy difference between points
R and P.
Thus, potential energy difference
∆U = U P − U R = WRP (2.2)
(Note here that this displacement is in an opposite sense to the electric
force and hence work done by electric field is negative, i.e., –WRP .)
Therefore, we can define electric potential energy difference between
two points as the work required to be done by an external force in moving
(without accelerating ) charge q from one point to another for electric field
of any arbitrary charge configuration.
Two important comments may be made at this stage:
(i) The right side of Eq. (2.2) depends only on the initial and final positions
of the charge. It means that the work done by an electrostatic field in
moving a charge from one point to another depends only on the initial
and the final points and is independent of the path taken to go from
one point to the other. This is the fundamental characteristic of a
conservative force. The concept of the potential energy would not be
meaningful if the work depended on the path. The path-independence
of work done by an electrostatic field can be proved using the
46 Coulomb’s law. We omit this proof here.

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 Electrostatic Potential
and Capacitance
(ii) Equation (2.2) defines potential energy difference in terms
of the physically meaningful quantity work. Clearly,
potential energy so defined is undetermined to within an
additive constant.What this means is that the actual value
of potential energy is not physically significant; it is only
the difference of potential energy that is significant. We can
always add an arbitrary constant a to potential energy at
every point, since this will not change the potential energy
difference:
(U P + α ) − (U R + α ) = U P − U R
Put it differently, there is a freedom in choosing the point
where potential energy is zero. A convenient choice is to have
electrostatic potential energy zero at infinity. With this choice,
if we take the point R at infinity, we get from Eq. (2.2) Count Alessandro Volta

COUNT ALESSANDRO VOLTA (1745 –1827)

(1745 – 1827) Italian
W ∞P = U P − U ∞ = U P (2.3) physicist, professor at
Since the point P is arbitrary, Eq. (2.3) provides us with a Pavia. Volta established
definition of potential energy of a charge q at any point. that the animal electri-
Potential energy of charge q at a point (in the presence of field city observed by Luigi
Galvani, 1737–1798, in
due to any charge configuration) is the work done by the
experiments with frog
external force (equal and opposite to the electric force) in
muscle tissue placed in
bringing the charge q from infinity to that point. contact with dissimilar
metals, was not due to
2.2 ELECTROSTATIC POTENTIAL any exceptional property
Consider any general static charge configuration. We define of animal tissues but
was also generated
potential energy of a test charge q in terms of the work done
whenever any wet body
on the charge q. This work is obviously proportional to q, since was sandwiched between
the force at any point is qE, where E is the electric field at that dissimilar metals. This
point due to the given charge configuration. It is, therefore, led him to develop the
convenient to divide the work by the amount of charge q, so first voltaic pile, or
that the resulting quantity is independent of q. In other words, battery, consisting of a
work done per unit test charge is characteristic of the electric large stack of moist disks
field associated with the charge configuration. This leads to of cardboard (electro-
lyte) sandwiched
the idea of electrostatic potential V due to a given charge
between disks of metal
configuration. From Eq. (2.1), we get: (electrodes).
Work done by external force in bringing a unit positive
charge from point R to P

 U −UR 
= VP – VR  = P  (2.4)
 q

where VP and VR are the electrostatic potentials at P and R, respectively.

Note, as before, that it is not the actual value of potential but the potential
difference that is physically significant. If, as before, we choose the
potential to be zero at infinity, Eq. (2.4) implies:
Work done by an external force in bringing a unit positive charge
from infinity to a point = electrostatic potential (V ) at that point. 47

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 Physics
In other words, the electrostatic potential (V )
at any point in a region with electrostatic field is
the work done in bringing a unit positive
charge (without acceleration) from infinity to
that point.
The qualifying remarks made earlier regarding
potential energy also apply to the definition of
potential. To obtain the work done per unit test
charge, we should take an infinitesimal test charge
FIGURE 2.2 Work done on a test charge q dq, obtain the work done dW in bringing it from
by the electrostatic field due to any given infinity to the point and determine the ratio
charge configuration is independent dW/dq. Also, the external force at every point of the
of the path, and depends only on
path is to be equal and opposite to the electrostatic
its initial and final positions.
force on the test charge at that point.

2.3 POTENTIAL DUE TO A POINT CHARGE

Consider a point charge Q at the origin (Fig. 2.3). For definiteness, take Q
to be positive. We wish to determine the potential at any point P with
position vector r from the origin. For that we must
calculate the work done in bringing a unit positive
test charge from infinity to the point P. For Q > 0,
the work done against the repulsive force on the
test charge is positive. Since work done is
independent of the path, we choose a convenient
path – along the radial direction from infinity to
the point P.
At some intermediate point P¢ on the path, the
electrostatic force on a unit positive charge is
FIGURE 2.3 Work done in bringing a unit
positive test charge from infinity to the Q ×1
rˆ ′ (2.5)
point P, against the repulsive force of 4 πε 0r '2
charge Q (Q > 0), is the potential at P due to
the charge Q. where rˆ ′ is the unit vector along OP¢. Work done
against this force from r¢ to r¢ + Dr¢ is
Q
∆W = − ∆r ′ (2.6)
4 πε 0r '2
The negative sign appears because for Dr ¢ < 0, DW is positive. Total
work done (W) by the external force is obtained by integrating Eq. (2.6)
from r¢ = ¥ to r¢ = r,

r r
Q Q Q
W = −∫ dr ′ = = (2.7)
∞
4 πε 0r ′ 2
4 πε 0r ′ ∞ 4 πε 0r

This, by definition is the potential at P due to the charge Q

Q
48 V (r ) = (2.8)
4 πε 0r

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 Electrostatic Potential
and Capacitance
Equation (2.8) is true for any
sign of the charge Q, though we
considered Q > 0 in its derivation.
For Q < 0, V < 0, i.e., work done (by
the external force) per unit positive
test charge in bringing it from
infinity to the point is negative. This
is equivalent to saying that work
done by the electrostatic force in
bringing the unit positive charge
form infinity to the point P is
positive. [This is as it should be,
since for Q < 0, the force on a unit
positive test charge is attractive, so
that the electrostatic force and the
displacement (from infinity to P) are FIGURE 2.4 Variation of potential V with r [in units of
in the same direction.] Finally, we (Q/4pe0) m-1] (blue curve) and field with r [in units
of (Q/4pe0) m-2] (black curve) for a point charge Q.
note that Eq. (2.8) is consistent with
the choice that potential at infinity
be zero.
Figure (2.4) shows how the electrostatic potential (  1/r ) and the
electrostatic field (  1/r 2 ) varies with r.

Example 2.1
(a) Calculate the potential at a point P due to a charge of 4 × 10–7C
located 9 cm away.
(b) Hence obtain the work done in bringing a charge of 2 × 10–9 C
from infinity to the point P. Does the answer depend on the path
along which the charge is brought?
Solution

(a)

= 4 × 104 V
(b) W = qV = 2 × 10–9C × 4 × 104V
= 8 × 10–5 J
EXAMPLE 2.1

No, work done will be path independent. Any arbitrary infinitesimal

path can be resolved into two perpendicular displacements: One along
r and another perpendicular to r. The work done corresponding to
the later will be zero.

2.4 POTENTIAL DUE TO AN ELECTRIC DIPOLE

As we learnt in the last chapter, an electric dipole consists of two charges
q and –q separated by a (small) distance 2a. Its total charge is zero. It is
characterised by a dipole moment vector p whose magnitude is q × 2a
and which points in the direction from –q to q (Fig. 2.5). We also saw that
the electric field of a dipole at a point with position vector r depends not
just on the magnitude r, but also on the angle between r and p. Further, 49

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 Physics
the field falls off, at large distance, not as
1/r 2 (typical of field due to a single charge)
but as 1/r 3. We, now, determine the electric
potential due to a dipole and contrast it
with the potential due to a single charge.
As before, we take the origin at the
centre of the dipole. Now we know that the
electric field obeys the superposition
principle. Since potential is related to the
work done by the field, electrostatic
potential also follows the superposition
principle. Thus, the potential due to the
dipole is the sum of potentials due to the
charges q and –q

1 q q
V = −
4 πε 0  r1 r2 
(2.9)
FIGURE 2.5 Quantities involved in the calculation
of potential due to a dipole.
where r1 and r2 are the distances of the
point P from q and –q, respectively.
Now, by geometry,
r12 = r 2 + a 2 − 2ar cosq

r22 = r 2 + a 2 + 2ar cosq (2.10)

We take r much greater than a ( r  a ) and retain terms only upto
the first order in a/r
 2a cos θ a 2 
r12 = r 2 1 − + 2 
 r r 

 2a cos θ 
≅ r 2 1 −  (2.11)
 r
Similarly,
 2a cos θ 
r22 ≅ r 2 1 +  (2.12)
 r
Using the Binomial theorem and retaining terms upto the first order
in a/r ; we obtain,
− 1/ 2
1 1 2a cos θ  1 a 
≅ 1 −  ≅ 1 + cos θ  [2.13(a)]
r1 r  r r r
− 1/ 2
1 1 2a cos θ  1 a 
≅ 1 +  ≅ 1 − cos θ  [2.13(b)]
r2 r  r r r
Using Eqs. (2.9) and (2.13) and p = 2qa, we get
q 2acosθ p cos θ
V = = (2.14)
4 πε 0 r2 4 πε 0r 2
50 Now, p cos q = p.r̂

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 Electrostatic Potential
and Capacitance
where r̂ is the unit vector along the position vector OP.
The electric potential of a dipole is then given by
1 p.r̂
V = ; (r >> a) (2.15)
4 πε 0 r 2
Equation (2.15) is, as indicated, approximately true only for distances
large compared to the size of the dipole, so that higher order terms in
a/r are negligible. For a point dipole p at the origin, Eq. (2.15) is, however,
exact.
From Eq. (2.15), potential on the dipole axis (q = 0, p ) is given by
1 p
V =± (2.16)
4 πε 0 r 2
(Positive sign for q = 0, negative sign for q = p.) The potential in the
equatorial plane (q = p/2) is zero.
The important contrasting features of electric potential of a dipole
from that due to a single charge are clear from Eqs. (2.8) and (2.15):
(i) The potential due to a dipole depends not just on r but also on the
angle between the position vector r and the dipole moment vector p.
(It is, however, axially symmetric about p. That is, if you rotate the
position vector r about p, keeping q fixed, the points corresponding
to P on the cone so generated will have the same potential as at P.)
(ii) The electric dipole potential falls off, at large distance, as 1/r 2, not as
1/r, characteristic of the potential due to a single charge. (You can
refer to the Fig. 2.5 for graphs of 1/r 2 versus r and 1/r versus r,
drawn there in another context.)

2.5 POTENTIAL DUE TO A SYSTEM OF CHARGES

Consider a system of charges q1, q2,…, qn with position vectors r1, r2,…,
rn relative to some origin (Fig. 2.6). The potential V1 at P due to the charge
q1 is
1 q1
V1 =
4 πε 0 r1P
where r1P is the distance between q1 and P.
Similarly, the potential V2 at P due to q2 and
V3 due to q3 are given by
1 q2 1 q3
V2 = , V3 =
4 πε 0 r2P 4 πε 0 r3P
where r2P and r3P are the distances of P from
charges q2 and q3, respectively; and so on for the
potential due to other charges. By the FIGURE 2.6 Potential at a point due to a
superposition principle, the potential V at P due system of charges is the sum of potentials
to the total charge configuration is the algebraic due to individual charges.
sum of the potentials due to the individual
charges
V = V1 + V2 + ... + Vn (2.17) 51

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 Physics
1  q1 q 2 q 
=  + + ...... + n  (2.18)
4 πε 0  r1P r2 P rnP 
If we have a continuous charge distribution characterised by a charge
density r (r), we divide it, as before, into small volume elements each of
size Dv and carrying a charge r Dv. We then determine the potential due
to each volume element and sum (strictly speaking , integrate) over all
such contributions, and thus determine the potential due to the entire
distribution.
We have seen in Chapter 1 that for a uniformly charged spherical shell,
the electric field outside the shell is as if the entire charge is concentrated
at the centre. Thus, the potential outside the shell is given by
1 q
V = (r ≥ R ) [2.19(a)]
4 πε0 r
where q is the total charge on the shell and R its radius. The electric field
inside the shell is zero. This implies (Section 2.6) that potential is constant
inside the shell (as no work is done in moving a charge inside the shell),
and, therefore, equals its value at the surface, which is
1 q
V = [2.19(b)]
4 πε 0 R

Example 2.2 Two charges 3 × 10–8 C and –2 × 10–8 C are located

15 cm apart. At what point on the line joining the two charges is the
electric potential zero? Take the potential at infinity to be zero.
Solution Let us take the origin O at the location of the positive charge.
The line joining the two charges is taken to be the x-axis; the negative
charge is taken to be on the right side of the origin (Fig. 2.7).

FIGURE 2.7

Let P be the required point on the x-axis where the potential is zero.
If x is the x-coordinate of P, obviously x must be positive. (There is no
possibility of potentials due to the two charges adding up to zero for
x < 0.) If x lies between O and A, we have

1  3 × 10 –8 2 × 10 –8 
− =0
4 πε 0  x × 10 (15 − x ) × 10 –2 
–2

where x is in cm. That is,

3 2
− =0
x 15 − x
EXAMPLE 2.2

which gives x = 9 cm.

If x lies on the extended line OA, the required condition is
3 2
− =0
x x − 15
52

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 Electrostatic Potential
and Capacitance

which gives

EXAMPLE 2.2
x = 45 cm
Thus, electric potential is zero at 9 cm and 45 cm away from the
positive charge on the side of the negative charge. Note that the
formula for potential used in the calculation required choosing
potential to be zero at infinity.

Example 2.3 Figures 2.8 (a) and (b) show the field lines of a positive
and negative point charge respectively.

equipotential-sufaces-12584/
http://video.mit.edu/watch/4-electrostatic-potential-elctric-energy-ev-conservative-field-
Electric potential, equipotential surfaces:
FIGURE 2.8

(a) Give the signs of the potential difference VP – VQ; VB – VA.

(b) Give the sign of the potential energy difference of a small negative
charge between the points Q and P; A and B.
(c) Give the sign of the work done by the field in moving a small
positive charge from Q to P.
(d) Give the sign of the work done by the external agency in moving
a small negative charge from B to A.
(e) Does the kinetic energy of a small negative charge increase or
decrease in going from B to A?
Solution
1
(a) As V ∝ , VP > VQ. Thus, (VP – VQ ) is positive. Also VB is less negative
r
than VA . Thus, VB > VA or (VB – VA) is positive.
(b) A small negative charge will be attracted towards positive charge.
The negative charge moves from higher potential energy to lower
potential energy. Therefore the sign of potential energy difference
of a small negative charge between Q and P is positive.
Similarly, (P.E.) A > (P.E.) B and hence sign of potential energy
differences is positive.
(c) In moving a small positive charge from Q to P, work has to be
done by an external agency against the electric field. Therefore,
work done by the field is negative.
EXAMPLE 2.3

(d) In moving a small negative charge from B to A work has to be

done by the external agency. It is positive.
(e) Due to force of repulsion on the negative charge, velocity decreases
and hence the kinetic energy decreases in going from B to A.
53

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2.6 EQUIPOTENTIAL SURFACES
An equipotential surface is a surface with a constant value of potential
at all points on the surface. For a single charge q, the potential is given
by Eq. (2.8):
1 q
V=
4 πεo r
This shows that V is a constant if r is constant. Thus, equipotential
surfaces of a single point charge are concentric spherical surfaces centred
at the charge.
Now the electric field lines for a single charge q are radial lines starting
from or ending at the charge, depending on whether q is positive or negative.
Clearly, the electric field at every point is normal to the equipotential surface
passing through that point. This is true in general: for any charge
configuration, equipotential surface through a point is normal to the
electric field at that point. The proof of this statement is simple.
If the field were not normal to the equipotential surface, it would
have non-zero component along the surface. To move a unit test charge
against the direction of the component of the field, work would have to
be done. But this is in contradiction to the definition of an equipotential
FIGURE 2.9 For a surface: there is no potential difference between any two points on the
single charge q surface and no work is required to move a test charge on the surface.
(a) equipotential The electric field must, therefore, be normal to the equipotential surface
surfaces are at every point. Equipotential surfaces offer an alternative visual picture
spherical surfaces in addition to the picture of electric field lines around a charge
centred at the
configuration.
charge, and
(b) electric field
lines are radial,
starting from the
charge if q > 0.

FIGURE 2.10 Equipotential surfaces for a uniform electric field.

For a uniform electric field E, say, along the x-axis, the equipotential
surfaces are planes normal to the x-axis, i.e., planes parallel to the y-z
plane (Fig. 2.10). Equipotential surfaces for (a) a dipole and (b) two
identical positive charges are shown in Fig. 2.11.

FIGURE 2.11 Some equipotential surfaces for (a) a dipole,

54 (b) two identical positive charges.

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 Electrostatic Potential
and Capacitance
2.6.1 Relation between field and potential
Consider two closely spaced equipotential surfaces A and B (Fig. 2.12)
with potential values V and V + d V, where d V is the change in V in the
direction of the electric field E. Let P be a point on the
surface B. d l is the perpendicular distance of the
surface A from P. Imagine that a unit positive charge
is moved along this perpendicular from the surface B
to surface A against the electric field. The work done
in this process is |E|d l.
This work equals the potential difference
VA–VB.
Thus,
|E|d l = V – (V + dV )= – dV

δV
i.e., |E|= − (2.20)
δl
Since dV is negative, dV = – |dV|. we can rewrite FIGURE 2.12 From the
Eq (2.20) as potential to the field.

δV δV
E =− =+ (2.21)
δl δl
We thus arrive at two important conclusions concerning the relation
between electric field and potential:
(i) Electric field is in the direction in which the potential decreases
steepest.
(ii) Its magnitude is given by the change in the magnitude of potential
per unit displacement normal to the equipotential surface at the point.

2.7 POTENTIAL ENERGY OF A SYSTEM OF CHARGES

Consider first the simple case of two charges q1and q2 with position vector
r1 and r2 relative to some origin. Let us calculate the work done
(externally) in building up this configuration. This means that we consider
the charges q1 and q2 initially at infinity and determine the work done by
an external agency to bring the charges to the given locations. Suppose,
first the charge q1 is brought from infinity to the point r1. There is no
external field against which work needs to be done, so work done in
bringing q1 from infinity to r1 is zero. This charge produces a potential in
space given by
1 q1
V1 =
4 πε 0 r1P
where r1P is the distance of a point P in space from the location of q1.
From the definition of potential, work done in bringing charge q2 from
infinity to the point r2 is q2 times the potential at r2 due to q1:
1 q1q2
work done on q2 =
4 πε 0 r12 55

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 Physics
where r12 is the distance between points 1 and 2.
Since electrostatic force is conservative, this work gets
stored in the form of potential energy of the system. Thus,
the potential energy of a system of two charges q1 and q2 is
FIGURE 2.13 Potential energy of a 1 q1q 2
U = (2.22)
system of charges q1 and q2 is 4 πε 0 r12
directly proportional to the product
of charges and inversely to the Obviously, if q2 was brought first to its present location and
distance between them. q1 brought later, the potential energy U would be the same.
More generally, the potential energy expression,
Eq. (2.22), is unaltered whatever way the charges are brought to the specified
locations, because of path-independence of work for electrostatic force.
Equation (2.22) is true for any sign of q1and q2. If q1q2 > 0, potential
energy is positive. This is as expected, since for like charges (q1q2 > 0),
electrostatic force is repulsive and a positive amount of work is needed to
be done against this force to bring the charges from infinity to a finite
distance apart. For unlike charges (q1 q2 < 0), the electrostatic force is
attractive. In that case, a positive amount of work is needed against this
force to take the charges from the given location to infinity. In other words,
a negative amount of work is needed for the reverse path (from infinity to
the present locations), so the potential energy is negative.
Equation (2.22) is easily generalised for a system of any number of
point charges. Let us calculate the potential energy of a system of three
charges q1, q2 and q3 located at r1, r2, r3, respectively. To bring q1 first
from infinity to r1, no work is required. Next we bring q2 from infinity to
r2. As before, work done in this step is
1 q1q2
q2V1( r2 ) = (2.23)
4 πε 0 r12
The charges q1 and q2 produce a potential, which at any point P is
given by

1  q1 q 2 
V1, 2 = + (2.24)
4 πε 0  r1P r2 P 
Work done next in bringing q3 from infinity to the point r3 is q3 times
V1, 2 at r3

1  q1q3 q 2q 3 
q3V1, 2 ( r3 ) = + (2.25)
4 πε 0  r13 r23 
The total work done in assembling the charges
at the given locations is obtained by adding the work
done in different steps [Eq. (2.23) and Eq. (2.25)],

1  q1q 2 q1q 3 q 2q 3 
U = + + (2.26)
FIGURE 2.14 Potential energy of a 4 πε 0  r12 r13 r23 
system of three charges is given by Again, because of the conservative nature of the
Eq. (2.26), with the notation given
electrostatic force (or equivalently, the path
in the figure.
independence of work done), the final expression for
U, Eq. (2.26), is independent of the manner in which
56 the configuration is assembled. The potential energy

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 Electrostatic Potential
and Capacitance
is characteristic of the present state of configuration, and not the way
the state is achieved.

Example 2.4 Four charges are arranged at the corners of a square

ABCD of side d, as shown in Fig. 2.15.(a) Find the work required to
put together this arrangement. (b) A charge q0 is brought to the centre
E of the square, the four charges being held fixed at its corners. How
much extra work is needed to do this?

FIGURE 2.15

Solution
(a) Since the work done depends on the final arrangement of the
charges, and not on how they are put together, we calculate work
needed for one way of putting the charges at A, B, C and D. Suppose,
first the charge +q is brought to A, and then the charges –q, +q, and
–q are brought to B, C and D, respectively. The total work needed can
be calculated in steps:
(i) Work needed to bring charge +q to A when no charge is present
elsewhere: this is zero.
(ii) Work needed to bring –q to B when +q is at A. This is given by
(charge at B) × (electrostatic potential at B due to charge +q at A)
 q  q2
= −q ×  = −
 4 πε 0d  4 πε 0d
(iii) Work needed to bring charge +q to C when +q is at A and –q is at
B. This is given by (charge at C) × (potential at C due to charges
at A and B)
 +q −q 
= +q  + 
 4 πε0d 2 4πε 0d 
−q 2  1 
= 1−
4πε 0d  
2
(iv) Work needed to bring –q to D when +q at A,–q at B, and +q at C.
This is given by (charge at D) × (potential at D due to charges at A,
B and C)
EXAMPLE 2.4

 +q −q q 
= −q  + + 
 4 πε 0d 4 πε0d 2 4πε 0d 
−q 2  1 
=  2 − 
4πε 0d 2 57

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 Physics
Add the work done in steps (i), (ii), (iii) and (iv). The total work
required is
−q 2   1   1 
= (0) + (1) + 1 −  + 2 − 
4 πε 0d   2  2

=
−q 2
4πε 0d
(
4− 2 )
The work done depends only on the arrangement of the charges, and
not how they are assembled. By definition, this is the total
electrostatic energy of the charges.
(Students may try calculating same work/energy by taking charges
in any other order they desire and convince themselves that the energy
will remain the same.)
(b) The extra work necessary to bring a charge q0 to the point E when
EXAMPLE 2.4

the four charges are at A, B, C and D is q0 × (electrostatic potential at

E due to the charges at A, B, C and D). The electrostatic potential at
E is clearly zero since potential due to A and C is cancelled by that
due to B and D. Hence, no work is required to bring any charge to
point E.

2.8 POTENTIAL ENERGY IN AN EXTERNAL FIELD

2.8.1 Potential energy of a single charge
In Section 2.7, the source of the electric field was specified – the charges
and their locations - and the potential energy of the system of those charges
was determined. In this section, we ask a related but a distinct question.
What is the potential energy of a charge q in a given field? This question
was, in fact, the starting point that led us to the notion of the electrostatic
potential (Sections 2.1 and 2.2). But here we address this question again
to clarify in what way it is different from the discussion in Section 2.7.
The main difference is that we are now concerned with the potential
energy of a charge (or charges) in an external field. The external field E is
not produced by the given charge(s) whose potential energy we wish to
calculate. E is produced by sources external to the given charge(s).The
external sources may be known, but often they are unknown or
unspecified; what is specified is the electric field E or the electrostatic
potential V due to the external sources. We assume that the charge q
does not significantly affect the sources producing the external field. This
is true if q is very small, or the external sources are held fixed by other
unspecified forces. Even if q is finite, its influence on the external sources
may still be ignored in the situation when very strong sources far away
at infinity produce a finite field E in the region of interest. Note again that
we are interested in determining the potential energy of a given charge q
(and later, a system of charges) in the external field; we are not interested
in the potential energy of the sources producing the external electric field.
The external electric field E and the corresponding external potential
V may vary from point to point. By definition, V at a point P is the work
58 done in bringing a unit positive charge from infinity to the point P.

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 Electrostatic Potential
and Capacitance
(We continue to take potential at infinity to be zero.) Thus, work done in
bringing a charge q from infinity to the point P in the external field is qV.
This work is stored in the form of potential energy of q. If the point P has
position vector r relative to some origin, we can write:
Potential energy of q at r in an external field
= qV (r) (2.27)
where V(r) is the external potential at the point r.
Thus, if an electron with charge q = e = 1.6×10–19 C is accelerated by
a potential difference of DV = 1 volt, it would gain energy of qDV = 1.6 ×
10–19J. This unit of energy is defined as 1 electron volt or 1eV, i.e.,
1 eV=1.6 × 10–19J. The units based on eV are most commonly used in
atomic, nuclear and particle physics, (1 keV = 103eV = 1.6 × 10–16J, 1 MeV
= 106eV = 1.6 × 10–13J, 1 GeV = 109eV = 1.6 × 10–10J and 1 TeV = 1012eV
= 1.6 × 10–7J). [This has already been defined on Page 117, XI Physics
Part I, Table 6.1.]

2.8.2 Potential energy of a system of two charges in an

external field
Next, we ask: what is the potential energy of a system of two charges q1
and q2 located at r1and r2, respectively, in an external field? First, we
calculate the work done in bringing the charge q1 from infinity to r1.
Work done in this step is q1 V(r1), using Eq. (2.27). Next, we consider the
work done in bringing q2 to r2. In this step, work is done not only against
the external field E but also against the field due to q1.
Work done on q2 against the external field
= q2 V (r2)
Work done on q2 against the field due to q1
q1q 2
=
4 πε or12
where r12 is the distance between q1 and q2. We have made use of Eqs.
(2.27) and (2.22). By the superposition principle for fields, we add up
the work done on q2 against the two fields (E and that due to q1):
Work done in bringing q2 to r2
q1q2
= q2V ( r2 ) + (2.28)
4 πεo r12
Thus,
Potential energy of the system
= the total work done in assembling the configuration
q1q2
= q1V ( r1 ) + q2V ( r2 ) + (2.29)
4 πε0r12

Example 2.5
EXAMPLE 2.5

(a) Determine the electrostatic potential energy of a system consisting

of two charges 7 mC and –2 mC (and with no external field) placed
at (–9 cm, 0, 0) and (9 cm, 0, 0) respectively.
(b) How much work is required to separate the two charges infinitely
away from each other? 59

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 Physics
(c) Suppose that the same system of charges is now placed in an
external electric field E = A (1/r 2); A = 9 × 105 NC–1 m2. What would
the electrostatic energy of the configuration be?
Solution
1 q1q 2 7 × ( −2) × 10 −12
(a) U = = 9 × 109 × = –0.7 J.
4 πε0 r 0.18
(b) W = U2 – U1 = 0 – U = 0 – (–0.7) = 0.7 J.
(c) The mutual interaction energy of the two charges remains
unchanged. In addition, there is the energy of interaction of the
two charges with the external electric field. We find,
7 µC −2 µ C
q1V ( r1 ) + q 2V ( r2 ) = A
+A
0.09m 0.09m
EXAMPLE 2.5

and the net electrostatic energy is

q1q2 7 µC −2 µC
q1V ( r1 ) + q2V ( r2 ) + =A +A − 0.7 J
4 πε 0r12 0.09 m 0.09 m
= 70 − 20 − 0.7 = 49.3 J

2.8.3 Potential energy of a dipole in an external field

Consider a dipole with charges q1 = +q and q2 = –q placed in a uniform
electric field E, as shown in Fig. 2.16.
As seen in the last chapter, in a uniform electric field,
the dipole experiences no net force; but experiences a
torque t given by
t=p×E (2.30)
which will tend to rotate it (unless p is parallel or
antiparallel to E). Suppose an external torque text is
applied in such a manner that it just neutralises this
torque and rotates it in the plane of paper from angle q0
to angle q1 at an infinitesimal angular speed and without
angular acceleration. The amount of work done by the
external torque will be given by

FIGURE 2.16 Potential energy of a

dipole in a uniform external field.
= pE (cos θ0 − cos θ1 ) (2.31)

This work is stored as the potential energy of the system. We can

then associate potential energy U(q ) with an inclination q of the dipole.
Similar to other potential energies, there is a freedom in choosing the
angle where the potential energy U is taken to be zero. A natural choice
is to take q0 = p / 2. (An explanation for it is provided towards the end of
discussion.) We can then write,

(2.32)
60

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 Electrostatic Potential
and Capacitance
This expression can alternately be understood also from Eq. (2.29).
We apply Eq. (2.29) to the present system of two charges +q and –q. The
potential energy expression then reads
q2
U ′ (θ ) = q [V ( r1 ) − V ( r2 )] − (2.33)
4 πε 0 × 2a
Here, r1 and r2 denote the position vectors of +q and –q. Now, the
potential difference between positions r1 and r2 equals the work done
in bringing a unit positive charge against field from r2 to r1. The
displacement parallel to the force is 2a cosq. Thus, [V(r1)–V (r2)] =
–E × 2a cosq . We thus obtain,
q2 q2
U ′ (θ ) = − pE cos θ − = − p.E − (2.34)
4πε 0 × 2a 4πε 0 × 2a
We note that U¢ (q ) differs from U(q ) by a quantity which is just a constant
for a given dipole. Since a constant is insignificant for potential energy, we
can drop the second term in Eq. (2.34) and it then reduces to Eq. (2.32).
We can now understand why we took q0=p/2. In this case, the work
done against the external field E in bringing +q and – q are equal and
opposite and cancel out, i.e., q [V (r1) – V (r2)]=0.

Example 2.6 A molecule of a substance has a permanent electric

dipole moment of magnitude 10–29 C m. A mole of this substance is
polarised (at low temperature) by applying a strong electrostatic field
of magnitude 106 V m–1. The direction of the field is suddenly changed
by an angle of 60º. Estimate the heat released by the substance in
aligning its dipoles along the new direction of the field. For simplicity,
assume 100% polarisation of the sample.
Solution Here, dipole moment of each molecules = 10–29 C m
As 1 mole of the substance contains 6 × 1023 molecules,
total dipole moment of all the molecules, p = 6 × 1023 × 10–29 C m
= 6 × 10–6 C m
EXAMPLE 2.6

Initial potential energy, Ui = –pE cos q = –6×10–6×106 cos 0° = –6 J

Final potential energy (when q = 60°), Uf = –6 × 10–6 × 106 cos 60° = –3 J
Change in potential energy = –3 J – (–6J) = 3 J
So, there is loss in potential energy. This must be the energy released
by the substance in the form of heat in aligning its dipoles.

2.9 ELECTROSTATICS OF CONDUCTORS

Conductors and insulators were described briefly in Chapter 1.
Conductors contain mobile charge carriers. In metallic conductors, these
charge carriers are electrons. In a metal, the outer (valence) electrons
part away from their atoms and are free to move. These electrons are free
within the metal but not free to leave the metal. The free electrons form a
kind of ‘gas’; they collide with each other and with the ions, and move
randomly in different directions. In an external electric field, they drift
against the direction of the field. The positive ions made up of the nuclei
and the bound electrons remain held in their fixed positions. In electrolytic
conductors, the charge carriers are both positive and negative ions; but 61

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the situation in this case is more involved – the movement of the charge
carriers is affected both by the external electric field as also by the
so-called chemical forces (see Chapter 3). We shall restrict our discussion
to metallic solid conductors. Let us note important results regarding
electrostatics of conductors.

1. Inside a conductor, electrostatic field is zero

Consider a conductor, neutral or charged. There may also be an external
electrostatic field. In the static situation, when there is no current inside
or on the surface of the conductor, the electric field is zero everywhere
inside the conductor. This fact can be taken as the defining property of a
conductor. A conductor has free electrons. As long as electric field is not
zero, the free charge carriers would experience force and drift. In the
static situation, the free charges have so distributed themselves that the
electric field is zero everywhere inside. Electrostatic field is zero inside a
conductor.

2. At the surface of a charged conductor, electrostatic field

must be normal to the surface at every point
If E were not normal to the surface, it would have some non-zero
component along the surface. Free charges on the surface of the conductor
would then experience force and move. In the static situation, therefore,
E should have no tangential component. Thus electrostatic field at the
surface of a charged conductor must be normal to the surface at every
point. (For a conductor without any surface charge density, field is zero
even at the surface.) See result 5.

3. The interior of a conductor can have no excess charge in

the static situation
A neutral conductor has equal amounts of positive and negative charges
in every small volume or surface element. When the conductor is charged,
the excess charge can reside only on the surface in the static situation.
This follows from the Gauss’s law. Consider any arbitrary volume element
v inside a conductor. On the closed surface S bounding the volume
element v, electrostatic field is zero. Thus the total electric flux through S
is zero. Hence, by Gauss’s law, there is no net charge enclosed by S. But
the surface S can be made as small as you like, i.e., the volume v can be
made vanishingly small. This means there is no net charge at any point
inside the conductor, and any excess charge must reside at the surface.

4. Electrostatic potential is constant throughout the volume

of the conductor and has the same value (as inside) on
its surface
This follows from results 1 and 2 above. Since E = 0 inside the conductor
and has no tangential component on the surface, no work is done in
moving a small test charge within the conductor and on its surface. That
is, there is no potential difference between any two points inside or on
62 the surface of the conductor. Hence, the result. If the conductor is charged,

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 Electrostatic Potential
and Capacitance
electric field normal to the surface exists; this means potential will be
different for the surface and a point just outside the surface.
In a system of conductors of arbitrary size, shape and
charge configuration, each conductor is characterised by a constant
value of potential, but this constant may differ from one conductor to
the other.

5. Electric field at the surface of a charged conductor

σ
E= ˆ
n (2.35)
ε0
where s is the surface charge density and n̂ is a unit vector normal
to the surface in the outward direction.
To derive the result, choose a pill box (a short cylinder) as the Gaussian
surface about any point P on the surface, as shown in Fig. 2.17. The pill
box is partly inside and partly outside the surface of the conductor. It
has a small area of cross section d S and negligible height.
Just inside the surface, the electrostatic field is zero; just outside, the
field is normal to the surface with magnitude E. Thus,
the contribution to the total flux through the pill box
comes only from the outside (circular) cross-section
of the pill box. This equals ± EdS (positive for s > 0,
negative for s < 0), since over the small area dS, E
may be considered constant and E and dS are parallel
or antiparallel. The charge enclosed by the pill box
is sdS.
By Gauss’s law
σ δS
EdS =
ε0

σ
E= (2.36)
ε0
Including the fact that electric field is normal to the FIGURE 2.17 The Gaussian surface
surface, we get the vector relation, Eq. (2.35), which (a pill box) chosen to derive Eq. (2.35)
is true for both signs of s. For s > 0, electric field is for electric field at the surface of a
normal to the surface outward; for s < 0, electric field charged conductor.
is normal to the surface inward.

6. Electrostatic shielding
Consider a conductor with a cavity, with no charges inside the cavity. A
remarkable result is that the electric field inside the cavity is zero, whatever
be the size and shape of the cavity and whatever be the charge on the
conductor and the external fields in which it might be placed. We have
proved a simple case of this result already: the electric field inside a charged
spherical shell is zero. The proof of the result for the shell makes use of
the spherical symmetry of the shell (see Chapter 1). But the vanishing of
electric field in the (charge-free) cavity of a conductor is, as mentioned
above, a very general result. A related result is that even if the conductor 63

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 Physics
is charged or charges are induced on a neutral
conductor by an external field, all charges reside
only on the outer surface of a conductor with cavity.
The proofs of the results noted in Fig. 2.18 are
omitted here, but we note their important
implication. Whatever be the charge and field
configuration outside, any cavity in a conductor
remains shielded from outside electric influence: the
field inside the cavity is always zero. This is known
as electrostatic shielding. The effect can be made
use of in protecting sensitive instruments from
FIGURE 2.18 The electric field inside a
outside electrical influence. Figure 2.19 gives a
cavity of any conductor is zero. All
summary of the important electrostatic properties
charges reside only on the outer surface
of a conductor with cavity. (There are no of a conductor.
charges placed in the cavity.)

FIGURE 2.19 Some important electrostatic properties of a conductor.

Example 2.7
(a) A comb run through one’s dry hair attracts small bits of paper.
Why?
What happens if the hair is wet or if it is a rainy day? (Remember,
a paper does not conduct electricity.)
(b) Ordinary rubber is an insulator. But special rubber tyres of
aircraft are made slightly conducting. Why is this necessary?
(c) Vehicles carrying inflammable materials usually have metallic
ropes touching the ground during motion. Why?
(d) A bird perches on a bare high power line, and nothing happens
to the bird. A man standing on the ground touches the same line
and gets a fatal shock. Why?
Solution
EXAMPLE 2.7

(a) This is because the comb gets charged by friction. The molecules
in the paper gets polarised by the charged comb, resulting in a
net force of attraction. If the hair is wet, or if it is rainy day, friction
between hair and the comb reduces. The comb does not get
charged and thus it will not attract small bits of paper.
64

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 Electrostatic Potential
and Capacitance

(b) To enable them to conduct charge (produced by friction) to the

EXAMPLE 2.7
ground; as too much of static electricity accumulated may result
in spark and result in fire.
(c) Reason similar to (b).
(d) Current passes only when there is difference in potential.

2.10 DIELECTRICS AND POLARISATION

Dielectrics are non-conducting substances. In contrast to conductors,
they have no (or negligible number of ) charge carriers. Recall from Section
2.9 what happens when a conductor is placed in an
external electric field. The free charge carriers move
and charge distribution in the conductor adjusts
itself in such a way that the electric field due to
induced charges opposes the external field within
the conductor. This happens until, in the static
situation, the two fields cancel each other and the
net electrostatic field in the conductor is zero. In a
dielectric, this free movement of charges is not
possible. It turns out that the external field induces
dipole moment by stretching or re-orienting
molecules of the dielectric. The collective effect of all
the molecular dipole moments is net charges on the
surface of the dielectric which produce a field that FIGURE 2.20 Difference in behaviour
of a conductor and a dielectric
opposes the external field. Unlike in a conductor,
in an external electric field.
however, the opposing field so induced does not
exactly cancel the external field. It only reduces it.
The extent of the effect depends on the
nature of the dielectric. To understand the
effect, we need to look at the charge
distribution of a dielectric at the
molecular level.
The molecules of a substance may be
polar or non-polar. In a non-polar
molecule, the centres of positive and
negative charges coincide. The molecule
then has no permanent (or intrinsic) dipole
moment. Examples of non-polar molecules
are oxygen (O 2 ) and hydrogen (H 2 )
molecules which, because of their
symmetry, have no dipole moment. On the
other hand, a polar molecule is one in which
the centres of positive and negative charges
are separated (even when there is no
FIGURE 2.21 Some examples of polar
external field). Such molecules have a
and non-polar molecules.
permanent dipole moment. An ionic
molecule such as HCl or a molecule of water
(H2O) are examples of polar molecules. 65

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 Physics
In an external electric field, the
positive and negative charges of a non-
polar molecule are displaced in opposite
directions. The displacement stops when
the external force on the constituent
charges of the molecule is balanced by
the restoring force (due to internal fields
in the molecule). The non-polar molecule
thus develops an induced dipole moment.
The dielectric is said to be polarised by
the external field. We consider only the
simple situation when the induced dipole
moment is in the direction of the field and
is proportional to the field strength.
(Substances for which this assumption
is true are called linear isotropic
dielectrics.) The induced dipole moments
of different molecules add up giving a net
dipole moment of the dielectric in the
presence of the external field.
A dielectric with polar molecules also
develops a net dipole moment in an
external field, but for a different reason.
FIGURE 2.22 A dielectric develops a net dipole In the absence of any external field, the
moment in an external electric field. (a) Non-polar different permanent dipoles are oriented
molecules, (b) Polar molecules.
randomly due to thermal agitation; so
the total dipole moment is zero. When
an external field is applied, the individual dipole moments tend to align
with the field. When summed overall the molecules, there is then a net
dipole moment in the direction of the external field, i.e., the dielectric is
polarised. The extent of polarisation depends on the relative strength of
two mutually opposite factors: the dipole potential energy in the external
field tending to align the dipoles with the field and thermal energy tending
to disrupt the alignment. There may be, in addition, the ‘induced dipole
moment’ effect as for non-polar molecules, but generally the alignment
effect is more important for polar molecules.
Thus in either case, whether polar or non-polar, a dielectric develops
a net dipole moment in the presence of an external field. The dipole
moment per unit volume is called polarisation and is denoted by P. For
linear isotropic dielectrics,
P = ε0 χe E (2.37)
where ce is a constant characteristic of the dielectric and is known as the
electric susceptibility of the dielectric medium.
It is possible to relate ce to the molecular properties of the substance,
but we shall not pursue that here.
The question is: how does the polarised dielectric modify the original
external field inside it? Let us consider, for simplicity, a rectangular
dielectric slab placed in a uniform external field E0 parallel to two of its
66 faces. The field causes a uniform polarisation P of the dielectric. Thus

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 Electrostatic Potential
and Capacitance
every volume element Dv of the slab has a dipole moment
P Dv in the direction of the field. The volume element Dv is
macroscopically small but contains a very large number of
molecular dipoles. Anywhere inside the dielectric, the
volume element Dv has no net charge (though it has net
dipole moment). This is, because, the positive charge of one
dipole sits close to the negative charge of the adjacent dipole.
However, at the surfaces of the dielectric normal to the
electric field, there is evidently a net charge density. As seen
in Fig 2.23, the positive ends of the dipoles remain
unneutralised at the right surface and the negative ends at
the left surface. The unbalanced charges are the induced
charges due to the external field.
Thus, the polarised dielectric is equivalent to two charged
surfaces with induced surface charge densities, say sp
and –sp. Clearly, the field produced by these surface charges
opposes the external field. The total field in the dielectric FIGURE 2.23 A uniformly
is, thereby, reduced from the case when no dielectric is polarised dielectric amounts
present. We should note that the surface charge density to induced surface charge
±sp arises from bound (not free charges) in the dielectric. density, but no volume
charge density.
2.11 CAPACITORS AND CAPACITANCE
A capacitor is a system of two conductors separated by an insulator
(Fig. 2.24). The conductors have charges, say Q1 and Q2, and potentials
V1 and V2. Usually, in practice, the two conductors have charges Q
and – Q, with potential difference V = V1 – V2 between them. We shall
consider only this kind of charge configuration of the capacitor. (Even a
single conductor can be used as a capacitor by assuming the other at
infinity.) The conductors may be so charged by connecting them to the
two terminals of a battery. Q is called the charge of the capacitor, though
this, in fact, is the charge on one of the conductors – the total charge of
the capacitor is zero.
The electric field in the region between the
conductors is proportional to the charge Q. That
is, if the charge on the capacitor is, say doubled,
the electric field will also be doubled at every point.
(This follows from the direct proportionality
between field and charge implied by Coulomb’s
law and the superposition principle.) Now,
potential difference V is the work done per unit
positive charge in taking a small test charge from
the conductor 2 to 1 against the field. FIGURE 2.24 A system of two conductors
Consequently, V is also proportional to Q, and the separated by an insulator forms a capacitor.
ratio Q/V is a constant:
Q
C= (2.38)
V
The constant C is called the capacitance of the capacitor. C is independent
of Q or V, as stated above. The capacitance C depends only on the 67

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geometrical configuration (shape, size, separation) of the system of two
conductors. [As we shall see later, it also depends on the nature of the
insulator (dielectric) separating the two conductors.] The SI unit of
capacitance is 1 farad (=1 coulomb volt-1) or 1 F = 1 C V –1. A capacitor
with fixed capacitance is symbolically shown as ---||---, while the one with
variable capacitance is shown as .
Equation (2.38) shows that for large C, V is small for a given Q. This
means a capacitor with large capacitance can hold large amount of charge
Q at a relatively small V. This is of practical importance. High potential
difference implies strong electric field around the conductors. A strong
electric field can ionise the surrounding air and accelerate the charges so
produced to the oppositely charged plates, thereby neutralising the charge
on the capacitor plates, at least partly. In other words, the charge of the
capacitor leaks away due to the reduction in insulating power of the
intervening medium.
The maximum electric field that a dielectric medium can withstand
without break-down (of its insulating property) is called its dielectric
strength; for air it is about 3 × 106 Vm–1. For a separation between
conductors of the order of 1 cm or so, this field corresponds to a potential
difference of 3 × 104 V between the conductors. Thus, for a capacitor to
store a large amount of charge without leaking, its capacitance should
be high enough so that the potential difference and hence the electric
field do not exceed the break-down limits. Put differently, there is a limit
to the amount of charge that can be stored on a given capacitor without
significant leaking. In practice, a farad is a very big unit; the most common
units are its sub-multiples 1 mF = 10–6 F, 1 nF = 10–9 F, 1 pF = 10–12 F,
etc. Besides its use in storing charge, a capacitor is a key element of most
ac circuits with important functions, as described in Chapter 7.

2.12 THE PARALLEL PLATE CAPACITOR

A parallel plate capacitor consists of two large plane parallel conducting
plates separated by a small distance (Fig. 2.25). We first take the
intervening medium between the plates to be
vacuum. The effect of a dielectric medium between
the plates is discussed in the next section. Let A be
the area of each plate and d the separation between
them. The two plates have charges Q and –Q. Since
d is much smaller than the linear dimension of the
plates (d2 << A), we can use the result on electric
field by an infinite plane sheet of uniform surface
charge density (Section 1.15). Plate 1 has surface
charge density s = Q/A and plate 2 has a surface
charge density –s. Using Eq. (1.33), the electric field
in different regions is:
Outer region I (region above the plate 1),
FIGURE 2.25 The parallel plate capacitor.
σ σ
68 E= − =0 (2.39)
2ε 0 2ε 0

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 Electrostatic Potential
and Capacitance
Outer region II (region below the plate 2),
σ σ
E= − =0 (2.40)
2ε 0 2ε 0
In the inner region between the plates 1 and 2, the electric fields due
to the two charged plates add up, giving
σ σ σ Q
E = + = = (2.41)
2ε 0 2ε 0 ε 0 ε 0 A
The direction of electric field is from the positive to the negative plate.
Thus, the electric field is localised between the two plates and is
uniform throughout. For plates with finite area, this will not be true near
the outer boundaries of the plates. The field lines bend outward at the
edges — an effect called ‘fringing of the field’. By the same token, s will

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Factors affecting capacitance, capacitors in action
not be strictly uniform on the entire plate. [E and s are related by Eq.
(2.35).] However, for d2 << A, these effects can be ignored in the regions
sufficiently far from the edges, and the field there is given by Eq. (2.41).
Now for uniform electric field, potential difference is simply the electric
field times the distance between the plates, that is,
1 Qd
V = Ed = (2.42)
ε0 A
The capacitance C of the parallel plate capacitor is then
Q ε0 A
C= = = (2.43)
V d
which, as expected, depends only on the geometry of the system. For
typical values like A = 1 m2, d = 1 mm, we get
8.85 × 10 −12 C2 N –1m –2 × 1m 2
C= = 8.85 × 10 −9 F (2.44)
10 −3 m
(You can check that if 1F= 1C V–1 = 1C (NC–1m)–1 = 1 C2 N–1m–1.)
This shows that 1F is too big a unit in practice, as remarked earlier.
Another way of seeing the ‘bigness’ of 1F is to calculate the area of the
plates needed to have C = 1F for a separation of, say 1 cm:

A=
Cd
= 1F × 10 −2 m
= 109 m 2 (2.45)
ε0 8.85 × 10 −12 C2 N –1m –2
which is a plate about 30 km in length and breadth!

2.13 EFFECT OF DIELECTRIC ON CAPACITANCE

With the understanding of the behaviour of dielectrics in an external
field developed in Section 2.10, let us see how the capacitance of a parallel
plate capacitor is modified when a dielectric is present. As before, we
have two large plates, each of area A, separated by a distance d. The
charge on the plates is ±Q, corresponding to the charge density ±s (with
s = Q/A). When there is vacuum between the plates,
σ
E0 =
ε0 69

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 Physics
and the potential difference V0 is
V0 = E0d
The capacitance C0 in this case is
Q A
C0 = = ε0 (2.46)
V0 d
Consider next a dielectric inserted between the plates fully occupying
the intervening region. The dielectric is polarised by the field and, as
explained in Section 2.10, the effect is equivalent to two charged sheets
(at the surfaces of the dielectric normal to the field) with surface charge
densities sp and –sp. The electric field in the dielectric then corresponds
to the case when the net surface charge density on the plates is ±(s – sp ).
That is,
σ − σP
E = (2.47)
ε0
so that the potential difference across the plates is
σ − σP
V = Ed = d (2.48)
ε0
For linear dielectrics, we expect sp to be proportional to E0, i.e., to s.
Thus, (s – sp ) is proportional to s and we can write
σ
σ − σP = (2.49)
K
where K is a constant characteristic of the dielectric. Clearly, K > 1. We
then have
σd Qd
V = = (2.50)
ε0 K Aε0 K
The capacitance C, with dielectric between the plates, is then
Q ε 0 KA
C= = (2.51)
V d
The product e0K is called the permittivity of the medium and is
denoted by e
e = e0 K (2.52)
For vacuum K = 1 and e = e0; e0 is called the permittivity of the vacuum.
The dimensionless ratio
ε
K = (2.53)
ε0
is called the dielectric constant of the substance. As remarked before,
from Eq. (2.49), it is clear that K is greater than 1. From Eqs. (2.46) and
(2. 51)
C
K = (2.54)
C0
Thus, the dielectric constant of a substance is the factor (>1) by which
the capacitance increases from its vacuum value, when the dielectric is
70 inserted fully between the plates of a capacitor. Though we arrived at

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 Electrostatic Potential
and Capacitance
Eq. (2.54) for the case of a parallel plate capacitor, it holds good for any
type of capacitor and can, in fact, be viewed in general as a definition of
the dielectric constant of a substance.

Example 2.8 A slab of material of dielectric constant K has the same

area as the plates of a parallel-plate capacitor but has a thickness
(3/4)d, where d is the separation of the plates. How is the capacitance
changed when the slab is inserted between the plates?
Solution Let E0 = V0/d be the electric field between the plates when
there is no dielectric and the potential difference is V0. If the dielectric
is now inserted, the electric field in the dielectric will be E = E0/K.
The potential difference will then be
1 E 3
V = E0 ( d ) + 0 ( d )
4 K 4
1 3 K +3
= E 0d ( + ) = V0
4 4K 4K
The potential difference decreases by the factor (K + 3)/4K while the

EXAMPLE 2.8
free charge Q0 on the plates remains unchanged. The capacitance
thus increases
Q 4 K Q0 4K
C= 0 = = C0
V K + 3 V0 K +3

2.14 COMBINATION OF CAPACITORS

We can combine several capacitors of
capacitance C1, C2,…, Cn to obtain a system with
some effective capacitance C. The effective
capacitance depends on the way the individual
capacitors are combined. Two simple
possibilities are discussed below.

2.14.1 Capacitors in series

Figure 2.26 shows capacitors C 1 and C 2 FIGURE 2.26 Combination of two
combined in series. capacitors in series.
The left plate of C1 and the right plate of C2
are connected to two terminals of a battery and
have charges Q and –Q , respectively. It then
follows that the right plate of C1 has charge –Q
and the left plate of C2 has charge Q. If this was
not so, the net charge on each capacitor would
not be zero. This would result in an electric field
in the conductor connecting C1and C2. Charge
would flow until the net charge on both C1 and
C2 is zero and there is no electric field in the
conductor connecting C1 and C2. Thus, in the
series combination, charges on the two plates FIGURE 2.27 Combination of n
(±Q) are the same on each capacitor. The total capacitors in series. 71

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 Physics
potential drop V across the combination is the sum of the potential drops
V1 and V2 across C1 and C2, respectively.
Q Q
V = V1 + V2 = C + C (2.55)
1 2

V 1 1
i.e., Q = C + C , (2.56)
1 2

Now we can regard the combination as an effective capacitor with

charge Q and potential difference V. The effective capacitance of the
combination is
Q
C= (2.57)
V
We compare Eq. (2.57) with Eq. (2.56), and obtain
1 1 1
= + (2.58)
C C1 C 2
The proof clearly goes through for any number of capacitors arranged
in a similar way. Equation (2.55), for n capacitors arranged in series,
generalises to
Q Q Q
V = V1 + V2 + ... + Vn = + + ... + (2.59)
C1 C2 Cn
Following the same steps as for the case of two
capacitors, we get the general formula for effective
capacitance of a series combination of n capacitors:
1 1 1 1 1
= + + + ... +
C C1 C 2 C3 C n (2.60)

2.14.2 Capacitors in parallel

Figure 2.28 (a) shows two capacitors arranged in
parallel. In this case, the same potential difference is
applied across both the capacitors. But the plate charges
(±Q1) on capacitor 1 and the plate charges (±Q2) on the
capacitor 2 are not necessarily the same:
Q1 = C1V, Q2 = C2V (2.61)
The equivalent capacitor is one with charge
Q = Q1 + Q2 (2.62)
and potential difference V.
Q = CV = C1V + C2V (2.63)
The effective capacitance C is, from Eq. (2.63),
C = C1 + C2 (2.64)
The general formula for effective capacitance C for
parallel combination of n capacitors [Fig. 2.28 (b)]
follows similarly,
Q = Q1 + Q2 + ... + Qn (2.65)
FIGURE 2.28 Parallel combination of i.e., CV = C1V + C2V + ... CnV(2.66)
(a) two capacitors, (b) n capacitors. which gives
C = C1 + C2 + ... Cn (2.67)
72

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 Electrostatic Potential
and Capacitance

Example 2.9 A network of four 10 mF capacitors is connected to a 500 V

supply, as shown in Fig. 2.29. Determine (a) the equivalent capacitance
of the network and (b) the charge on each capacitor. (Note, the charge on
a capacitor is the charge on the plate with higher potential, equal and
opposite to the charge on the plate with lower potential.)

FIGURE 2.29

Solution
(a) In the given network, C1, C2 and C3 are connected in series. The
effective capacitance C¢ of these three capacitors is given by
1 1 1 1
= + +
C ′ C1 C2 C3
For C1 = C2 = C3 = 10 mF, C¢ = (10/3) mF. The network has C¢ and C4
connected in parallel. Thus, the equivalent capacitance C of the
network is
 10 
C = C¢ + C4 =  + 10 mF =13.3mF
 3 
(b) Clearly, from the figure, the charge on each of the capacitors, C1,
C2 and C3 is the same, say Q. Let the charge on C4 be Q¢. Now, since
the potential difference across AB is Q/C1, across BC is Q/C2, across
CD is Q/C3 , we have
Q Q Q
+ + = 500 V .
C1 C2 C3
Also, Q¢/C4 = 500 V.
EXAMPLE 2.9

This gives for the given value of the capacitances,

10
Q = 500 V × µF = 1.7 × 10 −3 C and
3
Q ′ = 500 V × 10 µF = 5.0 × 10 −3 C

2.15 ENERGY STORED IN A CAPACITOR

A capacitor, as we have seen above, is a system of two conductors with
charge Q and –Q. To determine the energy stored in this configuration,
consider initially two uncharged conductors 1 and 2. Imagine next a
process of transferring charge from conductor 2 to conductor 1 bit by
73

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 Physics
bit, so that at the end, conductor 1 gets charge Q. By
charge conservation, conductor 2 has charge –Q at
the end (Fig 2.30 ).
In transferring positive charge from conductor 2
to conductor 1, work will be done externally, since at
any stage conductor 1 is at a higher potential than
conductor 2. To calculate the total work done, we first
calculate the work done in a small step involving
transfer of an infinitesimal (i.e., vanishingly small)
amount of charge. Consider the intermediate situation
when the conductors 1 and 2 have charges Q¢ and
FIGURE 2.30 (a) Work done in a small
–Q¢ respectively. At this stage, the potential difference
step of building charge on conductor 1
from Q¢ to Q¢ + d Q¢. (b) Total work done
V¢ between conductors 1 to 2 is Q¢/C, where C is the
in charging the capacitor may be capacitance of the system. Next imagine that a small
viewed as stored in the energy of charge d Q¢ is transferred from conductor 2 to 1. Work
electric field between the plates. done in this step (d W), resulting in charge Q¢ on
conductor 1 increasing to Q¢+ d Q¢, is given by
Q′
δ W = V ′δ Q ′ = δ Q′ (2.68)
C
Integrating eq. (2.68)
Q
Q′ 1 Q ′2
Q
Q2
W = ∫C δ Q ’ =
C 2
=
2C
0 0

We can write the final result, in different ways

Q2 1 1
W = = CV 2 = QV (2.69)
2C 2 2
Since electrostatic force is conservative, this work is stored in the form
of potential energy of the system. For the same reason, the final result for
potential energy [Eq. (2.69)] is independent of the manner in which the
charge configuration of the capacitor is built up. When the capacitor
discharges, this stored-up energy is released. It is possible to view the
potential energy of the capacitor as ‘stored’ in the electric field between
the plates. To see this, consider for simplicity, a parallel plate capacitor
[of area A (of each plate) and separation d between the plates].
Energy stored in the capacitor
1 Q 2 ( Aσ )2 d
= = × (2.70)
2 C 2 ε0 A
The surface charge density s is related to the electric field E between
the plates,
σ
E= (2.71)
ε0
From Eqs. (2.70) and (2.71) , we get
Energy stored in the capacitor

74 U = (1/ 2) ε 0 E 2 × A d (2.72)

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 Electrostatic Potential
and Capacitance
Note that Ad is the volume of the region between the plates (where
electric field alone exists). If we define energy density as energy stored
per unit volume of space, Eq (2.72) shows that
Energy density of electric field,
u =(1/2)e0E 2 (2.73)
Though we derived Eq. (2.73) for the case of a parallel plate
capacitor, the result on energy density of an electric field is, in fact,
very general and holds true for electric field due to any configuration
of charges.

Example 2.10 (a) A 900 pF capacitor is charged by 100 V battery

[Fig. 2.31(a)]. How much electrostatic energy is stored by the capacitor?
(b) The capacitor is disconnected from the battery and connected to
another 900 pF capacitor [Fig. 2.31(b)]. What is the electrostatic
energy stored by the system?

FIGURE 2.31

Solution
(a) The charge on the capacitor is
Q = CV = 900 × 10–12 F × 100 V = 9 × 10–8 C
The energy stored by the capacitor is
= (1/2) CV 2 = (1/2) QV
EXAMPLE 2.10

= (1/2) × 9 × 10–8C × 100 V = 4.5 × 10–6 J

(b) In the steady situation, the two capacitors have their positive
plates at the same potential, and their negative plates at the
same potential. Let the common potential difference be V¢. The
75

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 Physics
charge on each capacitor is then Q¢ = CV¢. By charge conservation,
Q¢ = Q/2. This implies V¢ = V/2. The total energy of the system is
1 1
=2× Q ' V ' = QV = 2.25 × 10 −6 J
2 4
Thus in going from (a) to (b), though no charge is lost; the final
energy is only half the initial energy. Where has the remaining energy
EXAMPLE 2.10

gone?
There is a transient period before the system settles to the
situation (b). During this period, a transient current flows from
the first capacitor to the second. Energy is lost during this time in
the form of heat and electromagnetic radiation.

SUMMARY

1. Electrostatic force is a conservative force. Work done by an external

force (equal and opposite to the electrostatic force) in bringing a charge
q from a point R to a point P is q(VP–VR), which is the difference in
potential energy of charge q between the final and initial points.
2. Potential at a point is the work done per unit charge (by an external
agency) in bringing a charge from infinity to that point. Potential at a
point is arbitrary to within an additive constant, since it is the potential
difference between two points which is physically significant. If potential
at infinity is chosen to be zero; potential at a point with position vector
r due to a point charge Q placed at the origin is given is given by
1 Q
V (r) =
4 πε o r
3. The electrostatic potential at a point with position vector r due to a
point dipole of dipole moment p placed at the origin is
1 p.rˆ
V (r ) =
4 πε o r 2
The result is true also for a dipole (with charges –q and q separated by
2a) for r >> a.

4. For a charge configuration q1 , q2 , ..., q n with position vectors r 1 ,

r2, ... rn, the potential at a point P is given by the superposition principle

1 q1 q2 q
V = ( + + ... + n )
4 πε 0 r1P r2P rnP
where r1P is the distance between q1 and P, as and so on.
5. An equipotential surface is a surface over which potential has a constant
value. For a point charge, concentric spheres centred at a location of the
charge are equipotential surfaces. The electric field E at a point is
perpendicular to the equipotential surface through the point. E is in the
direction of the steepest decrease of potential.

76

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 Electrostatic Potential
and Capacitance

6. Potential energy stored in a system of charges is the work done (by an

external agency) in assembling the charges at their locations. Potential
energy of two charges q1, q2 at r1, r2 is given by

1 q1 q 2
U =
4 πε 0 r12
where r12 is distance between q1 and q2.
7. The potential energy of a charge q in an external potential V(r) is qV(r).
The potential energy of a dipole moment p in a uniform electric field E
is –p.E.

8. Electrostatics field E is zero in the interior of a conductor; just outside

the surface of a charged conductor, E is normal to the surface given by
σ
E= ˆ where n̂ is the unit vector along the outward normal to the
n
ε0
surface and s is the surface charge density. Charges in a conductor can
reside only at its surface. Potential is constant within and on the surface
of a conductor. In a cavity within a conductor (with no charges), the
electric field is zero.

9. A capacitor is a system of two conductors separated by an insulator. Its

capacitance is defined by C = Q/V, where Q and –Q are the charges on the
two conductors and V is the potential difference between them. C is
determined purely geometrically, by the shapes, sizes and relative
positions of the two conductors. The unit of capacitance is farad:,
1 F = 1 C V –1. For a parallel plate capacitor (with vacuum between the
plates),
A
C= ε0
d
where A is the area of each plate and d the separation between them.
10. If the medium between the plates of a capacitor is filled with an insulating
substance (dielectric), the electric field due to the charged plates induces
a net dipole moment in the dielectric. This effect, called polarisation,
gives rise to a field in the opposite direction. The net electric field inside
the dielectric and hence the potential difference between the plates is
thus reduced. Consequently, the capacitance C increases from its value
C0 when there is no medium (vacuum),

C = KC0

where K is the dielectric constant of the insulating substance.

11. For capacitors in the series combination, the total capacitance C is given by

1 1 1 1
= + + + ...
C C1 C2 C3
In the parallel combination, the total capacitance C is:
C = C1 + C2 + C3 + ...

where C1, C2, C3... are individual capacitances.

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12. The energy U stored in a capacitor of capacitance C, with charge Q and
voltage V is

1 1 1 Q2
U = QV = CV 2 =
2 2 2 C
The electric energy density (energy per unit volume) in a region with
electric field is (1/2)e0E2.

Physical quantity Symbol Dimensions Unit Remark

Potential or V [M1 L2 T–3 A–1] V Potential difference is

physically significant
Capacitance C [M–1 L–2 T–4 A2] F
–2
Polarisation P [L AT] C m-2 Dipole moment per unit
volume
Dielectric constant K [Dimensionless]

POINTS TO PONDER

1. Electrostatics deals with forces between charges at rest. But if there is a

force on a charge, how can it be at rest? Thus, when we are talking of
electrostatic force between charges, it should be understood that each
charge is being kept at rest by some unspecified force that opposes the
net Coulomb force on the charge.
2. A capacitor is so configured that it confines the electric field lines within
a small region of space. Thus, even though field may have considerable
strength, the potential difference between the two conductors of a
capacitor is small.
3. Electric field is discontinuous across the surface of a spherical charged
σ
shell. It is zero inside and ε0 n̂ outside. Electric potential is, however
continuous across the surface, equal to q/4pe0R at the surface.
4. The torque p × E on a dipole causes it to oscillate about E. Only if there
is a dissipative mechanism, the oscillations are damped and the dipole
eventually aligns with E.
5. Potential due to a charge q at its own location is not defined – it is
infinite.
6. In the expression qV (r) for potential energy of a charge q, V (r) is the
potential due to external charges and not the potential due to q. As seen
in point 5, this expression will be ill-defined if V (r) includes potential
78 due to a charge q itself.

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 Electrostatic Potential
and Capacitance

7. A cavity inside a conductor is shielded from outside electrical influences.

It is worth noting that electrostatic shielding does not work the other
way round; that is, if you put charges inside the cavity, the exterior of
the conductor is not shielded from the fields by the inside charges.

EXERCISES

2.1 Two charges 5 × 10–8 C and –3 × 10–8 C are located 16 cm apart. At

what point(s) on the line joining the two charges is the electric
potential zero? Take the potential at infinity to be zero.
2.2 A regular hexagon of side 10 cm has a charge 5 mC at each of its
vertices. Calculate the potential at the centre of the hexagon.
2.3 Two charges 2 mC and –2 mC are placed at points A and B 6 cm
apart.
(a) Identify an equipotential surface of the system.
(b) What is the direction of the electric field at every point on this
surface?
2.4 A spherical conductor of radius 12 cm has a charge of 1.6 × 10–7C
distributed uniformly on its surface. What is the electric field
(a) inside the sphere
(b) just outside the sphere
(c) at a point 18 cm from the centre of the sphere?
2.5 A parallel plate capacitor with air between the plates has a
capacitance of 8 pF (1pF = 10–12 F). What will be the capacitance if
the distance between the plates is reduced by half, and the space
between them is filled with a substance of dielectric constant 6?
2.6 Three capacitors each of capacitance 9 pF are connected in series.
(a) What is the total capacitance of the combination?
(b) What is the potential difference across each capacitor if the
combination is connected to a 120 V supply?
2.7 Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected
in parallel.
(a) What is the total capacitance of the combination?
(b) Determine the charge on each capacitor if the combination is
connected to a 100 V supply.
2.8 In a parallel plate capacitor with air between the plates, each plate
has an area of 6 × 10–3 m2 and the distance between the plates is 3 mm.
Calculate the capacitance of the capacitor. If this capacitor is
connected to a 100 V supply, what is the charge on each plate of the
capacitor?
79

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2.9 Explain what would happen if in the capacitor given in Exercise
2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted
between the plates,
(a) while the voltage supply remained connected.
(b) after the supply was disconnected.
2.10 A 12pF capacitor is connected to a 50V battery. How much
electrostatic energy is stored in the capacitor?
2.11 A 600pF capacitor is charged by a 200V supply. It is then
disconnected from the supply and is connected to another
uncharged 600 pF capacitor. How much electrostatic energy is lost
in the process?

80

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Chapter Three

CURRENT
ELECTRICITY

3.1 INTRODUCTION
In Chapter 1, all charges whether free or bound, were considered to be at
rest. Charges in motion constitute an electric current. Such currents occur
naturally in many situations. Lightning is one such phenomenon in
which charges flow from the clouds to the earth through the atmosphere,
sometimes with disastrous results. The flow of charges in lightning is not
steady, but in our everyday life we see many devices where charges flow
in a steady manner, like water flowing smoothly in a river. A torch and a
cell-driven clock are examples of such devices. In the present chapter, we
shall study some of the basic laws concerning steady electric currents.

3.2 ELECTRIC CURRENT

Imagine a small area held normal to the direction of flow of charges. Both
the positive and the negative charges may flow forward and backward
across the area. In a given time interval t, let q+ be the net amount (i.e.,
forward minus backward) of positive charge that flows in the forward
direction across the area. Similarly, let q – be the net amount of negative
charge flowing across the area in the forward direction. The net amount
of charge flowing across the area in the forward direction in the time
interval t, then, is q = q+– q –. This is proportional to t for steady current

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 Physics
and the quotient
q
I= (3.1)
t
is defined to be the current across the area in the forward direction. (If it
turn out to be a negative number, it implies a current in the backward
direction.)
Currents are not always steady and hence more generally, we define
the current as follows. Let DQ be the net charge flowing across a cross-
section of a conductor during the time interval Dt [i.e., between times t
and (t + Dt)]. Then, the current at time t across the cross-section of the
conductor is defined as the value of the ratio of DQ to Dt in the limit of Dt
tending to zero,
∆Q
I (t ) ≡ lim (3.2)
∆t
∆t → 0

In SI units, the unit of current is ampere. An ampere is defined

through magnetic effects of currents that we will study in the following
chapter. An ampere is typically the order of magnitude of currents in
domestic appliances. An average lightning carries currents of the order
of tens of thousands of amperes and at the other extreme, currents in
our nerves are in microamperes.

3.3 ELECTRIC CURRENTS IN CONDUCTORS

An electric charge will experience a force if an electric field is applied. If it is
free to move, it will thus move contributing to a current. In nature, free
charged particles do exist like in upper strata of atmosphere called the
ionosphere. However, in atoms and molecules, the negatively charged
electrons and the positively charged nuclei are bound to each other and
are thus not free to move. Bulk matter is made up of many molecules, a
gram of water, for example, contains approximately 1022 molecules. These
molecules are so closely packed that the electrons are no longer attached
to individual nuclei. In some materials, the electrons will still be bound,
i.e., they will not accelerate even if an electric field is applied. In other
materials, notably metals, some of the electrons are practically free to move
within the bulk material. These materials, generally called conductors,
develop electric currents in them when an electric field is applied.
If we consider solid conductors, then of course the atoms are tightly
bound to each other so that the current is carried by the negatively
charged electrons. There are, however, other types of conductors like
electrolytic solutions where positive and negative charges both can move.
In our discussions, we will focus only on solid conductors so that the
current is carried by the negatively charged electrons in the background
of fixed positive ions.
Consider first the case when no electric field is present. The electrons
will be moving due to thermal motion during which they collide with the
fixed ions. An electron colliding with an ion emerges with the same speed
as before the collision. However, the direction of its velocity after the
collision is completely random. At a given time, there is no preferential
82 direction for the velocities of the electrons. Thus on the average, the

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 Current
Electricity
number of electrons travelling in any direction will be equal to the number
of electrons travelling in the opposite direction. So, there will be no net
electric current.
Let us now see what happens to such a
piece of conductor if an electric field is applied.
To focus our thoughts, imagine the conductor
in the shape of a cylinder of radius R (Fig. 3.1).
Suppose we now take two thin circular discs FIGURE 3.1 Charges +Q and –Q put at the ends
of a dielectric of the same radius and put of a metallic cylinder. The electrons will drift
positive charge +Q distributed over one disc because of the electric field created to
and similarly –Q at the other disc. We attach neutralise the charges. The current thus
the two discs on the two flat surfaces of the will stop after a while unless the charges +Q
cylinder. An electric field will be created and and –Q are continuously replenished.
is directed from the positive towards the
negative charge. The electrons will be accelerated due to this field towards
+Q. They will thus move to neutralise the charges. The electrons, as long
as they are moving, will constitute an electric current. Hence in the
situation considered, there will be a current for a very short while and no
current thereafter.
We can also imagine a mechanism where the ends of the cylinder are
supplied with fresh charges to make up for any charges neutralised by
electrons moving inside the conductor. In that case, there will be a steady
electric field in the body of the conductor. This will result in a continuous
current rather than a current for a short period of time. Mechanisms,
which maintain a steady electric field are cells or batteries that we shall
study later in this chapter. In the next sections, we shall study the steady
current that results from a steady electric field in conductors.

3.4 OHM’S LAW

A basic law regarding flow of currents was discovered by G.S. Ohm in
1828, long before the physical mechanism responsible for flow of currents
was discovered. Imagine a conductor through which a current I is flowing
and let V be the potential difference between the ends of the conductor.
Then Ohm’s law states that
VµI
or, V = R I (3.3)
where the constant of proportionality R is called the resistance of the
conductor. The SI units of resistance is ohm, and is denoted by the symbol
W. The resistance R not only depends on the material of the conductor
but also on the dimensions of the conductor. The dependence of R on the
dimensions of the conductor can easily be determined as follows. FIGURE 3.2
Consider a conductor satisfying Eq. (3.3) to be in the form of a slab of Illustrating the
length l and cross sectional area A [Fig. 3.2(a)]. Imagine placing two such relation R = rl/A for
identical slabs side by side [Fig. 3.2(b)], so that the length of the a rectangular slab
of length l and area
combination is 2l. The current flowing through the combination is the
of cross-section A.
same as that flowing through either of the slabs. If V is the potential
difference across the ends of the first slab, then V is also the potential
difference across the ends of the second slab since the second slab is 83

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 Physics
identical to the first and the same current I flows through
both. The potential difference across the ends of the
combination is clearly sum of the potential difference
across the two individual slabs and hence equals 2V. The
current through the combination is I and the resistance
GEORG SIMON OHM (1787–1854)

of the combination RC is [from Eq. (3.3)],

2V
RC = =2R (3.4)
I
since V/I = R, the resistance of either of the slabs. Thus,
doubling the length of a conductor doubles the
resistance. In general, then resistance is proportional to
length,
R ∝l (3.5)
Georg Simon Ohm (1787– Next, imagine dividing the slab into two by cutting it
1854) German physicist, lengthwise so that the slab can be considered as a
professor at Munich. Ohm combination of two identical slabs of length l , but each
was led to his law by an having a cross sectional area of A/2 [Fig. 3.2(c)].
analogy between the
For a given voltage V across the slab, if I is the current
conduction of heat: the
electric field is analogous to through the entire slab, then clearly the current flowing
the temperature gradient, through each of the two half-slabs is I/2. Since the
and the electric current is potential difference across the ends of the half-slabs is V,
analogous to the heat flow. i.e., the same as across the full slab, the resistance of each
of the half-slabs R1 is
V V
R1 = = 2 = 2R. (3.6)
( I /2) I
Thus, halving the area of the cross-section of a conductor doubles
the resistance. In general, then the resistance R is inversely proportional
to the cross-sectional area,
1
R ∝ (3.7)
A
Combining Eqs. (3.5) and (3.7), we have
l
R ∝ (3.8)
A
and hence for a given conductor
l
R=ρ (3.9)
A
where the constant of proportionality r depends on the material of the
conductor but not on its dimensions. r is called resistivity.
Using the last equation, Ohm’s law reads
I ρl
V =I ×R= (3.10)
A
Current per unit area (taken normal to the current), I/A, is called
current density and is denoted by j. The SI units of the current density
are A/m2. Further, if E is the magnitude of uniform electric field in the
conductor whose length is l, then the potential difference V across its
84 ends is El. Using these, the last equation reads

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 Current
Electricity
El=jrl
or, E = j r (3.11)
The above relation for magnitudes E and j can indeed be cast in a
vector form. The current density, (which we have defined as the current
through unit area normal to the current) is also directed along E, and is
also a vector j (º
º j E/E). Thus, the last equation can be written as,
E = jr (3.12)
or, j = s E (3.13)
where s º1/r is called the conductivity. Ohm’s law is often stated in an
equivalent form, Eq. (3.13) in addition to Eq.(3.3). In the next section, we
will try to understand the origin of the Ohm’s law as arising from the
characteristics of the drift of electrons.

3.5 DRIFT OF ELECTRONS AND THE ORIGIN

OF RESISTIVITY
As remarked before, an electron will suffer collisions with the heavy fixed
ions, but after collision, it will emerge with the same speed but in random
directions. If we consider all the electrons, their average velocity will be
zero since their directions are random. Thus, if there are N electrons and
the velocity of the ith electron (i = 1, 2, 3, ... N ) at a given time is vi , then
N
1
N
∑v i =0 (3.14)
i =1

Consider now the situation when an electric field is

present. Electrons will be accelerated due to this
field by
–e E
a = (3.15)
m
where –e is the charge and m is the mass of an electron.
Consider again the ith electron at a given time t. This
electron would have had its last collision some time
before t, and let ti be the time elapsed after its last
collision. If vi was its velocity immediately after the last
collision, then its velocity Vi at time t is
−eE
Vi = vi + ti (3.16)
m FIGURE 3.3 A schematic picture of
since starting with its last collision it was accelerated an electron moving from a point A to
(Fig. 3.3) with an acceleration given by Eq. (3.15) for a another point B through repeated
time interval ti . The average velocity of the electrons at collisions, and straight line travel
time t is the average of all the Vi’s. The average of vi’s is between collisions (full lines). If an
zero [Eq. (3.14)] since immediately after any collision, electric field is applied as shown, the
electron ends up at point B¢ (dotted
the direction of the velocity of an electron is completely
lines). A slight drift in a direction
random. The collisions of the electrons do not occur at opposite the electric field is visible.
regular intervals but at random times. Let us denote by
t, the average time between successive collisions. Then 85
at a given time, some of the electrons would have spent

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 Physics
time more than t and some less than t. In other words, the time ti in
Eq. (3.16) will be less than t for some and more than t for others as we go
through the values of i = 1, 2 ..... N. The average value of ti then is t
(known as relaxation time). Thus, averaging Eq. (3.16) over the
N-electrons at any given time t gives us for the average velocity vd
eE
v d ≡ ( Vi )average = ( vi )average − (t i )average
m
eE eE
=0– τ =− τ (3.17)
m m
This last result is surprising. It tells us that the
electrons move with an average velocity which is
independent of time, although electrons are
accelerated. This is the phenomenon of drift and the
velocity vd in Eq. (3.17) is called the drift velocity.
Because of the drift, there will be net transport of
charges across any area perpendicular to E. Consider
a planar area A, located inside the conductor such that
FIGURE 3.4 Current in a metallic the normal to the area is parallel to E (Fig. 3.4). Then
conductor. The magnitude of current because of the drift, in an infinitesimal amount of time
density in a metal is the magnitude of Dt, all electrons to the left of the area at distances upto
charge contained in a cylinder of unit |vd|Dt would have crossed the area. If n is the number
area and length vd. of free electrons per unit volume in the metal, then
there are n Dt |vd|A such electrons. Since each
electron carries a charge –e, the total charge transported across this area
A to the right in time Dt is –ne A|vd|Dt. E is directed towards the left and
hence the total charge transported along E across the area is negative of
this. The amount of charge crossing the area A in time Dt is by definition
[Eq. (3.2)] I Dt, where I is the magnitude of the current. Hence,

I ∆t = + n e A v d ∆t (3.18)
Substituting the value of |vd| from Eq. (3.17)
e2A
I ∆t = τ n ∆t E (3.19)
m
By definition I is related to the magnitude |j| of the current density by
I = |j|A (3.20)
Hence, from Eqs.(3.19) and (3.20),
ne 2
j= τE (3.21)
m
The vector j is parallel to E and hence we can write Eq. (3.21) in the
vector form
ne 2
j= τE (3.22)
m
Comparison with Eq. (3.13) shows that Eq. (3.22) is exactly the Ohm’s
86 law, if we identify the conductivity s as

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 Current
Electricity

ne 2
σ = τ (3.23)
m
We thus see that a very simple picture of electrical conduction
reproduces Ohm’s law. We have, of course, made assumptions that t
and n are constants, independent of E. We shall, in the next section,
discuss the limitations of Ohm’s law.

Example 3.1 (a) Estimate the average drift speed of conduction

electrons in a copper wire of cross-sectional area 1.0 × 10–7 m2 carrying
a current of 1.5 A. Assume that each copper atom contributes roughly
one conduction electron. The density of copper is 9.0 × 103 kg/m3,
and its atomic mass is 63.5 u. (b) Compare the drift speed obtained
above with, (i) thermal speeds of copper atoms at ordinary
temperatures, (ii) speed of propagation of electric field along the
conductor which causes the drift motion.
Solution
(a) The direction of drift velocity of conduction electrons is opposite
to the electric field direction, i.e., electrons drift in the direction
of increasing potential. The drift speed vd is given by Eq. (3.18)
vd = (I/neA)
Now, e = 1.6 × 10–19 C, A = 1.0 × 10–7m2, I = 1.5 A. The density of
conduction electrons, n is equal to the number of atoms per cubic
metre (assuming one conduction electron per Cu atom as is
reasonable from its valence electron count of one). A cubic metre
of copper has a mass of 9.0 × 103 kg. Since 6.0 × 1023 copper
atoms have a mass of 63.5 g,
6.0 × 1023
n= × 9.0 × 106
63.5
= 8.5 × 1028 m–3
which gives,
1.5
vd =
8.5 × 1028 × 1.6 × 10 –19 × 1.0 × 10 –7
= 1.1 × 10–3 m s–1 = 1.1 mm s–1
(b) (i) At a temperature T, the thermal speed* of a copper atom of
mass M is obtained from [<(1/2) Mv2 > = (3/2) kBT ] and is thus
typically of the order of k B T/M , where k B is the Boltzmann
constant. For copper at 300 K, this is about 2 × 102 m/s. This
figure indicates the random vibrational speeds of copper atoms
in a conductor. Note that the drift speed of electrons is much
smaller, about 10–5 times the typical thermal speed at ordinary
EXAMPLE 3.1

temperatures.
(ii) An electric field travelling along the conductor has a speed of
an electromagnetic wave, namely equal to 3.0 × 10 8 m s –1
(You will learn about this in Chapter 8). The drift speed is, in
comparison, extremely small; smaller by a factor of 10–11.

* See Eq. (12.23) of Chapter 12 from Class XI book. 87

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Example 3.2
(a) In Example 3.1, the electron drift speed is estimated to be only a
few mm s–1 for currents in the range of a few amperes? How then
is current established almost the instant a circuit is closed?
(b) The electron drift arises due to the force experienced by electrons
in the electric field inside the conductor. But force should cause
acceleration. Why then do the electrons acquire a steady average
drift speed?
(c) If the electron drift speed is so small, and the electron’s charge is
small, how can we still obtain large amounts of current in a
conductor?
(d) When electrons drift in a metal from lower to higher potential,
does it mean that all the ‘free’ electrons of the metal are moving
in the same direction?
(e) Are the paths of electrons straight lines between successive
collisions (with the positive ions of the metal) in the (i) absence of
electric field, (ii) presence of electric field?
Solution
(a) Electric field is established throughout the circuit, almost instantly
(with the speed of light) causing at every point a local electron
drift. Establishment of a current does not have to wait for electrons
from one end of the conductor travelling to the other end. However,
it does take a little while for the current to reach its steady value.
(b) Each ‘free’ electron does accelerate, increasing its drift speed until
it collides with a positive ion of the metal. It loses its drift speed
after collision but starts to accelerate and increases its drift speed
again only to suffer a collision again and so on. On the average,
therefore, electrons acquire only a drift speed.
(c) Simple, because the electron number density is enormous,
EXAMPLE 3.2

~1029 m–3.
(d) By no means. The drift velocity is superposed over the large
random velocities of electrons.
(e) In the absence of electric field, the paths are straight lines; in the
presence of electric field, the paths are, in general, curved.

3.5.1 Mobility
As we have seen, conductivity arises from mobile charge carriers. In
metals, these mobile charge carriers are electrons; in an ionised gas, they
are electrons and positive charged ions; in an electrolyte, these can be
both positive and negative ions.
An important quantity is the mobility m defined as the magnitude of
the drift velocity per unit electric field:

| vd |
µ= (3.24)
E
The SI unit of mobility is m2/Vs and is 104 of the mobility in practical
units (cm2/Vs). Mobility is positive. From Eq. (3.17), we have
e τE
88 vd =
m

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 Current
Electricity
Hence,
vd e τ
µ= = (3.25)
E m
where t is the average collision time for electrons.

3.6 LIMITATIONS OF OHM’S LAW

Although Ohm’s law has been found valid over a large class
of materials, there do exist materials and devices used in
electric circuits where the proportionality of V and I does not
hold. The deviations broadly are one or more of the following
FIGURE 3.5 The dashed line
types:
represents the linear Ohm’s
(a) V ceases to be proportional to I (Fig. 3.5). law. The solid line is the voltage
(b) The relation between V and I depends on the sign of V. In V versus current I for a good
other words, if I is the current for a certain V, then reversing conductor.
the direction of V keeping its magnitude fixed, does not
produce a current of the same magnitude as I in the opposite direction
(Fig. 3.6). This happens, for example, in a diode which we will study
in Chapter 14.

FIGURE 3.6 Characteristic curve FIGURE 3.7 Variation of current

of a diode. Note the different versus voltage for GaAs.
scales for negative and positive
values of the voltage and current.

(c) The relation between V and I is not unique, i.e., there is more than
one value of V for the same current I (Fig. 3.7). A material exhibiting
such behaviour is GaAs.
Materials and devices not obeying Ohm’s law in the form of Eq. (3.3)
are actually widely used in electronic circuits. In this and a few
subsequent chapters, however, we will study the electrical currents in
materials that obey Ohm’s law.

3.7 RESISTIVITY OF VARIOUS MATERIALS

The materials are classified as conductors, semiconductors and insulators
depending on their resistivities, in an increasing order of their values. 89

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Metals have low resistivities in the range of 10–8 Wm to 10–6 Wm. At the
other end are insulators like ceramic, rubber and plastics having
resistivities 1018 times greater than metals or more. In between the two
are the semiconductors. These, however, have resistivities
characteristically decreasing with a rise in temperature. The resistivities
of semiconductors can be decreased by adding small amount of suitable
impurities. This last feature is exploited in use of semiconductors for
electronic devices.

3.8 TEMPERATURE DEPENDENCE OF RESISTIVITY

The resistivity of a material is found to be dependent on the temperature.
Different materials do not exhibit the same dependence on temperatures.
Over a limited range of temperatures, that is not too large, the resistivity
of a metallic conductor is approximately given by,
rT = r0 [1 + a (T–T0)] (3.26)
where rT is the resistivity at a temperature T and r0 is the same at a
reference temperature T0. a is called the temperature co-efficient of
resistivity, and from Eq. (3.26), the dimension of a is (Temperature)–1.
For metals, a is positive.
The relation of Eq. (3.26) implies that a graph of rT plotted against T
would be a straight line. At temperatures much lower than 0°C, the graph,
however, deviates considerably from a straight line (Fig. 3.8).
Equation (3.26) thus, can be used approximately over a limited range
of T around any reference temperature T0, where the graph can be
approximated as a straight line.

FIGURE 3.8 FIGURE 3.9 Resistivity FIGURE 3.10

Resistivity rT of rT of nichrome as a Temperature dependence
copper as a function function of absolute of resistivity for a typical
of temperature T. temperature T. semiconductor.

Some materials like Nichrome (which is an alloy of nickel, iron and

chromium) exhibit a very weak dependence of resistivity with temperature
(Fig. 3.9). Manganin and constantan have similar properties. These
materials are thus widely used in wire bound standard resistors since
90 their resistance values would change very little with temperatures.

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Electricity
Unlike metals, the resistivities of semiconductors decrease with
increasing temperatures. A typical dependence is shown in Fig. 3.10.
We can qualitatively understand the temperature dependence of
resistivity, in the light of our derivation of Eq. (3.23). From this equation,
resistivity of a material is given by
1 m
ρ= = (3.27)
σ n e 2τ
r thus depends inversely both on the number n of free electrons per unit
volume and on the average time t between collisions. As we increase
temperature, average speed of the electrons, which act as the carriers of
current, increases resulting in more frequent collisions. The average time
of collisions t, thus decreases with temperature.
In a metal, n is not dependent on temperature to any appreciable
extent and thus the decrease in the value of t with rise in temperature
causes r to increase as we have observed.
For insulators and semiconductors, however, n increases with
temperature. This increase more than compensates any decrease in t in
Eq.(3.23) so that for such materials, r decreases with temperature.

Example 3.3 An electric toaster uses nichrome for its heating

element. When a negligibly small current passes through it, its
resistance at room temperature (27.0 °C) is found to be 75.3 W. When
the toaster is connected to a 230 V supply, the current settles, after
a few seconds, to a steady value of 2.68 A. What is the steady
temperature of the nichrome element? The temperature coefficient
of resistance of nichrome averaged over the temperature range
involved, is 1.70 × 10–4 °C–1.
Solution When the current through the element is very small, heating
effects can be ignored and the temperature T1 of the element is the
same as room temperature. When the toaster is connected to the
supply, its initial current will be slightly higher than its steady value
of 2.68 A. But due to heating effect of the current, the temperature
will rise. This will cause an increase in resistance and a slight
decrease in current. In a few seconds, a steady state will be reached
when temperature will rise no further, and both the resistance of the
element and the current drawn will achieve steady values. The
resistance R2 at the steady temperature T2 is
230 V
R2 = = 8 5 .8 Ω
2 .6 8 A
Using the relation
R2 = R1 [1 + a (T2 – T1)]
with a = 1.70 × 10–4 °C–1, we get

(85.8 – 75.3)
T2 – T1 = = 820 °C
EXAMPLE 3.3

(75.3) × 1.70 × 10 –4
that is, T2 = (820 + 27.0) °C = 847 °C
Thus, the steady temperature of the heating element (when heating
effect due to the current equals heat loss to the surroundings) is
91
847 °C.

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Example 3.4 The resistance of the platinum wire of a platinum

resistance thermometer at the ice point is 5 W and at steam point is
5.23 W. When the thermometer is inserted in a hot bath, the resistance
of the platinum wire is 5.795 W. Calculate the temperature of the
bath.
Solution R0 = 5 W, R100 = 5.23 W and Rt = 5.795 W
Rt − R0
Now, t = × 100, Rt = R0 (1 + α t )
R100 − R0
EXAMPLE 3.4

5.795 − 5
= × 100
5.23 − 5

0.795
= × 100 = 345.65 °C
0.23

3.9 ELECTRICAL ENERGY, POWER

Consider a conductor with end points A and B, in which a current I is
flowing from A to B. The electric potential at A and B are denoted by V (A)
and V (B) respectively. Since current is flowing from A to B, V (A) > V (B)
and the potential difference across AB is V = V(A) – V(B) > 0.
In a time interval Dt, an amount of charge DQ = I Dt travels from A to
B. The potential energy of the charge at A, by definition, was Q V (A) and
similarly at B, it is Q V(B). Thus, change in its potential energy DUpot is
DUpot = Final potential energy – Initial potential energy
= DQ[(V (B) – V (A)] = –DQ V
= –I VDt < 0 (3.28)
If charges moved without collisions through the conductor, their
kinetic energy would also change so that the total energy is unchanged.
Conservation of total energy would then imply that,
DK = –DUpot (3.29)
that is,
DK = I VDt > 0 (3.30)
Thus, in case charges were moving freely through the conductor under
the action of electric field, their kinetic energy would increase as they
move. We have, however, seen earlier that on the average, charge carriers
do not move with acceleration but with a steady drift velocity. This is
because of the collisions with ions and atoms during transit. During
collisions, the energy gained by the charges thus is shared with the atoms.
The atoms vibrate more vigorously, i.e., the conductor heats up. Thus,
in an actual conductor, an amount of energy dissipated as heat in the
conductor during the time interval Dt is,
DW = I VDt (3.31)
The energy dissipated per unit time is the power dissipated
P = DW/Dt and we have,
92 P=IV (3.32)

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Electricity

Using Ohm’s law V = IR, we get

P = I 2 R = V 2/R (3.33)
as the power loss (“ohmic loss”) in a conductor of resistance R carrying a
current I. It is this power which heats up, for example, the coil of an
electric bulb to incandescence, radiating out heat and light.
Where does the power come from? As we have
reasoned before, we need an external source to keep
a steady current through the conductor. It is clearly
this source which must supply this power. In the
simple circuit shown with a cell (Fig.3.11), it is the
chemical energy of the cell which supplies this power
for as long as it can.
The expressions for power, Eqs. (3.32) and (3.33),
show the dependence of the power dissipated in a
resistor R on the current through it and the voltage FIGURE 3.11 Heat is produced in the
across it. resistor R which is connected across
Equation (3.33) has an important application to the terminals of a cell. The energy
power transmission. Electrical power is transmitted dissipated in the resistor R comes from
from power stations to homes and factories, which the chemical energy of the electrolyte.
may be hundreds of miles away, via transmission
cables. One obviously wants to minimise the power
loss in the transmission cables connecting the power stations to homes
and factories. We shall see now how this can be achieved. Consider a
device R, to which a power P is to be delivered via transmission cables
having a resistance Rc to be dissipated by it finally. If V is the voltage
across R and I the current through it, then
P=VI (3.34)
The connecting wires from the power station to the device has a finite
resistance Rc. The power dissipated in the connecting wires, which is
wasted is Pc with
Pc = I 2 Rc
P 2 Rc
= (3.35)
V2
from Eq. (3.32). Thus, to drive a device of power P, the power wasted in the
connecting wires is inversely proportional to V 2. The transmission cables
from power stations are hundreds of miles long and their resistance Rc is
considerable. To reduce Pc, these wires carry current at enormous values
of V and this is the reason for the high voltage danger signs on transmission
lines — a common sight as we move away from populated areas. Using
electricity at such voltages is not safe and hence at the other end, a device
called a transformer lowers the voltage to a value suitable for use.

3.10 CELLS, EMF, INTERNAL RESISTANCE

We have already mentioned that a simple device to maintain a steady
current in an electric circuit is the electrolytic cell. Basically a cell has
two electrodes, called the positive (P) and the negative (N), as shown in 93

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Fig. 3.12. They are immersed in an electrolytic solution. Dipped in
the solution, the electrodes exchange charges with the electrolyte.
The positive electrode has a potential difference V+ (V+ > 0) between
itself and the electrolyte solution immediately adjacent to it marked
A in the figure. Similarly, the negative electrode develops a negative
potential – (V– ) (V– ≥ 0) relative to the electrolyte adjacent to it,
marked as B in the figure. When there is no current, the electrolyte
has the same potential throughout, so that the potential difference
between P and N is V+ – (–V–) = V+ + V– . This difference is called the
electromotive force (emf) of the cell and is denoted by e. Thus
e = V++V– > 0 (3.36)
Note that e is, actually, a potential difference and not a force. The
name emf, however, is used because of historical reasons, and was
given at a time when the phenomenon was not understood properly.
To understand the significance of e, consider a resistor R
connected across the cell (Fig. 3.12). A current I flows across R
FIGURE 3.12 (a) Sketch of from C to D. As explained before, a steady current is maintained
an electrolyte cell with because current flows from N to P through the electrolyte. Clearly,
positive terminal P and across the electrolyte the same current flows through the electrolyte
negative terminal N. The but from N to P, whereas through R, it flows from P to N.
gap between the electrodes The electrolyte through which a current flows has a finite
is exaggerated for clarity. A resistance r, called the internal resistance. Consider first the
and B are points in the
situation when R is infinite so that I = V/R = 0, where V is the
electrolyte typically close to
P and N. (b) the symbol for
potential difference between P and N. Now,
a cell, + referring to P and V = Potential difference between P and A
– referring to the N + Potential difference between A and B
electrode. Electrical + Potential difference between B and N
connections to the cell are =e (3.37)
made at P and N. Thus, emf e is the potential difference between the positive and
negative electrodes in an open circuit, i.e., when no current is
flowing through the cell.
If however R is finite, I is not zero. In that case the potential difference
between P and N is
V = V++ V– – I r
=e–Ir (3.38)
Note the negative sign in the expression (I r ) for the potential difference
between A and B. This is because the current I flows from B to A in the
electrolyte.
In practical calculations, internal resistances of cells in the circuit
may be neglected when the current I is such that e >> I r. The actual
values of the internal resistances of cells vary from cell to cell. The internal
resistance of dry cells, however, is much higher than the common
electrolytic cells.
We also observe that since V is the potential difference across R, we
have from Ohm’s law
V=I R (3.39)
94 Combining Eqs. (3.38) and (3.39), we get

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Electricity
I R = e–I r
ε
Or, I = (3.40)
R +r
The maximum current that can be drawn from a cell is for R = 0 and
it is Imax = e/r. However, in most cells the maximum allowed current is
much lower than this to prevent permanent damage to the cell.

3.11 CELLS IN SERIES AND IN PARALLEL

Like resistors, cells can be combined together in an electric circuit. And
like resistors, one can, for calculating currents and voltages in a circuit,
replace a combination of cells by an equivalent cell.

FIGURE 3.13 Two cells of emf’s e1 and e2 in the series. r1, r2 are their
internal resistances. For connections across A and C, the combination
can be considered as one cell of emf eeq and an internal resistance req.

Consider first two cells in series (Fig. 3.13), where one terminal of the
two cells is joined together leaving the other terminal in either cell free.
e1, e2 are the emf’s of the two cells and r1, r2 their internal resistances,
respectively.
Let V (A), V (B), V (C) be the potentials at points A, B and C shown in
Fig. 3.13. Then V (A) – V (B) is the potential difference between the positive
and negative terminals of the first cell. We have already calculated it in
Eq. (3.38) and hence,
V AB ≡ V ( A ) V (B) = ε1 I r1 (3.41)

Similarly,

VBC ≡ V (B) V (C) = ε 2 I r2 (3.42)

Hence, the potential difference between the terminals A and C of the
combination is

V AC ≡ V (A) – V (C) = V ( A ) – V ( B ) + V ( B ) – V (C )

= (ε1 + ε 2 ) – I (r1 + r2 ) (3.43)

If we wish to replace the combination by a single cell between A and
C of emf eeq and internal resistance req, we would have
VAC = eeq– I req (3.44)
Comparing the last two equations, we get
eeq = e1 + e2 (3.45)
and req = r1 + r2 (3.46)
In Fig.3.13, we had connected the negative electrode of the first to the
positive electrode of the second. If instead we connect the two negatives,
95

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Eq. (3.42) would change to VBC = –e2–Ir2 and we will get
eeq = e1 – e2 (e1 > e2) (3.47)
The rule for series combination clearly can be extended to any number
of cells:
(i) The equivalent emf of a series combination of n cells is just the sum of
their individual emf’s, and
(ii) The equivalent internal resistance of a series combination of n cells is
just the sum of their internal resistances.
This is so, when the current leaves each cell from the positive
electrode. If in the combination, the current leaves any cell from
the negative electrode, the emf of the cell enters the expression
for eeq with a negative sign, as in Eq. (3.47).
Next, consider a parallel combination of the cells (Fig. 3.14).
I1 and I2 are the currents leaving the positive electrodes of the
cells. At the point B1, I1 and I2 flow in whereas the current I flows
out. Since as much charge flows in as out, we have
FIGURE 3.14 Two cells in I = I1 + I2 (3.48)
parallel. For connections Let V (B1) and V (B2) be the potentials at B1 and B2, respectively.
across A and C, the Then, considering the first cell, the potential difference across its
combination can be terminals is V (B1) – V (B2). Hence, from Eq. (3.38)
replaced by one cell of emf
eeq and internal resistances V ≡ V ( B1 ) – V ( B2 ) = ε1 – I1r1 (3.49)
req whose values are given in Points B1 and B2 are connected exactly similarly to the second
Eqs. (3.54) and (3.55).
cell. Hence considering the second cell, we also have
V ≡ V ( B1 ) – V ( B2 ) = ε 2 – I 2r2 (3.50)
Combining the last three equations
I = I1 + I 2

ε1 – V ε 2 – V  ε1 ε 2  1 1
= + = +  –V  +  (3.51)
r1 r2  r1 r2   r1 r2 
Hence, V is given by,
ε1r2 + ε 2r1 r1r2
V = –I (3.52)
r1 + r2 r1 + r2
If we want to replace the combination by a single cell, between B1 and
B2, of emf eeq and internal resistance req, we would have
V = eeq – I req (3.53)
The last two equations should be the same and hence
ε1r2 + ε 2r1
ε eq = (3.54)
r1 + r2

r1r2
req = (3.55)
r1 + r2
We can put these equations in a simpler way,

96

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Electricity
1 1 1
= + (3.56)
req r1 r2

ε eq ε1 ε2
= + (3.57)
req r1 r2

GUSTAV ROBERT KIRCHHOFF (1824 –

In Fig. (3.14), we had joined the positive terminals
together and similarly the two negative ones, so that the
currents I1, I2 flow out of positive terminals. If the negative
terminal of the second is connected to positive terminal
of the first, Eqs. (3.56) and (3.57) would still be valid with
e 2 ® –e2
Equations (3.56) and (3.57) can be extended easily.
If there are n cells of emf e1, . . . en and of internal
resistances r1,... rn respectively, connected in parallel, the Gustav Robert Kirchhoff
combination is equivalent to a single cell of emf eeq and (1824 – 1887) German
internal resistance req, such that physicist, professor at
Heidelberg and at
1 1 1
= + ... + (3.58)
Berlin. Mainly known for
req r1 rn his development of
spectroscopy, he also
εeq ε1 ε made many important
= + ... + n (3.59) contributions to mathe-
req r1 rn
matical physics, among
them, his first and
3.12 KIRCHHOFF’S RULES second rules for circuits.

Electric circuits generally consist of a number of resistors

and cells interconnected sometimes in a complicated way.
The formulae we have derived earlier for series and parallel combinations
of resistors are not always sufficient to determine all the currents and
potential differences in the circuit. Two rules, called Kirchhoff’s rules,
are very useful for analysis of electric circuits.
Given a circuit, we start by labelling currents in each resistor by a
symbol, say I, and a directed arrow to indicate that a current I flows
along the resistor in the direction indicated. If ultimately I is determined
to be positive, the actual current in the resistor is in the direction of the
arrow. If I turns out to be negative, the current actually flows in a direction
opposite to the arrow. Similarly, for each source (i.e., cell or some other
source of electrical power) the positive and negative electrodes are labelled,
as well as, a directed arrow with a symbol for the current flowing through
the cell. This will tell us the potential difference, V = V (P) – V (N) = e – I r
[Eq. (3.38) between the positive terminal P and the negative terminal N; I
here is the current flowing from N to P through the cell]. If, while labelling
the current I through the cell one goes from P to N, then of course
V=e+Ir (3.60)
Having clarified labelling, we now state the rules and the proof:
(a) Junction rule: At any junction, the sum of the currents entering
the junction is equal to the sum of currents leaving the junction
(Fig. 3.15). 97

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This applies equally well if instead of a junction of
several lines, we consider a point in a line.
The proof of this rule follows from the fact that
when currents are steady, there is no accumulation
of charges at any junction or at any point in a line.
Thus, the total current flowing in, (which is the rate
at which charge flows into the junction), must equal
the total current flowing out.
(b) Loop rule: The algebraic sum of changes in
potential around any closed loop involving
resistors and cells in the loop is zero
FIGURE 3.15 At junction a the current (Fig. 3.15).
leaving is I1 + I2 and current entering is I3. This rule is also obvious, since electric potential is
The junction rule says I3 = I1 + I2. At point dependent on the location of the point. Thus
h current entering is I1. There is only one starting with any point if we come back to the same
current leaving h and by junction rule
point, the total change must be zero. In a closed
that will also be I1. For the loops ‘ahdcba’
loop, we do come back to the starting point and
and ‘ahdefga’, the loop rules give –30I1 –
41 I3 + 45 = 0 and –30I1 + 21 I2 – 80 = 0. hence the rule.

Example 3.5 A battery of 10 V and negligible internal resistance is

connected across the diagonally opposite corners of a cubical network
consisting of 12 resistors each of resistance 1 W (Fig. 3.16). Determine
the equivalent resistance of the network and the current along each
edge of the cube.

Z
EXAMPLE 3.5

FIGURE 3.16
98

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Electricity

Solution The network is not reducible to a simple series and parallel

combinations of resistors. There is, however, a clear symmetry in the
problem which we can exploit to obtain the equivalent resistance of
the network.
The paths AA¢, AD and AB are obviously symmetrically placed in the
network. Thus, the current in each must be the same, say, I. Further,
at the corners A¢, B and D, the incoming current I must split equally
into the two outgoing branches. In this manner, the current in all
the 12 edges of the cube are easily written down in terms of I, using
Kirchhoff’s first rule and the symmetry in the problem.
Next take a closed loop, say, ABCC¢EA, and apply Kirchhoff’s second
rule:
–IR – (1/2)IR – IR + e = 0
where R is the resistance of each edge and e the emf of battery. Thus,
5
e= IR
2

http://www.phys.hawaii.edu/~teb/optics/java/kirch3/
Similation for application of Kirchhoff ’s rules:
The equivalent resistance Req of the network is
ε 5
Req = = R
3I 6

EXAMPLE 3.5
For R = 1 W, Req = (5/6) W and for e = 10 V, the total current (= 3I ) in
the network is
3I = 10 V/(5/6) W = 12 A, i.e., I = 4 A
The current flowing in each edge can now be read off from the
Fig. 3.16.

It should be noted that because of the symmetry of the network, the

great power of Kirchhoff’s rules has not been very apparent in Example 3.5.
In a general network, there will be no such simplification due to symmetry,
and only by application of Kirchhoff’s rules to junctions and closed loops
(as many as necessary to solve the unknowns in the network) can we
handle the problem. This will be illustrated in Example 3.6.

Example 3.6 Determine the current in each branch of the network

shown in Fig. 3.17.
EXAMPLE 3.6

FIGURE 3.17
99

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Solution Each branch of the network is assigned an unknown current
to be determined by the application of Kirchhoff’s rules. To reduce
the number of unknowns at the outset, the first rule of Kirchhoff is
used at every junction to assign the unknown current in each branch.
We then have three unknowns I 1, I2 and I3 which can be found by
applying the second rule of Kirchhoff to three different closed loops.
Kirchhoff’s second rule for the closed loop ADCA gives,
10 – 4(I1– I2) + 2(I2 + I3 – I1) – I1 = 0 [3.61(a)]
that is, 7I1– 6I2 – 2I3 = 10
For the closed loop ABCA, we get
10 – 4I2– 2 (I2 + I3) – I1 = 0
that is, I1 + 6I2 + 2I3 =10 [3.61(b)]
For the closed loop BCDEB, we get
5 – 2 (I2 + I3 ) – 2 (I2 + I3 – I1) = 0
that is, 2I1 – 4I2 – 4I3 = –5 [3.61(c)]
Equations (3.61 a, b, c) are three simultaneous equations in three
unknowns. These can be solved by the usual method to give
5 7
I1 = 2.5A, I2 = A, I3 = 1 A
8 8
The currents in the various branches of the network are
5 1 7
AB : A, CA : 2 A, DEB : 1 A
8 2 8
7 1
AD : 1 A, CD : 0 A, BC : 2 A
8 2
It is easily verified that Kirchhoff’s second rule applied to the
remaining closed loops does not provide any additional independent
equation, that is, the above values of currents satisfy the second
EXAMPLE 3.6

rule for every closed loop of the network. For example, the total voltage
drop over the closed loop BADEB
5  15 
5 V+  × 4 V−  × 4 V
8   8 
equal to zero, as required by Kirchhoff’s second rule.

3.13 WHEATSTONE BRIDGE

As an application of Kirchhoff’s rules consider the circuit shown in
Fig. 3.18, which is called the Wheatstone bridge. The bridge has
four resistors R1, R2, R3 and R4. Across one pair of diagonally opposite
points (A and C in the figure) a source is connected. This (i.e., AC) is
called the battery arm. Between the other two vertices, B and D, a
galvanometer G (which is a device to detect currents) is connected. This
line, shown as BD in the figure, is called the galvanometer arm.
For simplicity, we assume that the cell has no internal resistance. In
general there will be currents flowing across all the resistors as well as a
current Ig through G. Of special interest, is the case of a balanced bridge
where the resistors are such that Ig = 0. We can easily get the balance
condition, such that there is no current through G. In this case, the
100 Kirchhoff’s junction rule applied to junctions D and B (see the figure)

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Electricity
immediately gives us the relations I1 = I3 and I2 = I4. Next, we apply
Kirchhoff’s loop rule to closed loops ADBA and CBDC. The first
loop gives
–I1 R1 + 0 + I2 R2 = 0 (Ig = 0) (3.62)
and the second loop gives, upon using I3 = I1, I4 = I2
I2 R4 + 0 – I1 R3 = 0 (3.63)
From Eq. (3.62), we obtain,
I1 R2
=
I 2 R1
whereas from Eq. (3.63), we obtain,
I1 R4
=
I 2 R3
Hence, we obtain the condition
R2 R FIGURE 3.18
= 4 [3.64(a)]
R1 R3
This last equation relating the four resistors is called the balance
condition for the galvanometer to give zero or null deflection.
The Wheatstone bridge and its balance condition provide a practical
method for determination of an unknown resistance. Let us suppose we
have an unknown resistance, which we insert in the fourth arm; R4 is
thus not known. Keeping known resistances R1 and R2 in the first and
second arm of the bridge, we go on varying R3 till the galvanometer shows
a null deflection. The bridge then is balanced, and from the balance
condition the value of the unknown resistance R4 is given by,
R
R4 = R3 2 [3.64(b)]
R1
A practical device using this principle is called the meter bridge.

Example 3.7 The four arms of a Wheatstone bridge (Fig. 3.19) have
the following resistances:
AB = 100W, BC = 10W, CD = 5W, and DA = 60W.
EXAMPLE 3.7

FIGURE 3.19 101

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A galvanometer of 15W resistance is connected across BD. Calculate
the current through the galvanometer when a potential difference of
10 V is maintained across AC.
Solution Considering the mesh BADB, we have
100I1 + 15Ig – 60I2 = 0
or 20I1 + 3Ig – 12I2= 0 [3.65(a)]
Considering the mesh BCDB, we have
10 (I1 – Ig) – 15Ig – 5 (I2 + Ig ) = 0
10I1 – 30Ig –5I2 = 0
2I1 – 6Ig – I2 = 0 [3.65(b)]
Considering the mesh ADCEA,
60I2 + 5 (I2 + Ig ) = 10
65I2 + 5Ig = 10
13I2 + Ig = 2 [3.65(c)]
Multiplying Eq. (3.65b) by 10
20I1 – 60Ig – 10I2 = 0 [3.65(d)]
From Eqs. (3.65d) and (3.65a) we have
63Ig – 2I2 = 0
I2 = 31.5Ig [3.65(e)]
EXAMPLE 3.7

Substituting the value of I2 into Eq. [3.65(c)], we get

13 (31.5Ig ) + Ig = 2
410.5 Ig = 2
Ig = 4.87 mA.

SUMMARY

1. Current through a given area of a conductor is the net charge passing

per unit time through the area.
2. To maintain a steady current, we must have a closed circuit in which
an external agency moves electric charge from lower to higher potential
energy. The work done per unit charge by the source in taking the
charge from lower to higher potential energy (i.e., from one terminal
of the source to the other) is called the electromotive force, or emf, of
the source. Note that the emf is not a force; it is the voltage difference
between the two terminals of a source in open circuit.
3. Ohm’s law: The electric current I flowing through a substance is
proportional to the voltage V across its ends, i.e., V µ I or V = RI,
where R is called the resistance of the substance. The unit of resistance
is ohm: 1W = 1 V A–1.

102

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 Current
Electricity

4. The resistance R of a conductor depends on its length l and

cross-sectional area A through the relation,
ρl
R=
A
where r, called resistivity is a property of the material and depends on
temperature and pressure.
5. Electrical resistivity of substances varies over a very wide range. Metals
have low resistivity, in the range of 10–8 W m to 10–6 W m. Insulators
like glass and rubber have 1022 to 1024 times greater resistivity.
Semiconductors like Si and Ge lie roughly in the middle range of
resistivity on a logarithmic scale.
6. In most substances, the carriers of current are electrons; in some
cases, for example, ionic crystals and electrolytic liquids, positive and
negative ions carry the electric current.
7. Current density j gives the amount of charge flowing per second per
unit area normal to the flow,
j = nq vd
where n is the number density (number per unit volume) of charge
carriers each of charge q, and vd is the drift velocity of the charge
carriers. For electrons q = – e. If j is normal to a cross-sectional area
A and is constant over the area, the magnitude of the current I through
the area is nevd A.
8. Using E = V/l, I = nevd A, and Ohm’s law, one obtains
eE ne 2
=ρ vd
m m
The proportionality between the force eE on the electrons in a metal
due to the external field E and the drift velocity vd (not acceleration)
can be understood, if we assume that the electrons suffer collisions
with ions in the metal, which deflect them randomly. If such collisions
occur on an average at a time interval t,
vd = at = eEt/m
where a is the acceleration of the electron. This gives
m
ρ=
ne 2τ
9. In the temperature range in which resistivity increases linearly with
temperature, the temperature coefficient of resistivity a is defined as
the fractional increase in resistivity per unit increase in temperature.
10. Ohm’s law is obeyed by many substances, but it is not a fundamental
law of nature. It fails if
(a) V depends on I non-linearly.
(b) the relation between V and I depends on the sign of V for the same
absolute value of V.
(c) The relation between V and I is non-unique.
An example of (a) is when r increases with I (even if temperature is
kept fixed). A rectifier combines features (a) and (b). GaAs shows the
feature (c).
11. When a source of emf e is connected to an external resistance R, the
voltage Vext across R is given by
ε
Vext = IR = R
R +r
where r is the internal resistance of the source. 103

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 Physics
12. Kirchhoff’s Rules –
(a) Junction Rule: At any junction of circuit elements, the sum of
currents entering the junction must equal the sum of currents
leaving it.
(b) Loop Rule: The algebraic sum of changes in potential around any
closed loop must be zero.
13. The Wheatstone bridge is an arrangement of four resistances – R1, R2,
R3, R4 as shown in the text. The null-point condition is given by
R1 R3
=
R2 R 4
using which the value of one resistance can be determined, knowing
the other three resistances.

Physical Quantity Symbol Dimensions Unit Remark

Electric current I [A] A SI base unit

Charge Q, q [T A] C
2 –3 –1
Voltage, Electric V [M L T A ] V Work/charge
potential difference
Electromotive force e [M L2 T –3 A–1] V Work/charge
2 –3 –2
Resistance R [M L T A ] W R = V/I
Resistivity r [M L3 T –3 A–2] Wm R = rl/A
–1 –3 3 2
Electrical s [M L T A] S s = 1/r
conductivity
–3 –1 –1 Electric force
Electric field E [M L T A ] Vm
charge
eEτ
vd =
–1 –1
Drift speed vd [L T ] ms
m
Relaxation time t [T] s
Current density j [L–2 A] A m–2 current/area
3 –4 –1 2 –1 –1
Mobility m [M L T A ] m V s vd / E

POINTS TO PONDER

1. Current is a scalar although we represent current with an arrow.

Currents do not obey the law of vector addition. That current is a
scalar also follows from it’s definition. The current I through an area
of cross-section is given by the scalar product of two vectors:
I = j . DS
104 where j and DS are vectors.

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 Current
Electricity

2. Refer to V-I curves of a resistor and a diode as drawn in the text. A

resistor obeys Ohm’s law while a diode does not. The assertion that
V = IR is a statement of Ohm’s law is not true. This equation defines
resistance and it may be applied to all conducting devices whether
they obey Ohm’s law or not. The Ohm’s law asserts that the plot of I
versus V is linear i.e., R is independent of V.
Equation E = r j leads to another statement of Ohm’s law, i.e., a
conducting material obeys Ohm’s law when the resistivity of the
material does not depend on the magnitude and direction of applied
electric field.
3. Homogeneous conductors like silver or semiconductors like pure
germanium or germanium containing impurities obey Ohm’s law within
some range of electric field values. If the field becomes too strong,
there are departures from Ohm’s law in all cases.
4. Motion of conduction electrons in electric field E is the sum of (i)
motion due to random collisions and (ii) that due to E. The motion
due to random collisions averages to zero and does not contribute to
vd (Chapter 11, Textbook of Class XI). vd , thus is only due to applied
electric field on the electron.
5. The relation j = r v should be applied to each type of charge carriers
separately. In a conducting wire, the total current and charge density
arises from both positive and negative charges:
j = r+ v+ + r– v–
r = r+ + r–
Now in a neutral wire carrying electric current,
r + = – r–
Further, v+ ~ 0 which gives
r=0
j = r– v
Thus, the relation j = r v does not apply to the total current charge
density.
6. Kirchhoff’s junction rule is based on conservation of charge and the
outgoing currents add up and are equal to incoming current at a
junction. Bending or reorienting the wire does not change the validity
of Kirchhoff’s junction rule.

EXERCISES
3.1 The storage battery of a car has an emf of 12 V. If the internal
resistance of the battery is 0.4 W, what is the maximum current
that can be drawn from the battery?
3.2 A battery of emf 10 V and internal resistance 3 W is connected to a
resistor. If the current in the circuit is 0.5 A, what is the resistance
of the resistor? What is the terminal voltage of the battery when the
circuit is closed?
3.3 At room temperature (27.0 °C) the resistance of a heating element
is 100 W. What is the temperature of the element if the resistance is
found to be 117 W, given that the temperature coefficient of the
material of the resistor is 1.70 × 10–4 °C–1. 105

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 Physics
3.4 A negligibly small current is passed through a wire of length 15 m
and uniform cross-section 6.0 × 10 –7 m 2, and its resistance is
measured to be 5.0 W. What is the resistivity of the material at the
temperature of the experiment?
3.5 A silver wire has a resistance of 2.1 W at 27.5 °C, and a resistance
of 2.7 W at 100 °C. Determine the temperature coefficient of
resistivity of silver.
3.6 A heating element using nichrome connected to a 230 V supply
draws an initial current of 3.2 A which settles after a few seconds to
a steady value of 2.8 A. What is the steady temperature of the heating
element if the room temperature is 27.0 °C? Temperature coefficient
of resistance of nichrome averaged over the temperature range
involved is 1.70 × 10–4 °C–1.
3.7 Determine the current in each branch of the network shown in
Fig. 3.20:

FIGURE 3.20

3.8 A storage battery of emf 8.0 V and internal resistance 0.5 W is being
charged by a 120 V dc supply using a series resistor of 15.5 W. What
is the terminal voltage of the battery during charging? What is the
purpose of having a series resistor in the charging circuit?
3.9 The number density of free electrons in a copper conductor
estimated in Example 3.1 is 8.5 × 1028 m–3. How long does an electron
take to drift from one end of a wire 3.0 m long to its other end? The
area of cross-section of the wire is 2.0 × 10–6 m2 and it is carrying a
current of 3.0 A.

106

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Chapter Four

MOVING CHARGES
AND MAGNETISM

4.1 INTRODUCTION
Both Electricity and Magnetism have been known for more than 2000
years. However, it was only about 200 years ago, in 1820, that it was
realised that they were intimately related. During a lecture demonstration
in the summer of 1820, Danish physicist Hans Christian Oersted noticed
that a current in a straight wire caused a noticeable deflection in a nearby
magnetic compass needle. He investigated this phenomenon. He found
that the alignment of the needle is tangential to an imaginary circle which
has the straight wire as its centre and has its plane perpendicular to the
wire. This situation is depicted in Fig.4.1(a). It is noticeable when the
current is large and the needle sufficiently close to the wire so that the
earth’s magnetic field may be ignored. Reversing the direction of the
current reverses the orientation of the needle [Fig. 4.1(b)]. The deflection
increases on increasing the current or bringing the needle closer to the
wire. Iron filings sprinkled around the wire arrange themselves in
concentric circles with the wire as the centre [Fig. 4.1(c)]. Oersted
concluded that moving charges or currents produced a magnetic field
in the surrounding space.
Following this, there was intense experimentation. In 1864, the laws
obeyed by electricity and magnetism were unified and formulated by

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 Physics
James Maxwell who then realised that light was electromagnetic waves.
Radio waves were discovered by Hertz, and produced by J.C.Bose and
G. Marconi by the end of the 19th century. A remarkable scientific and
technological progress took place in the 20th century. This was due to
our increased understanding of electromagnetism and the invention of
devices for production, amplification, transmission and detection of
electromagnetic waves.

FIGURE 4.1 The magnetic field due to a straight long current-carrying

wire. The wire is perpendicular to the plane of the paper. A ring of
compass needles surrounds the wire. The orientation of the needles is
shown when (a) the current emerges out of the plane of the paper,
(b) the current moves into the plane of the paper. (c) The arrangement of
iron filings around the wire. The darkened ends of the needle represent
north poles. The effect of the earth’s magnetic field is neglected.

In this chapter, we will see how magnetic field exerts

forces on moving charged particles, like electrons, protons,
HANS CHRISTIAN OERSTED (1777–1851)

and current-carrying wires. We shall also learn how

currents produce magnetic fields. We shall see how
particles can be accelerated to very high energies in a
cyclotron. We shall study how currents and voltages are
detected by a galvanometer.
In this and subsequent Chapter on magnetism,
we adopt the following convention: A current or a
field (electric or magnetic) emerging out of the plane of the
paper is depicted by a dot (¤). A current or a field going
into the plane of the paper is depicted by a cross (  )*.
Hans Christian Oersted Figures. 4.1(a) and 4.1(b) correspond to these two
(1777–1851) Danish situations, respectively.
physicist and chemist,
professor at Copenhagen.
He observed that a
4.2 MAGNETIC FORCE
compass needle suffers a 4.2.1 Sources and fields
deflection when placed
near a wire carrying an Before we introduce the concept of a magnetic field B, we
electric current. This shall recapitulate what we have learnt in Chapter 1 about
discovery gave the first the electric field E. We have seen that the interaction
empirical evidence of a between two charges can be considered in two stages.
connection between electric The charge Q, the source of the field, produces an electric
and magnetic phenomena.
field E, where

* A dot appears like the tip of an arrow pointed at you, a cross is like the feathered
108 tail of an arrow moving away from you.

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 Moving Charges and
Magnetism
E = Q r̂ / (4pe0)r2 (4.1)
where r̂ is unit vector along r, and the field E is a vector
field. A charge q interacts with this field and experiences
a force F given by
F = q E = q Q r̂ / (4pe0) r 2 (4.2)
As pointed out in the Chapter 1, the field E is not just
an artefact but has a physical role. It can convey energy
and momentum and is not established instantaneously
but takes finite time to propagate. The concept of a field
was specially stressed by Faraday and was incorporated
by Maxwell in his unification of electricity and magnetism.
In addition to depending on each point in space, it can
also vary with time, i.e., be a function of time. In our

HENDRIK ANTOON LORENTZ (1853 – 1928)

Hendrik Antoon Lorentz
discussions in this chapter, we will assume that the fields (1853 – 1928) Dutch
do not change with time. theoretical physicist,
The field at a particular point can be due to one or professor at Leiden. He
investigated the
more charges. If there are more charges the fields add
relationship between
vectorially. You have already learnt in Chapter 1 that this electricity, magnetism, and
is called the principle of superposition. Once the field is mechanics. In order to
known, the force on a test charge is given by Eq. (4.2). explain the observed effect
Just as static charges produce an electric field, the of magnetic fields on
currents or moving charges produce (in addition) a emitters of light (Zeeman
magnetic field, denoted by B (r), again a vector field. It effect), he postulated the
has several basic properties identical to the electric field. existence of electric charges
in the atom, for which he
It is defined at each point in space (and can in addition
was awarded the Nobel Prize
depend on time). Experimentally, it is found to obey the in 1902. He derived a set of
principle of superposition: the magnetic field of several transformation equations
sources is the vector addition of magnetic field of each (known after him, as
individual source. Lorentz transformation
equations) by some tangled
4.2.2 Magnetic Field, Lorentz Force mathematical arguments,
but he was not aware that
Let us suppose that there is a point charge q (moving
these equations hinge on a
with a velocity v and, located at r at a given time t) in
new concept of space and
presence of both the electric field E (r) and the magnetic time.
field B (r). The force on an electric charge q due to both
of them can be written as
F = q [ E (r) + v × B (r)] º Felectric +Fmagnetic (4.3)
This force was given first by H.A. Lorentz based on the extensive
experiments of Ampere and others. It is called the Lorentz force. You
have already studied in detail the force due to the electric field. If we
look at the interaction with the magnetic field, we find the following
features.
(i) It depends on q, v and B (charge of the particle, the velocity and the
magnetic field). Force on a negative charge is opposite to that on a
positive charge.
(ii) The magnetic force q [ v × B ] includes a vector product of velocity
and magnetic field. The vector product makes the force due to magnetic 109

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 Physics
field vanish (become zero) if velocity and magnetic field are parallel
or anti-parallel. The force acts in a (sideways) direction perpendicular
to both the velocity and the magnetic field. Its
direction is given by the screw rule or right hand
rule for vector (or cross) product as illustrated
in Fig. 4.2.
(iii) The magnetic force is zero if charge is not
moving (as then |v|= 0). Only a moving
charge feels the magnetic force.
The expression for the magnetic force helps
us to define the unit of the magnetic field, if one
FIGURE 4.2 The direction of the magnetic takes q, F and v, all to be unity in the force
force acting on a charged particle. (a) The equation F = q [ v × B] =q v B sin q n̂ , where q is
force on a positively charged particle with the angle between v and B [see Fig. 4.2 (a)]. The
velocity v and making an angle q with the magnitude of magnetic field B is 1 SI unit, when
magnetic field B is given by the right-hand
the force acting on a unit charge (1 C), moving
rule. (b) A moving charged particle q is
perpendicular to B with a speed 1m/s, is one
deflected in an opposite sense to –q in the
presence of magnetic field. newton.
Dimensionally, we have [B] = [F/qv] and the unit
of B are Newton second / (coulomb metre). This unit is called tesla ( T )
named after Nikola Tesla (1856 – 1943). Tesla is a rather large unit. A
smaller unit (non-SI) called gauss (=10–4 tesla) is also often used. The
earth’s magnetic field is about 3.6 × 10–5 T.

4.2.3 Magnetic force on a current-carrying conductor

We can extend the analysis for force due to magnetic field on a single
moving charge to a straight rod carrying current. Consider a rod of a
uniform cross-sectional area A and length l. We shall assume one kind
of mobile carriers as in a conductor (here electrons). Let the number
density of these mobile charge carriers in it be n. Then the total number
of mobile charge carriers in it is nlA. For a steady current I in this
conducting rod, we may assume that each mobile carrier has an average
drift velocity vd (see Chapter 3). In the presence of an external magnetic
field B, the force on these carriers is:
F = (nlA)q vd ´ B
where q is the value of the charge on a carrier. Now nq vd is the current
density j and |(nq vd )|A is the current I (see Chapter 3 for the discussion
of current and current density). Thus,
F = [(nq vd )l A] × B = [ jAl ] ´ B
= Il ´ B (4.4)
where l is a vector of magnitude l, the length of the rod, and with a direction
identical to the current I. Note that the current I is not a vector. In the last
step leading to Eq. (4.4), we have transferred the vector sign from j to l.
Equation (4.4) holds for a straight rod. In this equation, B is the
external magnetic field. It is not the field produced by the current-carrying
rod. If the wire has an arbitrary shape we can calculate the Lorentz force
on it by considering it as a collection of linear strips dlj and summing
F   Idl j × B
j
110
This summation can be converted to an integral in most cases.

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 Moving Charges and
Magnetism

Example 4.1 A straight wire of mass 200 g and length 1.5 m carries
a current of 2 A. It is suspended in mid-air by a uniform horizontal
magnetic field B (Fig. 4.3). What is the magnitude of the magnetic
field?

FIGURE 4.3

Solution From Eq. (4.4), we find that there is an upward force F, of

magnitude IlB,. For mid-air suspension, this must be balanced by
the force due to gravity:
m g = I lB

http://www.phys.hawaii.edu/~teb/optics/java/partmagn/index.html
Interactive demonstration:
Charged particles moving in a magnetic field.
mg
B=
Il

EXAMPLE 4.1
0.2 × 9.8
= = 0.65 T
2 × 1.5
Note that it would have been sufficient to specify m/l, the mass per
unit length of the wire. The earth’s magnetic field is approximately
4 × 10–5 T and we have ignored it.

Example 4.2 If the magnetic field is parallel to the positive y-axis

and the charged particle is moving along the positive x-axis (Fig. 4.4),
which way would the Lorentz force be for (a) an electron (negative
charge), (b) a proton (positive charge).

FIGURE 4.4
EXAMPLE 4.2

Solution The velocity v of particle is along the x-axis, while B, the

magnetic field is along the y-axis, so v × B is along the z-axis (screw
rule or right-hand thumb rule). So, (a) for electron it will be along –z
axis. (b) for a positive charge (proton) the force is along +z axis.

111

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 Physics
4.3 MOTION IN A MAGNETIC FIELD
We will now consider, in greater detail, the motion of a charge moving in
a magnetic field. We have learnt in Mechanics (see Class XI book, Chapter
6) that a force on a particle does work if the force has a component along
(or opposed to) the direction of motion of the particle. In the case of motion
of a charge in a magnetic field, the magnetic force is perpendicular to the
velocity of the particle. So no work is done and no change in the magnitude
of the velocity is produced (though the direction of momentum may be
changed). [Notice that this is unlike the force due to an electric field, q E,
which can have a component parallel (or antiparallel) to motion and thus
can transfer energy in addition to momentum.]
We shall consider motion of a charged particle in a uniform magnetic
field. First consider the case of v perpendicular to B. The
perpendicular force, q v × B, acts as a centripetal force and
produces a circular motion perpendicular to the magnetic field.
The particle will describe a circle if v and B are perpendicular
to each other (Fig. 4.5).
If velocity has a component along B, this component
remains unchanged as the motion along the magnetic field will
not be affected by the magnetic field. The motion in a plane
perpendicular to B is as before a circular one, thereby producing
a helical motion (Fig. 4.6).
You have already learnt in earlier classes (See Class XI,
Chapter 4) that if r is the radius of the circular path of a particle,
then a force of m v2 / r, acts perpendicular to the path towards
the centre of the circle, and is called the centripetal force. If the
FIGURE 4.5 Circular motion velocity v is perpendicular to the magnetic field B, the magnetic
force is perpendicular to both v and B and acts
like a centripetal force. It has a magnitude q v
B. Equating the two expressions for centripetal
force,
m v 2/r = q v B, which gives
r = m v / qB (4.5)
for the radius of the circle described by the
charged particle. The larger the momentum, the
larger is the radius and bigger the circle
described. If w is the angular frequency, then v
= w r. So,
w = 2p n = q B/ m [4.6(a)]
which is independent of the velocity or energy .
Here n is the frequency of rotation. The
independence of n from energy has important
application in the design of a cyclotron (see
Section 4.4.2).
The time taken for one revolution is T= 2p/w
FIGURE 4.6 Helical motion º 1/n. If there is a component of the velocity
112 parallel to the magnetic field (denoted by v||), it will make the particle

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 Moving Charges and
Magnetism
move along the field and the path of the particle would be a helical one
(Fig. 4.6). The distance moved along the magnetic field in one rotation is
called pitch p. Using Eq. [4.6 (a)], we have
p = v||T = 2pm v|| / q B [4.6(b)]
The radius of the circular component of motion is called the radius of
the helix.

Example 4.3 What is the radius of the path of an electron (mass

9 × 10-31 kg and charge 1.6 × 10–19 C) moving at a speed of 3 ×107 m/s in
a magnetic field of 6 × 10–4 T perpendicular to it? What is its
frequency? Calculate its energy in keV. ( 1 eV = 1.6 × 10–19 J).
Solution Using Eq. (4.5) we find

EXAMPLE 4.3
r = m v / (qB ) = 9 ×10–31 kg × 3 × 107 m s–1 / ( 1.6 × 10–19 C × 6 × 10–4 T )
= 28 × 10–2 m = 28 cm
n = v / (2 pr) = 17×106 s–1 = 17×106 Hz =17 MHz.
E = (½ )mv 2 = (½ ) 9 × 10–31 kg × 9 × 1014 m2/s2 = 40.5 ×10–17 J
≈ 4×10 J = 2.5 keV.
–16

4.4 MAGNETIC FIELD DUE TO A CURRENT ELEMENT,

BIOT-SAVART LAW
All magnetic fields that we know are due to currents (or moving
charges) and due to intrinsic magnetic moments of particles.
Here, we shall study the relation between current and the
magnetic field it produces. It is given by the Biot-Savart’s law.
Fig. 4.7 shows a finite conductor XY carrying current I. Consider
an infinitesimal element dl of the conductor. The magnetic field
dB due to this element is to be determined at a point P which is at
a distance r from it. Let q be the angle between dl and the
displacement vector r. According to Biot-Savart’s law, the
magnitude of the magnetic field dB is proportional to the current
I, the element length |dl|, and inversely proportional to the square
of the distance r. Its direction* is perpendicular to the plane
containing dl and r . Thus, in vector notation,
I dl × r FIGURE 4.7 Illustration of
dB ∝ the Biot-Savart law. The
r3
current element I dl
µ I dl × r produces a field dB at a
= 0 [4.11(a)] distance r. The Ä sign
4π r 3 indicates that the
where m0/4p is a constant of proportionality. The above expression field is perpendicular
holds when the medium is vacuum. to the plane of this
page and directed
into it.
* The sense of dl × r is also given by the Right Hand Screw rule : Look at the
plane containing vectors dl and r. Imagine moving from the first vector towards
second vector. If the movement is anticlockwise, the resultant is towards you.
If it is clockwise, the resultant is away from you. 113

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 Physics
The magnitude of this field is,
µ0 I dl sin θ
dB = [4.11(b)]
4π r2
where we have used the property of cross-product. Equation [4.11 (a)]
constitutes our basic equation for the magnetic field. The proportionality
constant in SI units has the exact value,
µ0
= 10 − 7 Tm/A [4.11(c)]
4π
We call m0 the permeability of free space (or vacuum).
The Biot-Savart law for the magnetic field has certain similarities, as
well as, differences with the Coulomb’s law for the electrostatic field. Some
of these are:
(i) Both are long range, since both depend inversely on the square of
distance from the source to the point of interest. The principle of
superposition applies to both fields. [In this connection, note that
the magnetic field is linear in the source I dl just as the electrostatic
field is linear in its source: the electric charge.]
(ii) The electrostatic field is produced by a scalar source, namely, the electric
charge. The magnetic field is produced by a vector source I dl.
(iii) The electrostatic field is along the displacement vector joining the
source and the field point. The magnetic field is perpendicular to the
plane containing the displacement vector r and the current element I dl.
(iv) There is an angle dependence in the Biot-Savart law which is not
present in the electrostatic case. In Fig. 4.7, the magnetic field at any
point in the direction of dl (the dashed line) is zero. Along this line,
q = 0, sin q = 0 and from Eq. [4.11(a)], |dB| = 0.
There is an interesting relation between e0, the permittivity of free
space; m , the permeability of free space; and c, the speed of light in vacuum:
0

µ0
ε 0 µ 0 = ( 4 πε 0 )
4π
=
1
9 × 109
(10 ) = (3 × 10
−7 1
8 2
)
=
1
c2
We will discuss this connection further in Chapter 8 on the
electromagnetic waves. Since the speed of light in vacuum is constant,
the product m0e0 is fixed in magnitude. Choosing the value of either e0 or
m0, fixes the value of the other. In SI units, m0 is fixed to be equal to
4p × 10–7 in magnitude.

Example 4.5 An element  l  x ˆi is placed at the origin and carries

a large current I = 10 A (Fig. 4.8). What is the magnetic field on the y-
axis at a distance of 0.5 m. Dx = 1 cm.
EXAMPLE 4.5

114 FIGURE 4.8

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 Moving Charges and
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Solution
µ 0 I dl sin θ
|dB | = [using Eq. (4.11)]
4π r2
Tm
dl = ∆x = 10 −2 m , I = 10 A, r = 0.5 m = y, µ0 / 4 π = 10 −7
A
q = 90° ; sin q = 1
10−7 × 10 × 10 −2
dB = = 4 × 10–8 T
25 × 10 −2
The direction of the field is in the +z-direction. This is so since,

( )

EXAMPLE 4.5
dl × r  x ˆi × y ˆj = y ∆x ˆi × ˆj = y ∆x k
ˆ
We remind you of the following cyclic property of cross-products,
ˆi × ˆj = k
ˆ ; ˆj × k
ˆ = ˆi ; k
ˆ × ˆi = ˆj
Note that the field is small in magnitude.

In the next section, we shall use the Biot-Savart law to calculate the
magnetic field due to a circular loop.

4.5 MAGNETIC FIELD ON THE AXIS OF A CIRCULAR

CURRENT LOOP
In this section, we shall evaluate the magnetic field due
to a circular coil along its axis. The evaluation entails
summing up the effect of infinitesimal current elements
(I dl) mentioned in the previous section. We assume
that the current I is steady and that the evaluation is
carried out in free space (i.e., vacuum).
Fig. 4.9 depicts a circular loop carrying a steady
current I. The loop is placed in the y-z plane with its
centre at the origin O and has a radius R. The x-axis is
the axis of the loop. We wish to calculate the magnetic
field at the point P on this axis. Let x be the distance of
P from the centre O of the loop.
Consider a conducting element dl of the loop. This is
FIGURE 4.9 Magnetic field on the
shown in Fig. 4.9. The magnitude dB of the magnetic
axis of a current carrying circular
field due to dl is given by the Biot-Savart law [Eq. 4.11(a)], loop of radius R. Shown are the
0 I dl × r magnetic field dB (due to a line
dB  (4.12) element dl ) and its
4 r3 components along and
Now r 2 = x 2 + R 2 . Further, any element of the loop perpendicular to the axis.
will be perpendicular to the displacement vector from
the element to the axial point. For example, the element dl in Fig. 4.9 is
in the y-z plane, whereas, the displacement vector r from dl to the axial
point P is in the x-y plane. Hence |dl × r|=r dl. Thus,
µ0 Idl
dB =
(
4π x + R 2
2
) (4.13) 115

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The direction of dB is shown in Fig. 4.9. It is perpendicular to the
plane formed by dl and r. It has an x-component dBx and a component
perpendicular to x-axis, dB^. When the components perpendicular to
the x-axis are summed over, they cancel out and we obtain a null result.
For example, the dB^ component due to dl is cancelled by the contribution
due to the diametrically opposite dl element, shown in
Fig. 4.9. Thus, only the x-component survives. The net contribution along
x-direction can be obtained by integrating dBx = dB cos q over the loop.
For Fig. 4.9,
R
cos θ = (4.14)
(x 2 + R 2 )1/ 2
From Eqs. (4.13) and (4.14),
µ0 Idl R
dB x =
(x )
3/2
4π 2
+ R2
The summation of elements dl over the loop yields 2pR, the
circumference of the loop. Thus, the magnetic field at P due to entire
circular loop is
µ0 I R 2
B = Bx ˆi = ˆi
( )
2 3/2 (4.15)
2 x +R 2

As a special case of the above result, we may obtain the field at the centre
of the loop. Here x = 0, and we obtain,
µ0 I ˆ
B0 = i (4.16)
2R
The magnetic field lines due to a circular wire form closed loops and
are shown in Fig. 4.10. The direction of the magnetic field is given by
(another) right-hand thumb rule stated below:
Curl the palm of your right hand around the circular wire with the
fingers pointing in the direction of the current. The right-hand thumb
gives the direction of the magnetic field.

FIGURE 4.10 The magnetic field lines for a current loop. The direction of
the field is given by the right-hand thumb rule described in the text. The
upper side of the loop may be thought of as the north pole and the lower
116 side as the south pole of a magnet.

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Example 4.6 A straight wire carrying a current of 12 A is bent into a

semi-circular arc of radius 2.0 cm as shown in Fig. 4.11(a). Consider
the magnetic field B at the centre of the arc. (a) What is the magnetic
field due to the straight segments? (b) In what way the contribution
to B from the semicircle differs from that of a circular loop and in
what way does it resemble? (c) Would your answer be different if the
wire were bent into a semi-circular arc of the same radius but in the
opposite way as shown in Fig. 4.11(b)?

FIGURE 4.11

Solution
(a) dl and r for each element of the straight segments are parallel.
Therefore, dl × r = 0. Straight segments do not contribute to
|B|.
(b) For all segments of the semicircular arc, dl × r are all parallel to
each other (into the plane of the paper). All such contributions

EXAMPLE 4.6
add up in magnitude. Hence direction of B for a semicircular arc
is given by the right-hand rule and magnitude is half that of a
circular loop. Thus B is 1.9 × 10–4 T normal to the plane of the
paper going into it.
(c) Same magnitude of B but opposite in direction to that in (b).

Example 4.7 Consider a tightly wound 100 turn coil of radius 10 cm,
carrying a current of 1 A. What is the magnitude of the magnetic
field at the centre of the coil?
Solution Since the coil is tightly wound, we may take each circular
EXAMPLE 4.7

element to have the same radius R = 10 cm = 0.1 m. The number of

turns N = 100. The magnitude of the magnetic field is,

µ0 NI 4 π × 10 –7 × 102 × 1
B= = −4
= 2π × 10 −4 = 6.28 × 10 T
2R 2 × 10 –1

4.6 AMPERE’S CIRCUITAL LAW

There is an alternative and appealing way in which the
Biot-Savart law may be expressed. Ampere’s circuital law
considers an open surface with a boundary (Fig. 4.14).
The surface has current passing through it. We consider
the boundary to be made up of a number of small line
elements. Consider one such element of length dl. We
take the value of the tangential component of the
magnetic field, Bt, at this element and multiply it by the FIGURE 4.12 117

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length of that element dl [Note: Btdl=B.d l]. All such
products are added together. We consider the limit as the
lengths of elements get smaller and their number gets
larger. The sum then tends to an integral. Ampere’s law
states that this integral is equal to m0 times the total
current passing through the surface, i.e.,
“ B.dl = m I 0
[4.17(a)]
where I is the total current through the surface. The
integral is taken over the closed loop coinciding with the
boundary C of the surface. The relation above involves a
sign-convention, given by the right-hand rule. Let the
fingers of the right-hand be curled in the sense the
“
boundary is traversed in the loop integral B.dl. Then
the direction of the thumb gives the sense in which the
Andre Ampere (1775 –
1836) Andre Marie Ampere
current I is regarded as positive.
was a French physicist, For several applications, a much simplified version of
mathematician and chemist Eq. [4.17(a)] proves sufficient. We shall assume that, in
who founded the science of such cases, it is possible to choose the loop (called
electrodynamics. Ampere an amperian loop) such that at each point of the
was a child prodigy loop, either
who mastered advanced (i) B is tangential to the loop and is a non-zero constant
mathematics by the age of
B, or
12. Ampere grasped the
significance of Oersted’s (ii) B is normal to the loop, or
discovery. He carried out a (iii) B vanishes.
large series of experiments Now, let L be the length (part) of the loop for which B
to explore the relationship is tangential. Let Ie be the current enclosed by the loop.
between current electricity
Then, Eq. (4.17) reduces to,
and magnetism. These
investigations culminated BL =m0Ie [4.17(b)]
ANDRE AMPERE (1775 –1836)

in 1827 with the

When there is a system with a symmetry such as for
publication of the
‘Mathematical Theory of
a straight infinite current-carrying wire in Fig. 4.13, the
Electrodynamic Pheno- Ampere’s law enables an easy evaluation of the magnetic
mena Deduced Solely from field, much the same way Gauss’ law helps in
Experiments’. He hypo- determination of the electric field. This is exhibited in the
thesised that all magnetic Example 4.9 below. The boundary of the loop chosen is
phenomena are due to a circle and magnetic field is tangential to the
circulating electric circumference of the circle. The law gives, for the left hand
currents. Ampere was side of Eq. [4.17 (b)], B. 2pr. We find that the magnetic
humble and absent-
field at a distance r outside the wire is tangential and
minded. He once forgot an
given by
invitation to dine with the
Emperor Napoleon. He died B × 2pr = m0 I,
of pneumonia at the age of
61. His gravestone bears B = m0 I/ (2pr) (4.18)
the epitaph: Tandem Felix The above result for the infinite wire is interesting
(Happy at last). from several points of view.
(i) It implies that the field at every point on a circle of
radius r, (with the wire along the axis), is same in
118 magnitude. In other words, the magnetic field

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 Moving Charges and
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possesses what is called a cylindrical symmetry. The field that
normally can depend on three coordinates depends only on one: r.
Whenever there is symmetry, the solutions simplify.
(ii) The field direction at any point on this circle is tangential to it.
Thus, the lines of constant magnitude of magnetic field form
concentric circles. Notice now, in Fig. 4.1(c), the iron filings form
concentric circles. These lines called magnetic field lines form closed
loops. This is unlike the electrostatic field lines which originate
from positive charges and end at negative charges. The expression
for the magnetic field of a straight wire provides a theoretical
justification to Oersted’s experiments.
(iii) Another interesting point to note is that even though the wire is
infinite, the field due to it at a non-zero distance is not infinite. It
tends to blow up only when we come very close to the wire. The
field is directly proportional to the current and inversely
proportional to the distance from the (infinitely long) current source.
(iv) There exists a simple rule to determine the direction of the magnetic
field due to a long wire. This rule, called the right-hand rule*, is:
Grasp the wire in your right hand with your extended thumb pointing
in the direction of the current. Your fingers will curl around in the
direction of the magnetic field.
Ampere’s circuital law is not new in content from Biot-Savart law.
Both relate the magnetic field and the current, and both express the same
physical consequences of a steady electrical current. Ampere’s law is to
Biot-Savart law, what Gauss’s law is to Coulomb’s law. Both, Ampere’s
and Gauss’s law relate a physical quantity on the periphery or boundary
(magnetic or electric field) to another physical quantity, namely, the source,
in the interior (current or charge). We also note that Ampere’s circuital
law holds for steady currents which do not fluctuate with time. The
following example will help us understand what is meant by the term
enclosed current.
Example 4.8 Figure 4.13 shows a long straight wire of a circular
cross-section (radius a) carrying steady current I. The current I is
uniformly distributed across this cross-section. Calculate the
magnetic field in the region r < a and r > a.
EXAMPLE 4.8

FIGURE 4.13

* Note that there are two distinct right-hand rules: One which gives the direction
of B on the axis of current-loop and the other which gives direction of B
for a straight conducting wire. Fingers and thumb play different roles in
the two.
119

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Solution (a) Consider the case r > a . The Amperian loop, labelled 2,
is a circle concentric with the cross-section. For this loop,
L =2pr
Ie = Current enclosed by the loop = I
The result is the familiar expression for a long straight wire
B (2p r) = m0I
µ0 I
B= [4.19(a)]
2πr

1
B∝ (r > a)
r
Now the current enclosed I e is not I, but is less than this value.
Since the current distribution is uniform, the current enclosed is,
 πr 2  Ir 2
Ie = I  2 =
 πa  a2

I r2
Using Ampere’s law, B (2 π r ) = µ0
a2

 µ I 
B =  0 2r [4.19(b)]
 2 pa 
Bµr (r < a)

FIGURE 4.14

Figure (4.14) shows a plot of the magnitude of B with distance r

from the centre of the wire. The direction of the field is tangential to
EXAMPLE 4.8

the respective circular loop (1 or 2) and given by the right-hand

rule described earlier in this section.
This example possesses the required symmetry so that Ampere’s
law can be applied readily.

It should be noted that while Ampere’s circuital law holds for any
loop, it may not always facilitate an evaluation of the magnetic field in
every case. For example, for the case of the circular loop discussed in
Section 4.6, it cannot be applied to extract the simple expression
B = m0I/2R [Eq. (4.16)] for the field at the centre of the loop. However,
there exists a large number of situations of high symmetry where the law
120 can be conveniently applied. We shall use it in the next section to calculate

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the magnetic field produced by two commonly used and very useful
magnetic systems: the solenoid and the toroid.

4.7 THE SOLENOID

We shall discuss a long solenoid. By long solenoid we mean that the
solenoid’s length is large compared to its radius. It consists of a long wire
wound in the form of a helix where the neighbouring turns are closely
spaced. So each turn can be regarded as a circular loop. The net magnetic
field is the vector sum of the fields due to all the turns. Enamelled wires
are used for winding so that turns are insulated from each other.

FIGURE 4.15 (a) The magnetic field due to a section of the solenoid which has been
stretched out for clarity. Only the exterior semi-circular part is shown. Notice
how the circular loops between neighbouring turns tend to cancel.
(b) The magnetic field of a finite solenoid.

Figure 4.15 displays the magnetic field lines for a finite solenoid. We
show a section of this solenoid in an enlarged manner in Fig. 4.15(a).
Figure 4.15(b) shows the entire finite solenoid with its magnetic field. In
Fig. 4.15(a), it is clear from the circular loops that the field between two
neighbouring turns vanishes. In Fig. 4.15(b), we see that the field at the
interior mid-point P is uniform, strong and along the axis of the solenoid.
The field at the exterior mid-point Q is weak and moreover is along the
axis of the solenoid with no perpendicular or normal component. As the

FIGURE 4.16 The magnetic field of a very long solenoid. We consider a

rectangular Amperian loop abcd to determine the field. 121

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solenoid is made longer it appears like a long cylindrical metal sheet.
Figure 4.16 represents this idealised picture. The field outside the solenoid
approaches zero. We shall assume that the field outside is zero. The field
inside becomes everywhere parallel to the axis.
Consider a rectangular Amperian loop abcd. Along cd the field is zero
as argued above. Along transverse sections bc and ad, the field component
is zero. Thus, these two sections make no contribution. Let the field along
ab be B. Thus, the relevant length of the Amperian loop is, L = h.
Let n be the number of turns per unit length, then the total number
of turns is nh. The enclosed current is, Ie = I (n h), where I is the current
in the solenoid. From Ampere’s circuital law [Eq. 4.17 (b)]
BL = m0Ie, B h = m0I (n h)
B = m0 n I (4.20)
The direction of the field is given by the right-hand rule. The solenoid
is commonly used to obtain a uniform magnetic field. We shall see in the
next chapter that a large field is possible by inserting a soft iron core
inside the solenoid.

Example 4.9 A solenoid of length 0.5 m has a radius of 1 cm and is

made up of 500 turns. It carries a current of 5 A. What is the
magnitude of the magnetic field inside the solenoid?
Solution The number of turns per unit length is,
500
n = = 1000 turns/m
0.5
EXAMPLE 4.9

The length l = 0.5 m and radius r = 0.01 m. Thus, l/a = 50 i.e., l >> a .
Hence, we can use the long solenoid formula, namely, Eq. (4.20)
B = m0n I
= 4p × 10–7 × 103 × 5
= 6.28 × 10–3 T

4.8 FORCE BETWEEN TWO PARALLEL

CURRENTS, THE AMPERE
We have learnt that there exists a magnetic field due to a
conductor carrying a current which obeys the Biot-Savart
law. Further, we have learnt that an external magnetic field
will exert a force on a current-carrying conductor. This
follows from the Lorentz force formula. Thus, it is logical
to expect that two current-carrying conductors placed near
each other will exert (magnetic) forces on each other. In
the period 1820-25, Ampere studied the nature of this
FIGURE 4.17 Two long straight magnetic force and its dependence on the magnitude of
parallel conductors carrying steady the current, on the shape and size of the conductors, as
currents Ia and Ib and separated by a
well as, the distances between the conductors. In this
distance d. Ba is the magnetic field set
up by conductor ‘a’ at conductor ‘b’. section, we shall take the simple example of two parallel
current- carrying conductors, which will perhaps help us
122 to appreciate Ampere’s painstaking work.

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 Moving Charges and
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Figure 4.17 shows two long parallel conductors a and b separated
by a distance d and carrying (parallel) currents Ia and Ib, respectively.
The conductor ‘a’ produces, the same magnetic field Ba at all points
along the conductor ‘b’. The right-hand rule tells us that the direction of
this field is downwards (when the conductors are placed horizontally).
Its magnitude is given by Eq. [4.19(a)] or from Ampere’s circuital law,

µ0 I a
Ba =
2πd

The conductor ‘b’ carrying a current Ib will experience a sideways

force due to the field Ba. The direction of this force is towards the
conductor ‘a’ (Verify this). We label this force as Fba, the force on a
segment L of ‘b’ due to ‘a’. The magnitude of this force is given by
Eq. (4.4),
Fba = Ib L Ba
µ0 I a I b
= L (4.23)
2 πd
It is of course possible to compute the force on ‘a’ due to ‘b’. From
considerations similar to above we can find the force Fab, on a segment of
length L of ‘a’ due to the current in ‘b’. It is equal in magnitude to Fba,
and directed towards ‘b’. Thus,
Fba = –Fab (4.24)
Note that this is consistent with Newton’s third Law. Thus, at least for
parallel conductors and steady currents, we have shown that the
Biot-Savart law and the Lorentz force yield results in accordance with
Newton’s third Law*.
We have seen from above that currents flowing in the same direction
attract each other. One can show that oppositely directed currents repel
each other. Thus,
Parallel currents attract, and antiparallel currents repel.
This rule is the opposite of what we find in electrostatics. Like (same
sign) charges repel each other, but like (parallel) currents attract each
other.
Let fba represent the magnitude of the force Fba per unit length. Then,
from Eq. (4.23),
µ0 I a I b
f ba = (4.25)
2πd
The above expression is used to define the ampere (A), which is one of
the seven SI base units.

* It turns out that when we have time-dependent currents and/or charges in

motion, Newton’s third law may not hold for forces between charges and/or
conductors. An essential consequence of the Newton’s third law in mechanics
is conservation of momentum of an isolated system. This, however, holds even
for the case of time-dependent situations with electromagnetic fields, provided 123
the momentum carried by fields is also taken into account.

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The ampere is the value of that steady current which, when maintained
in each of the two very long, straight, parallel conductors of negligible
cross-section, and placed one metre apart in vacuum, would produce
on each of these conductors a force equal to 2 × 10–7 newtons per metre
of length.
This definition of the ampere was adopted in 1946. It is a theoretical
definition. In practice, one must eliminate the effect of the earth’s magnetic
field and substitute very long wires by multiturn coils of appropriate
geometries. An instrument called the current balance is used to measure
this mechanical force.
The SI unit of charge, namely, the coulomb, can now be defined in
terms of the ampere.
When a steady current of 1A is set up in a conductor, the quantity of
charge that flows through its cross-section in 1s is one coulomb (1C).

Example 4.10 The horizontal component of the earth’s magnetic field

at a certain place is 3.0 ×10–5 T and the direction of the field is from
the geographic south to the geographic north. A very long straight
conductor is carrying a steady current of 1A. What is the force per
unit length on it when it is placed on a horizontal table and the
direction of the current is (a) east to west; (b) south to north?
Solution F = I l × B
F = IlB sinq
The force per unit length is
f = F/l = I B sinq
(a) When the current is flowing from east to west,
q = 90°
Hence,
f=IB
= 1 × 3 × 10–5 = 3 × 10–5 N m–1
This is larger than the value 2×10–7 Nm–1 quoted in the definition
of the ampere. Hence it is important to eliminate the effect of the
earth’s magnetic field and other stray fields while standardising
the ampere.
EXAMPLE 4.10

The direction of the force is downwards. This direction may be

obtained by the directional property of cross product of vectors.
(b) When the current is flowing from south to north,
q = 0o
f=0
Hence there is no force on the conductor.

4.9 TORQUE ON CURRENT LOOP, MAGNETIC DIPOLE

4.9.1 Torque on a rectangular current loop in a uniform
magnetic field
We now show that a rectangular loop carrying a steady current I and
placed in a uniform magnetic field experiences a torque. It does not
experience a net force. This behaviour is analogous to that of electric
124 dipole in a uniform electric field (Section 1.12).

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We first consider the simple case when the
rectangular loop is placed such that the uniform
magnetic field B is in the plane of the loop. This is
illustrated in Fig. 4.18(a).
The field exerts no force on the two arms AD and BC
of the loop. It is perpendicular to the arm AB of the loop
and exerts a force F1 on it which is directed into the
plane of the loop. Its magnitude is,
F1 = I b B
Similarly, it exerts a force F2 on the arm CD and F2
is directed out of the plane of the paper.
F2 = I b B = F1
Thus, the net force on the loop is zero. There is a
torque on the loop due to the pair of forces F1 and F2.
Figure 4.18(b) shows a view of the loop from the AD
end. It shows that the torque on the loop tends to rotate
it anticlockwise. This torque is (in magnitude),
a a
τ = F1 + F2
2 2
a a
= IbB + IbB = I (ab ) B
2 2
=IAB (4.26) FIGURE 4.18 (a) A rectangular
current-carrying coil in uniform
where A = ab is the area of the rectangle. magnetic field. The magnetic moment
We next consider the case when the plane of the loop, m points downwards. The torque t is
is not along the magnetic field, but makes an angle with along the axis and tends to rotate the
it. We take the angle between the field and the normal to coil anticlockwise. (b) The couple
acting on the coil.
the coil to be angle q (The previous case corresponds to
q = p/2). Figure 4.19 illustrates this general case.
The forces on the arms BC and DA are equal, opposite, and act along
the axis of the coil, which connects the centres of mass of BC and DA.
Being collinear along the axis they cancel each other, resulting in no net
force or torque. The forces on arms AB and CD are F1 and F2. They too
are equal and opposite, with magnitude,
F1 = F2 = I b B
But they are not collinear! This results in a couple as before. The
torque is, however, less than the earlier case when plane of loop was
along the magnetic field. This is because the perpendicular distance
between the forces of the couple has decreased. Figure 4.19(b) is a view
of the arrangement from the AD end and it illustrates these two forces
constituting a couple. The magnitude of the torque on the loop is,
a a
τ = F1 sin θ + F2 sin θ
2 2
= I ab B sin q
= I A B sin q (4.27)
125

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As q à 0, the perpendicular distance between
the forces of the couple also approaches zero. This
makes the forces collinear and the net force and
torque zero. The torques in Eqs. (4.26) and (4.27)
can be expressed as vector product of the magnetic
moment of the coil and the magnetic field.
We define the magnetic moment of the current
loop as,
m=IA (4.28)
where the direction of the area vector A is given by
the right-hand thumb rule and is directed into
the plane of the paper in Fig. 4.18. Then as the
angle between m and B is q , Eqs. (4.26) and (4.27)
can be expressed by one expression
(4.29)
This is analogous to the electrostatic case
(Electric dipole of dipole moment pe in an electric
field E).
τ = pe × E
As is clear from Eq. (4.28), the dimensions of the
magnetic moment are [A][L2] and its unit is Am2.
FIGURE 4.19 (a) The area vector of the loop From Eq. (4.29), we see that the torque t
ABCD makes an arbitrary angle q with vanishes when m is either parallel or antiparallel
the magnetic field. (b) Top view of to the magnetic field B. This indicates a state of
the loop. The forces F1 and F2 acting equilibrium as there is no torque on the coil (this
on the arms AB and CD
also applies to any object with a magnetic moment
are indicated.
m). When m and B are parallel the equilibrium is
a stable one. Any small rotation of the coil
produces a torque which brings it back to its original position. When
they are antiparallel, the equilibrium is unstable as any rotation produces
a torque which increases with the amount of rotation. The presence of
this torque is also the reason why a small magnet or any magnetic dipole
aligns itself with the external magnetic field.
If the loop has N closely wound turns, the expression for torque, Eq.
(4.29), still holds, with
m=NIA (4.30)

Example 4.11 A 100 turn closely wound circular coil of radius 10 cm

carries a current of 3.2 A. (a) What is the field at the centre of the
coil? (b) What is the magnetic moment of this coil?
The coil is placed in a vertical plane and is free to rotate about a
horizontal axis which coincides with its diameter. A uniform magnetic
EXAMPLE 4.11

field of 2T in the horizontal direction exists such that initially the axis
of the coil is in the direction of the field. The coil rotates through an
angle of 90° under the influence of the magnetic field. (c) What are the
magnitudes of the torques on the coil in the initial and final position?
(d) What is the angular speed acquired by the coil when it has rotated
126 by 90°? The moment of inertia of the coil is 0.1 kg m2.

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 Moving Charges and
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Solution
(a) From Eq. (4.16)

µ0 NI
B=
2R
Here, N = 100; I = 3.2 A, and R = 0.1 m. Hence,
4 × 10 −5 × 10
= (using p ´ 3.2 = 10)
2 × 10 −1
= 2 × 10–3 T
The direction is given by the right-hand thumb rule.
(b) The magnetic moment is given by Eq. (4.30),
m = N I A = N I p r2 = 100 × 3.2 × 3.14 × 10–2 = 10 A m2
The direction is once again given by the right-hand thumb rule.
(c) t = m × B [from Eq. (4.29)]
= m B sin q

Initially, q = 0. Thus, initial torque ti = 0. Finally, q = p/2 (or 90º).

Thus, final torque tf = m B = 10 ´ 2 = 20 N m.
(d) From Newton’s second law,

where I is the moment of inertia of the coil. From chain rule,

d d d d 
  
dt d dt d 
Using this,
I  d  m B sin  d
Integrating from q = 0 to q = p/2,
EXAMPLE 4.11

Example 4.12
(a) A current-carrying circular loop lies on a smooth horizontal plane.
EXAMPLE 4.12

Can a uniform magnetic field be set up in such a manner that

the loop turns around itself (i.e., turns about the vertical axis).
(b) A current-carrying circular loop is located in a uniform external
magnetic field. If the loop is free to turn, what is its orientation
of stable equilibrium? Show that in this orientation, the flux of
127

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the total field (external field + field produced by the loop) is
maximum.
(c) A loop of irregular shape carrying current is located in an external
magnetic field. If the wire is flexible, why does it change to a
circular shape?
Solution
(a) No, because that would require t to be in the vertical direction.
But t = I A × B, and since A of the horizontal loop is in the vertical
direction, t would be in the plane of the loop for any B.
(b) Orientation of stable equilibrium is one where the area vector A
of the loop is in the direction of external magnetic field. In this
EXAMPLE 4.12

orientation, the magnetic field produced by the loop is in the same

direction as external field, both normal to the plane of the loop,
thus giving rise to maximum flux of the total field.
(c) It assumes circular shape with its plane normal to the field to
maximise flux, since for a given perimeter, a circle encloses greater
area than any other shape.

4.9.2 Circular current loop as a magnetic dipole

In this section, we shall consider the elementary magnetic element: the
current loop. We shall show that the magnetic field (at large distances)
due to current in a circular current loop is very similar in behaviour to
the electric field of an electric dipole. In Section 4.6, we have evaluated
the magnetic field on the axis of a circular loop, of a radius R, carrying a
steady current I. The magnitude of this field is [(Eq. (4.15)],
µ0 I R 2
B=
( )
3/2
2 x 2 + R2
and its direction is along the axis and given by the right-hand thumb
rule (Fig. 4.12). Here, x is the distance along the axis from the centre of
the loop. For x >> R, we may drop the R 2 term in the denominator. Thus,
µ0 IR 2
B=
2x 3
Note that the area of the loop A = pR2. Thus,
µ0 IA
B=
2 πx 3
As earlier, we define the magnetic moment m to have a magnitude IA,
m = I A. Hence,
µ m
B≃ 0 3
2πx
µ0 2 m
= [4.31(a)]
4π x 3
The expression of Eq. [4.31(a)] is very similar to an expression obtained
earlier for the electric field of a dipole. The similarity may be seen if we
substitute,
128
µ0 → 1/ ε 0

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 Moving Charges and
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m → pe (electrostatic dipole)
B → E (electrostatic field)
We then obtain,
2pe
E=
4 π ε0 x 3
which is precisely the field for an electric dipole at a point on its axis.
considered in Chapter 1, Section 1.10 [Eq. (1.20)].
It can be shown that the above analogy can be carried further. We
had found in Chapter 1 that the electric field on the perpendicular bisector
of the dipole is given by [See Eq.(1.21)],
pe
E≃
4πε0 x 3
where x is the distance from the dipole. If we replace p à m and µ0 → 1/ ε 0
in the above expression, we obtain the result for B for a point in the
plane of the loop at a distance x from the centre. For x >>R,
µ m
B≃ 0 3; x >> R [4.31(b)]
4π x
The results given by Eqs. [4.31(a)] and [4.31(b)] become exact for a
point magnetic dipole.
The results obtained above can be shown to apply to any planar loop:
a planar current loop is equivalent to a magnetic dipole of dipole moment
m = I A, which is the analogue of electric dipole moment p. Note, however,
a fundamental difference: an electric dipole is built up of two elementary
units — the charges (or electric monopoles). In magnetism, a magnetic
dipole (or a current loop) is the most elementary element. The equivalent
of electric charges, i.e., magnetic monopoles, are not known to exist.
We have shown that a current loop (i) produces a magnetic field and
behaves like a magnetic dipole at large distances, and
(ii) is subject to torque like a magnetic needle. This led Ampere to suggest
that all magnetism is due to circulating currents. This seems to be partly
true and no magnetic monopoles have been seen so far. However,
elementary particles such as an electron or a proton also carry an intrinsic
magnetic moment, not accounted by circulating currents.

4.10 THE MOVING COIL GALVANOMETER

Currents and voltages in circuits have been discussed extensively in
Chapters 3. But how do we measure them? How do we claim that
current in a circuit is 1.5 A or the voltage drop across a resistor is 1.2 V?
Figure 4.20 exhibits a very useful instrument for this purpose: the moving
coil galvanometer (MCG). It is a device whose principle can be understood
on the basis of our discussion in Section 4.10.
The galvanometer consists of a coil, with many turns, free to rotate
about a fixed axis (Fig. 4.20), in a uniform radial magnetic field. There is
a cylindrical soft iron core which not only makes the field radial but also
increases the strength of the magnetic field. When a current flows through
the coil, a torque acts on it. This torque is given by Eq. (4.26) to be
129
t = NI AB

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 Physics
where the symbols have their usual meaning. Since
the field is radial by design, we have taken sin q = 1 in
the above expression for the torque. The magnetic
torque NIAB tends to rotate the coil. A spring Sp
provides a counter torque k f that balances the
magnetic torque NIAB; resulting in a steady angular
deflection f. In equilibrium
kf = NI AB
where k is the torsional constant of the spring; i.e. the
restoring torque per unit twist. The deflection f is
indicated on the scale by a pointer attached to the
spring. We have
 NAB 
φ=
 k 
I (4.38)
The quantity in brackets is a constant for a given
galvanometer.
The galvanometer can be used in a number of ways.
It can be used as a detector to check if a current is
FIGURE 4.20 The moving coil flowing in the circuit. We have come across this usage
galvanometer. Its elements are in the Wheatstone’s bridge arrangement. In this usage
described in the text. Depending on the neutral position of the pointer (when no current is
the requirement, this device can be flowing through the galvanometer) is in the middle of
used as a current detector or for the scale and not at the left end as shown in Fig.4.20.
measuring the value of the current Depending on the direction of the current, the pointer’s
(ammeter) or voltage (voltmeter). deflection is either to the right or the left.
The galvanometer cannot as such be used as an
ammeter to measure the value of the current in a given circuit. This is for
two reasons: (i) Galvanometer is a very sensitive device, it gives a full-
scale deflection for a current of the order of mA. (ii) For measuring currents,
the galvanometer has to be connected in series, and as it has a large
resistance, this will change the value of the current in the circuit. To
overcome these difficulties, one attaches a small resistance rs, called shunt
resistance, in parallel with the galvanometer coil; so that most of the
current passes through the shunt. The resistance of this arrangement is,
RG rs / (RG + rs ) ≃ rs if RG >> rs
If rs has small value, in relation to the resistance of the rest of the
circuit Rc, the effect of introducing the measuring instrument is also small
and negligible. This arrangement is schematically shown in Fig. 4.21.
FIGURE 4.21 The scale of this ammeter is calibrated and then graduated to read off
Conversion of a the current value with ease. We define the current sensitivity of the
galvanometer (G) to galvanometer as the deflection per unit current. From Eq. (4.38) this
an ammeter by the current sensitivity is,
introduction of a
shunt resistance rs of φ NAB
= (4.39)
very small value in I k
parallel. A convenient way for the manufacturer to increase the sensitivity is
to increase the number of turns N. We choose galvanometers having
130 sensitivities of value, required by our experiment.

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The galvanometer can also be used as a voltmeter to measure the
voltage across a given section of the circuit. For this it must be connected
in parallel with that section of the circuit. Further, it must draw a very
small current, otherwise the voltage measurement will disturb the original
set up by an amount which is very large. Usually we like to keep the
disturbance due to the measuring device below one per cent. To ensure
this, a large resistance R is connected in series with the galvanometer.
This arrangement is schematically depicted in Fig.4.22. Note that the
resistance of the voltmeter is now,
FIGURE 4.22
RG + R ≃ R : large Conversion of a
The scale of the voltmeter is calibrated to read off the voltage value galvanometer (G) to a
with ease. We define the voltage sensitivity as the deflection per unit voltmeter by the
voltage. From Eq. (4.38), introduction of a
resistance R of large
φ  NAB  I  NAB  1 value in series.
=  =  (4.40)
V  k V  k R
An interesting point to note is that increasing the current sensitivity
may not necessarily increase the voltage sensitivity. Let us take Eq. (4.39)
which provides a measure of current sensitivity. If N ® 2N, i.e., we double
the number of turns, then
φ φ
→2
I I
Thus, the current sensitivity doubles. However, the resistance of the
galvanometer is also likely to double, since it is proportional to the length
of the wire. In Eq. (4.40), N ®2N, and R ®2R, thus the voltage sensitivity,
φ φ
→
V V
remains unchanged. So in general, the modification needed for conversion
of a galvanometer to an ammeter will be different from what is needed for
converting it into a voltmeter.
Example 4.13 In the circuit (Fig. 4.23) the current is to be
measured. What is the value of the current if the ammeter shown
(a) is a galvanometer with a resistance R G = 60.00 W; (b) is a
galvanometer described in (a) but converted to an ammeter by a
shunt resistance r s = 0.02 W; (c) is an ideal ammeter with zero
resistance?
EXAMPLE 4.13

FIGURE 4.23 131

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 Physics
Solution
(a) Total resistance in the circuit is,
RG + 3 = 63 Ω . Hence, I = 3/63 = 0.048 A.
(b) Resistance of the galvanometer converted to an ammeter is,
RG rs 60 Ω × 0.02Ω
=
EXAMPLE 4.13

RG + rs (60 + 0.02)Ω ≃ 0.02W

Total resistance in the circuit is,
0.02 Ω + 3 Ω = 3.02 Ω . Hence, I = 3/3.02 = 0.99 A.
(c) For the ideal ammeter with zero resistance,
I = 3/3 = 1.00 A

SUMMARY

1. The total force on a charge q moving with velocity v in the presence of

magnetic and electric fields B and E, respectively is called the Lorentz
force. It is given by the expression:
F = q (v × B + E)
The magnetic force q (v × B) is normal to v and work done by it is zero.
2. A straight conductor of length l and carrying a steady current I
experiences a force F in a uniform external magnetic field B,
F=Il×B
where|l| = l and the direction of l is given by the direction of the
current.
3. In a uniform magnetic field B, a charge q executes a circular orbit in
a plane normal to B. Its frequency of uniform circular motion is called
the cyclotron frequency and is given by:
qB
νc =
2 πm
This frequency is independent of the particle’s speed and radius. This
fact is exploited in a machine, the cyclotron, which is used to
accelerate charged particles.
4. The Biot-Savart law asserts that the magnetic field dB due to an
element dl carrying a steady current I at a point P at a distance r from
the current element is:
µ0 dl × r
dB = I
4π r3
To obtain the total field at P, we must integrate this vector expression
over the entire length of the conductor.
5. The magnitude of the magnetic field due to a circular coil of radius R
carrying a current I at an axial distance x from the centre is

132

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 Moving Charges and
Magnetism

µ0 IR 2
B=
2( x + R 2 )3 / 2
2

At the centre this reduces to

µ0 I
B=
2R
6. Ampere’s Circuital Law: Let an open surface S be bounded by a loop
C. Then the Ampere’s law states that
∫
Ñ B.d l = µ I where I refers to
C
0

the current passing through S. The sign of I is determined from the

right-hand rule. We have discussed a simplified form of this law. If B
is directed along the tangent to every point on the perimeter L of a
closed curve and is constant in magnitude along perimeter then,
BL = m0 Ie
where Ie is the net current enclosed by the closed circuit.
7. The magnitude of the magnetic field at a distance R from a long,
straight wire carrying a current I is given by:
µ0 I
B=
2πR
The field lines are circles concentric with the wire.
8. The magnitude of the field B inside a long solenoid carrying a current
I is
B = m0nI
where n is the number of turns per unit length.
where N is the total number of turns and r is the average radius.
9. Parallel currents attract and anti-parallel currents repel.
10. A planar loop carrying a current I, having N closely wound turns, and
an area A possesses a magnetic moment m where,
m=NIA
and the direction of m is given by the right-hand thumb rule : curl
the palm of your right hand along the loop with the fingers pointing
in the direction of the current. The thumb sticking out gives the
direction of m (and A)
When this loop is placed in a uniform magnetic field B, the force F on
it is: F = 0
And the torque on it is,
t=m×B
In a moving coil galvanometer, this torque is balanced by a counter-
torque due to a spring, yielding
kf = NI AB
where f is the equilibrium deflection and k the torsion constant of
the spring.
11. A moving coil galvanometer can be converted into a ammeter by
introducing a shunt resistance rs, of small value in parallel. It can be
converted into a voltmeter by introducing a resistance of a large value
in series.

133

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Physical Quantity Symbol Nature Dimensions Units Remarks

Permeability of free m0 Scalar [MLT –2A–2] T m A–1 4p ´ 10–7 T m A–1

space

Magnetic Field B Vector [M T –2A–1] T (telsa)

Magnetic Moment m Vector [L2A] A m2 or J/T

Torsion Constant k Scalar [M L2T –2] N m rad–1 Appears in MCG

POINTS TO PONDER

1. Electrostatic field lines originate at a positive charge and terminate at a

negative charge or fade at infinity. Magnetic field lines always form
closed loops.
2. The discussion in this Chapter holds only for steady currents which do
not vary with time.
When currents vary with time Newton’s third law is valid only if momentum
carried by the electromagnetic field is taken into account.
3. Recall the expression for the Lorentz force,
F = q (v × B + E)
This velocity dependent force has occupied the attention of some of the
greatest scientific thinkers. If one switches to a frame with instantaneous
velocity v, the magnetic part of the force vanishes. The motion of the
charged particle is then explained by arguing that there exists an
appropriate electric field in the new frame. We shall not discuss the
details of this mechanism. However, we stress that the resolution of this
paradox implies that electricity and magnetism are linked phenomena
(electromagnetism) and that the Lorentz force expression does not imply
a universal preferred frame of reference in nature.
4. Ampere’s Circuital law is not independent of the Biot-Savart law. It
can be derived from the Biot-Savart law. Its relationship to the
Biot-Savart law is similar to the relationship between Gauss’s law and
Coulomb’s law.

EXERCISES
4.1 A circular coil of wire consisting of 100 turns, each of radius 8.0 cm
carries a current of 0.40 A. What is the magnitude of the magnetic
field B at the centre of the coil?
4.2 A long straight wire carries a current of 35 A. What is the magnitude
of the field B at a point 20 cm from the wire?
4.3 A long straight wire in the horizontal plane carries a current of 50 A
in north to south direction. Give the magnitude and direction of B
134 at a point 2.5 m east of the wire.

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4.4 A horizontal overhead power line carries a current of 90 A in east to
west direction. What is the magnitude and direction of the magnetic
field due to the current 1.5 m below the line?
4.5 What is the magnitude of magnetic force per unit length on a wire
carrying a current of 8 A and making an angle of 30º with the
direction of a uniform magnetic field of 0.15 T ?
4.6 A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid
perpendicular to its axis. The magnetic field inside the solenoid is
given to be 0.27 T. What is the magnetic force on the wire?
4.7 Two long and parallel straight wires A and B carrying currents of
8.0 A and 5.0 A in the same direction are separated by a distance of
4.0 cm. Estimate the force on a 10 cm section of wire A.
4.8 A closely wound solenoid 80 cm long has 5 layers of windings of 400
turns each. The diameter of the solenoid is 1.8 cm. If the current
carried is 8.0 A, estimate the magnitude of B inside the solenoid
near its centre.
4.9 A square coil of side 10 cm consists of 20 turns and carries a current
of 12 A. The coil is suspended vertically and the normal to the plane
of the coil makes an angle of 30º with the direction of a uniform
horizontal magnetic field of magnitude 0.80 T. What is the magnitude
of torque experienced by the coil?
4.10 Two moving coil meters, M1 and M2 have the following particulars:
R1 = 10 W, N1 = 30,
A1 = 3.6 × 10–3 m2, B1 = 0.25 T
R2 = 14 W, N2 = 42,
A2 = 1.8 × 10–3 m2, B2 = 0.50 T
(The spring constants are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage
sensitivity of M2 and M1.
4.11 In a chamber, a uniform magnetic field of 6.5 G (1 G = 10 –4 T ) is
maintained. An electron is shot into the field with a speed of
4.8 × 106 m s–1 normal to the field. Explain why the path of the
electron is a circle. Determine the radius of the circular orbit.
(e = 1.5 × 10–19 C, me = 9.1×10–31 kg )
4.12 In Exercise 4.11 obtain the frequency of revolution of the electron in
its circular orbit. Does the answer depend on the speed of the
electron? Explain.
4.13 (a) A circular coil of 30 turns and radius 8.0 cm carrying a current
of 6.0 A is suspended vertically in a uniform horizontal magnetic
field of magnitude 1.0 T. The field lines make an angle of 60°
with the normal of the coil. Calculate the magnitude of the
counter torque that must be applied to prevent the coil from
turning.
(b) Would your answer change, if the circular coil in (a) were replaced
by a planar coil of some irregular shape that encloses the same
area? (All other particulars are also unaltered.)

135

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Chapter Five

MAGNETISM AND
MATTER

5.1 INTRODUCTION
Magnetic phenomena are universal in nature. Vast, distant galaxies, the
tiny invisible atoms, humans and beasts all are permeated through and
through with a host of magnetic fields from a variety of sources. The earth’s
magnetism predates human evolution. The word magnet is derived from
the name of an island in Greece called magnesia where magnetic ore
deposits were found, as early as 600 BC.
In the previous chapter we have learned that moving charges or electric
currents produce magnetic fields. This discovery, which was made in the
early part of the nineteenth century is credited to Oersted, Ampere, Biot
and Savart, among others.
In the present chapter, we take a look at magnetism as a subject in its
own right.
Some of the commonly known ideas regarding magnetism are:
(i) The earth behaves as a magnet with the magnetic field pointing
approximately from the geographic south to the north.
(ii) When a bar magnet is freely suspended, it points in the north-south
direction. The tip which points to the geographic north is called the
north pole and the tip which points to the geographic south is called
136 the south pole of the magnet.

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 Magnetism and
Matter
(iii) There is a repulsive force when north poles ( or south poles ) of two
magnets are brought close together. Conversely, there is an attractive
force between the north pole of one magnet and the south pole of
the other.
(iv) We cannot isolate the north, or south pole of a magnet. If a bar magnet
is broken into two halves, we get two similar bar magnets with
somewhat weaker properties. Unlike electric charges, isolated magnetic
north and south poles known as magnetic monopoles do not exist.
(v) It is possible to make magnets out of iron and its alloys.
We begin with a description of a bar magnet and its behaviour in an
external magnetic field. We describe Gauss’s law of magnetism. We then
follow it up with an account of the earth’s magnetic field. We next describe
how materials can be classified on the basis of their magnetic properties.
We describe para-, dia-, and ferromagnetism. We conclude with a section
on electromagnets and permanent magnets.

5.2 THE BAR MAGNET

One of the earliest childhood memories of the famous physicist Albert
Einstein was that of a magnet gifted to him by a relative. Einstein was
fascinated, and played endlessly with it. He wondered how the magnet
could affect objects such as nails or pins placed away from it and not in
any way connected to it by a spring or string.
We begin our study by examining iron filings sprinkled on a sheet of
glass placed over a short bar magnet. The arrangement of iron filings is
shown in Fig. 5.1.
The pattern of iron filings suggests that the magnet has two poles
similar to the positive and negative charge of an electric dipole. As
FIGURE 5.1 The
mentioned in the introductory section, one pole is designated the North
arrangement of iron
pole and the other, the South pole. When suspended freely, these poles filings surrounding a
point approximately towards the geographic north and south poles, bar magnet. The
respectively. A similar pattern of iron filings is observed around a current pattern mimics
carrying solenoid. magnetic field lines.
The pattern suggests
5.2.1 The magnetic field lines that the bar magnet
The pattern of iron filings permits us to plot the magnetic field lines*. This is is a magnetic dipole.
shown both for the bar-magnet and the current-carrying solenoid in
Fig. 5.2. For comparison refer to the Chapter 1, Figure 1.17(d). Electric field
lines of an electric dipole are also displayed in Fig. 5.2(c). The magnetic field
lines are a visual and intuitive realisation of the magnetic field. Their
properties are:
(i) The magnetic field lines of a magnet (or a solenoid) form continuous
closed loops. This is unlike the electric dipole where these field lines
begin from a positive charge and end on the negative charge or escape
to infinity.

* In some textbooks the magnetic field lines are called magnetic lines of force.
This nomenclature is avoided since it can be confusing. Unlike electrostatics
the field lines in magnetism do not indicate the direction of the force on a 137
(moving) charge.

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FIGURE 5.2 The field lines of (a) a bar magnet, (b) a current-carrying finite solenoid and
(c) electric dipole. At large distances, the field lines are very similar. The curves
labelled i and ii are closed Gaussian surfaces.

(ii) The tangent to the field line at a given

point represents the direction of the net
magnetic field B at that point.
(iii) The larger the number of field lines
crossing per unit area, the stronger is
the magnitude of the magnetic field B.
In Fig. 5.2(a), B is larger around region
ii than in region i .
(iv) The magnetic field lines do not
intersect, for if they did, the direction
of the magnetic field would not be
unique at the point of intersection.
One can plot the magnetic field lines
in a variety of ways. One way is to place a
small magnetic compass needle at various
positions and note its orientation. This
gives us an idea of the magnetic field
direction at various points in space.

5.2.2 Bar magnet as an equivalent

FIGURE 5.3 Calculation of (a) The axial field of a
finite solenoid in order to demonstrate its similarity solenoid
to that of a bar magnet. (b) A magnetic needle In the previous chapter, we have explained
in a uniform magnetic field B. The how a current loop acts as a magnetic
arrangement may be used to dipole (Section 4.10). We mentioned
determine either B or the magnetic
Ampere’s hypothesis that all magnetic
moment m of the needle.
phenomena can be explained in terms of
138 circulating currents.

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 Magnetism and
Matter
The resemblance of magnetic field lines for a bar magnet and a solenoid
suggest that a bar magnet may be thought of as a large number of
circulating currents in analogy with a solenoid. Cutting a bar magnet in
half is like cutting a solenoid. We get two smaller solenoids with weaker
magnetic properties. The field lines remain continuous, emerging from
one face of the solenoid and entering into the other face. One can test this
analogy by moving a small compass needle in the neighbourhood of a
bar magnet and a current-carrying finite solenoid and noting that the
deflections of the needle are similar in both cases.
To make this analogy more firm we calculate the axial field of a finite
solenoid depicted in Fig. 5.3 (a). We shall demonstrate that at large
distances this axial field resembles that of a bar magnet.
µ 0 2m
B = (5.1)
4π r 3
This is also the far axial magnetic field of a bar magnet which one may
obtain experimentally. Thus, a bar magnet and a solenoid produce similar
magnetic fields. The magnetic moment of a bar magnet is thus equal to
the magnetic moment of an equivalent solenoid that produces the same
magnetic field.

5.2.3 The dipole in a uniform magnetic field

Let’s place a small compass needle of known magnetic moment m and
allowing it to oscillate in the magnetic field. This arrangement is shown in
Fig. 5.3(b).
The torque on the needle is [see Eq. (4.23)],
t=m×B (5.2)
In magnitude t = mB sinq
Here t is restoring torque and q is the angle between m and B.
An expression for magnetic potential energy can also be obtained on
lines similar to electrostatic potential energy.
The magnetic potential energy Um is given by
U m = ∫ τ (θ )dθ

= ∫ mB sin θ dθ = −mB cos θ

= –m.B (5.3)
We have emphasised in Chapter 2 that the zero of potential energy
can be fixed at one’s convenience. Taking the constant of integration to be
zero means fixing the zero of potential energy at q = 90°, i.e., when the
needle is perpendicular to the field. Equation (5.6) shows that potential
energy is minimum (= –mB) at q = 0° (most stable position) and maximum
(= +mB) at q = 180° (most unstable position).

Example 5.1
EXAMPLE 5.1

(a) What happens if a bar magnet is cut into two pieces: (i) transverse
to its length, (ii) along its length?
(b) A magnetised needle in a uniform magnetic field experiences a
torque but no net force. An iron nail near a bar magnet, however,
experiences a force of attraction in addition to a torque. Why? 139

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(c) Must every magnetic configuration have a north pole and a south
pole? What about the field due to a toroid?
(d) Two identical looking iron bars A and B are given, one of which
is definitely known to be magnetised. (We do not know which
one.) How would one ascertain whether or not both are
magnetised? If only one is magnetised, how does one ascertain
which one? [Use nothing else but the bars A and B.]
Solution
(a) In either case, one gets two magnets, each with a north and
south pole.
(b) No force if the field is uniform. The iron nail experiences a non-
uniform field due to the bar magnet. There is induced magnetic
moment in the nail, therefore, it experiences both force and
torque. The net force is attractive because the induced south
pole (say) in the nail is closer to the north pole of magnet than
induced north pole.
(c) Not necessarily. True only if the source of the field has a net
non-zero magnetic moment. This is not so for a toroid or even for
a straight infinite conductor.
(d) Try to bring different ends of the bars closer. A repulsive force in
some situation establishes that both are magnetised. If it is
always attractive, then one of them is not magnetised. In a bar
magnet the intensity of the magnetic field is the strongest at the
two ends (poles) and weakest at the central region. This fact
may be used to determine whether A or B is the magnet. In this
case, to see which one of the two bars is a magnet, pick up one,
EXAMPLE 5.1

(say, A) and lower one of its ends; first on one of the ends of the
other (say, B), and then on the middle of B. If you notice that in
the middle of B, A experiences no force, then B is magnetised. If
you do not notice any change from the end to the middle of B,
then A is magnetised.

5.2.4 The electrostatic analog

Comparison of Eqs. (5.2), (5.3) and (5.6) with the corresponding equations
for electric dipole (Chapter 1), suggests that magnetic field at large
distances due to a bar magnet of magnetic moment m can be obtained
from the equation for electric field due to an electric dipole of dipole moment
p, by making the following replacements:
µ
→

1
E →B , p m, → 0
4 πε 0 4π
In particular, we can write down the equatorial field (BE) of a bar magnet
at a distance r, for r >> l, where l is the size of the magnet:
µ0 m
BE = − (5.4)
4 πr 3
Likewise, the axial field (BA) of a bar magnet for r >> l is:
µ0 2m
140 BA = (5.5)
4 π r3

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 Magnetism and
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Equation (5.8) is just Eq. (5.2) in the vector form. Table 5.1 summarises
the analogy between electric and magnetic dipoles.

TABLE 5.1 THE DIPOLE ANALOGY

Electrostatics Magnetism
1/e0 m0
Dipole moment p m
Equatorial Field for a short dipole –p/4pe0r 3 – m0 m / 4p r 3
Axial Field for a short dipole 2p/4pe0r 3 m0 2m / 4p r 3
External Field: torque p×E m×B
External Field: Energy –p.E –m.B

Example 5.2 Figure 5.4 shows a small magnetised needle P placed at

a point O. The arrow shows the direction of its magnetic moment. The
other arrows show different positions (and orientations of the magnetic
moment) of another identical magnetised needle Q.
(a) In which configuration the system is not in equilibrium?
(b) In which configuration is the system in (i) stable, and (ii) unstable
equilibrium?
(c) Which configuration corresponds to the lowest potential energy
among all the configurations shown?

FIGURE 5.4
Solution
Potential energy of the configuration arises due to the potential energy of
one dipole (say, Q) in the magnetic field due to other (P). Use the result
that the field due to P is given by the expression [Eqs. (5.7) and (5.8)]:
µ0 m P
BP = − (on the normal bisector)
4π r 3
µ0 2 mP
EXAMPLE 5.2

BP = (on the axis)

4π r 3
where mP is the magnetic moment of the dipole P.
Equilibrium is stable when mQ is parallel to BP, and unstable when it
is anti-parallel to BP.
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For instance for the configuration Q 3 for which Q is along the
perpendicular bisector of the dipole P, the magnetic moment of Q is
parallel to the magnetic field at the position 3. Hence Q3 is stable.
EXAMPLE 5.2 Thus,
(a) PQ1 and PQ2
(b) (i) PQ3, PQ6 (stable); (ii) PQ5, PQ4 (unstable)
(c) PQ6

5.3 MAGNETISM AND GAUSS’S LAW

In Chapter 1, we studied Gauss’s law for electrostatics.
In Fig 5.3(c), we see that for a closed surface represented
by i , the number of lines leaving the surface is equal to
the number of lines entering it. This is consistent with the
fact that no net charge is enclosed by the surface. However,
KARL FRIEDRICH GAUSS (1777 – 1855)

in the same figure, for the closed surface ii , there is a net

outward flux, since it does include a net (positive) charge.
The situation is radically different for magnetic fields
which are continuous and form closed loops. Examine the
Gaussian surfaces represented by i or ii in Fig 5.3(a) or
Karl Friedrich Gauss Fig. 5.3(b). Both cases visually demonstrate that the
(1777 – 1855) He was a number of magnetic field lines leaving the surface is
child prodigy and was gifted balanced by the number of lines entering it. The net
in mathematics, physics, magnetic flux is zero for both the surfaces. This is true
engineering, astronomy for any closed surface.
and even land surveying.
The properties of numbers
fascinated him, and in his
work he anticipated major
mathematical development
of later times. Along with
Wilhelm Welser, he built the
first electric telegraph in
1833. His mathematical
theory of curved surface
laid the foundation for the
later work of Riemann.
FIGURE 5.5
Consider a small vector area element DS of a closed surface S as in
Fig. 5.5. The magnetic flux through ÄS is defined as DfB = B.DS, where B
is the field at DS. We divide S into many small area elements and calculate
the individual flux through each. Then, the net flux fB is,
φB = ∑ ∆φ B = ∑ B.∆S = 0 (5.6)
’ all ’ ’ all ’

where ‘all’ stands for ‘all area elements DS¢. Compare this with the Gauss’s
law of electrostatics. The flux through a closed surface in that case is
given by
142 q
∑ E.∆S = ε
0

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where q is the electric charge enclosed by the surface.
The difference between the Gauss’s law of magnetism and that for
electrostatics is a reflection of the fact that isolated magnetic poles (also
called monopoles) are not known to exist. There are no sources or sinks
of B; the simplest magnetic element is a dipole or a current loop. All
magnetic phenomena can be explained in terms of an arrangement of
dipoles and/or current loops.
Thus, Gauss’s law for magnetism is:
The net magnetic flux through any closed surface is zero.

Example 5.3 Many of the diagrams given in Fig. 5.7 show magnetic
field lines (thick lines in the figure) wrongly. Point out what is wrong
with them. Some of them may describe electrostatic field lines correctly.
Point out which ones.

EXAMPLE 5.3

FIGURE 5.6
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Solution
(a) Wrong. Magnetic field lines can never emanate from a point, as
shown in figure. Over any closed surface, the net flux of B must
always be zero, i.e., pictorially as many field lines should seem to
enter the surface as the number of lines leaving it. The field lines
shown, in fact, represent electric field of a long positively charged
wire. The correct magnetic field lines are circling the straight
conductor, as described in Chapter 4.
(b) Wrong. Magnetic field lines (like electric field lines) can never cross
each other, because otherwise the direction of field at the point of
intersection is ambiguous. There is further error in the figure.
Magnetostatic field lines can never form closed loops around empty
space. A closed loop of static magnetic field line must enclose a
region across which a current is passing. By contrast, electrostatic
field lines can never form closed loops, neither in empty space,
nor when the loop encloses charges.
(c) Right. Magnetic lines are completely confined within a toroid.
Nothing wrong here in field lines forming closed loops, since each
loop encloses a region across which a current passes. Note, for
clarity of figure, only a few field lines within the toroid have been
shown. Actually, the entire region enclosed by the windings
contains magnetic field.
(d) Wrong. Field lines due to a solenoid at its ends and outside cannot
be so completely straight and confined; such a thing violates
Ampere’s law. The lines should curve out at both ends, and meet
eventually to form closed loops.
(e) Right. These are field lines outside and inside a bar magnet. Note
carefully the direction of field lines inside. Not all field lines emanate
out of a north pole (or converge into a south pole). Around both
the N-pole, and the S-pole, the net flux of the field is zero.
(f ) Wrong. These field lines cannot possibly represent a magnetic field.
Look at the upper region. All the field lines seem to emanate out of
the shaded plate. The net flux through a surface surrounding the
shaded plate is not zero. This is impossible for a magnetic field.
The given field lines, in fact, show the electrostatic field lines
around a positively charged upper plate and a negatively charged
lower plate. The difference between Fig. [5.7(e) and (f )] should be
EXAMPLE 5.3

carefully grasped.
(g) Wrong. Magnetic field lines between two pole pieces cannot be
precisely straight at the ends. Some fringing of lines is inevitable.
Otherwise, Ampere’s law is violated. This is also true for electric
field lines.

Example 5.4
(a) Magnetic field lines show the direction (at every point) along which
a small magnetised needle aligns (at the point). Do the magnetic
field lines also represent the lines of force on a moving charged
particle at every point?
EXAMPLE 5.4

(b) If magnetic monopoles existed, how would the Gauss’s law of

magnetism be modified?
(c) Does a bar magnet exert a torque on itself due to its own field?
Does one element of a current-carrying wire exert a force on another
element of the same wire?
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(d) Magnetic field arises due to charges in motion. Can a system
have magnetic moments even though its net charge is zero?
Solution
(a) No. The magnetic force is always normal to B (remember magnetic
force = qv × B). It is misleading to call magnetic field lines as lines
of force.
(b) Gauss’s law of magnetism states that the flux of B through any
closed surface is always zero ∫s B .∆s = 0 .
If monopoles existed, the right hand side would be equal to the
monopole (magnetic charge) qm enclosed by S. [Analogous to

Gauss’s law of electrostatics, ∫ B.∆s = µ q

S
0 m where qm is the
(monopole) magnetic charge enclosed by S .]
(c) No. There is no force or torque on an element due to the field
produced by that element itself. But there is a force (or torque)
on an element of the same wire. (For the special case of a straight
wire, this force is zero.)

EXAMPLE 5.4
(d) Yes. The average of the charge in the system may be zero. Yet,
the mean of the magnetic moments due to various current loops
may not be zero. We will come across such examples in connection
with paramagnetic material where atoms have net dipole moment
through their net charge is zero.

5.4 MAGNETISATION AND MAGNETIC INTENSITY

The earth abounds with a bewildering variety of elements and compounds.
In addition, we have been synthesising new alloys, compounds and even
elements. One would like to classify the magnetic properties of these
substances. In the present section, we define and explain certain terms
which will help us to carry out this exercise.
We have seen that a circulating electron in an atom has a magnetic
moment. In a bulk material, these moments add up vectorially and they
can give a net magnetic moment which is non-zero. We define
magnetisation M of a sample to be equal to its net magnetic moment per
unit volume:
mnet
M= (5.7)
V
M is a vector with dimensions L–1 A and is measured in a units of A m–1.
Consider a long solenoid of n turns per unit length and carrying a
current I. The magnetic field in the interior of the solenoid was shown to
be given by
B0 = m0 nI (5.8)
If the interior of the solenoid is filled with a material with non-zero
magnetisation, the field inside the solenoid will be greater than B0. The
net B field in the interior of the solenoid may be expressed as
B = B0 + Bm (5.9) 145

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where Bm is the field contributed by the material core. It turns out that
this additional field Bm is proportional to the magnetisation M of the
material and is expressed as
Bm = m0M (5.10)
where m0 is the same constant (permittivity of vacuum) that appears in
Biot-Savart’s law.
It is convenient to introduce another vector field H, called the magnetic
intensity, which is defined by
B
H= –M (5.11)
µ0
where H has the same dimensions as M and is measured in units of A m–1.
Thus, the total magnetic field B is written as
B = m0 (H + M) (5.12)
We repeat our defining procedure. We have partitioned the contribution
to the total magnetic field inside the sample into two parts: one, due to
external factors such as the current in the solenoid. This is represented
by H. The other is due to the specific nature of the magnetic material,
namely M. The latter quantity can be influenced by external factors. This
influence is mathematically expressed as
M = χH (5.13)
where c , a dimensionless quantity, is appropriately called the magnetic
susceptibility. It is a measure of how a magnetic material responds to an
external field. c is small and positive for materials, which are called
paramagnetic. It is small and negative for materials, which are termed
diamagnetic. In the latter case M and H are opposite in direction. From
Eqs. (5.12) and (5.13) we obtain,
B = µ0 (1 + χ )H (5.14)

= m0 mr H

= mH (5.15)
where mr= 1 + c, is a dimensionless quantity called the relative magnetic
permeability of the substance. It is the analog of the dielectric constant in
electrostatics. The magnetic permeability of the substance is m and it has
the same dimensions and units as m0;
m = m0mr = m0 (1+c).
The three quantities c, mr and m are interrelated and only one of
them is independent. Given one, the other two may be easily determined.
EXAMPLE 5.5

Example 5.5 A solenoid has a core of a material with relative

permeability 400. The windings of the solenoid are insulated from the
core and carry a current of 2A. If the number of turns is 1000 per
metre, calculate (a) H, (b) M, (c) B and (d) the magnetising current Im.
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Solution
(a) The field H is dependent of the material of the core, and is
H = nI = 1000 × 2.0 = 2 ×103 A/m.
(b) The magnetic field B is given by
B = mr m0 H
= 400 × 4p ×10–7 (N/A2) × 2 × 103 (A/m)
= 1.0 T
(c) Magnetisation is given by
M = (B– m0 H )/ m0
= (mr m0 H–m0 H )/m0 = (mr – 1)H = 399 × H
@ 8 × 105 A/m

EXAMPLE 5.5
(d) The magnetising current IM is the additional current that needs to
be passed through the windings of the solenoid in the absence of
the core which would give a B value as in the presence of the core.
Thus B = mr n (I + IM). Using I = 2A, B = 1 T, we get IM = 794 A.

5.5 MAGNETIC PROPERTIES OF MATERIALS

The discussion in the previous section helps us to classify materials as
diamagnetic, paramagnetic or ferromagnetic. In terms of the susceptibility
c, a material is diamagnetic if c is negative, para- if c is positive and small,
and ferro- if c is large and positive.
A glance at Table 5.3 gives one a better feeling for these materials.
Here e is a small positive number introduced to quantify paramagnetic
materials. Next, we describe these materials in some detail.

TABLE 5.3
Diamagnetic Paramagnetic Ferromagnetic

–1 £ c < 0 0 < c< e c >> 1

0 £ mr < 1 1< mr < 1+ e mr >> 1
m < m0 m > m0 m >> m0

5.5.1 Diamagnetism
Diamagnetic substances are those which have tendency to move from FIGURE 5.7
stronger to the weaker part of the external magnetic field. In other words, Behaviour of
unlike the way a magnet attracts metals like iron, it would repel a magnetic field lines
diamagnetic substance. near a
Figure 5.7(a) shows a bar of diamagnetic material placed in an external (a) diamagnetic,
magnetic field. The field lines are repelled or expelled and the field inside (b) paramagnetic
the material is reduced. In most cases, this reduction is slight, being one substance.
part in 105. When placed in a non-uniform magnetic field, the bar will tend
to move from high to low field.
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The simplest explanation for diamagnetism is as follows. Electrons in
an atom orbiting around nucleus possess orbital angular momentum.
These orbiting electrons are equivalent to current-carrying loop and thus
possess orbital magnetic moment. Diamagnetic substances are the ones
in which resultant magnetic moment in an atom is zero. When magnetic
field is applied, those electrons having orbital magnetic moment in the
same direction slow down and those in the opposite direction speed up.
This happens due to induced current in accordance with Lenz’s law which
you will study in Chapter 6. Thus, the substance develops a net magnetic
moment in direction opposite to that of the applied field and hence repulsion.
Some diamagnetic materials are bismuth, copper, lead, silicon,
nitrogen (at STP), water and sodium chloride. Diamagnetism is present
in all the substances. However, the effect is so weak in most cases that it
gets shifted by other effects like paramagnetism, ferromagnetism, etc.
The most exotic diamagnetic materials are superconductors. These
are metals, cooled to very low temperatures which exhibits both perfect
conductivity and perfect diamagnetism. Here the field lines are completely
expelled! c = –1 and mr = 0. A superconductor repels a magnet and (by
Newton’s third law) is repelled by the magnet. The phenomenon of perfect
diamagnetism in superconductors is called the Meissner effect, after the
name of its discoverer. Superconducting magnets can be gainfully
exploited in variety of situations, for example, for running magnetically
levitated superfast trains.

5.5.2 Paramagnetism
Paramagnetic substances are those which get weakly magnetised when
placed in an external magnetic field. They have tendency to move from a
region of weak magnetic field to strong magnetic field, i.e., they get weakly
attracted to a magnet.
The individual atoms (or ions or molecules) of a paramagnetic material
possess a permanent magnetic dipole moment of their own. On account
of the ceaseless random thermal motion of the atoms, no net magnetisation
is seen. In the presence of an external field B0, which is strong enough,
and at low temperatures, the individual atomic dipole moment can be
made to align and point in the same direction as B0. Figure 5.7(b) shows
a bar of paramagnetic material placed in an external field. The field lines
gets concentrated inside the material, and the field inside is enhanced. In
most cases, this enhancement is slight, being one part in 105. When placed
in a non-uniform magnetic field, the bar will tend to move from weak field
to strong.
Some paramagnetic materials are aluminium, sodium, calcium,
oxygen (at STP) and copper chloride. For a paramagnetic material both c
and mr depend not only on the material, but also (in a simple fashion) on
the sample temperature. As the field is increased or the temperature is
lowered, the magnetisation increases until it reaches the saturation value
at which point all the dipoles are perfectly aligned with the field.

5.5.3 Ferromagnetism
Ferromagnetic substances are those which gets strongly magnetised when
148 placed in an external magnetic field. They have strong tendency to move

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from a region of weak magnetic field to strong magnetic field, i.e., they get
strongly attracted to a magnet.
The individual atoms (or ions or molecules) in a ferromagnetic material
possess a dipole moment as in a paramagnetic material. However, they
interact with one another in such a way that they spontaneously align
themselves in a common direction over a macroscopic volume called
domain. The explanation of this cooperative effect requires quantum
mechanics and is beyond the scope of this textbook. Each domain has a
net magnetisation. Typical domain size is 1mm and the domain contains
about 1011 atoms. In the first instant, the magnetisation varies randomly
from domain to domain and there is no bulk magnetisation. This is shown
in Fig. 5.8(a). When we apply an external magnetic field B0, the domains
orient themselves in the direction of B0 and simultaneously the domain
oriented in the direction of B0 grow in size. This existence of domains and
their motion in B0 are not speculations. One may observe this under a
microscope after sprinkling a liquid suspension of powdered FIGURE 5.8
ferromagnetic substance of samples. This motion of suspension can be (a) Randomly
observed. Fig. 5.8(b) shows the situation when the domains have aligned oriented domains,
and amalgamated to form a single ‘giant’ domain. (b) Aligned domains.
Thus, in a ferromagnetic material the field lines are highly
concentrated. In non-uniform magnetic field, the sample tends to move
towards the region of high field. We may wonder as to what happens
when the external field is removed. In some ferromagnetic materials the
magnetisation persists. Such materials are called hard magnetic materials
or hard ferromagnets. Alnico, an alloy of iron, aluminium, nickel, cobalt
and copper, is one such material. The naturally occurring lodestone is
another. Such materials form permanent magnets to be used among other
things as a compass needle. On the other hand, there is a class of
ferromagnetic materials in which the magnetisation disappears on removal
of the external field. Soft iron is one such material. Appropriately enough,
such materials are called soft ferromagnetic materials. There are a number
of elements, which are ferromagnetic: iron, cobalt, nickel, gadolinium,
etc. The relative magnetic permeability is >1000!
The ferromagnetic property depends on temperature. At high enough
temperature, a ferromagnet becomes a paramagnet. The domain structure
disintegrates with temperature. This disappearance of magnetisation with
temperature is gradual.

SUMMARY

1. The science of magnetism is old. It has been known since ancient times
that magnetic materials tend to point in the north-south direction; like
magnetic poles repel and unlike ones attract; and cutting a bar magnet
in two leads to two smaller magnets. Magnetic poles cannot be isolated.
2. When a bar magnet of dipole moment m is placed in a uniform magnetic
field B,
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(a) the force on it is zero,
(b) the torque on it is m × B,
(c) its potential energy is –m.B, where we choose the zero of energy at
the orientation when m is perpendicular to B.
3. Consider a bar magnet of size l and magnetic moment m, at a distance
r from its mid-point, where r >>l, the magnetic field B due to this bar
is,
µ0 m
B= (along axis)
2 πr 3

µ0 m
=– (along equator)
4 πr3
4. Gauss’s law for magnetism states that the net magnetic flux through
any closed surface is zero

B Bi S 0
all area
elements S

5. Consider a material placed in an external magnetic field B0. The

magnetic intensity is defined as,
B
H= 0
µ0
The magnetisation M of the material is its dipole moment per unit volume.
The magnetic field B in the material is,
B = m0 (H + M)
6. For a linear material M = c H. So that B = m H and c is called the
magnetic susceptibility of the material. The three quantities, c, the
relative magnetic permeability mr, and the magnetic permeability m are
related as follows:
m = m0 mr
mr = 1+ c
7. Magnetic materials are broadly classified as: diamagnetic, paramagnetic,
and ferromagnetic. For diamagnetic materials c is negative and small
and for paramagnetic materials it is positive and small. Ferromagnetic
materials have large c and are characterised by non-linear relation
between B and H.
8. Substances, which at room temperature, retain their ferromagnetic
property for a long period of time are called permanent magnets.

Physical quantity Symbol Nature Dimensions Units Remarks

Permeability of m0 Scalar [MLT–2 A–2] T m A–1 m0/4p = 10–7

free space
Magnetic field, B Vector [MT–2 A–1] T (tesla) 104 G (gauss) = 1 T
Magnetic induction,
Magnetic flux density
Magnetic moment m Vector [L–2 A] A m2

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Magnetic flux fB Scalar [ML2T–2 A–1] W (weber) W = T m2

Magnetisation M Vector [L–1 A] A m–1 Magnetic moment

Volume
Magnetic intensity H Vector [L–1 A] A m–1 B = m0 (H + M)
Magnetic field
strength
Magnetic c Scalar - - M = cH
susceptibility
Relative magnetic mr Scalar - - B = m0 mr H
permeability
Magnetic permeability m Scalar [MLT–2 A–2] T m A–1 m = m0 mr
N A–2 B=mH

POINTS TO PONDER

1. A satisfactory understanding of magnetic phenomenon in terms of moving

charges/currents was arrived at after 1800 AD. But technological
exploitation of the directional properties of magnets predates this scientific
understanding by two thousand years. Thus, scientific understanding is
not a necessary condition for engineering applications. Ideally, science
and engineering go hand-in-hand, one leading and assisting the other in
tandem.
2. Magnetic monopoles do not exist. If you slice a magnet in half, you get
two smaller magnets. On the other hand, isolated positive and negative
charges exist. There exists a smallest unit of charge, for example, the
electronic charge with value |e| = 1.6 ×10–19 C. All other charges are
integral multiples of this smallest unit charge. In other words, charge is
quantised. We do not know why magnetic monopoles do not exist or why
electric charge is quantised.
3. A consequence of the fact that magnetic monopoles do not exist is that
the magnetic field lines are continuous and form closed loops. In contrast,
the electrostatic lines of force begin on a positive charge and terminate
on the negative charge (or fade out at infinity).
4. A miniscule difference in the value of c, the magnetic susceptibility, yields
radically different behaviour: diamagnetic versus paramagnetic. For
diamagnetic materials c = –10–5 whereas c = +10–5 for paramagnetic
materials.
5. There exists a perfect diamagnet, namely, a superconductor. This is a
metal at very low temperatures. In this case c = –1, mr = 0, m = 0. The
external magnetic field is totally expelled. Interestingly, this material is
also a perfect conductor. However, there exists no classical theory which
ties these two properties together. A quantum-mechanical theory by
Bardeen, Cooper, and Schrieffer (BCS theory) explains these effects. The
BCS theory was proposed in1957 and was eventually recognised by a Nobel
Prize in physics in 1970.

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6. Diamagnetism is universal. It is present in all materials. But it
is weak and hard to detect if the substance is para- or ferromagnetic.
7. We have classified materials as diamagnetic, paramagnetic, and
ferromagnetic. However, there exist additional types of magnetic material
such as ferrimagnetic, anti-ferromagnetic, spin glass, etc. with properties
which are exotic and mysterious.

EXERCISES
5.1 A short bar magnet placed with its axis at 30° with a uniform external
magnetic field of 0.25 T experiences a torque of magnitude equal to
4.5 × 10–2 J. What is the magnitude of magnetic moment of the magnet?
5.2 A short bar magnet of magnetic moment m = 0.32 J T –1 is placed in a
uniform magnetic field of 0.15 T. If the bar is free to rotate in the
plane of the field, which orientation would correspond to its (a) stable,
and (b) unstable equilibrium? What is the potential energy of the
magnet in each case?
5.3 A closely wound solenoid of 800 turns and area of cross section
2.5 × 10–4 m2 carries a current of 3.0 A. Explain the sense in which
the solenoid acts like a bar magnet. What is its associated magnetic
moment?
5.4 If the solenoid in Exercise 5.5 is free to turn about the vertical
direction and a uniform horizontal magnetic field of 0.25 T is applied,
what is the magnitude of torque on the solenoid when its axis makes
an angle of 30° with the direction of applied field?
5.5 A bar magnet of magnetic moment 1.5 J T –1 lies aligned with the
direction of a uniform magnetic field of 0.22 T.
(a) What is the amount of work required by an external torque to
turn the magnet so as to align its magnetic moment: (i) normal
to the field direction, (ii) opposite to the field direction?
(b) What is the torque on the magnet in cases (i) and (ii)?
5.6 A closely wound solenoid of 2000 turns and area of cross-section
1.6 × 10 –4 m2, carrying a current of 4.0 A, is suspended through its
centre allowing it to turn in a horizontal plane.
(a) What is the magnetic moment associated with the solenoid?
(b) What is the force and torque on the solenoid if a uniform
horizontal magnetic field of 7.5 × 10–2 T is set up at an angle of
30° with the axis of the solenoid?
5.7 A short bar magnet has a magnetic moment of 0.48 J T –1. Give the
direction and magnitude of the magnetic field produced by the magnet
at a distance of 10 cm from the centre of the magnet on (a) the axis,
(b) the equatorial lines (normal bisector) of the magnet.

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5.8 A short bar magnet placed in a horizontal plane has its axis aligned
along the magnetic north-south direction. Null points are found on
the axis of the magnet at 14 cm from the centre of the magnet. The
earth’s magnetic field at the place is 0.36 G and the angle of dip is
zero. What is the total magnetic field on the normal bisector of the
magnet at the same distance as the null–point (i.e., 14 cm) from the
centre of the magnet? (At null points, field due to a magnet is equal
and opposite to the horizontal component of earth’s magnetic field.)
5.9 If the bar magnet in exercise 5.13 is turned around by 180°, where
will the new null points be located?

153

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Chapter Six

ELECTROMAGNETIC
INDUCTION

6.1 INTRODUCTION
Electricity and magnetism were considered separate and unrelated
phenomena for a long time. In the early decades of the nineteenth century,
experiments on electric current by Oersted, Ampere and a few others
established the fact that electricity and magnetism are inter-related. They
found that moving electric charges produce magnetic fields. For example,
an electric current deflects a magnetic compass needle placed in its vicinity.
This naturally raises the questions like: Is the converse effect possible?
Can moving magnets produce electric currents? Does the nature permit
such a relation between electricity and magnetism? The answer is
resounding yes! The experiments of Michael Faraday in England and
Joseph Henry in USA, conducted around 1830, demonstrated
conclusively that electric currents were induced in closed coils when
subjected to changing magnetic fields. In this chapter, we will study the
phenomena associated with changing magnetic fields and understand
the underlying principles. The phenomenon in which electric current is
generated by varying magnetic fields is appropriately called
electromagnetic induction.
When Faraday first made public his discovery that relative motion
between a bar magnet and a wire loop produced a small current in the
latter, he was asked, “What is the use of it?” His reply was: “What is the
154 use of a new born baby?” The phenomenon of electromagnetic induction

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Induction
is not merely of theoretical or academic interest but also
of practical utility. Imagine a world where there is no
electricity – no electric lights, no trains, no telephones and
no personal computers. The pioneering experiments of
Faraday and Henry have led directly to the development
of modern day generators and transformers. Today’s
civilisation owes its progress to a great extent to the
discovery of electromagnetic induction.

6.2 THE EXPERIMENTS OF FARADAY AND

HENRY

JOSEPH HENRY (1797 – 1878)

The discovery and understanding of electromagnetic
induction are based on a long series of experiments carried Josheph Henry [1797 –
out by Faraday and Henry. We shall now describe some 1878] American experimental
of these experiments. physicist, professor at
Princeton University and first
Experiment 6.1 director of the Smithsonian
Institution. He made important
Figure 6.1 shows a coil C1* connected to a galvanometer improvements in electro-
G. When the North-pole of a bar magnet is pushed magnets by winding coils of
insulated wire around iron
towards the coil, the pointer in the galvanometer deflects,
pole pieces and invented an
indicating the presence of electric current in the coil. The electromagnetic motor and a
deflection lasts as long as the bar magnet is in motion. new, efficient telegraph. He
The galvanometer does not show any deflection when the discoverd self-induction and
magnet is held stationary. When the magnet is pulled investigated how currents in
away from the coil, the galvanometer shows deflection in one circuit induce currents in
another.
the opposite direction, which indicates reversal of the
current’s direction. Moreover, when the South-pole of
the bar magnet is moved towards or away from the
coil, the deflections in the galvanometer are opposite
to that observed with the North-pole for similar
movements. Further, the deflection (and hence current)
is found to be larger when the magnet is pushed
towards or pulled away from the coil faster. Instead,
when the bar magnet is held fixed and the coil C1 is
moved towards or away from the magnet, the same
effects are observed. It shows that it is the relative
motion between the magnet and the coil that is
responsible for generation (induction) of electric
current in the coil.

Experiment 6.2
FIGURE 6.1 When the bar magnet is
In Fig. 6.2 the bar magnet is replaced by a second coil pushed towards the coil, the pointer in
C2 connected to a battery. The steady current in the the galvanometer G deflects.
coil C2 produces a steady magnetic field. As coil C2 is

* Wherever the term ‘coil’ or ‘loop’ is used, it is assumed that they are made up of
conducting material and are prepared using wires which are coated with insulating
material. 155

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moved towards the coil C 1, the galvanometer shows a
deflection. This indicates that electric current is induced in
coil C1. When C2 is moved away, the galvanometer shows a
deflection again, but this time in the opposite direction. The
deflection lasts as long as coil C2 is in motion. When the coil
C2 is held fixed and C1 is moved, the same effects are observed.
Again, it is the relative motion between the coils that induces
the electric current.

Experiment 6.3
The above two experiments involved relative motion between
a magnet and a coil and between two coils, respectively.
Through another experiment, Faraday showed that this
FIGURE 6.2 Current is
relative motion is not an absolute requirement. Figure 6.3
induced in coil C1 due to motion shows two coils C1 and C2 held stationary. Coil C1 is connected
of the current carrying coil C2. to galvanometer G while the second coil C2 is connected to a
battery through a tapping key K.

FIGURE 6.3 Experimental set-up for Experiment 6.3.

It is observed that the galvanometer shows a momentary deflection

when the tapping key K is pressed. The pointer in the galvanometer returns
to zero immediately. If the key is held pressed continuously, there is no
deflection in the galvanometer. When the key is released, a momentory
deflection is observed again, but in the opposite direction. It is also observed
that the deflection increases dramatically when an iron rod is inserted
into the coils along their axis.

6.3 MAGNETIC FLUX

Faraday’s great insight lay in discovering a simple mathematical relation
to explain the series of experiments he carried out on electromagnetic
induction. However, before we state and appreciate his laws, we must get
familiar with the notion of magnetic flux, F B. Magnetic flux is defined in
156 the same way as electric flux is defined in Chapter 1. Magnetic flux through

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a plane of area A placed in a uniform magnetic field B (Fig. 6.4) can
be written as
F B = B . A = BA cos q (6.1)
where q is angle between B and A. The notion of the area as a vector
has been discussed earlier in Chapter 1. Equation (6.1) can be
extended to curved surfaces and nonuniform fields.
If the magnetic field has different magnitudes and directions at
various parts of a surface as shown in Fig. 6.5, then the magnetic
flux through the surface is given by
Φ = B . dA + B . dA + ... =
B 1 1 2 2 ∑ Bi . dA i
all
(6.2)
FIGURE 6.4 A plane of
where ‘all’ stands for summation over all the area elements dAi
surface area A placed in a
comprising the surface and Bi is the magnetic field at the area element uniform magnetic field B.
dAi. The SI unit of magnetic flux is weber (Wb) or tesla meter
squared (T m2). Magnetic flux is a scalar quantity.

6.4 FARADAY’S LAW OF INDUCTION

From the experimental observations, Faraday arrived at a
conclusion that an emf is induced in a coil when magnetic flux
through the coil changes with time. Experimental observations
discussed in Section 6.2 can be explained using this concept.
The motion of a magnet towards or away from coil C1 in
Experiment 6.1 and moving a current-carrying coil C2 towards
or away from coil C1 in Experiment 6.2, change the magnetic
flux associated with coil C1. The change in magnetic flux induces
emf in coil C1. It was this induced emf which caused electric
current to flow in coil C1 and through the galvanometer. A FIGURE 6.5 Magnetic field Bi
plausible explanation for the observations of Experiment 6.3 is at the i th area element. dAi
as follows: When the tapping key K is pressed, the current in represents area vector of the
i th area element.
coil C2 (and the resulting magnetic field) rises from zero to a
maximum value in a short time. Consequently, the magnetic
flux through the neighbouring coil C1 also increases. It is the change in
magnetic flux through coil C1 that produces an induced emf in coil C1.
When the key is held pressed, current in coil C2 is constant. Therefore,
there is no change in the magnetic flux through coil C1 and the current in
coil C1 drops to zero. When the key is released, the current in C2 and the
resulting magnetic field decreases from the maximum value to zero in a
short time. This results in a decrease in magnetic flux through coil C1
and hence again induces an electric current in coil C1*. The common
point in all these observations is that the time rate of change of magnetic
flux through a circuit induces emf in it. Faraday stated experimental
observations in the form of a law called Faraday’s law of electromagnetic
induction. The law is stated below.

* Note that sensitive electrical instruments in the vicinity of an electromagnet

can be damaged due to the induced emfs (and the resulting currents) when the
electromagnet is turned on or off. 157

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The magnitude of the induced emf in a circuit is equal
to the time rate of change of magnetic flux through the
circuit.
Mathematically, the induced emf is given by
dΦB
ε=– (6.3)
dt
The negative sign indicates the direction of e and hence
the direction of current in a closed loop. This will be
discussed in detail in the next section.
MICHAEL FARADAY (1791–1867)

In the case of a closely wound coil of N turns, change

of flux associated with each turn, is the same. Therefore,
Michael Faraday [1791–
the expression for the total induced emf is given by
1867] Faraday made
numerous contributions to dΦB
science, viz., the discovery ε = –N (6.4)
dt
of electromagnetic
induction, the laws of The induced emf can be increased by increasing the
electrolysis, benzene, and number of turns N of a closed coil.
the fact that the plane of
polarisation is rotated in an
From Eqs. (6.1) and (6.2), we see that the flux can be
electric field. He is also varied by changing any one or more of the terms B, A and
credited with the invention q. In Experiments 6.1 and 6.2 in Section 6.2, the flux is
of the electric motor, the changed by varying B. The flux can also be altered by
electric generator and the changing the shape of a coil (that is, by shrinking it or
transformer. He is widely stretching it) in a magnetic field, or rotating a coil in a
regarded as the greatest magnetic field such that the angle q between B and A
experimental scientist of
changes. In these cases too, an emf is induced in the
the nineteenth century.
respective coils.

Example 6.1 Consider Experiment 6.2. (a) What would you do to obtain
a large deflection of the galvanometer? (b) How would you demonstrate
the presence of an induced current in the absence of a galvanometer?
Solution
(a) To obtain a large deflection, one or more of the following steps can
be taken: (i) Use a rod made of soft iron inside the coil C2, (ii) Connect
the coil to a powerful battery, and (iii) Move the arrangement rapidly
towards the test coil C1.
(b) Replace the galvanometer by a small bulb, the kind one finds in a
small torch light. The relative motion between the two coils will cause
EXAMPLE 6.1

the bulb to glow and thus demonstrate the presence of an induced

current.
In experimental physics one must learn to innovate. Michael Faraday
who is ranked as one of the best experimentalists ever, was legendary
for his innovative skills.
EXAMPLE 6.2

Example 6.2 A square loop of side 10 cm and resistance 0.5 W is

placed vertically in the east-west plane. A uniform magnetic field of
0.10 T is set up across the plane in the north-east direction. The
magnetic field is decreased to zero in 0.70 s at a steady rate. Determine
the magnitudes of induced emf and current during this time-interval.
158

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Induction

Solution The angle q made by the area vector of the coil with the
magnetic field is 45°. From Eq. (6.1), the initial magnetic flux is
F = BA cos q

0.1 × 10 –2
= Wb
2
Final flux, Fmin = 0
The change in flux is brought about in 0.70 s. From Eq. (6.3), the
magnitude of the induced emf is given by

∆ΦB (Φ – 0) 10 –3
ε= = = = 1.0 mV
∆t ∆t 2 × 0.7
And the magnitude of the current is

EXAMPLE 6.2
10 –3 V
I = = = 2 mA
R 0.5Ω
Note that the earth’s magnetic field also produces a flux through the
loop. But it is a steady field (which does not change within the time
span of the experiment) and hence does not induce any emf.

Example 6.3
A circular coil of radius 10 cm, 500 turns and resistance 2 W is placed
with its plane perpendicular to the horizontal component of the earth’s
magnetic field. It is rotated about its vertical diameter through 180°
in 0.25 s. Estimate the magnitudes of the emf and current induced in
the coil. Horizontal component of the earth’s magnetic field at the
place is 3.0 × 10–5 T.
Solution
Initial flux through the coil,
FB (initial) = BA cos q
= 3.0 × 10–5 × (p ×10–2) × cos 0°
= 3p × 10–7 Wb
Final flux after the rotation,
FB (final) = 3.0 × 10–5 × (p ×10–2) × cos 180°
= –3p × 10–7 Wb
Therefore, estimated value of the induced emf is,

∆Φ
ε=N
∆t
= 500 × (6p × 10–7)/0.25
= 3.8 × 10–3 V
EXAMPLE 6.3

I = e/R = 1.9 × 10–3 A

Note that the magnitudes of e and I are the estimated values. Their
instantaneous values are different and depend upon the speed of
rotation at the particular instant.
159

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6.5 LENZ’S LAW AND CONSERVATION OF ENERGY
In 1834, German physicist Heinrich Friedrich Lenz (1804-1865) deduced
a rule, known as Lenz’s law which gives the polarity of the induced emf
in a clear and concise fashion. The statement of the law is:
The polarity of induced emf is such that it tends to produce a current
which opposes the change in magnetic flux that produced it.
The negative sign shown in Eq. (6.3) represents this effect. We can
understand Lenz’s law by examining Experiment 6.1 in Section 6.2.1. In
Fig. 6.1, we see that the North-pole of a bar magnet is being pushed
towards the closed coil. As the North-pole of the bar magnet moves towards
the coil, the magnetic flux through the coil increases. Hence current is
induced in the coil in such a direction that it opposes the increase in flux.
This is possible only if the current in the coil is in a counter-clockwise
direction with respect to an observer situated on the side of the magnet.
Note that magnetic moment associated with this current has North polarity
towards the North-pole of the approaching magnet. Similarly, if the North-
pole of the magnet is being withdrawn from the coil, the magnetic flux
through the coil will decrease. To counter this decrease in magnetic flux,
the induced current in the coil flows in clockwise direction and its South-
pole faces the receding North-pole of the bar magnet. This would result in
an attractive force which opposes the motion of the magnet and the
corresponding decrease in flux.
What will happen if an open circuit is used in place of the closed loop
in the above example? In this case too, an emf is induced across the open
ends of the circuit. The direction of the induced emf can be found
using Lenz’s law. Consider Figs. 6.6 (a) and (b). They provide an easier
way to understand the direction of induced currents. Note that the
direction shown by and indicate the directions of the induced
currents.
A little reflection on this matter should convince us on the
correctness of Lenz’s law. Suppose that the induced current was in
the direction opposite to the one depicted in Fig. 6.6(a). In that case,
the South-pole due to the induced current will face the approaching
North-pole of the magnet. The bar magnet will then be attracted
towards the coil at an ever increasing acceleration. A gentle push on
the magnet will initiate the process and its velocity and kinetic energy
will continuously increase without expending any energy. If this can
happen, one could construct a perpetual-motion machine by a
suitable arrangement. This violates the law of conservation of energy
and hence can not happen.
FIGURE 6.6 Now consider the correct case shown in Fig. 6.6(a). In this situation,
Illustration of the bar magnet experiences a repulsive force due to the induced
Lenz’s law. current. Therefore, a person has to do work in moving the magnet.
Where does the energy spent by the person go? This energy is
160 dissipated by Joule heating produced by the induced current.

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Example 6.4
Figure 6.7 shows planar loops of different shapes moving out of or
into a region of a magnetic field which is directed normal to the plane
of the loop away from the reader. Determine the direction of induced
current in each loop using Lenz’s law.

FIGURE 6.7

Solution
(i) The magnetic flux through the rectangular loop abcd increases,
due to the motion of the loop into the region of magnetic field, The
induced current must flow along the path bcdab so that it opposes
the increasing flux.
(ii) Due to the outward motion, magnetic flux through the triangular
loop abc decreases due to which the induced current flows along
bacb, so as to oppose the change in flux.
EXAMPLE 6.4

(iii) As the magnetic flux decreases due to motion of the irregular

shaped loop abcd out of the region of magnetic field, the induced
current flows along cdabc, so as to oppose change in flux.
Note that there are no induced current as long as the loops are
completely inside or outside the region of the magnetic field.

Example 6.5
(a) A closed loop is held stationary in the magnetic field between the
north and south poles of two permanent magnets held fixed. Can
we hope to generate current in the loop by using very strong
magnets?
(b) A closed loop moves normal to the constant electric field between
the plates of a large capacitor. Is a current induced in the loop
(i) when it is wholly inside the region between the capacitor plates
(ii) when it is partially outside the plates of the capacitor? The
electric field is normal to the plane of the loop.
EXAMPLE 6.5

(c) A rectangular loop and a circular loop are moving out of a uniform
magnetic field region (Fig. 6.8) to a field-free region with a constant
velocity v. In which loop do you expect the induced emf to be
constant during the passage out of the field region? The field is
normal to the loops. 161

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FIGURE 6.8

(d) Predict the polarity of the capacitor in the situation described by

Fig. 6.9.

FIGURE 6.9
Solution
(a) No. However strong the magnet may be, current can be induced
only by changing the magnetic flux through the loop.
(b) No current is induced in either case. Current can not be induced
by changing the electric flux.
EXAMPLE 6.5

(c) The induced emf is expected to be constant only in the case of the
rectangular loop. In the case of circular loop, the rate of change of
area of the loop during its passage out of the field region is not
constant, hence induced emf will vary accordingly.
(d) The polarity of plate ‘A’ will be positive with respect to plate ‘B’ in
the capacitor.

6.6 MOTIONAL ELECTROMOTIVE FORCE

Let us consider a straight conductor moving in a uniform and time-
independent magnetic field. Figure 6.10 shows a rectangular conductor
PQRS in which the conductor PQ is free to move. The rod PQ is moved
towards the left with a constant velocity v as
shown in the figure. Assume that there is no
loss of energy due to friction. PQRS forms a
closed circuit enclosing an area that changes
as PQ moves. It is placed in a uniform magnetic
field B which is perpendicular to the plane of
this system. If the length RQ = x and RS = l, the
magnetic flux FB enclosed by the loop PQRS
will be
FB = Blx
Since x is changing with time, the rate of change
of flux FB will induce an emf given by:
FIGURE 6.10 The arm PQ is moved to the left – dΦB d
side, thus decreasing the area of the ε= = – ( Blx )
rectangular loop. This movement dt dt
induces a current I as shown. dx
162 = – Bl = Blv (6.5)
dt

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 Electromagnetic
Induction
where we have used dx/dt = –v which is the speed of the conductor PQ.
The induced emf Blv is called motional emf. Thus, we are able to produce
induced emf by moving a conductor instead of varying the magnetic field,
that is, by changing the magnetic flux enclosed by the circuit.
It is also possible to explain the motional emf expression in Eq. (6.5)
by invoking the Lorentz force acting on the free charge carriers of conductor
PQ. Consider any arbitrary charge q in the conductor PQ. When the rod
moves with speed v, the charge will also be moving with speed v in the
magnetic field B. The Lorentz force on this charge is qvB in magnitude,
and its direction is towards Q. All charges experience the same force, in
magnitude and direction, irrespective of their position in the rod PQ.
The work done in moving the charge from P to Q is,
W = qvBl
Since emf is the work done per unit charge,
W
ε =
q
= Blv
This equation gives emf induced across the rod PQ and is identical
to Eq. (6.5). We stress that our presentation is not wholly rigorous. But
it does help us to understand the basis of Faraday’s law when
the conductor is moving in a uniform and time-independent
magnetic field.
On the other hand, it is not obvious how an emf is induced when a
conductor is stationary and the magnetic field is changing – a fact which
Faraday verified by numerous experiments. In the case of a stationary
conductor, the force on its charges is given by
F = q (E + v ´ B) = qE (6.6)
since v = 0. Thus, any force on the charge must arise from the electric
field term E alone. Therefore, to explain the existence of induced emf or
induced current, we must assume that a time-varying magnetic field
generates an electric field. However, we hasten to add that electric fields
produced by static electric charges have properties different from those
produced by time-varying magnetic fields. In Chapter 4, we learnt that
charges in motion (current) can exert force/torque on a stationary magnet.
Conversely, a bar magnet in motion (or more generally, a changing
magnetic field) can exert a force on the stationary charge. This is the
fundamental significance of the Faraday’s discovery. Electricity and
magnetism are related.

Example 6.6 A metallic rod of 1 m length is rotated with a frequency

of 50 rev/s, with one end hinged at the centre and the other end at the
EXAMPLE 6.6

circumference of a circular metallic ring of radius 1 m, about an axis

passing through the centre and perpendicular to the plane of the ring
(Fig. 6.11). A constant and uniform magnetic field of 1 T parallel to the
axis is present everywhere. What is the emf between the centre and
the metallic ring? 163

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FIGURE 6.11
Solution
Method I
As the rod is rotated, free electrons in the rod move towards the outer
end due to Lorentz force and get distributed over the ring. Thus, the
resulting separation of charges produces an emf across the ends of
the rod. At a certain value of emf, there is no more flow of electrons
and a steady state is reached. Using Eq. (6.5), the magnitude of the
emf generated across a length dr of the rod as it moves at right angles
to the magnetic field is given by
dε = Bv dr . Hence,
R R
B ωR2
ε =∫ dε = ∫ Bv dr = ∫ B ωr dr =
0 0
2
Note that we have used v = w r. This gives
1
e = × 1.0 × 2 π × 50 × (12 )
2
= 157 V
Method II
To calculate the emf, we can imagine a closed loop OPQ in which
point O and P are connected with a resistor R and OQ is the rotating
rod. The potential difference across the resistor is then equal to the
induced emf and equals B × (rate of change of area of loop). If q is the
angle between the rod and the radius of the circle at P at time t, the
area of the sector OPQ is given by
θ 1 2
π R2 × R θ
=
2π 2
where R is the radius of the circle. Hence, the induced emf is
d 1 2  1 2 dθ Bω R 2
e =B × R θ BR =
dt  2  =
EXAMPLE 6.6

2 dt 2
dθ
[Note: = ω = 2πν ]
dt
This expression is identical to the expression obtained by Method I
and we get the same value of e.
164

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Example 6.7
A wheel with 10 metallic spokes each 0.5 m long is rotated with a
speed of 120 rev/min in a plane normal to the horizontal component
of earth’s magnetic field HE at a place. If HE = 0.4 G at the place, what
is the induced emf between the axle and the rim of the wheel? Note
that 1 G = 10–4 T.
Solution
Induced emf = (1/2) w B R2

EXAMPLE 6.7
= (1/2) × 4p × 0.4 × 10–4 × (0.5)2
= 6.28 × 10–5 V
The number of spokes is immaterial because the emf’s across the
spokes are in parallel.

6.7 INDUCTANCE
An electric current can be induced in a coil by flux change produced by
another coil in its vicinity or flux change produced by the same coil. These
two situations are described separately in the next two sub-sections.
However, in both the cases, the flux through a coil is proportional to the
current. That is, FB a I.
Further, if the geometry of the coil does not vary with time then,
dΦB dI
∝
dt dt
For a closely wound coil of N turns, the same magnetic flux is linked
with all the turns. When the flux FB through the coil changes, each turn
contributes to the induced emf. Therefore, a term called flux linkage is
used which is equal to NFB for a closely wound coil and in such a case
NF B ∝ I
The constant of proportionality, in this relation, is called inductance.
We shall see that inductance depends only on the geometry of the coil
and intrinsic material properties. This aspect is akin to capacitance which
for a parallel plate capacitor depends on the plate area and plate separation
(geometry) and the dielectric constant K of the intervening medium
(intrinsic material property).
Inductance is a scalar quantity. It has the dimensions of [M L2 T –2 A–2]
given by the dimensions of flux divided by the dimensions of current. The
SI unit of inductance is henry and is denoted by H. It is named in honour
of Joseph Henry who discovered electromagnetic induction in USA,
independently of Faraday in England.

6.7.1 Mutual inductance

Consider Fig. 6.11 which shows two long co-axial solenoids each of length
l. We denote the radius of the inner solenoid S1 by r1 and the number of
turns per unit length by n1. The corresponding quantities for the outer
solenoid S2 are r2 and n2, respectively. Let N1 and N2 be the total number
of turns of coils S1 and S2, respectively. 165

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When a current I2 is set up through S2, it in turn sets
up a magnetic flux through S1. Let us denote it by F1.
The corresponding flux linkage with solenoid S1 is
N1 Φ1 = M12 I 2 (6.7)
M12 is called the mutual inductance of solenoid S1 with
respect to solenoid S2. It is also referred to as the
coefficient of mutual induction.
For these simple co-axial solenoids it is possible to
calculate M12. The magnetic field due to the current I2 in
S2 is m0n2I2. The resulting flux linkage with coil S1 is,

( ) (µ n I )
N1Φ1 = (n1l ) πr12 0 2 2

= µ0n1n 2 πr12l I 2 (6.8)

where n1l is the total number of turns in solenoid S1. Thus,
FIGURE 6.12 Two long co-axial from Eq. (6.9) and Eq. (6.10),
solenoids of same
M12 = m0n1n2pr 12l (6.9)
length l.
Note that we neglected the edge effects and considered
the magnetic field m0n2I2 to be uniform throughout the
length and width of the solenoid S2. This is a good approximation keeping
in mind that the solenoid is long, implying l >> r2.
We now consider the reverse case. A current I1 is passed through the
solenoid S1 and the flux linkage with coil S2 is,
N2F2 = M21 I1 (6.10)
M21 is called the mutual inductance of solenoid S2 with respect to
solenoid S1.
The flux due to the current I1 in S1 can be assumed to be confined
solely inside S1 since the solenoids are very long. Thus, flux linkage with
solenoid S2 is

( ) (µ n I )
N 2Φ2 = (n 2l ) πr12 0 1 1

where n2l is the total number of turns of S2. From Eq. (6.12),
M21 = m0n1n2pr 21 l (6.11)
Using Eq. (6.11) and Eq. (6.12), we get
M12 = M21= M (say) (6.12)
We have demonstrated this equality for long co-axial solenoids.
However, the relation is far more general. Note that if the inner solenoid
was much shorter than (and placed well inside) the outer solenoid, then
we could still have calculated the flux linkage N1F1 because the inner
solenoid is effectively immersed in a uniform magnetic field due to the
outer solenoid. In this case, the calculation of M12 would be easy. However,
it would be extremely difficult to calculate the flux linkage with the outer
solenoid as the magnetic field due to the inner solenoid would vary across
the length as well as cross section of the outer solenoid. Therefore, the
calculation of M21 would also be extremely difficult in this case. The
166 equality M12=M21 is very useful in such situations.

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 Electromagnetic
Induction
We explained the above example with air as the medium within the
solenoids. Instead, if a medium of relative permeability mr had been present,
the mutual inductance would be
M =mr m0 n1n2p r12 l
It is also important to know that the mutual inductance of a pair of
coils, solenoids, etc., depends on their separation as well as their relative
orientation.

Example 6.8 Two concentric circular coils, one of small radius r1 and
the other of large radius r2, such that r1 << r2, are placed co-axially
with centres coinciding. Obtain the mutual inductance of the
arrangement.
Solution Let a current I2 flow through the outer circular coil. The
field at the centre of the coil is B 2 = m 0I 2 / 2r2. Since the other
co-axially placed coil has a very small radius, B2 may be considered
constant over its cross-sectional area. Hence,
F1 = pr 12B2
µ0 πr12
= I2
2r2
= M12 I2
Thus,
µ0 πr12
M12 =
2r2
From Eq. (6.12)
µ0 π r12
M12 = M 21 =
EXAMPLE 6.8

2 r2
Note that we calculated M12 from an approximate value of F1, assuming
the magnetic field B2 to be uniform over the area p r12. However, we
can accept this value because r1 << r2.

Now, let us recollect Experiment 6.3 in Section 6.2. In that experiment,

emf is induced in coil C1 wherever there was any change in current through
coil C2. Let F1 be the flux through coil C1 (say of N1 turns) when current in
coil C2 is I2.
Then, from Eq. (6.7), we have
N1F1 = MI2
For currents varrying with time,
d ( N 1Φ1 ) d (M I 2 )
=
dt dt
Since induced emf in coil C1 is given by
d ( N 1Φ1 )
ε1 = –
dt
We get,
dI 2
ε1 = – M 167
dt

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 Physics
It shows that varying current in a coil can induce emf in a neighbouring
coil. The magnitude of the induced emf depends upon the rate of change
of current and mutual inductance of the two coils.

6.7.2 Self-inductance
In the previous sub-section, we considered the flux in one solenoid due
to the current in the other. It is also possible that emf is induced in a
single isolated coil due to change of flux through the coil by means of
varying the current through the same coil. This phenomenon is called
self-induction. In this case, flux linkage through a coil of N turns is
proportional to the current through the coil and is expressed as
N ΦB ∝ I
N ΦB = L I (6.13)
where constant of proportionality L is called self-inductance of the coil. It
is also called the coefficient of self-induction of the coil. When the current
is varied, the flux linked with the coil also changes and an emf is induced
in the coil. Using Eq. (6.13), the induced emf is given by
d ( N ΦB )
ε=–
dt
dI
ε = –L (6.14)
dt
Thus, the self-induced emf always opposes any change (increase or
decrease) of current in the coil.
It is possible to calculate the self-inductance for circuits with simple
geometries. Let us calculate the self-inductance of a long solenoid of cross-
sectional area A and length l, having n turns per unit length. The magnetic
field due to a current I flowing in the solenoid is B = m0 n I (neglecting edge
effects, as before). The total flux linked with the solenoid is
N ΦB = (nl ) ( µ0n I ) ( A )

  0n 2 Al I
where nl is the total number of turns. Thus, the self-inductance is,
ΝΦΒ
L =
I
= µ0n 2 Al (6.15)
If we fill the inside of the solenoid with a material of relative permeability
mr (for example soft iron, which has a high value of relative permeability),
then,
L = µr µ0 n 2 Al (6.16)
The self-inductance of the coil depends on its geometry and on the
permeability of the medium.
The self-induced emf is also called the back emf as it opposes any
168 change in the current in a circuit. Physically, the self-inductance plays

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 Electromagnetic
Induction
the role of inertia. It is the electromagnetic analogue of mass in mechanics.
So, work needs to be done against the back emf (e ) in establishing the
current. This work done is stored as magnetic potential energy. For the
current I at an instant in a circuit, the rate of work done is
dW
= εI
dt
If we ignore the resistive losses and consider only inductive effect,
then using Eq. (6.16),
dW dI
=L I
dt dt
Total amount of work done in establishing the current I is
I
W = ∫ dW = ∫ L I dI
0

Thus, the energy required to build up the current I is,

1 2
W = LI (6.17)
2
This expression reminds us of mv 2/2 for the (mechanical) kinetic energy
of a particle of mass m, and shows that L is analogous to m (i.e., L is
electrical inertia and opposes growth and decay of current in the circuit).
Consider the general case of currents flowing simultaneously in two
nearby coils. The flux linked with one coil will be the sum of two fluxes
which exist independently. Equation (6.7) would be modified into
N1 Φ1 = M11 I1 + M12 I 2
where M11 represents inductance due to the same coil.
Therefore, using Faraday’s law,
dI 1 dI
ε1 = − M 11 − M 12 2
dt dt
M11 is the self-inductance and is written as L1. Therefore,
d I1 dI
ε1 = − L1 − M 12 2
dt dt

Example 6.9 (a) Obtain the expression for the magnetic energy stored
in a solenoid in terms of magnetic field B, area A and length l of the
solenoid. (b) How does this magnetic energy compare with the
electrostatic energy stored in a capacitor?
Solution
(a) From Eq. (6.17), the magnetic energy is
1 2
EXAMPLE 6.9

UB = LI
2
2
1  B 
= L (since B = µ0 nI , for a solenoid)
2  µ0n 
169

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 Physics
2
http://micro.magnet.fsu.edu/electromag/java/generator/ac.html
1  B 
= ( µ0n 2 Al )  [from Eq. (6.15)]
2  µ0n 

1
= B 2 Al
2 µ0
(b) The magnetic energy per unit volume is,
UB
uB = (where V is volume that contains flux)
V
Interactive animation on ac generator:

UB
=
Al
B2
= (6.18)
2 µ0
We have already obtained the relation for the electrostatic energy
stored per unit volume in a parallel plate capacitor (refer to Chapter 2,
Eq. 2.73),
1
uΕ = ε0 E 2 (2.73)
2
EXAMPLE 6.9

In both the cases energy is proportional to the square of the field

strength. Equations (6.18) and (2.73) have been derived for special
cases: a solenoid and a parallel plate capacitor, respectively. But they
are general and valid for any region of space in which a magnetic field
or/and an electric field exist.

6.8 AC GENERATOR
The phenomenon of electromagnetic induction
has been technologically exploited in many ways.
An exceptionally important application is the
generation of alternating currents (ac). The
modern ac generator with a typical output
capacity of 100 MW is a highly evolved machine.
In this section, we shall describe the basic
principles behind this machine. The Yugoslav
inventor Nicola Tesla is credited with the
development of the machine. As was pointed out
in Section 6.3, one method to induce an emf or
current in a loop is through a change in the
loop’s orientation or a change in its effective area.
As the coil rotates in a magnetic field B, the
effective area of the loop (the face perpendicular
to the field) is A cos q, where q is the angle
between A and B. This method of producing a
FIGURE 6.13 AC Generator flux change is the principle of operation of a

170

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 Electromagnetic
Induction
simple ac generator. An ac generator converts mechanical energy into
electrical energy.
The basic elements of an ac generator are shown in Fig. 6.13. It consists
of a coil mounted on a rotor shaft. The axis of rotation of the coil is
perpendicular to the direction of the magnetic field. The coil (called
armature) is mechanically rotated in the uniform magnetic field by some
external means. The rotation of the coil causes the magnetic flux through
it to change, so an emf is induced in the coil. The ends of the
coil are connected to an external circuit by means of slip rings
and brushes.
When the coil is rotated with a constant angular speed w, the angle q
between the magnetic field vector B and the area vector A of the coil at any
instant t is q = wt (assuming q = 0° at t = 0). As a result, the effective area
of the coil exposed to the magnetic field lines changes with time, and from
Eq. (6.1), the flux at any time t is
FB = BA cos q = BA cos wt
From Faraday’s law, the induced emf for the rotating coil of N turns
is then,
dΦB d
ε=–N = – NBA (cos ω t )
dt dt
Thus, the instantaneous value of the emf is
ε = NBA ω sin ωt (6.19)
where NBAw is the maximum value of the emf, which occurs when
sin wt = ±1. If we denote NBAw as e0, then
e = e0 sin wt (6.20)
Since the value of the sine fuction varies between +1 and –1, the sign, or
polarity of the emf changes with time. Note from Fig. 6.14 that the emf
has its extremum value when q = 90° or q = 270°, as the change of flux is
greatest at these points.
The direction of the current changes periodically and therefore the current
is called alternating current (ac). Since w = 2pn, Eq (6.20) can be written as
e = e0sin 2p n t (6.21)
where n is the frequency of revolution of the generator’s coil.
Note that Eq. (6.20) and (6.21) give the instantaneous value of the emf
and e varies between +e0 and –e0 periodically. We shall learn how to
determine the time-averaged value for the alternating voltage and current
in the next chapter.
In commercial generators, the mechanical energy required for
rotation of the armature is provided by water falling from a height, for
example, from dams. These are called hydro-electric generators.
Alternatively, water is heated to produce steam using coal or other
sources. The steam at high pressure produces the rotation of the
armature. These are called thermal generators. Instead of coal, if a
nuclear fuel is used, we get nuclear power generators. Modern day
generators produce electric power as high as 500 MW, i.e., one can light

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 Physics

FIGURE 6.14 An alternating emf is generated by a loop of wire rotating in a magnetic field.

up 5 million 100 W bulbs! In most generators, the coils are held

stationary and it is the electromagnets which are rotated. The frequency
of rotation is 50 Hz in India. In certain countries such as USA, it is
60 Hz.

Example 6.10 Kamla peddles a stationary bicycle. The pedals of the

bicycle are attached to a 100 turn coil of area 0.10 m2. The coil rotates
at half a revolution per second and it is placed in a uniform magnetic
field of 0.01 T perpendicular to the axis of rotation of the coil. What is
the maximum voltage generated in the coil?
Solution Here n = 0.5 Hz; N =100, A = 0.1 m2 and B = 0.01 T. Employing
Eq. (6.19)
e0 = NBA (2 p n)
= 100 × 0.01 × 0.1 × 2 × 3.14 × 0.5
EXAMPLE 6.10

= 0.314 V
The maximum voltage is 0.314 V.
We urge you to explore such alternative possibilities for power
generation.

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 Electromagnetic
Induction

SUMMARY

1. The magnetic flux through a surface of area A placed in a uniform magnetic

field B is defined as,
FB = B.A = BA cos q
where q is the angle between B and A.
2. Faraday’s laws of induction imply that the emf induced in a coil of N
turns is directly related to the rate of change of flux through it,
dΦB
ε = −N
dt
Here FB is the flux linked with one turn of the coil. If the circuit is
closed, a current I = e/R is set up in it, where R is the resistance of the
circuit.
3. Lenz’s law states that the polarity of the induced emf is such that it
tends to produce a current which opposes the change in magnetic flux
that produces it. The negative sign in the expression for Faraday’s law
indicates this fact.
4. When a metal rod of length l is placed normal to a uniform magnetic
field B and moved with a velocity v perpendicular to the field, the
induced emf (called motional emf ) across its ends is
e = Bl v
5. Inductance is the ratio of the flux-linkage to current. It is equal to NF/I.
6. A changing current in a coil (coil 2) can induce an emf in a nearby coil
(coil 1). This relation is given by,
dI 2
ε1 = − M12
dt
The quantity M12 is called mutual inductance of coil 1 with respect to
coil 2. One can similarly define M21. There exists a general equality,
M12 = M21
7. When a current in a coil changes, it induces a back emf in the same
coil. The self-induced emf is given by,
dI
ε = −L
dt
L is the self-inductance of the coil. It is a measure of the inertia of the
coil against the change of current through it.
8. The self-inductance of a long solenoid, the core of which consists of a
magnetic material of relative permeability mr, is given by
L = mr m0 n2 A l
where A is the area of cross-section of the solenoid, l its length and n
the number of turns per unit length.
9. In an ac generator, mechanical energy is converted to electrical energy
by virtue of electromagnetic induction. If coil of N turn and area A is
rotated at n revolutions per second in a uniform magnetic field B, then
the motional emf produced is
e = NBA ( 2pn) sin (2p nt)
where we have assumed that at time t = 0 s, the coil is perpendicular to
the field.

173

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 Physics
Quantity Symbol Units Dimensions Equations

Magnetic Flux FB Wb (weber) [M L2 T –2 A–1] FB = B i A

EMF e V (volt) [M L2 T –3 A–1] e = − d( N ΦB )/ dt
Mutual Inductance M H (henry) 2 –2
[M L T A ] –2
e1 = −M12 (dI 2 /dt )

Self Inductance L H (henry) [M L2 T –2 A–2] ε = − L (d I / d t )

POINTS TO PONDER

1. Electricity and magnetism are intimately related. In the early part of the
nineteenth century, the experiments of Oersted, Ampere and others
established that moving charges (currents) produce a magnetic field.
Somewhat later, around 1830, the experiments of Faraday and Henry
demonstrated that a moving magnet can induce electric current.
2. In a closed circuit, electric currents are induced so as to oppose the
changing magnetic flux. It is as per the law of conservation of energy.
However, in case of an open circuit, an emf is induced across its ends.
How is it related to the flux change?
3. The motional emf discussed in Section 6.5 can be argued independently
from Faraday’s law using the Lorentz force on moving charges. However,
even if the charges are stationary [and the q (v × B) term of the Lorentz
force is not operative], an emf is nevertheless induced in the presence of a
time-varying magnetic field. Thus, moving charges in static field and static
charges in a time-varying field seem to be symmetric situation for Faraday’s
law. This gives a tantalising hint on the relevance of the principle of
relativity for Faraday’s law.

EXERCISES
6.1 Predict the direction of induced current in the situations described
by the following Figs. 6.15(a) to (f ).

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 Electromagnetic
Induction

FIGURE 6.15

6.2 Use Lenz’s law to determine the direction of induced current in the
situations described by Fig. 6.16:
(a) A wire of irregular shape turning into a circular shape;

175

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 Physics
(b) A circular loop being deformed into a narrow straight wire.

FIGURE 6.16

6.3 A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2
placed inside the solenoid normal to its axis. If the current carried
by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is
the induced emf in the loop while the current is changing?
6.4 A rectangular wire loop of sides 8 cm and 2 cm with a small cut is
moving out of a region of uniform magnetic field of magnitude 0.3 T
directed normal to the loop. What is the emf developed across the
cut if the velocity of the loop is 1 cm s–1 in a direction normal to the
(a) longer side, (b) shorter side of the loop? For how long does the
induced voltage last in each case?
6.5 A 1.0 m long metallic rod is rotated with an angular frequency of
400 rad s–1 about an axis normal to the rod passing through its one
end. The other end of the rod is in contact with a circular metallic
ring. A constant and uniform magnetic field of 0.5 T parallel to the
axis exists everywhere. Calculate the emf developed between the
centre and the ring.
6.6 A horizontal straight wire 10 m long extending from east to west is
falling with a speed of 5.0 m s–1, at right angles to the horizontal
component of the earth’s magnetic field, 0.30 ´ 10–4 Wb m–2.
(a) What is the instantaneous value of the emf induced in the wire?
(b) What is the direction of the emf?
(c) Which end of the wire is at the higher electrical potential?
6.7 Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf
of 200 V induced, give an estimate of the self-inductance of the circuit.
6.8 A pair of adjacent coils has a mutual inductance of 1.5 H. If the
current in one coil changes from 0 to 20 A in 0.5 s, what is the
change of flux linkage with the other coil?

176

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Chapter Seven

ALTERNATING
CURRENT

7.1 INTRODUCTION
We have so far considered direct current (dc) sources and circuits with dc
sources. These currents do not change direction with time. But voltages
and currents that vary with time are very common. The electric mains
supply in our homes and offices is a voltage that varies like a sine function
with time. Such a voltage is called alternating voltage (ac voltage) and
the current driven by it in a circuit is called the alternating current (ac
current)*. Today, most of the electrical devices we use require ac voltage.
This is mainly because most of the electrical energy sold by power
companies is transmitted and distributed as alternating current. The main
reason for preferring use of ac voltage over dc voltage is that ac voltages
can be easily and efficiently converted from one voltage to the other by
means of transformers. Further, electrical energy can also be transmitted
economically over long distances. AC circuits exhibit characteristics which
are exploited in many devices of daily use. For example, whenever we
tune our radio to a favourite station, we are taking advantage of a special
property of ac circuits – one of many that you will study in this chapter.

* The phrases ac voltage and ac current are contradictory and redundant,

respectively, since they mean, literally, alternating current voltage and alternating
current current. Still, the abbreviation ac to designate an electrical quantity
displaying simple harmonic time dependance has become so universally accepted
that we follow others in its use. Further, voltage – another phrase commonly
used means potential difference between two points.

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 Physics
7.2 AC VOLTAGE APPLIED TO A RESISTOR
Figure 7.1 shows a resistor connected to a source e of
ac voltage. The symbol for an ac source in a circuit
diagram is . We consider a source which produces
sinusoidally varying potential difference across its
terminals. Let this potential difference, also called ac
voltage, be given by
v = vm sin ω t (7.1)
where vm is the amplitude of the oscillating potential
difference and w is its angular frequency.

Nicola Tesla (1856 –

1943) Serbian-American
scientist, inventor and
NICOLA TESLA (1856 – 1943)

genius. He conceived the

idea of the rotating
magnetic field, which is the
basis of practically all
alternating current
machinery, and which
helped usher in the age of FIGURE 7.1 AC voltage applied to a resistor.
electric power. He also
invented among other To find the value of current through the resistor, we
things the induction motor,
the polyphase system of ac apply Kirchhoff’s loop rule ∑ ε(t ) = 0 (refer to Section
power, and the high 3.13), to the circuit shown in Fig. 7.1 to get
frequency induction coil
(the Tesla coil) used in radio
vm sin ω t = i R
and television sets and vm
other electronic equipment. or i = sin ω t
R
The SI unit of magnetic field
is named in his honour. Since R is a constant, we can write this equation as
i = i m sin ω t (7.2)
where the current amplitude im is given by
vm
im = (7.3)
R
Equation (7.3) is Ohm’s law, which for resistors, works equally
well for both ac and dc voltages. The voltage across a pure resistor
and the current through it, given by Eqs. (7.1) and (7.2) are
plotted as a function of time in Fig. 7.2. Note, in particular that
both v and i reach zero, minimum and maximum values at the
FIGURE 7.2 In a pure same time. Clearly, the voltage and current are in phase with
resistor, the voltage and each other.
current are in phase. The We see that, like the applied voltage, the current varies
minima, zero and maxima sinusoidally and has corresponding positive and negative values
occur at the same
respective times. during each cycle. Thus, the sum of the instantaneous current
values over one complete cycle is zero, and the average current
178 is zero. The fact that the average current is zero, however, does

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 Alternating Current
not mean that the average power consumed is zero and
that there is no dissipation of electrical energy. As you
know, Joule heating is given by i 2R and depends on i 2
(which is always positive whether i is positive or negative)
and not on i. Thus, there is Joule heating and
dissipation of electrical energy when an
ac current passes through a resistor.
The instantaneous power dissipated in the resistor is
p = i 2 R = i m2 R sin 2 ω t (7.4)
The average value of p over a cycle is*
p = < i 2 R > = < i m2 R sin 2 ω t > [7.5(a)]
where the bar over a letter (here, p) denotes its average
Geor ge Westinghouse

GEORGE WESTINGHOUSE (1846 – 1914)

value and <......> denotes taking average of the quantity
(1846 – 1914) A leading
inside the bracket. Since, i 2m and R are constants, proponent of the use of
p = i m2 R < sin2 ωt > [7.5(b)] alternating current over
direct current. Thus,
Using the trigonometric identity, sin w t = 2
he came into conflict
1/2 (1– cos 2wt ), we have < sin2 wt > = (1/2) (1– < cos 2wt >) with Thomas Alva Edison,
and since < cos2wt > = 0**, we have, an advocate of direct
1 current. Westinghouse
< sin2 ω t > = was convinced that the
2
technology of alternating
Thus, current was the key to
1 2 the electrical future.
p= im R [7.5(c)] He founded the famous
2
Company named after him
To express ac power in the same form as dc power and enlisted the services
(P = I2R), a special value of current is defined and used. of Nicola Tesla and
It is called, root mean square (rms) or effective current other inventors in the
(Fig. 7.3) and is denoted by Irms or I. development of alternating
current motors and
apparatus for the
transmission of high
tension current, pioneering
in large scale lighting.

FIGURE 7.3 The rms current I is related to the

peak current im by I = i m / 2 = 0.707 im.
T
1
T ∫0
* The average value of a function F (t ) over a period T is given by F (t ) = F (t ) dt

1 T
1  sin 2ω t  T 1
** < cos 2ω t > = T ∫ cos 2ω t dt = T  =
2ω  0 2ω T
[sin 2ω T − 0] = 0
0 179

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 Physics
It is defined by
1 2 i
I = i2 = im = m
2 2
= 0.707 im (7.6)
In terms of I, the average power, denoted by P is
1 2
P = p= im R = I 2 R (7.7)
2
Similarly, we define the rms voltage or effective voltage by
vm
V= = 0.707 vm (7.8)
2
From Eq. (7.3), we have
v m = i mR
vm im
or, = R
2 2
or, V = IR (7.9)
Equation (7.9) gives the relation between ac current and ac voltage
and is similar to that in the dc case. This shows the advantage of
introducing the concept of rms values. In terms of rms values, the equation
for power [Eq. (7.7)] and relation between current and voltage in ac circuits
are essentially the same as those for the dc case.
It is customary to measure and specify rms values for ac quantities. For
example, the household line voltage of 220 V is an rms value with a peak
voltage of
vm = 2 V = (1.414)(220 V) = 311 V
In fact, the I or rms current is the equivalent dc current that would
produce the same average power loss as the alternating current. Equation
(7.7) can also be written as
P = V2 / R = I V (since V = I R )

Example 7.1 A light bulb is rated at 100W for a 220 V supply. Find
(a) the resistance of the bulb; (b) the peak voltage of the source; and
(c) the rms current through the bulb.

Solution
(a) We are given P = 100 W and V = 220 V. The resistance of the
bulb is
V 2 (220 V )
2

R= = = 484 Ω
P 100 W
(b) The peak voltage of the source is
EXAMPLE 7.1

v m = 2V = 311 V
(c) Since, P = I V
P 100 W
I 0.454A
V 220 V
180

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 Alternating Current

7.3 REPRESENTATION OF AC CURRENT AND VOLTAGE

BY ROTATING VECTORS — PHASORS
In the previous section, we learnt that the current through a resistor is
in phase with the ac voltage. But this is not so in the case of an inductor,
a capacitor or a combination of these circuit elements. In order to show
phase relationship between voltage and current
in an ac circuit, we use the notion of phasors.
The analysis of an ac circuit is facilitated by the
use of a phasor diagram. A phasor* is a vector
which rotates about the origin with angular
speed w, as shown in Fig. 7.4. The vertical
components of phasors V and I represent the
sinusoidally varying quantities v and i. The
magnitudes of phasors V and I represent the
amplitudes or the peak values vm and im of these
oscillating quantities. Figure 7.4(a) shows the FIGURE 7.4 (a) A phasor diagram for the
voltage and current phasors and their circuit in Fig 7.1. (b) Graph of v and
relationship at time t1 for the case of an ac source i versus wt.
connected to a resistor i.e., corresponding to the
circuit shown in Fig. 7.1. The projection of
voltage and current phasors on vertical axis, i.e., vm sinw t and im sinw t,
respectively represent the value of voltage and current at that instant. As
they rotate with frequency w, curves in Fig. 7.4(b) are generated.
From Fig. 7.4(a) we see that phasors V and I for the case of a resistor are
in the same direction. This is so for all times. This means that the phase
angle between the voltage and the current is zero.

7.4 AC VOLTAGE APPLIED TO AN INDUCTOR

Figure 7.5 shows an ac source connected to an inductor. Usually,
inductors have appreciable resistance in their windings, but we shall
assume that this inductor has negligible resistance.
Thus, the circuit is a purely inductive ac circuit. Let
the voltage across the source be v = vm sinw t. Using
the Kirchhoff’s loop rule, ∑ ε (t ) = 0 , and since there
is no resistor in the circuit,
di
v −L =0 (7.10)
dt
where the second term is the self-induced Faraday FIGURE 7.5 An ac source
emf in the inductor; and L is the self-inductance of connected to an inductor.

* Though voltage and current in ac circuit are represented by phasors – rotating

vectors, they are not vectors themselves. They are scalar quantities. It so happens
that the amplitudes and phases of harmonically varying scalars combine
mathematically in the same way as do the projections of rotating vectors of
corresponding magnitudes and directions. The rotating vectors that represent
harmonically varying scalar quantities are introduced only to provide us with a
simple way of adding these quantities using a rule that we already know. 181

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 Physics
the inductor. The negative sign follows from Lenz’s law (Chapter 6).
Combining Eqs. (7.1) and (7.10), we have
di v v
= = m sin ω t (7.11)
dt L L
Equation (7.11) implies that the equation for i(t), the current as a
function of time, must be such that its slope di/dt is a sinusoidally varying
quantity, with the same phase as the source voltage and an amplitude
Interactive animation on Phasor diagrams of ac circuits containing, R, L, C and RLC series circuits:

given by vm/L. To obtain the current, we integrate di/dt with respect to

time:
di vm
∫ dt dt = L ∫ sin(ωt )dt
and get,
vm
i =− cos(ωt ) + constant
ωL
The integration constant has the dimension of current and is time-
independent. Since the source has an emf which oscillates symmetrically
about zero, the current it sustains also oscillates symmetrically about
zero, so that no constant or time-independent component of the current
exists. Therefore, the integration constant is zero.
http://www.animations.physics.unsw.edu.au//jw/AC.html

Using
 π
− cos(ω t ) = sin  ω t −  , we have
 2

 π
i = i m sin  ωt −  (7.12)
 2
vm
where im = is the amplitude of the current. The quantity w L is
ωL
analogous to the resistance and is called inductive reactance, denoted
by XL:
XL = w L (7.13)
The amplitude of the current is, then
vm
im = (7.14)
XL
The dimension of inductive reactance is the same as that of resistance
and its SI unit is ohm (W). The inductive reactance limits the current in a
purely inductive circuit in the same way as the resistance limits the
current in a purely resistive circuit. The inductive reactance is directly
proportional to the inductance and to the frequency of the current.
A comparison of Eqs. (7.1) and (7.12) for the source voltage and the
current in an inductor shows that the current lags the voltage by p/2 or
one-quarter (1/4) cycle. Figure 7.6 (a) shows the voltage and the current
phasors in the present case at instant t1. The current phasor I is p/2
behind the voltage phasor V. When rotated with frequency w counter-
clockwise, they generate the voltage and current given by Eqs. (7.1) and
182 (7.12), respectively and as shown in Fig. 7.6(b).

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 Alternating Current

FIGURE 7.6 (a) A Phasor diagram for the circuit in Fig. 7.5.
(b) Graph of v and i versus wt.

We see that the current reaches its maximum value later than the
T π/2
voltage by one-fourth of a period  =
ω 
. You have seen that an
4
inductor has reactance that limits current similar to resistance in a
dc circuit. Does it also consume power like a resistance? Let us try to
find out.
The instantaneous power supplied to the inductor is
 π
p L = i v = im sin  ω t −  ×vm sin (ωt )
 2
= −i m vm cos (ωt ) sin (ωt )
i m vm
=− sin (2ωt )
2
So, the average power over a complete cycle is
i m vm
PL = − sin (2ω t )
2

i m vm
=− sin (2ω t ) = 0,
2
since the average of sin (2wt) over a complete cycle is zero.
Thus, the average power supplied to an inductor over one complete
cycle is zero.

Example 7.2 A pure inductor of 25.0 mH is connected to a source of

220 V. Find the inductive reactance and rms current in the circuit if
the frequency of the source is 50 Hz.
Solution The inductive reactance,
X L = 2 π ν L = 2 × 3.14 × 50 × 25 × 10–3 Ω
EXAMPLE 7.2

= 7.85W
The rms current in the circuit is
V 220 V
I = = = 28A
X L 7.85 Ω 183

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 Physics
7.5 AC VOLTAGE APPLIED TO A CAPACITOR
Figure 7.7 shows an ac source e generating ac voltage v = vm sin wt
connected to a capacitor only, a purely capacitive ac circuit.
When a capacitor is connected to a voltage source
in a dc circuit, current will flow for the short time
required to charge the capacitor. As charge
accumulates on the capacitor plates, the voltage
across them increases, opposing the current. That is,
a capacitor in a dc circuit will limit or oppose the
current as it charges. When the capacitor is fully
charged, the current in the circuit falls to zero.
When the capacitor is connected to an ac source,
as in Fig. 7.7, it limits or regulates the current, but
FIGURE 7.7 An ac source does not completely prevent the flow of charge. The
connected to a capacitor. capacitor is alternately charged and discharged as
the current reverses each half cycle. Let q be the
charge on the capacitor at any time t. The instantaneous voltage v across
the capacitor is
q
v= (7.15)
C
From the Kirchhoff’s loop rule, the voltage across the source and the
capacitor are equal,
q
vm sin ω t =
C
dq
To find the current, we use the relation i =
dt
d
i =
dt
(vm C sin ω t ) = ω C vm cos(ω t )
π
Using the relation, cos(ω t ) = sin  ω t +  , we have
 2

 π
i = im sin  ω t +  (7.16)
 2
where the amplitude of the oscillating current is im = wCvm. We can rewrite
it as
vm
im =
(1/ ω C )
Comparing it to im= vm/R for a purely resistive circuit, we find that
(1/wC) plays the role of resistance. It is called capacitive reactance and
is denoted by Xc,
Xc= 1/wC (7.17)
so that the amplitude of the current is
vm
184 im = (7.18)
XC

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 Alternating Current
The dimension of capacitive reactance is the
same as that of resistance and its SI unit is
ohm (W). The capacitive reactance limits the
amplitude of the current in a purely capacitive
circuit in the same way as the resistance limits
the current in a purely resistive circuit. But it
is inversely proportional to the frequency and
the capacitance.
A comparison of Eq. (7.16) with the FIGURE 7.8 (a) A Phasor diagram for the circuit
equation of source voltage, Eq. (7.1) shows that in Fig. 7.8. (b) Graph of v and i versus wt.
the current is p/2 ahead of voltage.
Figure 7.8(a) shows the phasor diagram at an instant t1. Here the current
phasor I is p/2 ahead of the voltage phasor V as they rotate
counterclockwise. Figure 7.8(b) shows the variation of voltage and current
with time. We see that the current reaches its maximum value earlier than
the voltage by one-fourth of a period.
The instantaneous power supplied to the capacitor is
pc = i v = im cos(wt)vm sin(wt)
= imvm cos(wt) sin(wt)
i m vm
= sin(2ωt ) (7.19)
2
So, as in the case of an inductor, the average power
i m vm i v
PC = sin(2ωt ) = m m sin(2ωt ) = 0
2 2
since <sin (2wt)> = 0 over a complete cycle.
Thus, we see that in the case of an inductor, the current lags the voltage
by p/2 and in the case of a capacitor, the current leads the voltage by p/2.

Example 7.3 A lamp is connected in series with a capacitor. Predict

your observations for dc and ac connections. What happens in each
case if the capacitance of the capacitor is reduced?
Solution When a dc source is connected to a capacitor, the capacitor
gets charged and after charging no current flows in the circuit and
EXAMPLE 7.3

the lamp will not glow. There will be no change even if C is reduced.
With ac source, the capacitor offers capacitative reactance (1/w C )
and the current flows in the circuit. Consequently, the lamp will shine.
Reducing C will increase reactance and the lamp will shine less brightly
than before.

Example 7.4 A 15.0 mF capacitor is connected to a 220 V, 50 Hz source.

Find the capacitive reactance and the current (rms and peak) in the
circuit. If the frequency is doubled, what happens to the capacitive
reactance and the current?
EXAMPLE 7.4

Solution The capacitive reactance is

1 1
XC = = = 212 Ω
2 π ν C 2π (50Hz)(15.0 × 10−6 F)
The rms current is
185

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 Physics
V 220 V
I = = = 1.04 A
X C 212 Ω
The peak current is

i m = 2I = (1.41)(1.04 A ) = 1.47 A
EXAMPLE 7.4

This current oscillates between +1.47A and –1.47 A, and is ahead of

the voltage by p/2.
If the frequency is doubled, the capacitive reactance is halved and
consequently, the current is doubled.

Example 7.5 A light bulb and an open coil inductor are connected to
an ac source through a key as shown in Fig. 7.9.

FIGURE 7.9
The switch is closed and after sometime, an iron rod is inserted into
the interior of the inductor. The glow of the light bulb (a) increases; (b)
decreases; (c) is unchanged, as the iron rod is inserted. Give your
answer with reasons.
Solution As the iron rod is inserted, the magnetic field inside the coil
EXAMPLE 7.5

magnetizes the iron increasing the magnetic field inside it. Hence,
the inductance of the coil increases. Consequently, the inductive
reactance of the coil increases. As a result, a larger fraction of the
applied ac voltage appears across the inductor, leaving less voltage
across the bulb. Therefore, the glow of the light bulb decreases.

7.6 AC VOLTAGE APPLIED TO A SERIES LCR CIRCUIT

Figure 7.10 shows a series LCR circuit connected to an ac source e. As
usual, we take the voltage of the source to be v = vm sin wt.
If q is the charge on the capacitor and i the
current, at time t, we have, from Kirchhoff’s loop
rule:
di q
L +iR + =v (7.20)
dt C
We want to determine the instantaneous
current i and its phase relationship to the applied
alternating voltage v. We shall solve this problem
by two methods. First, we use the technique of
FIGURE 7.10 A series LCR circuit phasors and in the second method, we solve
connected to an ac source. Eq. (7.20) analytically to obtain the time–
186 dependence of i .

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 Alternating Current
7.6.1 Phasor-diagram solution
From the circuit shown in Fig. 7.10, we see that the resistor, inductor
and capacitor are in series. Therefore, the ac current in each element is
the same at any time, having the same amplitude and phase. Let it be
i = im sin(wt+f ) (7.21)
where f is the phase difference between the voltage across the source and
the current in the circuit. On the basis of what we have learnt in the previous
sections, we shall construct a phasor diagram for the present case.
Let I be the phasor representing the current in the circuit as given by
Eq. (7.21). Further, let VL, VR, VC, and V represent the voltage across the
inductor, resistor, capacitor and the source, respectively. From previous
section, we know that VR is parallel to I, VC is p/2
behind I and VL is p/2 ahead of I. VL, VR, VC and I
are shown in Fig. 7.11(a) with apppropriate phase-
relations.
The length of these phasors or the amplitude
of VR, VC and VL are:
vRm = im R, vCm = im XC, vLm = im XL (7.22)
The voltage Equation (7.20) for the circuit can
be written as
vL + vR + vC = v (7.23)
The phasor relation whose vertical component
gives the above equation is FIGURE 7.11 (a) Relation between the
phasors VL, VR, VC, and I, (b) Relation
VL + VR + VC = V (7.24) between the phasors VL, VR, and (VL + VC)
This relation is represented in Fig. 7.11(b). Since for the circuit in Fig. 7.10.
VC and VL are always along the same line and in
opposite directions, they can be combined into a single phasor (VC + VL)
which has a magnitude ½vCm – vLm½. Since V is represented as the
hypotenuse of a right-triangle whose sides are VR and (VC + VL), the
pythagorean theorem gives:
+ (vCm − v Lm )
2
vm2 = v Rm
2

Substituting the values of vRm, vCm, and vLm from Eq. (7.22) into the above
equation, we have
vm2 = (im R )2 + (i m X C − im X L )2

= im2 R 2 + ( X C − X L )2 

vm
or, i m = [7.25(a)]
R + ( X C − X L )2
2

By analogy to the resistance in a circuit, we introduce the impedance Z

in an ac circuit:
vm
im = [7.25(b)]
Z

where Z = R 2 + ( X C − X L )2 (7.26) 187

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 Physics
Since phasor I is always parallel to phasor VR, the phase angle f
is the angle between VR and V and can be determined from
Fig. 7.12:
vCm − v Lm
tan φ =
v Rm
Using Eq. (7.22), we have
XC − X L
tan φ = (7.27)
R
Equations (7.26) and (7.27) are graphically shown in Fig. (7.12).
FIGURE 7.12 Impedance This is called Impedance diagram which is a right-triangle with
diagram. Z as its hypotenuse.
Equation 7.25(a) gives the amplitude of the current and Eq. (7.27)
gives the phase angle. With these, Eq. (7.21) is completely specified.
If XC > XL, f is positive and the circuit is predominantly capacitive.
Consequently, the current in the circuit leads the source voltage. If
XC < X L, f is negative and the circuit is predominantly inductive.
Consequently, the current in the circuit lags the source voltage.
Figure 7.13 shows the phasor diagram and variation of v and i with w t
for the case XC > XL.
Thus, we have obtained the amplitude
and phase of current for an LCR series circuit
using the technique of phasors. But this
method of analysing ac circuits suffers from
certain disadvantages. First, the phasor
diagram say nothing about the initial
condition. One can take any arbitrary value
of t (say, t1, as done throughout this chapter)
and draw different phasors which show the
relative angle between different phasors.
The solution so obtained is called the
steady-state solution. This is not a general
FIGURE 7.13 (a) Phasor diagram of V and I. solution. Additionally, we do have a
(b) Graphs of v and i versus w t for a series LCR transient solution which exists even for
circuit where XC > XL. v = 0. The general solution is the sum of the
transient solution and the steady-state
solution. After a sufficiently long time, the effects of the transient solution
die out and the behaviour of the circuit is described by the steady-state
solution.

7.6.2 Resonance
An interesting characteristic of the series RLC circuit is the phenomenon
of resonance. The phenomenon of resonance is common among systems
that have a tendency to oscillate at a particular frequency. This frequency
is called the system’s natural frequency. If such a system is driven by an
energy source at a frequency that is near the natural frequency, the
amplitude of oscillation is found to be large. A familiar example of this
phenomenon is a child on a swing. The swing has a natural frequency
188
for swinging back and forth like a pendulum. If the child pulls on the

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 Alternating Current
rope at regular intervals and the frequency of the pulls is almost the
same as the frequency of swinging, the amplitude of the swinging will be
large (Chapter 13, Class XI).
For an RLC circuit driven with voltage of amplitude vm and frequency
w, we found that the current amplitude is given by
vm vm
im = =
Z R + ( X C − X L )2
2

with Xc = 1/wC and XL = w L . So if w is varied, then at a particular frequency

(
w0, Xc = XL, and the impedance is minimum Z = R + 0 = R . This
2 2
)
frequency is called the resonant frequency:
1
X c = X L or = ω0 L
ω0 C

1
or ω 0 = (7.28)
LC
At resonant frequency, the current amplitude
is maximum; im = vm/R.
Figure 7.16 shows the variation of im with w
in a RLC series circuit with L = 1.00 mH, C =
1.00 nF for two values of R: (i) R = 100 W
and (ii) R = 200 W. For the source applied vm =
1
100 V. w0 for this case is = 1.00×106
LC
rad/s. FIGURE 7.14 Variation of im with w for two
We see that the current amplitude is cases: (i) R = 100 W, (ii) R = 200 W,
L = 1.00 mH.
maximum at the resonant frequency. Since im =
vm / R at resonance, the current amplitude for
case (i) is twice to that for case (ii).
Resonant circuits have a variety of applications, for example, in the
tuning mechanism of a radio or a TV set. The antenna of a radio accepts
signals from many broadcasting stations. The signals picked up in the
antenna acts as a source in the tuning circuit of the radio, so the circuit
can be driven at many frequencies. But to hear one particular radio
station, we tune the radio. In tuning, we vary the capacitance of a
capacitor in the tuning circuit such that the resonant frequency of the
circuit becomes nearly equal to the frequency of the radio signal received.
When this happens, the amplitude of the current with the frequency of
the signal of the particular radio station in the circuit is maximum.
It is important to note that resonance phenomenon is exhibited by a
circuit only if both L and C are present in the circuit. Only then do the
voltages across L and C cancel each other (both being out of phase)
and the current amplitude is vm/R, the total source voltage appearing
across R. This means that we cannot have resonance in a RL or
RC circuit. 189

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 Physics
Example 7.6 A resistor of 200 W and a capacitor of 15.0 mF are
connected in series to a 220 V, 50 Hz ac source. (a) Calculate the
current in the circuit; (b) Calculate the voltage (rms) across the
resistor and the capacitor. Is the algebraic sum of these voltages
more than the source voltage? If yes, resolve the paradox.
Solution
Given
R = 200 Ω, C = 15.0 µF = 15.0 × 10−6 F
V = 220 V, ν = 50 Hz
(a) In order to calculate the current, we need the impedance of
the circuit. It is

Z = R 2 + X C2 = R 2 + (2π ν C )−2

= (200 Ω )2 + (2 × 3.14 × 50 × 15.0 × 10−6 F)−2

= (200 Ω)2 + (212.3 Ω)2

= 291.67 Ω
Therefore, the current in the circuit is
V 220 V
I = = = 0.755 A
Z 291.5 Ω
(b) Since the current is the same throughout the circuit, we have
V R = I R = (0.755 A)(200 Ω) = 151 V
VC = I X C = (0.755 A)(212.3 Ω ) = 160.3 V
The algebraic sum of the two voltages, VR and VC is 311.3 V which
is more than the source voltage of 220 V. How to resolve this
paradox? As you have learnt in the text, the two voltages are not
in the same phase. Therefore, they cannot be added like ordinary
numbers. The two voltages are out of phase by ninety degrees.
Therefore, the total of these voltages must be obtained using the
Pythagorean theorem:
EXAMPLE 7.6

V R +C = VR2 + VC2
= 220 V
Thus, if the phase difference between two voltages is properly taken
into account, the total voltage across the resistor and the capacitor
is equal to the voltage of the source.

7.7 POWER IN AC CIRCUIT: THE POWER FACTOR

We have seen that a voltage v = vm sinwt applied to a series RLC circuit
drives a current in the circuit given by i = im sin(wt + f) where
vm  X − XL 
im = and φ = tan −1  C 
Z  R
190 Therefore, the instantaneous power p supplied by the source is

Rationalised 2023-24
 Alternating Current

p = v i = (vm sin ω t ) × [im sin(ω t + φ )]

vm i m
=
2
[cos φ − cos(2ω t + φ )] (7.29)
The average power over a cycle is given by the average of the two terms in
R.H.S. of Eq. (7.37). It is only the second term which is time-dependent.
Its average is zero (the positive half of the cosine cancels the negative
half). Therefore,
vm i m v i
P = cos φ = m m cos φ
2 2 2

= V I cos φ [7.30(a)]
This can also be written as,

P = I 2 Z cos φ [7.30(b)]
So, the average power dissipated depends not only on the voltage and
current but also on the cosine of the phase angle f between them. The
quantity cosf is called the power factor. Let us discuss the following
cases:
Case (i) Resistive circuit: If the circuit contains only pure R, it is called
resistive. In that case f = 0, cos f = 1. There is maximum power dissipation.
Case (ii) Purely inductive or capacitive circuit: If the circuit contains
only an inductor or capacitor, we know that the phase difference between
voltage and current is p/2. Therefore, cos f = 0, and no power is dissipated
even though a current is flowing in the circuit. This current is sometimes
referred to as wattless current.
Case (iii) LCR series circuit: In an LCR series circuit, power dissipated is
given by Eq. (7.30) where f = tan–1 (Xc – XL )/ R. So, f may be non-zero in
a RL or RC or RCL circuit. Even in such cases, power is dissipated only in
the resistor.
Case (iv) Power dissipated at resonance in LCR circuit: At resonance
Xc – XL= 0, and f = 0. Therefore, cosf = 1 and P = I 2Z = I 2 R. That is,
maximum power is dissipated in a circuit (through R) at resonance.

Example 7.7 (a) For circuits used for transporting electric power, a
low power factor implies large power loss in transmission. Explain.
(b) Power factor can often be improved by the use of a capacitor of
appropriate capacitance in the circuit. Explain.
Solution (a) We know that P = I V cosf where cosf is the power factor.
To supply a given power at a given voltage, if cosf is small, we have to
increase current accordingly. But this will lead to large power loss
(I2R) in transmission.
EXAMPLE 7.7

(b)Suppose in a circuit, current I lags the voltage by an angle f. Then

power factor cosf =R/Z.
We can improve the power factor (tending to 1) by making Z tend to
R. Let us understand, with the help of a phasor diagram (Fig. 7.15) 191

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 Physics

FIGURE 7.15

how this can be achieved. Let us resolve I into two components. Ip

along the applied voltage V and Iq perpendicular to the applied
voltage. Iq as you have learnt in Section 7.7, is called the wattless
component since corresponding to this component of current, there
is no power loss. IP is known as the power component because it is
in phase with the voltage and corresponds to power loss in the circuit.
EXAMPLE 7.7

It’s clear from this analysis that if we want to improve power factor,
we must completely neutralize the lagging wattless current Iq by an
equal leading wattless current I¢ q. This can be done by connecting
a capacitor of appropriate value in parallel so that Iq and I¢q cancel
each other and P is effectively Ip V.

Example 7.8 A sinusoidal voltage of peak value 283 V and frequency

50 Hz is applied to a series LCR circuit in which
R = 3 W, L = 25.48 mH, and C = 796 mF. Find (a) the impedance of the
circuit; (b) the phase difference between the voltage across the source
and the current; (c) the power dissipated in the circuit; and (d) the
power factor.
Solution
(a) To find the impedance of the circuit, we first calculate XL and XC.
XL = 2 pnL
= 2 × 3.14 × 50 × 25.48 × 10–3 W = 8 W
1
XC =
2 πν C
1
= = 4Ω
2 × 3.14 × 50 × 796 × 10−6
Therefore,
Z = R 2 + ( X L − X C )2 = 32 + (8 − 4)2
=5W
EXAMPLE 7.8

XC − X L
(b) Phase difference, f = tan–1
R

 4 − 8
= tan −1  = −53.1°
192  3 

Rationalised 2023-24
 Alternating Current

Since f is negative, the current in the circuit lags the voltage

across the source.
(c) The power dissipated in the circuit is
P = I 2R

EXAMPLE 7.8
im 1  283 
Now, I = =   = 40A
2 2 5 
Therefore, P = (40A )2 × 3 Ω = 4800 W
(d) Power factor = cos cos –53.1 0.6

Example 7.9 Suppose the frequency of the source in the previous

example can be varied. (a) What is the frequency of the source at
which resonance occurs? (b) Calculate the impedance, the current,
and the power dissipated at the resonant condition.
Solution
(a) The frequency at which the resonance occurs is

1 1
ω0 = =
LC 25.48 × 10 −3 × 796 × 10 −6

= 222.1rad/s

ω0 221.1
νr = = Hz = 35.4Hz
2π 2 × 3.14

(b) The impedance Z at resonant condition is equal to the resistance:

Z = R = 3Ω

The rms current at resonance is

V V  283  1
= = = = 66.7 A
Z R  2  3
EXAMPLE 7.9

The power dissipated at resonance is

P = I 2 × R = (66.7)2 × 3 = 13.35 kW
You can see that in the present case, power dissipated
at resonance is more than the power dissipated in Example 7.8.

Example 7.10 At an airport, a person is made to walk through the

doorway of a metal detector, for security reasons. If she/he is carrying
anything made of metal, the metal detector emits a sound. On what
principle does this detector work?
Solution The metal detector works on the principle of resonance in
ac circuits. When you walk through a metal detector, you are,
in fact, walking through a coil of many turns. The coil is connected to
EXAMPLE 7.10

a capacitor tuned so that the circuit is in resonance. When

you walk through with metal in your pocket, the impedance of the
circuit changes – resulting in significant change in current in the
circuit. This change in current is detected and the electronic circuitry
causes a sound to be emitted as an alarm. 193

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 Physics
7.8 TRANSFORMERS
For many purposes, it is necessary to change (or transform) an alternating
voltage from one to another of greater or smaller value. This is done with
a device called transformer using the principle of mutual induction.
A transformer consists of two sets of coils, insulated from each other.
They are wound on a soft-iron core, either one on top of the other as in
Fig. 7.16(a) or on separate limbs of the core as in Fig. 7.16(b). One of the
coils called the primary coil has Np turns. The other coil is called the
secondary coil; it has Ns turns. Often the primary coil is the input coil
and the secondary coil is the output coil of the transformer.

FIGURE 7.16 Two arrangements for winding of primary and secondary coil in a transformer:
(a) two coils on top of each other, (b) two coils on separate limbs of the core.

When an alternating voltage is applied to the primary, the resulting

current produces an alternating magnetic flux which links the secondary
and induces an emf in it. The value of this emf depends on the number of
turns in the secondary. We consider an ideal transformer in which the
primary has negligible resistance and all the flux in the core links both
primary and secondary windings. Let f be the flux in each turn in the core
at time t due to current in the primary when a voltage vp is applied to it.
Then the induced emf or voltage es, in the secondary with Ns turns is
dφ
εs = − N s (7.31)
dt
The alternating flux f also induces an emf, called back emf in the
primary. This is
dφ
ε p = −N p (7.32)
dt
But ep = vp. If this were not so, the primary current would be infinite
since the primary has zero resistance (as assumed). If the secondary is
an open circuit or the current taken from it is small, then to a good
approximation
194 es = vs

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 Alternating Current
where vs is the voltage across the secondary. Therefore, Eqs. (7.31) and
(7.32) can be written as
dφ
vs = − N s [7.31(a)]
dt

dφ
v p = −N p [7.32(a)]
dt
From Eqs. [7.31 (a)] and [7.32 (a)], we have
vs N
= s (7.33)
vp N p
Note that the above relation has been obtained using three
assumptions: (i) the primary resistance and current are small; (ii) the
same flux links both the primary and the secondary as very little flux
escapes from the core, and (iii) the secondary current is small.
If the transformer is assumed to be 100% efficient (no energy losses),
the power input is equal to the power output, and since p = i v,
ipvp = isvs (7.34)
Although some energy is always lost, this is a good approximation,
since a well designed transformer may have an efficiency of more than
95%. Combining Eqs. (7.33) and (7.34), we have
i p vs N
= = s (7.35)
is v p N p

Since i and v both oscillate with the same frequency as the ac source,
Eq. (7.35) also gives the ratio of the amplitudes or rms values of
corresponding quantities.
Now, we can see how a transformer affects the voltage and current.
We have:
N   Np 
Vs =  s  V p and I s =  Ip
 N s 
(7.36)
 Np 
That is, if the secondary coil has a greater number of turns than the
primary (Ns > Np), the voltage is stepped up (Vs > Vp). This type of
arrangement is called a step-up transformer. However, in this arrangement,
there is less current in the secondary than in the primary (Np/Ns < 1 and Is
< Ip). For example, if the primary coil of a transformer has 100 turns and
the secondary has 200 turns, Ns/Np = 2 and Np/Ns=1/2. Thus, a 220V
input at 10A will step-up to 440 V output at 5.0 A.
If the secondary coil has less turns than the primary (Ns < Np),
we have a step-down transformer. In this case, Vs < Vp and Is > Ip. That
is, the voltage is stepped down, or reduced, and the current
is increased.
The equations obtained above apply to ideal transformers (without
any energy losses). But in actual transformers, small energy losses do
occur due to the following reasons:
(i) Flux Leakage: There is always some flux leakage; that is, not all of
the flux due to primary passes through the secondary due to poor 195

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 Physics
design of the core or the air gaps in the core. It can be reduced by
winding the primary and secondary coils one over the other.
(ii) Resistance of the windings: The wire used for the windings has some
resistance and so, energy is lost due to heat produced in the wire
(I 2R). In high current, low voltage windings, these are minimised by
using thick wire.
(iii) Eddy currents: The alternating magnetic flux induces eddy currents
in the iron core and causes heating. The effect is reduced by using a
laminated core.
(iv) Hysteresis: The magnetisation of the core is repeatedly reversed by
the alternating magnetic field. The resulting expenditure of energy in
the core appears as heat and is kept to a minimum by using a magnetic
material which has a low hysteresis loss.
The large scale transmission and distribution of electrical energy over
long distances is done with the use of transformers. The voltage output
of the generator is stepped-up (so that current is reduced and
consequently, the I 2R loss is cut down). It is then transmitted over long
distances to an area sub-station near the consumers. There the voltage
is stepped down. It is further stepped down at distributing sub-stations
and utility poles before a power supply of 240 V reaches our homes.

SUMMARY

1. An alternating voltage v = vm sin ω t applied to a resistor R drives a

vm
current i = im sinwt in the resistor, im = . The current is in phase with
R
the applied voltage.
2. For an alternating current i = im sin wt passing through a resistor R, the
average power loss P (averaged over a cycle) due to joule heating is
( 1/2 )i 2mR. To express it in the same form as the dc power (P = I 2R), a
special value of current is used. It is called root mean square (rms)
current and is donoted by I:
im
I = = 0.707 im
2
Similarly, the rms voltage is defined by

vm
V = = 0.707 vm
2
We have P = IV = I 2R
3. An ac voltage v = vm sin wt applied to a pure inductor L, drives a current
in the inductor i = im sin (wt – p/2), where im = vm/XL. XL = wL is called
inductive reactance. The current in the inductor lags the voltage by
p/2. The average power supplied to an inductor over one complete cycle
is zero.

196

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 Alternating Current

4. An ac voltage v = vm sinwt applied to a capacitor drives a current in the

capacitor: i = im sin (wt + p/2). Here,
vm 1
im = , XC =
XC ωC is called capacitive reactance.
The current through the capacitor is p/2 ahead of the applied voltage.
As in the case of inductor, the average power supplied to a capacitor
over one complete cycle is zero.
5. For a series RLC circuit driven by voltage v = vm sin wt, the current is
given by i = im sin (wt + f )
vm
where im =
R + ( XC − X L )
2 2

XC − X L
and φ = tan −1
R

Z = R2 + ( X C − X L )
2
is called the impedance of the circuit.
The average power loss over a complete cycle is given by
P = V I cosf
The term cosf is called the power factor.
6. In a purely inductive or capacitive circuit, cosf = 0 and no power is
dissipated even though a current is flowing in the circuit. In such cases,
current is referred to as a wattless current.
7. The phase relationship between current and voltage in an ac circuit
can be shown conveniently by representing voltage and current by
rotating vectors called phasors. A phasor is a vector which rotates
about the origin with angular speed w. The magnitude of a phasor
represents the amplitude or peak value of the quantity (voltage or
current) represented by the phasor.
The analysis of an ac circuit is facilitated by the use of a phasor
diagram.
8. A transformer consists of an iron core on which are bound a primary
coil of Np turns and a secondary coil of Ns turns. If the primary coil is
connected to an ac source, the primary and secondary voltages are
related by

N 
Vs =  s  V p
 Np 
and the currents are related by

 Np 
Is =   I p
 Ns
If the secondary coil has a greater number of turns than the primary, the
voltage is stepped-up (Vs > Vp). This type of arrangement is called a step-
up transformer. If the secondary coil has turns less than the primary, we
have a step-down transformer.

197

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 Physics
Physical quantity Symbol Dimensions Unit Remarks

2 –3 –1 vm
rms voltage V [M L T A ] V V = , vm is the
2
amplitude of the ac voltage.

im
rms current I [ A] A I= , im is the amplitude of
2
the ac current.

Reactance:
XL =  L
2 –3 –2
Inductive XL [M L T A ]
XC = 1/  C
2 –3 –2
Capacitive XC [M L T A ]
2 –3 –2
Impedance Z [M L T A ] Depends on elements
present in the circuit.

1
Resonant wr or w0 [T ]
–1
Hz w0  for a
frequency LC
series RLC circuit

ω0 L 1
Quality factor Q Dimensionless Q= = for a series
R ω0 C R
RLC circuit.
Power factor Dimensionless = cos f , f is the phase
difference between voltage
applied and current in
the circuit.

POINTS TO PONDER

1. When a value is given for ac voltage or current, it is ordinarily the rms

value. The voltage across the terminals of an outlet in your room is
normally 240 V. This refers to the rms value of the voltage. The amplitude
of this voltage is

vm = 2V = 2(240) = 340 V
2. The power rating of an element used in ac circuits refers to its average
power rating.
3. The power consumed in an ac circuit is never negative.
4. Both alternating current and direct current are measured in amperes.
But how is the ampere defined for an alternating current? It cannot be
derived from the mutual attraction of two parallel wires carrying ac
198 currents, as the dc ampere is derived. An ac current changes direction

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 Alternating Current

with the source frequency and the attractive force would average to
zero. Thus, the ac ampere must be defined in terms of some property
that is independent of the direction of the current. Joule heating
is such a property, and there is one ampere of rms value of
alternating current in a circuit if the current produces the same
average heating effect as one ampere of dc current would produce
under the same conditions.
5. In an ac circuit, while adding voltages across different elements, one
should take care of their phases properly. For example, if V R and VC
are voltages across R and C, respectively in an RC circuit, then the

total voltage across RC combination is VRC = VR2 + VC2 and not

VR + VC since V C is p/2 out of phase of V R.
6. Though in a phasor diagram, voltage and current are represented by
vectors, these quantities are not really vectors themselves. They are
scalar quantities. It so happens that the amplitudes and phases of
harmonically varying scalars combine mathematically in the same
way as do the projections of rotating vectors of corresponding
magnitudes and directions. The ‘rotating vectors’ that represent
harmonically varying scalar quantities are introduced only to provide
us with a simple way of adding these quantities using a rule that
we already know as the law of vector addition.
7. There are no power losses associated with pure capacitances and pure
inductances in an ac circuit. The only element that dissipates energy
in an ac circuit is the resistive element.
8. In a RLC circuit, resonance phenomenon occur when XL = X C or
1
ω0 = . For resonance to occur, the presence of both L and C
LC
elements in the circuit is a must. With only one of these (L or C )
elements, there is no possibility of voltage cancellation and hence,
no resonance is possible.
9. The power factor in a RLC circuit is a measure of how close the
circuit is to expending the maximum power.
10. In generators and motors, the roles of input and output are
reversed. In a motor, electric energy is the input and mechanical
energy is the output. In a generator, mechanical energy is the
input and electric energy is the output. Both devices simply
transfor m energy from one form to another.
11. A transformer (step-up) changes a low-voltage into a high-voltage.
This does not violate the law of conservation of energy. The
current is reduced by the same proportion.

199

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 Physics
EXERCISES
7.1 A 100 W resistor is connected to a 220 V, 50 Hz ac supply.
(a) What is the rms value of current in the circuit?
(b) What is the net power consumed over a full cycle?
7.2 (a) The peak voltage of an ac supply is 300 V. What is the rms voltage?
(b) The rms value of current in an ac circuit is 10 A. What is the
peak current?
7.3 A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine
the rms value of the current in the circuit.
7.4 A 60 mF capacitor is connected to a 110 V, 60 Hz ac supply. Determine
the rms value of the current in the circuit.
7.5 In Exercises 7.3 and 7.4, what is the net power absorbed by each
circuit over a complete cycle. Explain your answer.
7.6 A charged 30 mF capacitor is connected to a 27 mH inductor. What is
the angular frequency of free oscillations of the circuit?
7.7 A series LCR circuit with R = 20 W, L = 1.5 H and C = 35 mF is connected
to a variable-frequency 200 V ac supply. When the frequency of the
supply equals the natural frequency of the circuit, what is the average
power transferred to the circuit in one complete cycle?
7.8 Figure 7.17 shows a series LCR circuit connected to a variable
frequency 230 V source. L = 5.0 H, C = 80mF, R = 40 W.

FIGURE 7.17

(a) Determine the source frequency which drives the circuit in

resonance.
(b) Obtain the impedance of the circuit and the amplitude of current
at the resonating frequency.
(c) Determine the rms potential drops across the three elements of
the circuit. Show that the potential drop across the LC
combination is zero at the resonating frequency.

200

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Chapter Eight

ELECTROMAGNETIC
WAVES

8.1 INTRODUCTION
In Chapter 4, we learnt that an electric current produces magnetic field
and that two current-carrying wires exert a magnetic force on each other.
Further, in Chapter 6, we have seen that a magnetic field changing with
time gives rise to an electric field. Is the converse also true? Does an
electric field changing with time give rise to a magnetic field? James Clerk
Maxwell (1831-1879), argued that this was indeed the case – not only
an electric current but also a time-varying electric field generates magnetic
field. While applying the Ampere’s circuital law to find magnetic field at a
point outside a capacitor connected to a time-varying current, Maxwell
noticed an inconsistency in the Ampere’s circuital law. He suggested the
existence of an additional current, called by him, the displacement
current to remove this inconsistency.
Maxwell formulated a set of equations involving electric and magnetic
fields, and their sources, the charge and current densities. These
equations are known as Maxwell’s equations. Together with the Lorentz
force formula (Chapter 4), they mathematically express all the basic laws
of electromagnetism.
The most important prediction to emerge from Maxwell’s equations
is the existence of electromagnetic waves, which are (coupled) time-
varying electric and magnetic fields that propagate in space. The speed
of the waves, according to these equations, turned out to be very close to

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 Physics
the speed of light( 3 ×108 m/s), obtained from optical
measurements. This led to the remarkable conclusion
that light is an electromagnetic wave. Maxwell’s work
thus unified the domain of electricity, magnetism and
light. Hertz, in 1885, experimentally demonstrated the
existence of electromagnetic waves. Its technological use
by Marconi and others led in due course to the
revolution in communication that we are witnessing
today.
In this chapter, we first discuss the need for
displacement current and its consequences. Then we
present a descriptive account of electromagnetic waves.
James Clerk Maxwell The broad spectrum of electromagnetic waves,
(1831 – 1879) Born in stretching from g rays (wavelength ~10–12 m) to long
Edinburgh, Scotland,
radio waves (wavelength ~106 m) is described.
was among the greatest
physicists of the
nineteenth century. He 8.2 DISPLACEMENT CURRENT
derived the thermal We have seen in Chapter 4 that an electrical current
velocity distribution of
produces a magnetic field around it. Maxwell showed
molecules in a gas and
was among the first to that for logical consistency, a changing electric field must
obtain reliable also produce a magnetic field. This effect is of great
estimates of molecular importance because it explains the existence of radio
parameters from waves, gamma rays and visible light, as well as all other
measurable quantities forms of electromagnetic waves.
like viscosity, etc.
To see how a changing electric field gives rise to
JAMES CLERK MAXWELL (1831–1879)

Maxwell’s greatest
acheivement was the a magnetic field, let us consider the process of
unification of the laws of charging of a capacitor and apply Ampere’s circuital
electricity and law given by (Chapter 4)
magnetism (discovered
by Coulomb, Oersted, “B.dl = m0 i (t ) (8.1)
Ampere and Faraday)
to find magnetic field at a point outside the capacitor.
into a consistent set of
equations now called Figure 8.1(a) shows a parallel plate capacitor C which
Maxwell’s equations. is a part of circuit through which a time-dependent
From these he arrived at current i (t ) flows . Let us find the magnetic field at a
the most important point such as P, in a region outside the parallel plate
conclusion that light is capacitor. For this, we consider a plane circular loop of
an electromagnetic radius r whose plane is perpendicular to the direction
wave. Interestingly,
of the current-carrying wire, and which is centred
Maxwell did not agree
with the idea (strongly symmetrically with respect to the wire [Fig. 8.1(a)]. From
suggested by the symmetry, the magnetic field is directed along the
Faraday’s laws of circumference of the circular loop and is the same in
electrolysis) that magnitude at all points on the loop so that if B is the
electricity was magnitude of the field, the left side of Eq. (8.1) is B (2p r).
particulate in nature. So we have
B (2pr) = m0i (t ) (8 .2)
202

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 Electromagnetic
Waves
Now, consider a different surface, which has the same boundary. This
is a pot like surface [Fig. 8.1(b)] which nowhere touches the current, but
has its bottom between the capacitor plates; its mouth is the circular
loop mentioned above. Another such surface is shaped like a tiffin box
(without the lid) [Fig. 8.1(c)]. On applying Ampere’s circuital law to such
surfaces with the same perimeter, we find that the left hand side of
Eq. (8.1) has not changed but the right hand side is zero and not m0 i,
since no current passes through the surface of Fig. 8.1(b) and (c). So we
have a contradiction; calculated one way, there is a magnetic field at a
point P; calculated another way, the magnetic field at P is zero.
Since the contradiction arises from our use of Ampere’s circuital law,
this law must be missing something. The missing term must be such
that one gets the same magnetic field at point P, no matter what surface
is used.
We can actually guess the missing term by looking carefully at
Fig. 8.1(c). Is there anything passing through the surface S between the
plates of the capacitor? Yes, of course, the electric field! If the plates of the
capacitor have an area A, and a total charge Q, the magnitude of the
electric field E between the plates is (Q/A)/e0 (see Eq. 2.41). The field is
perpendicular to the surface S of Fig. 8.1(c). It has the same magnitude
over the area A of the capacitor plates, and vanishes outside it. So what
is the electric flux FE through the surface S ? Using Gauss’s law, it is
1 Q Q
ΦE = E A = A= (8.3)
ε0 A ε0
Now if the charge Q on the capacitor plates changes with time, there is a
current i = (dQ/dt), so that using Eq. (8.3), we have
dΦE d  Q  1 dQ FIGURE 8.1 A
= =
dt dt  ε 0  ε0 dt parallel plate
capacitor C, as part of
This implies that for consistency, a circuit through
which a time
 dΦ  dependent current
ε0  E  = i (8.4) i (t) flows, (a) a loop of
 dt 
radius r, to determine
This is the missing term in Ampere’s circuital law. If we generalise magnetic field at a
this law by adding to the total current carried by conductors through point P on the loop;
the surface, another term which is e0 times the rate of change of electric (b) a pot-shaped
surface passing
flux through the same surface, the total has the same value of current i through the interior
for all surfaces. If this is done, there is no contradiction in the value of B between the capacitor
obtained anywhere using the generalised Ampere’s law. B at the point P plates with the loop
is non-zero no matter which surface is used for calculating it. B at a shown in (a) as its
rim; (c) a tiffin-
point P outside the plates [Fig. 8.1(a)] is the same as at a point M just shaped surface with
inside, as it should be. The current carried by conductors due to flow of the circular loop as
charges is called conduction current. The current, given by Eq. (8.4), is a its rim and a flat
circular bottom S
new term, and is due to changing electric field (or electric displacement, between the capacitor
an old term still used sometimes). It is, therefore, called displacement plates. The arrows
current or Maxwell’s displacement current. Figure 8.2 shows the electric show uniform electric
field between the
and magnetic fields inside the parallel plate capacitor discussed above. capacitor plates.
The generalisation made by Maxwell then is the following. The source
of a magnetic field is not just the conduction electric current due to flowing 203

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 Physics
charges, but also the time rate of change of electric field. More
precisely, the total current i is the sum of the conduction current
denoted by ic, and the displacement current denoted by id (= e0 (dFE/
dt)). So we have
dΦE
i = ic + id = ic + ε0 (8.5)
dt
In explicit terms, this means that outside the capacitor plates,
we have only conduction current ic = i, and no displacement
current, i.e., id = 0. On the other hand, inside the capacitor, there is
no conduction current, i.e., ic = 0, and there is only displacement
current, so that id = i.
The generalised (and correct) Ampere’s circuital law has the same
form as Eq. (8.1), with one difference: “the total current passing
through any surface of which the closed loop is the perimeter” is
the sum of the conduction current and the displacement current.
The generalised law is
dΦE
∫ Bgdl = µ0 i c + µ0 ε0
dt
(8.6)
and is known as Ampere-Maxwell law.
In all respects, the displacement current has the same physical
effects as the conduction current. In some cases, for example, steady
electric fields in a conducting wire, the displacement current may
be zero since the electric field E does not change with time. In other
FIGURE 8.2 (a) The cases, for example, the charging capacitor above, both conduction
electric and magnetic and displacement currents may be present in different regions of
fields E and B between space. In most of the cases, they both may be present in the same
the capacitor plates, at region of space, as there exist no perfectly conducting or perfectly
the point M. (b) A cross
insulating medium. Most interestingly, there may be large regions
sectional view of Fig. (a).
of space where there is no conduction current, but there is only a
displacement current due to time-varying electric fields. In such a
region, we expect a magnetic field, though there is no (conduction)
current source nearby! The prediction of such a displacement current
can be verified experimentally. For example, a magnetic field (say at point
M) between the plates of the capacitor in Fig. 8.2(a) can be measured and
is seen to be the same as that just outside (at P).
The displacement current has (literally) far reaching consequences.
One thing we immediately notice is that the laws of electricity and
magnetism are now more symmetrical*. Faraday’s law of induction states
that there is an induced emf equal to the rate of change of magnetic flux.
Now, since the emf between two points 1 and 2 is the work done per unit
charge in taking it from 1 to 2, the existence of an emf implies the existence
of an electric field. So, we can rephrase Faraday’s law of electromagnetic
induction by saying that a magnetic field, changing with time, gives rise
to an electric field. Then, the fact that an electric field changing with
time gives rise to a magnetic field, is the symmetrical counterpart, and is

* They are still not perfectly symmetrical; there are no known sources of magnetic
field (magnetic monopoles) analogous to electric charges which are sources of
204 electric field.

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 Electromagnetic
Waves
a consequence of the displacement current being a source of a magnetic
field. Thus, time- dependent electric and magnetic fields give rise to each
other! Faraday’s law of electromagnetic induction and Ampere-Maxwell
law give a quantitative expression of this statement, with the current
being the total current, as in Eq. (8.5). One very important consequence
of this symmetry is the existence of electromagnetic waves, which we
discuss qualitatively in the next section.

MAXWELL’S EQUATIONS IN VACUUM

1. “E.dA = Q/✒ 0
(Gauss’s Law for electricity)

2. “B.dA = 0 (Gauss’s Law for magnetism)

3. “E.dll == –ddΦt B
(Faraday’s Law)

dΦE
4. “B.dl == µ i + µ0 ε 0
0 c
dt
(Ampere – Maxwell Law)

8.3 ELECTROMAGNETIC WAVES

8.3.1 Sources of electromagnetic waves
How are electromagnetic waves produced? Neither stationary charges
nor charges in uniform motion (steady currents) can be sources of
electromagnetic waves. The former produces only electrostatic fields, while
the latter produces magnetic fields that, however, do not vary with time.
It is an important result of Maxwell’s theory that accelerated charges
radiate electromagnetic waves. The proof of this basic result is beyond
the scope of this book, but we can accept it on the basis of rough,
qualitative reasoning. Consider a charge oscillating with some frequency.
(An oscillating charge is an example of accelerating charge.) This
produces an oscillating electric field in space, which produces an
oscillating magnetic field, which in turn, is a source of oscillating electric
field, and so on. The oscillating electric and magnetic fields thus
regenerate each other, so to speak, as the wave propagates through the
space. The frequency of the electromagnetic wave naturally equals the
frequency of oscillation of the charge. The energy associated with the
propagating wave comes at the expense of the energy of the source – the
accelerated charge.
From the preceding discussion, it might appear easy to test the
prediction that light is an electromagnetic wave. We might think that all
we needed to do was to set up an ac circuit in which the current oscillate
at the frequency of visible light, say, yellow light. But, alas, that is not
possible. The frequency of yellow light is about 6 × 1014 Hz, while the
frequency that we get even with modern electronic circuits is hardly about
1011 Hz. This is why the experimental demonstration of electromagnetic 205

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 Physics
wave had to come in the low frequency region (the radio
wave region), as in the Hertz’s experiment (1887).
Hertz’s successful experimental test of Maxwell’s
theory created a sensation and sparked off other
important works in this field. Two important
achievements in this connection deserve mention. Seven
years after Hertz, Jagdish Chandra Bose, working at
Calcutta (now Kolkata), succeeded in producing and
observing electromagnetic waves of much shorter
EXAMPLE 8.1

wavelength (25 mm to 5 mm). His experiment, like that

of Hertz’s, was confined to the laboratory.
At around the same time, Guglielmo Marconi in Italy
followed Hertz’s work and succeeded in transmitting
electromagnetic waves over distances of many kilometres.
HEINRICH RUDOLF HERTZ (1857–1894)

Heinrich Rudolf Hertz Marconi’s experiment marks the beginning of the field of
(1857 – 1894) German communication using electromagnetic waves.
physicist who was the
first to broadcast and 8.3.2 Nature of electromagnetic waves
receive radio waves. He
It can be shown from Maxwell’s equations that electric
produced electro-
magnetic waves, sent and magnetic fields in an electromagnetic wave are
them through space, and perpendicular to each other, and to the direction of
measured their wave- propagation. It appears reasonable, say from our
length and speed. He discussion of the displacement current. Consider
showed that the nature Fig. 8.2. The electric field inside the plates of the capacitor
of their vibration, is directed perpendicular to the plates. The magnetic
reflection and refraction
field this gives rise to via the displacement current is
was the same as that of
light and heat waves,
along the perimeter of a circle parallel to the capacitor
establishing their plates. So B and E are perpendicular in this case. This
identity for the first time. is a general feature.
He also pioneered In Fig. 8.3, we show a typical example of a plane
research on discharge of electromagnetic wave propagating along the z direction
electricity through gases, (the fields are shown as a function of the z coordinate, at
and discovered the a given time t). The electric field Ex is along the x-axis,
photoelectric effect.
and varies sinusoidally with z, at a given time. The
magnetic field By is along the y-axis, and again varies
sinusoidally with z. The electric and magnetic fields Ex
and By are perpendicular to each
other, and to the direction z of
propagation. We can write Ex and
By as follows:
Ex= E0 sin (kz–wt ) [8.7(a)]
By= B0 sin (kz–wt ) [8.7(b)]
Here k is related to the wave length
FIGURE 8.3 A linearly polarised electromagnetic wave, l of the wave by the usual
propagating in the z-direction with the oscillating electric field E
equation
along the x-direction and the oscillating magnetic field B along
the y-direction. 2π
k= (8.8)
206 λ

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 Electromagnetic
Waves
and w is the angular frequency. k is the magnitude of the wave vector (or
propagation vector) k and its direction describes the direction of
propagation of the wave. The speed of propagation of the wave is (w/k ).
Using Eqs. [8.7(a) and (b)] for Ex and By and Maxwell’s equations, one
finds that
w = ck, where, c = 1/ µ0 ε 0 [8.9(a)]
The relation w = ck is the standard one for waves (see for example,
Section 15.4 of class XI Physics textbook). This relation is often written
in terms of frequency, n (=w/2p) and wavelength, l (=2p/k) as
 2π 
2πν = c   or
 λ 
nl = c [8.9(b)]
It is also seen from Maxwell’s equations that the magnitude of the
electric and the magnetic fields in an electromagnetic wave are related as
B0 = (E0/c) (8.10)
We here make remarks on some features of electromagnetic waves.
They are self-sustaining oscillations of electric and magnetic fields in
free space, or vacuum. They differ from all the other waves we have
studied so far, in respect that no material medium is involved in the
vibrations of the electric and magnetic fields.
But what if a material medium is actually there? We know that light,
an electromagnetic wave, does propagate through glass, for example. We
have seen earlier that the total electric and magnetic fields inside a
medium are described in terms of a permittivity e and a magnetic
permeability m (these describe the factors by which the total fields differ
from the external fields). These replace e0 and m0 in the description to
electric and magnetic fields in Maxwell’s equations with the result that in
a material medium of permittivity e and magnetic permeability m, the
velocity of light becomes,
1
v= (8.11)
µε
Thus, the velocity of light depends on electric and magnetic properties of
the medium. We shall see in the next chapter that the refractive index of
one medium with respect to the other is equal to the ratio of velocities of
light in the two media.
The velocity of electromagnetic waves in free space or vacuum is an
important fundamental constant. It has been shown by experiments on
electromagnetic waves of different wavelengths that this velocity is the
same (independent of wavelength) to within a few metres per second, out
of a value of 3×108 m/s. The constancy of the velocity of em waves in
vacuum is so strongly supported by experiments and the actual value is
so well known now that this is used to define a standard of length.
The great technological importance of electromagnetic waves stems
from their capability to carry energy from one place to another. The
radio and TV signals from broadcasting stations carry energy. Light
carries energy from the sun to the earth, thus making life possible on
the earth. 207

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 Physics
Example 8.1 A plane electromagnetic wave of frequency
25 MHz travels in free space along the x-direction. At a particular
point in space and time, E = 6.3 ĵ V/m. What is B at this point?
Solution Using Eq. (8.10), the magnitude of B is
E
B=
c
6.3 V/m
= = 2.1 × 10 –8 T
3 × 108 m/s
EXAMPLE 8.1

To find the direction, we note that E is along y-direction and the

wave propagates along x-axis. Therefore, B should be in a direction
perpendicular to both x- and y-axes. Using vector algebra, E × B should
be along x-direction. Since, (+ ĵ ) × (+ k̂ ) = î , B is along the z-direction.
Thus, B = 2.1 × 10–8 k̂ T

Example 8.2 The magnetic field in a plane electromagnetic wave is

http://www.fnal.gov/pub/inquiring/more/light

given by By = (2 × 10–7) T sin (0.5×103x+1.5×1011t).

http://imagine.gsfc.nasa.gov/docs/science/

(a) What is the wavelength and frequency of the wave?

(b) Write an expression for the electric field.
Solution
(a) Comparing the given equation with
Electromagnetic spectrum

  x t 
By = B0 sin 2π  +  
 λ T 
2π
We get, λ = m = 1.26 cm,
0.5 × 103
1
and
T
( )
= ν = 1.5 × 1011 /2π = 23.9 GHz
EXAMPLE 8.2

(b) E0 = B0c = 2×10–7 T × 3 × 108 m/s = 6 × 101 V/m

The electric field component is perpendicular to the direction of
propagation and the direction of magnetic field. Therefore, the
electric field component along the z-axis is obtained as
Ez = 60 sin (0.5 × 103x + 1.5 × 1011 t) V/m

8.4 ELECTROMAGNETIC SPECTRUM

At the time Maxwell predicted the existence of electromagnetic waves, the
only familiar electromagnetic waves were the visible light waves. The existence
of ultraviolet and infrared waves was barely established. By the end of the
nineteenth century, X-rays and gamma rays had also been discovered. We
now know that, electromagnetic waves include visible light waves, X-rays,
gamma rays, radio waves, microwaves, ultraviolet and infrared waves. The
classification of em waves according to frequency is the electromagnetic
spectrum (Fig. 8.5). There is no sharp division between one kind of wave
and the next. The classification is based roughly on how the waves are
produced and/or detected.
We briefly describe these different types of electromagnetic waves, in
208 order of decreasing wavelengths.

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 Electromagnetic
Waves

FIGURE 8.5 The electromagnetic spectrum, with common names for various
part of it. The various regions do not have sharply defined boundaries.

8.4.1 Radio waves

Radio waves are produced by the accelerated motion of charges in conducting
wires. They are used in radio and television communication systems. They
are generally in the frequency range from 500 kHz to about 1000 MHz.
The AM (amplitude modulated) band is from 530 kHz to 1710 kHz. Higher
frequencies upto 54 MHz are used for short wave bands. TV waves range
from 54 MHz to 890 MHz. The FM (frequency modulated) radio band
extends from 88 MHz to 108 MHz. Cellular phones use radio waves to
transmit voice communication in the ultrahigh frequency (UHF) band. How
these waves are transmitted and received is described in Chapter 15.
8.4.2 Microwaves
Microwaves (short-wavelength radio waves), with frequencies in the
gigahertz (GHz) range, are produced by special vacuum tubes (called
klystrons, magnetrons and Gunn diodes). Due to their short wavelengths,
they are suitable for the radar systems used in aircraft navigation. Radar
also provides the basis for the speed guns used to time fast balls, tennis-
serves, and automobiles. Microwave ovens are an interesting domestic
application of these waves. In such ovens, the frequency of the microwaves
is selected to match the resonant frequency of water molecules so that
energy from the waves is transferred efficiently to the kinetic energy of
the molecules. This raises the temperature of any food containing water. 209

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 Physics
8.4.3 Infrared waves
Infrared waves are produced by hot bodies and molecules. This band
lies adjacent to the low-frequency or long-wave length end of the visible
spectrum. Infrared waves are sometimes referred to as heat waves. This
is because water molecules present in most materials readily absorb
infrared waves (many other molecules, for example, CO2, NH3, also absorb
infrared waves). After absorption, their thermal motion increases, that is,
they heat up and heat their surroundings. Infrared lamps are used in
physical therapy. Infrared radiation also plays an important role in
maintaining the earth’s warmth or average temperature through the
greenhouse effect. Incoming visible light (which passes relatively easily
through the atmosphere) is absorbed by the earth’s surface and re-
radiated as infrared (longer wavelength) radiations. This radiation is
trapped by greenhouse gases such as carbon dioxide and water vapour.
Infrared detectors are used in Earth satellites, both for military purposes
and to observe growth of crops. Electronic devices (for example
semiconductor light emitting diodes) also emit infrared and are widely
used in the remote switches of household electronic systems such as TV
sets, video recorders and hi-fi systems.

8.4.4 Visible rays

It is the most familiar form of electromagnetic waves. It is the part of the
spectrum that is detected by the human eye. It runs from about
4 × 1014 Hz to about 7 × 1014 Hz or a wavelength range of about 700 –
400 nm. Visible light emitted or reflected from objects around us provides
us information about the world. Our eyes are sensitive to this range of
wavelengths. Different animals are sensitive to different range of
wavelengths. For example, snakes can detect infrared waves, and the
‘visible’ range of many insects extends well into the utraviolet.

8.4.5 Ultraviolet rays

It covers wavelengths ranging from about 4 × 10–7 m (400 nm) down to
6 × 10–10m (0.6 nm). Ultraviolet (UV) radiation is produced by special
lamps and very hot bodies. The sun is an important source of ultraviolet
light. But fortunately, most of it is absorbed in the ozone layer in the
atmosphere at an altitude of about 40 – 50 km. UV light in large quantities
has harmful effects on humans. Exposure to UV radiation induces the
production of more melanin, causing tanning of the skin. UV radiation is
absorbed by ordinary glass. Hence, one cannot get tanned or sunburn
through glass windows.
Welders wear special glass goggles or face masks with glass windows
to protect their eyes from large amount of UV produced by welding arcs.
Due to its shorter wavelengths, UV radiations can be focussed into very
narrow beams for high precision applications such as LASIK (Laser-
assisted in situ keratomileusis) eye surgery. UV lamps are used to kill
germs in water purifiers.
Ozone layer in the atmosphere plays a protective role, and hence its
depletion by chlorofluorocarbons (CFCs) gas (such as freon) is a matter
210
of international concern.

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 Electromagnetic
Waves
8.4.6 X-rays
Beyond the UV region of the electromagnetic spectrum lies the X-ray
region. We are familiar with X-rays because of its medical applications. It
covers wavelengths from about 10–8 m (10 nm) down to 10–13 m
(10–4 nm). One common way to generate X-rays is to bombard a metal
target by high energy electrons. X-rays are used as a diagnostic tool in
medicine and as a treatment for certain forms of cancer. Because X-rays
damage or destroy living tissues and organisms, care must be taken to
avoid unnecessary or over exposure.

8.4.7 Gamma rays

They lie in the upper frequency range of the electromagnetic spectrum
and have wavelengths of from about 10–10m to less than 10–14m. This
high frequency radiation is produced in nuclear reactions and
also emitted by radioactive nuclei. They are used in medicine to destroy
cancer cells.
Table 8.1 summarises different types of electromagnetic waves, their
production and detections. As mentioned earlier, the demarcation between
different regions is not sharp and there are overlaps.

TABLE 8.1 DIFFERENT TYPES OF ELECTROMAGNETIC WAVES

Type Wavelength range Production Detection

Radio > 0.1 m Rapid acceleration and Receiver’s aerials

decelerations of electrons
in aerials
Microwave 0.1m to 1 mm Klystron valve or Point contact diodes
magnetron valve
Infra-red 1mm to 700 nm Vibration of atoms Thermopiles
and molecules Bolometer, Infrared
photographic film
Light 700 nm to 400 nm Electrons in atoms emit The eye
light when they move from Photocells
one energy level to a Photographic film
lower energy level
Ultraviolet 400 nm to 1nm Inner shell electrons in Photocells
atoms moving from one Photographic film
energy level to a lower level
X-rays 1nm to 10–3 nm X-ray tubes or inner shell Photographic film
electrons Geiger tubes
Ionisation chamber
Gamma rays <10–3 nm Radioactive decay of the -do-
nucleus

211

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 Physics
SUMMARY

1. Maxwell found an inconsistency in the Ampere’s law and suggested the

existence of an additional current, called displacement current, to remove
this inconsistency. This displacement current is due to time-varying electric
field and is given by
dΦΕ
id = ε0
dt
and acts as a source of magnetic field in exactly the same way as conduction
current.
2. An accelerating charge produces electromagnetic waves. An electric charge
oscillating harmonically with frequency n, produces electromagnetic waves
of the same frequency n . An electric dipole is a basic source of
electromagnetic waves.
3. Electromagnetic waves with wavelength of the order of a few metres were
first produced and detected in the laboratory by Hertz in 1887. He thus
verified a basic prediction of Maxwell’s equations.
4. Electric and magnetic fields oscillate sinusoidally in space and time in an
electromagnetic wave. The oscillating electric and magnetic fields, E and
B are perpendicular to each other, and to the direction of propagation of
the electromagnetic wave. For a wave of frequency n, wavelength l ,
propagating along z-direction, we have
E = Ex (t) = E0 sin (kz – w t )

 z    z t 
= E0 sin 2π  λ − νt   = E 0 sin 2π  λ − T  
   
B = By(t) = B0 sin (kz – w t)

 z    z t 
= B0 sin 2π  − νt   = B0 sin 2π  −  
  λ     λ T 
They are related by E0/B0 = c.
5. The speed c of electromagnetic wave in vacuum is related to m0 and e0 (the
free space permeability and permittivity constants) as follows:

c = 1/ µ0 ε 0 . The value of c equals the speed of light obtained from

optical measurements.
Light is an electromagnetic wave; c is, therefore, also the speed of light.
Electromagnetic waves other than light also have the same velocity c in
free space.
The speed of light, or of electromagnetic waves in a material medium is
given by v = 1/ µ ε
where m is the permeability of the medium and e its permittivity.
6. The spectrum of electromagnetic waves stretches, in principle, over an
infinite range of wavelengths. Different regions are known by different
names; g-rays, X-rays, ultraviolet rays, visible rays, infrared rays,
microwaves and radio waves in order of increasing wavelength from 10–2 Å
or 10–12 m to 106 m.
They interact with matter via their electric and magnetic fields which set
in oscillation charges present in all matter. The detailed interaction and
so the mechanism of absorption, scattering, etc., depend on the wavelength
of the electromagnetic wave, and the nature of the atoms and molecules
212 in the medium.

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 Electromagnetic
Waves

POINTS TO PONDER

1. The basic difference between various types of electromagnetic waves

lies in their wavelengths or frequencies since all of them travel through
vacuum with the same speed. Consequently, the waves differ
considerably in their mode of interaction with matter.
2. Accelerated charged particles radiate electromagnetic waves. The
wavelength of the electromagnetic wave is often correlated with the
characteristic size of the system that radiates. Thus, gamma radiation,
having wavelength of 10–14 m to 10–15 m, typically originate from an
atomic nucleus. X-rays are emitted from heavy atoms. Radio waves
are produced by accelerating electrons in a circuit. A transmitting
antenna can most efficiently radiate waves having a wavelength of
about the same size as the antenna. Visible radiation emitted by atoms
is, however, much longer in wavelength than atomic size.
3. Infrared waves, with frequencies lower than those of visible light,
vibrate not only the electrons, but entire atoms or molecules of a
substance. This vibration increases the internal energy and
consequently, the temperature of the substance. This is why infrared
waves are often called heat waves.
4. The centre of sensitivity of our eyes coincides with the centre of the
wavelength distribution of the sun. It is because humans have evolved
with visions most sensitive to the strongest wavelengths from
the sun.

EXERCISES
8.1 Figure 8.5 shows a capacitor made of two circular plates each of
radius 12 cm, and separated by 5.0 cm. The capacitor is being
charged by an external source (not shown in the figure). The
charging current is constant and equal to 0.15A.
(a) Calculate the capacitance and the rate of change of potential
difference between the plates.
(b) Obtain the displacement current across the plates.
(c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the
capacitor? Explain.

FIGURE 8.5

8.2 A parallel plate capacitor (Fig. 8.6) made of circular plates each of radius
R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to
a 230 V ac supply with a (angular) frequency of 300 rad s–1. 213

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 Physics
(a) What is the rms value of the conduction current?
(b) Is the conduction current equal to the displacement current?
(c) Determine the amplitude of B at a point 3.0 cm from the axis
between the plates.

FIGURE 8.6
8.3 What physical quantity is the same for X-rays of wavelength
10–10 m, red light of wavelength 6800 Å and radiowaves of wavelength
500m?
8.4 A plane electromagnetic wave travels in vacuum along z-direction.
What can you say about the directions of its electric and magnetic
field vectors? If the frequency of the wave is 30 MHz, what is its
wavelength?
8.5 A radio can tune in to any station in the 7.5 MHz to 12 MHz band.
What is the corresponding wavelength band?
8.6 A charged particle oscillates about its mean equilibrium position
with a frequency of 10 9 Hz. What is the frequency of the
electromagnetic waves produced by the oscillator?
8.7 The amplitude of the magnetic field part of a harmonic
electromagnetic wave in vacuum is B 0 = 510 nT. What is the
amplitude of the electric field part of the wave?
8.8 Suppose that the electric field amplitude of an electromagnetic wave
is E0 = 120 N/C and that its frequency is n = 50.0 MHz. (a) Determine,
B0,w, k, and l. (b) Find expressions for E and B.
8.9 The terminology of different parts of the electromagnetic spectrum
is given in the text. Use the formula E = hn (for energy of a quantum
of radiation: photon) and obtain the photon energy in units of eV for
different parts of the electromagnetic spectrum. In what way are
the different scales of photon energies that you obtain related to the
sources of electromagnetic radiation?
8.10 In a plane electromagnetic wave, the electric field oscillates
sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V m–1.
(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average energy density of the E field equals the
average energy density of the B field. [c = 3 × 108 m s–1.]

214

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 PHYSICS
PART – II

TEXTBOOK FOR CLASS XII

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Rationalised 2023-24
 PHYSICS
PART – II

TEXTBOOK FOR CLASS XII

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12090 – PHYSICS PART-II ISBN 81-7450-631-4 (Part I)
Textbook for Class XII ISBN 81-7450-671-3 (Part II)

First Edition ALL RIGHTS RESERVED

March 2007 Chaitra 1928 q No part of this publication may be reproduced, stored in a retrieval system
or transmitted, in any form or by any means, electronic, mechanical,
photocopying, recording or otherwise without the prior permission of the
Reprinted publisher.
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December 2009, January 2011, be lent, re-sold, hired out or otherwise disposed of without the publisher’s
consent, in any form of binding or cover other than that in which it is
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and should be unacceptable.
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Revised Edition OFFICES OF THE PUBLICATION

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Industrial Area, Phase-II, New Delhi 110 028

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 FOREWORD

The National Curriculum Framework (NCF), 2005 recommends that children’s life at school must
be linked to their life outside the school. This principle marks a departure from the legacy of bookish
learning which continues to shape our system and causes a gap between the school, home and
community. The syllabi and textbooks developed on the basis of NCF signify an attempt to implement
this basic idea. They also attempt to discourage rote learning and the maintenance of sharp
boundaries between different subject areas. We hope these measures will take us significantly
further in the direction of a child-centred system of education outlined in the National Policy on
Education (NPE), 1986.
The success of this effort depends on the steps that school principals and teachers will take to
encourage children to reflect on their own learning and to pursue imaginative activities and questions.
We must recognise that, given space, time and freedom, children generate new knowledge by engaging
with the information passed on to them by adults. Treating the prescribed textbook as the sole basis
of examination is one of the key reasons why other resources and sites of learning are ignored.
Inculcating creativity and initiative is possible if we perceive and treat children as participants in
learning, not as receivers of a fixed body of knowledge.
These aims imply considerable change in school routines and mode of functioning. Flexibility in
the daily time-table is as necessary as rigour in implementing the annual calendar so that the
required number of teaching days are actually devoted to teaching. The methods used for teaching
and evaluation will also determine how effective this textbook proves for making children’s life at
school a happy experience, rather than a source of stress or boredom. Syllabus designers have tried
to address the problem of curricular burden by restructuring and reorienting knowledge at different
stages with greater consideration for child psychology and the time available for teaching. The textbook
attempts to enhance this endeavour by giving higher priority and space to opportunities for
contemplation and wondering, discussion in small groups, and activities requiring hands-on
experience.
The National Council of Educational Research and Training (NCERT) appreciates the hard
work done by the textbook development committee responsible for this book. We wish to thank the
Chairperson of the advisory group in science and mathematics, Professor J.V. Narlikar and the
Chief Advisor for this book, Professor A.W. Joshi for guiding the work of this committee. Several
teachers contributed to the development of this textbook; we are grateful to their principals for
making this possible. We are indebted to the institutions and organisations which have generously
permitted us to draw upon their resources, material and personnel. We are especially grateful to
the members of the National Monitoring Committee, appointed by the Department of Secondary
and Higher Education, Ministry of Human Resource Development under the Chairpersonship of
Professor Mrinal Miri and Professor G.P. Deshpande, for their valuable time and contribution. As
an organisation committed to systemic reform and continuous improvement in the quality of its
products, NCERT welcomes comments and suggestions which will enable us to undertake further
revision and refinement.

Director
New Delhi National Council of Educational
20 November 2006 Research and Training

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Rationalised 2023-24
 RATIONALISATION OF CONTENT IN THE TEXTBOOKS

In view of the COVID-19 pandemic, it is imperative to reduce content load on

students. The National Education Policy 2020, also emphasises reducing the
content load and providing opportunities for experiential learning with creative
mindset. In this background, the NCERT has undertaken the exercise to rationalise
the textbooks across all classes. Learning Outcomes already developed by the NCERT
across classes have been taken into consideration in this exercise.
Contents of the textbooks have been rationalised in view of the following:
• Overlapping with similar content included in other subject areas in the same
class
• Similar content included in the lower or higher class in the same subject
• Difficulty level
• Content, which is easily accessible to students without much interventions
from teachers and can be learned by children through self-learning or peer-
learning
• Content, which is irrelevant in the present context

This present edition, is a reformatted version after carrying out the changes
given above.

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Rationalised 2023-24
 PREFACE

It gives me pleasure to place this book in the hands of the students, teachers and the
public at large (whose role cannot be overlooked). It is a natural sequel to the Class XI
textbook which was brought out in 2006. This book is also a trimmed version of the
textbooks which existed so far. The chapter on thermal and chemical effects of current
has been cut out. This topic has also been dropped from the CBSE syllabus. Similarly,
the chapter on communications has been substantially curtailed. It has been rewritten
in an easily comprehensible form.
Although most other chapters have been based on the earlier versions, several parts
and sections in them have been rewritten. The Development Team has been guided by
the feedback received from innumerable teachers across the country.
In producing these books, Class XI as well as Class XII, there has been a basic
change of emphasis. Both the books present physics to students without assuming
that they would pursue this subject beyond the higher secondary level. This new view
has been prompted by the various observations and suggestions made in the National
Curriculum Framework (NCF), 2005. Similarly, in today’s educational scenario where
students can opt for various combinations of subjects, we cannot assume that a physics
student is also studying mathematics. Therefore, physics has to be presented, so to
say, in a stand-alone form.
As in Class XI textbook, some interesting box items have been inserted in many
chapters. They are not meant for teaching or examinations. Their purpose is to catch
the attention of the reader, to show some applications in daily life or in other areas of
science and technology, to suggest a simple experiment, to show connection of concepts
in different areas of physics, and in general, to break the monotony and enliven the
book.
Features like Summary, Points to Ponder, Exercises and Additional Exercises at
the end of each chapter, and Examples have been retained. Several concept-based
Exercises have been transferred from end-of-chapter Exercises to Examples with
Solutions in the text. It is hoped that this will make the concepts discussed in the
chapter more comprehensible. Several new examples and exercises have been added.
Students wishing to pursue physics further would find Points to Ponder and Additional
Exercises very useful and thoughtful. To provide resources beyond the textbook and
to encourage eLearning, each chapter has been provided with some relevant website
addresses under the title ePhysics. These sites provide additional materials on specific
topics and also provide learners the opportunites for interactive demonstrations/
experiments.
The intricate concepts of physics must be understood, comprehended and
appreciated. Students must learn to ask questions like ‘why’, ‘how’, ‘how do we know
it’. They will find almost always that the question ‘why’ has no answer within the domain
of physics and science in general. But that itself is a learning experience, is it not? On
the other hand, the question ‘how’ has been reasonably well answered by physicists in
the case of most natural phenomena. In fact, with the understanding of how things
happen, it has been possible to make use of many phenomena to create technological
applications for the use of humans.

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 For example, consider statements in a book, like ‘A negatively charged electron is
attracted by the positively charged plate’, or ‘In this experiment, light (or electron)
behaves like a wave’. You will realise that it is not possible to answer ‘why’. This question
belongs to the domain of philosophy or metaphysics. But we can answer ‘how’, we can
find the force acting, we can find the wavelength of the photon (or electron), we can
determine how things behave under different conditions, and we can develop instruments
which will use these phenomena to our advantage.
It has been a pleasure to work for these books at the higher secondary level, along
with a team of members. The Textbook Development Team, the Review Team and Editing
Teams involved college and university teachers, teachers from Indian Institutes of
Technology, scientists from national institutes and laboratories, as well as higher
secondary teachers. The feedback and critical look provided by higher secondary
teachers in the various teams are highly laudable. Most box items were generated by
members of one or the other team, but three of them were generated by friends and
well-wishers not part of any team. We are thankful to Dr P.N. Sen of Pune, Professor
Roopmanjari Ghosh of Delhi and Dr Rajesh B Khaparde of Mumbai for allowing us to
use their box items, respectively in Chapters 3, 4 (Part I) and 9 (Part II). We are very
thankful to the members of the Review and Editing Workshops to discuss and refine
the first draft of the textbook. We also express our gratitude to Prof. Krishna Kumar,
Director, NCERT, for entrusting us with the task of presenting this textbook as a part of
the national effort for improving science education. I also thank Prof. G. Ravindra, Joint
Director, NCERT, for his help from time-to-time. Prof. Hukum Singh, Head, Department
of Education in Science and Mathematics, NCERT, was always willing to help us in our
endeavour in every possible way.
We welcome suggestions and comments from our valued users, especially students
and teachers. We wish our young readers a happy journey into the exciting realm of
physics.

A. W. JOSHI
Chief Advisor
Textbook Development Committee

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 TEXTBOOK DEVELOPMENT COMMITTEE

CHAIRPERSON, ADVISORY COMMITTEE FOR TEXTBOOKS IN SCIENCE AND MATHEMATICS

J.V. Narlikar, Emeritus Professor, Inter-University Centre for Astronomy and Astrophysics
(IUCAA), Ganeshkhind, Pune University Campus, Pune

CHIEF ADVISOR
A.W. Joshi, Honorary Visiting Scientist, National Centre for Radio Astrophysics (NCRA), Pune
University Campus, Pune (Formerly Professor at Department of Physics, University of Pune)

MEMBERS
A.K. Ghatak, Emeritus Professor, Department of Physics, Indian Institute of Technology,
New Delhi
Alika Khare, Professor, Department of Physics, Indian Institute of Technology, Guwahati
Anjali Kshirsagar, Reader, Department of Physics, University of Pune, Pune
Anuradha Mathur, PGT , Modern School, Vasant Vihar, New Delhi
Atul Mody, Lecturer (S.G.), VES College of Arts, Science and Commerce, Mumbai
B.K. Sharma, Professor, DESM, NCERT, New Delhi
Chitra Goel, PGT, Rajkiya Pratibha Vikas Vidyalaya, Tyagraj Nagar, New Delhi
Gagan Gupta, Reader, DESM, NCERT, New Delhi
H.C. Pradhan, Professor, Homi Bhabha Centre of Science Education (TIFR), Mumbai
N. Panchapakesan, Professor (Retd.), Department of Physics and Astrophysics, University of
Delhi, Delhi
R. Joshi, Lecturer (S.G.), DESM, NCERT, New Delhi
S.K. Dash, Reader, DESM, NCERT, New Delhi
S. Rai Choudhary, Professor, Department of Physics and Astrophysics, University of Delhi, Delhi
S.K. Upadhyay, PGT, Jawahar Navodaya Vidyalaya, Muzaffar Nagar
S.N. Prabhakara, PGT, DM School, Regional Institute of Education (NCERT), Mysore
V.H. Raybagkar, Reader, Nowrosjee Wadia College, Pune
Vishwajeet Kulkarni, Teacher (Grade I ), Higher Secondary Section, Smt. Parvatibai Chowgule
College, Margao, Goa

MEMBER-COORDINATOR
V.P. Srivastava, Reader, DESM, NCERT, New Delhi

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Rationalised 2023-24
 ACKNOWLEDGEMENTS

The National Council of Educational Research and Training acknowledges the valuable
contribution of the individuals and organisations involved in the development of Physics Textbook
for Class XII. The Council also acknowledges the valuable contribution of the following academics
for reviewing and refining the manuscripts of this book:
Anu Venugopalan, Lecturer, School of Basic and Applied Sciences, GGSIP University, Delhi; A.K.
Das, PGT, St. Xavier’s Senior Secondary School, Delhi; Bharati Kukkal, PGT, Kendriya Vidyalaya,
Pushp Vihar, New Delhi; D.A. Desai, Lecturer (Retd.), Ruparel College, Mumbai; Devendra Kumar,
PGT, Rajkiya Pratibha Vikas Vidyalaya, Yamuna Vihar, Delhi; I.K. Gogia, PGT, Kendriya Vidyalaya,
Gole Market, New Delhi; K.C. Sharma, Reader, Regional Institute of Education (NCERT), Ajmer;
M.K. Nandy, Associate Professor, Department of Physics, Indian Institute of Technology, Guwahati;
M.N. Bapat, Reader, Regional Institute of Education (NCERT), Mysore; R. Bhattacharjee, Asstt.
Professor, Department of Electronics and Communication Engineering, Indian Institute of Technology,
Guwahati; R.S. Das, Vice-Principal (Retd.), Balwant Ray Mehta Senior Secondary School, Lajpat
Nagar, New Delhi; Sangeeta D. Gadre, Reader, Kirori Mal College, Delhi; Suresh Kumar, PGT, Delhi
Public School, Dwarka, New Delhi; Sushma Jaireth, Reader, Department of Women’s Studies, NCERT,
New Delhi; Shyama Rath, Reader, Department of Physics and Astrophysics, University of Delhi,
Delhi; Yashu Kumar, PGT, Kulachi Hans Raj Model School, Ashok Vihar, Delhi.
The Council also gratefully acknowledges the valuable contribution of the following academics
for the editing and finalisation of this book: B.B. Tripathi, Professor (Retd.), Department of Physics,
Indian Institute of Technology, New Delhi; Dipan K. Ghosh, Professor, Department of Physics,
Indian Institute of Technology, Mumbai; Dipanjan Mitra, Scientist, National Centre for Radio
Astrophysics (TIFR), Pune; G.K. Mehta, Raja Ramanna Fellow, Inter-University Accelerator Centre,
New Delhi; G.S. Visweswaran, Professor, Department of Electrical Engineering, Indian Institute of
Technology, New Delhi; H.C. Kandpal, Head, Optical Radiation Standards, National Physical
Laboratory, New Delhi; H.S. Mani, Raja Ramanna Fellow, Institute of Mathematical Sciences,
Chennai; K. Thyagarajan, Professor, Department of Physics, Indian Institute of Technology, New
Delhi; P.C. Vinod Kumar, Professor, Department of Physics, Sardar Patel University, Vallabh
Vidyanagar, Gujarat; S. Annapoorni, Professor, Department of Physics and Astrophysics, University
of Delhi, Delhi; S.C. Dutta Roy, Emeritus Professor, Department of Electrical Engineering, Indian
Institute of Technology, New Delhi; S.D. Joglekar, Professor, Department of Physics, Indian Institute
of Technology, Kanpur; V. Sundara Raja, Professor, Sri Venkateswara University, Tirupati.
The Council also acknowledges the valuable contributions of the following academics for
refining the text in 2017: A.K. Srivastava, Assistant Professor, DESM, NCERT, New Delhi; Arnab
Sen, Assistant Professor, NERIE, Shillong; L.S. Chauhan, Assistant Professor, RIE, Bhopal;
O.N. Awasthi, Professor (Retd.), RIE, Bhopal; Rachna Garg, Professor, DESM, NCERT, New
Delhi; Raman Namboodiri, Assistant Professor, RIE, Mysuru; R.R. Koireng, Assistant Professor,
DCS, NCERT, New Delhi; Shashi Prabha, Professor, DESM, NCERT, New Delhi; and S.V. Sharma,
Professor, RIE, Ajmer.
Special thanks are due to Hukum Singh, Professor and Head, DESM, NCERT for his support.
The Council also acknowledges the support provided by the APC office and the administrative
staff of the DESM; Deepak Kapoor, Incharge, Computer Station; Inder Kumar, DTP Operator;
Mohd. Qamar Tabrez and Hari Darshan Lodhi Copy Editor ; Rishi Pal Singh, Sr. Proof Reader,
NCERT and Ashima Srivastava, Proof Reader in shaping this book.
The contributions of the Publication Department in bringing out this book are also duly
acknowledged.

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 Contents of Physics Part I
Class XII

CHAPTER ONE
ELECTRIC CHARGES AND FIELDS 1
CHAPTER TWO
ELECTROSTATIC POTENTIAL AND CAPACITANCE 45
CHAPTER THREE
CURRENT ELECTRICITY 81
CHAPTER FOUR
MOVING CHARGES AND MAGNETISM 107
CHAPTER FIVE
MAGNETISM AND MATTER 136
CHAPTER SIX
ELECTROMAGNETIC INDUCTION 154
CHAPTER SEVEN
ALTERNATING CURRENT 177
CHAPTER EIGHT
ELECTROMAGNETIC WAVES 201
ANSWERS 217

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 CONTENTS

FOREWORD v
RATIONALISATION OF CONTENT IN THE TEXTBOOK vii
PREFACE ix

CHAPTER NINE
RAY OPTICS AND OPTICAL INSTRUMENTS
9.1 Introduction 221
9.2 Reflection of Light by Spherical Mirrors 222
9.3 Refraction 228
9.4 Total Internal Reflection 229
9.5 Refraction at Spherical Surfaces and by Lenses 232
9.6 Refraction through a Prism 239
9.7 Optical Instruments 240

CHAPTER TEN
WAVE OPTICS
10.1 Introduction 255
10.2 Huygens Principle 257
10.3 Refraction and Reflection of Plane Waves using Huygens Principle 258
10.4 Coherent and Incoherent Addition of Waves 262
10.5 Interference of Light Waves and Young’s Experiment 265
10.6 Diffraction 266
10.7 Polarisation 269

CHAPTER ELEVEN
DUAL NATURE OF RADIATION AND MATTER
11.1 Introduction 274
11.2 Electron Emission 275

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11.3 Photoelectric Effect 276
11.4 Experimental Study of Photoelectric Effect 277
11.5 Photoelectric Effect and Wave Theory of Light 280
11.6 Einstein’s Photoelectric Equation: Energy Quantum of Radiation 281
11.7 Particle Nature of Light: The Photon 283
11.8 Wave Nature of Matter 284

CHAPTER TWELVE
ATOMS
12.1 Introduction 290
12.2 Alpha-particle Scattering and Rutherford’s Nuclear Model of Atom 291
12.3 Atomic Spectra 296
12.4 Bohr Model of the Hydrogen Atom 297
12.5 The Line Spectra of the Hydrogen Atom 300
12.6 DE Broglie’s Explanation of Bohr’s Second Postulate of Quantisation 301

CHAPTER THIRTEEN
NUCLEI
13.1 Introduction 306
13.2 Atomic Masses and Composition of Nucleus 306
13.3 Size of the Nucleus 309
13.4 Mass-Energy and Nuclear Binding Energy 310
13.5 Nuclear Force 313
13.6 Radioactivity 314
13.7 Nuclear Energy 314

CHAPTER FOURTEEN
SEMICONDUCTOR ELECTRONICS: MATERIALS, DEVICES AND SIMPLE CIRCUITS
14.1 Introduction 323
14.2 Classification of Metals, Conductors and Semiconductors 324
14.3 Intrinsic Semiconductor 327
14.4 Extrinsic Semiconductor 329

xvi

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14.5 p-n Junction 333
14.6 Semiconductor Diode 334
14.7 Application of Junction Diode as a Rectifier 338

APPENDICES 344

ANSWERS 346

BIBLIOGRAPHY 353

xvii

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 COVER DESIGN
(Adapted from http://nobelprize.org and
the Nobel Prize in Physics 2006)

Different stages in the evolution of

the universe.

BACK COVER
(Adapted from http://www.iter.org and
http://www.dae.gov.in)

Cut away view of International Thermonuclear Experimental Reactor (ITER)

device. The man in the bottom shows the scale.
ITER is a joint international research and development project that
aims to demonstrate the scientific and technical feasibility of fusion power.
India is one of the seven full partners in the project, the others being
the European Union (represented by EURATOM), Japan, the People’s
Republic of China, the Republic of Korea, the Russian Federation and the
USA. ITER will be constructed in Europe, at Cadarache in the South of
France and will provide 500 MW of fusion power.
Fusion is the energy source of the sun and the stars. On earth, fusion
research is aimed at demonstrating that this energy source can be used to
produce electricity in a safe and environmentally benign way, with
abundant fuel resources, to meet the needs of a growing world population.
For details of India’s role, see Nuclear India, Vol. 39, No. 11-12/
May-June 2006, issue available at Department of Atomic Energy (DAE)
website mentioned above.

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Chapter Nine

RAY OPTICS
AND OPTICAL
INSTRUMENTS

9.1 INTRODUCTION
Nature has endowed the human eye (retina) with the sensitivity to detect
electromagnetic waves within a small range of the electromagnetic
spectrum. Electromagnetic radiation belonging to this region of the
spectrum (wavelength of about 400 nm to 750 nm) is called light. It is
mainly through light and the sense of vision that we know and interpret
the world around us.
There are two things that we can intuitively mention about light from
common experience. First, that it travels with enormous speed and second,
that it travels in a straight line. It took some time for people to realise that
the speed of light is finite and measurable. Its presently accepted value
in vacuum is c = 2.99792458 × 108 m s–1. For many purposes, it suffices
to take c = 3 × 108 m s–1. The speed of light in vacuum is the highest
speed attainable in nature.
The intuitive notion that light travels in a straight line seems to
contradict what we have learnt in Chapter 8, that light is an
electromagnetic wave of wavelength belonging to the visible part of the
spectrum. How to reconcile the two facts? The answer is that the
wavelength of light is very small compared to the size of ordinary objects
that we encounter commonly (generally of the order of a few cm or larger).
In this situation, as you will learn in Chapter 10, a light wave can be
considered to travel from one point to another, along a straight line joining

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 Physics
them. The path is called a ray of light, and a bundle of such rays
constitutes a beam of light.
In this chapter, we consider the phenomena of reflection, refraction
and dispersion of light, using the ray picture of light. Using the basic
laws of reflection and refraction, we shall study the image formation by
plane and spherical reflecting and refracting surfaces. We then go on to
describe the construction and working of some important optical
instruments, including the human eye.

9.2 REFLECTION OF LIGHT BY SPHERICAL MIRRORS

We are familiar with the laws of reflection. The
angle of reflection (i.e., the angle between reflected
ray and the normal to the reflecting surface or
the mirror) equals the angle of incidence (angle
between incident ray and the normal). Also that
the incident ray, reflected ray and the normal to
the reflecting surface at the point of incidence lie
in the same plane (Fig. 9.1). These laws are valid
at each point on any reflecting surface whether
plane or curved. However, we shall restrict our
discussion to the special case of curved surfaces,
FIGURE 9.1 The incident ray, reflected ray that is, spherical surfaces. The normal in this case
and the normal to the reflecting surface lie is to be taken as normal to the tangent to surface
in the same plane. at the point of incidence. That is, the normal is
along the radius, the line joining the centre of curvature of the mirror to
the point of incidence.
We have already studied that the geometric centre of a spherical mirror
is called its pole while that of a spherical lens is called its optical centre.
The line joining the pole and the centre of curvature of the spherical
mirror is known as the principal axis. In the case of spherical lenses, the
principal axis is the line joining the optical centre with its principal focus
as you will see later.

9.2.1 Sign convention

To derive the relevant formulae for
reflection by spherical mirrors and
refraction by spherical lenses, we must
first adopt a sign convention for
measuring distances. In this book, we
shall follow the Cartesian sign
convention. According to this
convention, all distances are measured
from the pole of the mirror or the optical
centre of the lens. The distances
measured in the same direction as the
incident light are taken as positive and
FIGURE 9.2 The Cartesian Sign Convention.
those measured in the direction
opposite to the direction of incident light are taken as negative (Fig. 9.2).
222 The heights measured upwards with respect to x-axis and normal to the

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 Ray Optics and
Optical Instruments
principal axis (x-axis) of the mirror/lens are taken as positive (Fig. 9.2).
The heights measured downwards are taken as negative.
With a common accepted convention, it turns out that a single formula
for spherical mirrors and a single formula for spherical lenses can handle
all different cases.

9.2.2 Focal length of spherical mirrors

Figure 9.3 shows what happens when a parallel beam of light is incident
on (a) a concave mirror, and (b) a convex mirror. We assume that the rays
are paraxial, i.e., they are incident at points close to the pole P of the mirror
and make small angles with the principal axis. The reflected rays converge
at a point F on the principal axis of a concave mirror [Fig. 9.3(a)].
For a convex mirror, the reflected rays appear to diverge from a point F
on its principal axis [Fig. 9.3(b)]. The point F is called the principal focus
of the mirror. If the parallel paraxial beam of light were incident, making
some angle with the principal axis, the reflected rays would converge (or
appear to diverge) from a point in a plane through F normal to the principal
axis. This is called the focal plane of the mirror [Fig. 9.3(c)].

FIGURE 9.3 Focus of a concave and convex mirror.

The distance between the focus F and the pole P of the mirror is called
the focal length of the mirror, denoted by f. We now show that f = R/2,
where R is the radius of curvature of the mirror. The geometry of reflection
of an incident ray is shown in Fig. 9.4.
Let C be the centre of curvature of the mirror. Consider a ray parallel
to the principal axis striking the mirror at M. Then CM will be
perpendicular to the mirror at M. Let q be the angle of incidence, and MD 223

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 Physics
be the perpendicular from M on the principal axis. Then,
ÐMCP = q and ÐMFP = 2q
Now,
MD MD
tanq = and tan 2q = (9.1)
CD FD
For small q, which is true for paraxial rays, tanq » q,
tan 2q » 2q. Therefore, Eq. (9.1) gives
MD MD
=2
FD CD
CD
or, FD = (9.2)
2
Now, for small q, the point D is very close to the point P.
Therefore, FD = f and CD = R. Equation (9.2) then gives
f = R/2 (9.3)

9.2.3 The mirror equation

If rays emanating from a point actually meet at another point
FIGURE 9.4 Geometry of after reflection and/or refraction, that point is called the image
reflection of an incident ray on
of the first point. The image is real if the rays actually converge
(a) concave spherical mirror,
and (b) convex spherical mirror.
to the point; it is virtual if the rays do not actually meet but
appear to diverge from the point when produced
backwards. An image is thus a point-to-point
correspondence with the object established through
reflection and/or refraction.
In principle, we can take any two rays emanating
from a point on an object, trace their paths, find their
point of intersection and thus, obtain the image of
the point due to reflection at a spherical mirror. In
practice, however, it is convenient to choose any two
of the following rays:
(i) The ray from the point which is parallel to the
principal axis. The reflected ray goes through
the focus of the mirror.
(ii) The ray passing through the centre of
curvature of a concave mirror or appearing to
FIGURE 9.5 Ray diagram for image
formation by a concave mirror. pass through it for a convex mirror. The
reflected ray simply retraces the path.
(iii) The ray passing through (or directed towards) the focus of the concave
mirror or appearing to pass through (or directed towards) the focus
of a convex mirror. The reflected ray is parallel to the principal axis.
(iv) The ray incident at any angle at the pole. The reflected ray follows
laws of reflection.
Figure 9.5 shows the ray diagram considering three rays. It shows
the image A¢B¢ (in this case, real) of an object A B formed by a concave
mirror. It does not mean that only three rays emanate from the point A.
An infinite number of rays emanate from any source, in all directions.
Thus, point A¢ is image point of A if every ray originating at point A and
224 falling on the concave mirror after reflection passes through the point A¢.

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 Ray Optics and
Optical Instruments
We now derive the mirror equation or the relation between the object
distance (u), image distance (v) and the focal length ( f ).
From Fig. 9.5, the two right-angled triangles A¢B¢F and MPF are
similar. (For paraxial rays, MP can be considered to be a straight line
perpendicular to CP.) Therefore,
B ′A ′ B ′F
=
PM FP
B ′A ′ B ′F
or = ( ∵ PM = AB) (9.4)
BA FP
Since Ð APB = Ð A¢PB¢, the right angled triangles A¢B¢P and ABP are
also similar. Therefore,
B ′A ′ B ′ P
= (9.5)
B A BP
Comparing Eqs. (9.4) and (9.5), we get
B ′F B ′P – FP B ′P
= = (9.6)
FP FP BP
Equation (9.6) is a relation involving magnitude of distances. We now
apply the sign convention. We note that light travels from the object to
the mirror MPN. Hence this is taken as the positive direction. To reach
the object AB, image A¢B¢ as well as the focus F from the pole P, we have
to travel opposite to the direction of incident light. Hence, all the three
will have negative signs. Thus,
B¢ P = –v, FP = –f, BP = –u
Using these in Eq. (9.6), we get
–v + f –v
=
–f –u
v– f v
or =
f u
v v
= 1+
f u
Dividing it by v, we get
1 1 1
+ =
v u f (9.7)
This relation is known as the mirror equation.
The size of the image relative to the size of the object is another
important quantity to consider. We define linear magnification (m ) as the
ratio of the height of the image (h¢) to the height of the object (h):
h′
m= (9.8)
h
h and h¢ will be taken positive or negative in accordance with the accepted
sign convention. In triangles A¢B¢P and ABP, we have,
B ′A ′ B ′P
=
BA BP
With the sign convention, this becomes 225

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 Physics
– h′ –v
=
h –u
so that
h′ v
m= = – (9.9)
h u
We have derived here the mirror equation, Eq. (9.7), and the
magnification formula, Eq. (9.9), for the case of real, inverted image formed
by a concave mirror. With the proper use of sign convention, these are,
in fact, valid for all the cases of reflection by a spherical mirror (concave
or convex) whether the image formed is real or virtual. Figure 9.6 shows
the ray diagrams for virtual image formed by a concave and convex mirror.
You should verify that Eqs. (9.7) and (9.9) are valid for these cases as
well.

FIGURE 9.6 Image formation by (a) a concave mirror with object between
P and F, and (b) a convex mirror.

Example 9.1 Suppose that the lower half of the concave mirror’s
reflecting surface in Fig. 9.6 is covered with an opaque (non-reflective)
material. What effect will this have on the image of an object placed
in front of the mirror?
EXAMPLE 9.1

Solution You may think that the image will now show only half of the
object, but taking the laws of reflection to be true for all points of the
remaining part of the mirror, the image will be that of the whole object.
However, as the area of the reflecting surface has been reduced, the
intensity of the image will be low (in this case, half).

Example 9.2 A mobile phone lies along the principal axis of a concave
mirror, as shown in Fig. 9.7. Show by suitable diagram, the formation
of its image. Explain why the magnification is not uniform. Will the
distortion of image depend on the location of the phone with respect
to the mirror?
EXAMPLE 9.2

226 FIGURE 9.7

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 Ray Optics and
Optical Instruments

EXAMPLE 9.2
Solution
The ray diagram for the formation of the image of the phone is shown
in Fig. 9.7. The image of the part which is on the plane perpendicular
to principal axis will be on the same plane. It will be of the same size,
i.e., B¢C = BC. You can yourself realise why the image is distorted.

Example 9.3 An object is placed at (i) 10 cm, (ii) 5 cm in front of a

concave mirror of radius of curvature 15 cm. Find the position, nature,
and magnification of the image in each case.
Solution
The focal length f = –15/2 cm = –7.5 cm
(i) The object distance u = –10 cm. Then Eq. (9.7) gives
1 1 1
+ =
v – 10 – 7 .5

10 × 7.5
or v= = – 30 cm
−2.5
The image is 30 cm from the mirror on the same side as the object.
v ( −30)
Also, magnification m = – =– =–3
u ( −10)
The image is magnified, real and inverted.
(ii) The object distance u = –5 cm. Then from Eq. (9.7),
1 1 1
+ =
v −5 −7.5
5 × 7.5
or v = = 15 cm
(7.5 – 5)
EXAMPLE 9.3

This image is formed at 15 cm behind the mirror. It is a virtual image.

v 15
Magnification m = – =– =3
u ( −5)
The image is magnified, virtual and erect.

Example 9.4 Suppose while sitting in a parked car, you notice a

jogger approaching towards you in the side view mirror of R = 2 m. If
the jogger is running at a speed of 5 m s–1, how fast the image of the
jogger appear to move when the jogger is (a) 39 m, (b) 29 m, (c) 19 m,
and (d) 9 m away.
Solution
From the mirror equation, Eq. (9.7), we get
fu
v=
u− f
For convex mirror, since R = 2 m, f = 1 m. Then
EXAMPLE 9.4

( −39) × 1 39
for u = –39 m, v = = m
−39 − 1 40
Since the jogger moves at a constant speed of 5 m s–1, after 1 s the
position of the image v (for u = –39 + 5 = –34) is (34/35 )m.
227

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 Physics
The shift in the position of image in 1 s is
39 34 1365 − 1360 5 1
− = = = m
40 35 1400 1400 280
Therefore, the average speed of the image when the jogger is between
39 m and 34 m from the mirror, is (1/280) m s–1
Similarly, it can be seen that for u = –29 m, –19 m and –9 m, the
speed with which the image appears to move is
1 1 1
m s –1 , m s –1 and m s –1 , respectively.
150 60 10
Although the jogger has been moving with a constant speed, the speed
EXAMPLE 9.4

of his/her image appears to increase substantially as he/she moves

closer to the mirror. This phenomenon can be noticed by any person
sitting in a stationary car or a bus. In case of moving vehicles, a
similar phenomenon could be observed if the vehicle in the rear is
moving closer with a constant speed.

9.3 REFRACTION
When a beam of light encounters another transparent medium, a part of
light gets reflected back into the first medium while the rest enters the
other. A ray of light represents a beam. The direction of propagation of an
obliquely incident (0°< i < 90°) ray of light that enters the other medium,
changes at the interface of the two media. This phenomenon is called
refraction of light. Snell experimentally obtained the following laws
of refraction:

(i) The incident ray, the refracted ray and the

normal to the interface at the point of
incidence, all lie in the same plane.
(ii) The ratio of the sine of the angle of incidence
to the sine of angle of refraction is constant.
Remember that the angles of incidence (i ) and
refraction (r ) are the angles that the incident
and its refracted ray make with the normal,
respectively. We have
sin i
= n 21 (9.10)
sin r
where n 21 is a constant, called the refractive
index of the second medium with respect to the
FIGURE 9.8 Refraction and reflection of light. first medium. Equation (9.10) is the well-known
Snell’s law of refraction. We note that n 21 is a
characteristic of the pair of media (and also depends on the wavelength
of light), but is independent of the angle of incidence.
From Eq. (9.10), if n 21 > 1, r < i , i.e., the refracted ray bends towards
the normal. In such a case medium 2 is said to be optically denser (or
228 denser, in short) than medium 1. On the other hand, if n 21 <1, r > i, the

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 Ray Optics and
Optical Instruments
refracted ray bends away from the normal. This
is the case when incident ray in a denser
medium refracts into a rarer medium.
Note: Optical density should not be
confused with mass density, which is mass
per unit volume. It is possible that mass
density of an optically denser medium may
be less than that of an optically rarer
medium (optical density is the ratio of the
speed of light in two media). For example, FIGURE 9.9 Lateral shift of a ray refracted
turpentine and water. Mass density of through a parallel-sided slab.
turpentine is less than that of water but
its optical density is higher.
If n 21 is the refractive index of medium 2 with
respect to medium 1 and n12 the refractive index
of medium 1 with respect to medium 2, then it
should be clear that
1
n12 = (9.11)
n 21
It also follows that if n 32 is the refractive index
of medium 3 with respect to medium 2 then n 32 =
n 31 × n 12, where n 31 is the refractive index of
medium 3 with respect to medium 1.
Some elementary results based on the laws of
refraction follow immediately. For a rectangular
slab, refraction takes place at two interfaces (air-
glass and glass-air). It is easily seen from Fig. 9.9
that r2 = i1, i.e., the emergent ray is parallel to the
incident ray—there is no deviation, but it does
suffer lateral displacement/shift with respect to the
incident ray. Another familiar observation is that FIGURE 9.10 Apparent depth for
the bottom of a tank filled with water appears to be (a) normal, and (b) oblique viewing.
raised (Fig. 9.10). For viewing near the normal direction, it can be shown
that the apparent depth (h1) is real depth (h 2) divided by the refractive
index of the medium (water).

9.4 TOTAL INTERNAL REFLECTION

When light travels from an optically denser medium to a rarer medium
at the interface, it is partly reflected back into the same medium and
partly refracted to the second medium. This reflection is called the internal
reflection.
When a ray of light enters from a denser medium to a rarer medium,
it bends away from the normal, for example, the ray AO1 B in Fig. 9.11.
The incident ray AO1 is partially reflected (O1C) and partially transmitted
(O1B) or refracted, the angle of refraction (r ) being larger than the angle of
incidence (i ). As the angle of incidence increases, so does the angle of 229

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refraction, till for the ray AO3, the angle of
refraction is p/2. The refracted ray is bent
so much away from the normal that it
grazes the surface at the interface between
the two media. This is shown by the ray
AO3 D in Fig. 9.11. If the angle of incidence
is increased still further (e.g., the ray AO4),
refraction is not possible, and the incident
ray is totally reflected. This is called total
internal reflection. When light gets
reflected by a surface, normally some
FIGURE 9.11 Refraction and internal reflection fraction of it gets transmitted. The
of rays from a point A in the denser medium reflected ray, therefore, is always less
(water) incident at different angles at the interface intense than the incident ray, howsoever
with a rarer medium (air). smooth the reflecting surface may be. In
total internal reflection, on the other hand,
no transmission of light takes place.
The angle of incidence corresponding to an angle of refraction 90°,
say ÐAO3N, is called the critical angle (ic ) for the given pair of media. We
see from Snell’s law [Eq. (9.10)] that if the relative refractive index of the
refracting medium is less than one then, since the maximum value of sin
r is unity, there is an upper limit to the value of sin i for which the law
can be satisfied, that is, i = ic such that
sin ic = n 21 (9.12)
For values of i larger than ic, Snell’s law of refraction cannot be
satisfied, and hence no refraction is possible.
The refractive index of denser medium 1 with respect to rarer medium
2 will be n12 = 1/sin ic. Some typical critical angles are listed in Table 9.1.

TABLE 9.1 CRITICAL ANGLE OF SOME TRANSPARENT MEDIA WITH RESPECT TO AIR
Substance medium Refractive index Critical angle

Water 1.33 48.75

Crown glass 1.52 41.14
Dense flint glass 1.62 37.31
Diamond 2.42 24.41

A demonstration for total internal reflection

All optical phenomena can be demonstrated very easily with the use of a
laser torch or pointer, which is easily available nowadays. Take a glass
beaker with clear water in it. Add a few drops of milk or any other
suspension to water and stir so that water becomes a little turbid. Take
a laser pointer and shine its beam through the turbid water. You will
find that the path of the beam inside the water shines brightly.
230

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Shine the beam from below the beaker such that it strikes at the
upper water surface at the other end. Do you find that it undergoes partial
reflection (which is seen as a spot on the table below) and partial refraction
[which comes out in the air and is seen as a spot on the roof; Fig. 9.12(a)]?
Now direct the laser beam from one side of the beaker such that it strikes
the upper surface of water more obliquely [Fig. 9.12(b)]. Adjust the
direction of laser beam until you find the angle for which the refraction
above the water surface is totally absent and the beam is totally reflected
back to water. This is total internal reflection at its simplest.
Pour this water in a long test tube and shine the laser light from top,
as shown in Fig. 9.12(c). Adjust the direction of the laser beam such that
it is totally internally reflected every time it strikes the walls of the tube.
This is similar to what happens in optical fibres.
Take care not to look into the laser beam directly and not to point it
at anybody’s face.

9.4.1 Total internal reflection in nature and

its technelogical applications
(i) Prism : Prisms designed to bend light by 90° or by 180° make use of
total internal reflection [Fig. 9.13(a) and (b)]. Such a prism is also
used to invert images without chxanging their size [Fig. 9.13(c)].
In the first two cases, the critical angle ic for the material of the prism FIGURE 9.12
must be less than 45°. We see from Table 9.1 that this is true for both Observing total internal
crown glass and dense flint glass. reflection in water with
(ii) Optical fibres: Nowadays optical fibres are extensively used for a laser beam (refraction
transmitting audio and video signals through long distances. Optical due to glass of beaker
fibres too make use of the phenomenon of total internal reflection. neglected being very
Optical fibres are fabricated with high quality composite glass/quartz thin).
fibres. Each fibre consists of a core and
cladding. The refractive index of the
material of the core is higher than that
of the cladding.
When a signal in the form of light is
directed at one end of the fibre at a suitable
angle, it undergoes repeated total internal
reflections along the length of the fibre and
finally comes out at the other end (Fig.
9.14). Since light undergoes total internal
reflection at each stage, there is no
appreciable loss in the intensity of the light
signal. Optical fibres are fabricated such
that light reflected at one side of inner
surface strikes the other at an angle larger
than the critical angle. Even if the fibre is
bent, light can easily travel along its length.
Thus, an optical fibre can be used to act as
an optical pipe. FIGURE 9.13 Prisms designed to bend rays by
A bundle of optical fibres can be put to 90° and 180° or to invert image without changing
several uses. Optical fibres are extensively its size make use of total internal reflection.
used for transmitting and receiving 231

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electrical signals which are converted to light
by suitable transducers. Obviously, optical
fibres can also be used for transmission of
optical signals. For example, these are used
as a ‘light pipe’ to facilitate visual examination
of internal organs like esophagus, stomach
and intestines. You might have seen a
commonly available decorative lamp with fine
FIGURE 9.14 Light undergoes successive total plastic fibres with their free ends forming a
internal reflections as it moves through an fountain like structure. The other end of the
optical fibre.
fibres is fixed over an electric lamp. When the
lamp is switched on, the light travels from the bottom of each fibre and
appears at the tip of its free end as a dot of light. The fibres in such
decorative lamps are optical fibres.
The main requirement in fabricating optical fibres is that there should
be very little absorption of light as it travels for long distances inside
them. This has been achieved by purification and special preparation of
materials such as quartz. In silica glass fibres, it is possible to transmit
more than 95% of the light over a fibre length of 1 km. (Compare with
what you expect for a block of ordinary window glass 1 km thick.)

9.5 REFRACTION AT SPHERICAL SURFACES

AND BY LENSES
We have so far considered refraction at a plane interface. We shall now
consider refraction at a spherical interface between two transparent media.
An infinitesimal part of a spherical surface can be regarded as planar
and the same laws of refraction can be applied at every point on the
surface. Just as for reflection by a spherical mirror, the normal at the
point of incidence is perpendicular to the tangent plane to the spherical
surface at that point and, therefore, passes through its centre of
curvature. We first consider refraction by a single spherical surface and
follow it by thin lenses. A thin lens is a transparent optical medium
bounded by two surfaces; at least one of which should be spherical.
Applying the formula for image formation by a single spherical surface
successively at the two surfaces of a lens, we shall obtain the lens maker’s
formula and then the lens formula.

9.5.1 Refraction at a spherical surface

Figure 9.15 shows the geometry of formation of image I of an object O on
the principal axis of a spherical surface with centre of curvature C, and
radius of curvature R. The rays are incident from a medium of refractive
index n1, to another of refractive index n 2. As before, we take the aperture
(or the lateral size) of the surface to be small compared to other distances
involved, so that small angle approximation can be made. In particular,
NM will be taken to be nearly equal to the length of the perpendicular
from the point N on the principal axis. We have, for small angles,
MN
232 tan ÐNOM =
OM

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MN
tan ÐNCM =
MC

MN
tan ÐNIM =
MI
Now, for DNOC, i is the exterior angle. Therefore, i
= ÐNOM + ÐNCM
MN MN
i= + (9.13)
OM MC
Similarly, FIGURE 9.15 Refraction at a spherical
r = ÐNCM – ÐNIM surface separating two media.
MN MN
i.e., r = − (9.14)
MC MI
Now, by Snell’s law
n1 sin i = n 2 sin r
or for small angles
n1i = n 2r
Substituting i and r from Eqs. (9.13) and (9.14), we get
n1 n 2 n 2 − n1
+ = (9.15)
OM MI MC
Here, OM, MI and MC represent magnitudes of distances. Applying the
Cartesian sign convention,
OM = –u, MI = +v, MC = +R
Substituting these in Eq. (9.15), we get
n 2 n1 n 2 − n1
− = (9.16)
v u R
Equation (9.16) gives us a relation between object and image distance
in terms of refractive index of the medium and the radius of
curvature of the curved spherical surface. It holds for any curved
spherical surface.

Example 9.5 Light from a point source in air falls on a spherical

glass surface (n = 1.5 and radius of curvature = 20 cm). The distance
of the light source from the glass surface is 100 cm. At what position
the image is formed?
Solution
We use the relation given by Eq. (9.16). Here
u = – 100 cm, v = ?, R = + 20 cm, n1 = 1, and n 2 = 1.5.
We then have
1.5 1 0.5
+ =
EXAMPLE 9.5

v 100 20
or v = +100 cm
The image is formed at a distance of 100 cm from the glass surface,
in the direction of incident light.
233

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9.5.2 Refraction by a lens
Figure 9.16(a) shows the geometry of image formation by a double convex
lens. The image formation can be seen in terms of two steps:
(i) The first refracting surface forms the image I 1 of the object O
[Fig. 9.16(b)]. The image I1 acts as a virtual object for the second surface
that forms the image at I [Fig. 9.16(c)]. Applying Eq. (9.15) to the first
interface ABC, we get
n 1 n 2 n 2 − n1
+ = (9.17)
OB BI1 BC1
A similar procedure applied to the second
interface* ADC gives,
n 2 n1 n 2 − n 1
− + = (9.18)
DI1 DI DC2
For a thin lens, BI1 = DI1 . Adding
Eqs. (9.17) and (9.18), we get

n1 n1  1 1 
+ = (n 2 − n1 )  +
 BC1 DC2 
(9.19)
OB DI
Suppose the object is at infinity, i.e.,
OB ® ¥ and DI = f, Eq. (9.19) gives

n1  1 1 
= (n 2 − n1 )  +
 BC1 DC2 
(9.20)
f
The point where image of an object
placed at infinity is formed is called the
focus F, of the lens and the distance f gives
its focal length. A lens has two foci, F and
F¢, on either side of it (Fig. 9.16). By the
sign convention,
BC1 = + R1,
DC2 = –R 2
So Eq. (9.20) can be written as

(9.21)

Equation (9.21) is known as the lens

maker’s formula. It is useful to design
lenses of desired focal length using surfaces
FIGURE 9.16 (a) The position of object, and the of suitable radii of curvature. Note that the
image formed by a double convex lens, formula is true for a concave lens also. In
(b) Refraction at the first spherical surface and that case R1is negative, R 2 positive and
(c) Refraction at the second spherical surface.
therefore, f is negative.

* Note that now the refractive index of the medium on the right side of ADC is n1
while on its left it is n 2. Further DI1 is negative as the distance is measured
234 against the direction of incident light.

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From Eqs. (9.19) and (9.20), we get
n1 n1 n
+ = 1 (9.22)
OB DI f
Again, in the thin lens approximation, B and D are both close to the
optical centre of the lens. Applying the sign convention,
BO = – u, DI = +v, we get
1 1 1
− = (9.23)
v u f
Equation (9.23) is the familiar thin lens formula. Though we derived
it for a real image formed by a convex lens, the formula is valid for both
convex as well as concave lenses and for both real and virtual images.
It is worth mentioning that the two foci, F and F¢, of a double convex
or concave lens are equidistant from the optical centre. The focus on the
side of the (original) source of light is called the first focal point, whereas
the other is called the second focal point.
To find the image of an object by a lens, we can, in principle, take any
two rays emanating from a point on an object; trace their paths using the
laws of refraction and find the point where the refracted rays meet (or
appear to meet). In practice, however, it is convenient to choose any two
of the following rays:
(i) A ray emanating from the object parallel to the principal axis of the
lens after refraction passes through the second principal focus F¢ (in
a convex lens) or appears to diverge (in a concave lens) from the first
principal focus F.
(ii) A ray of light, passing through the optical
centre of the lens, emerges without any
deviation after refraction.
(iii) (a) A ray of light passing through the first
principal focus of a convex lens [Fig. 9.17(a)]
emerges parallel to the principal axis after
refraction.
(b) A ray of light incident on a concave lens
appearing to meet the principal axis at
second focus point emerges parallel to the
principal axis after refraction [Fig. 9.17(b)].
Figures 9.17(a) and (b) illustrate these rules
for a convex and a concave lens, respectively.
You should practice drawing similar ray
diagrams for different positions of the object with
respect to the lens and also verify that the lens
formula, Eq. (9.23), holds good for all cases.
Here again it must be remembered that each
point on an object gives out infinite number of
rays. All these rays will pass through the same
image point after refraction at the lens. FIGURE 9.17 Tracing rays through (a)
Magnification (m ) produced by a lens is convex lens (b) concave lens.
defined, like that for a mirror, as the ratio of the
235
size of the image to that of the object. Proceeding

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in the same way as for spherical mirrors, it is easily seen that
for a lens
h′ v
m= = (9.24)
h u
When we apply the sign convention, we see that, for erect (and virtual)
image formed by a convex or concave lens, m is positive, while for an
inverted (and real) image, m is negative.

Example 9.6 A magician during a show makes a glass lens with

n = 1.47 disappear in a trough of liquid. What is the refractive index
of the liquid? Could the liquid be water?
EXAMPLE 9.6

Solution
The refractive index of the liquid must be equal to 1.47 in order to
make the lens disappear. This means n1 = n 2. This gives 1/f = 0 or
f ® ¥. The lens in the liquid will act like a plane sheet of glass. No,
the liquid is not water. It could be glycerine.

9.5.3 Power of a lens

Power of a lens is a measure of the convergence or
divergence, which a lens introduces in the light falling on
it. Clearly, a lens of shorter focal length bends the incident
light more, while converging it in case of a convex lens
and diverging it in case of a concave lens. The power P of
a lens is defined as the tangent of the angle by which it
converges or diverges a beam of light parallel to the
principal axis falling at unit distance from the optical
centre (Fig. 9.18).
FIGURE 9.18 Power of a lens.
h 1 1
tan δ = ; if h = 1, tan δ = or δ = for small
f f f
value of d. Thus,
1
P= (9.25)
f
The SI unit for power of a lens is dioptre (D): 1D = 1m–1. The power of
a lens of focal length of 1 metre is one dioptre. Power of a lens is positive
for a converging lens and negative for a diverging lens. Thus, when an
optician prescribes a corrective lens of power + 2.5 D, the required lens is
a convex lens of focal length + 40 cm. A lens of power of – 4.0 D means a
concave lens of focal length – 25 cm.

Example 9.7 (i) If f = 0.5 m for a glass lens, what is the power of the
EXAMPLE 9.7

lens? (ii) The radii of curvature of the faces of a double convex lens
are 10 cm and 15 cm. Its focal length is 12 cm. What is the refractive
index of glass? (iii) A convex lens has 20 cm focal length in air. What
is focal length in water? (Refractive index of air-water = 1.33, refractive
index for air -glass = 1.5.)
236

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Solution
(i) Power = +2 dioptre.
(ii) Here, we have f = +12 cm, R1 = +10 cm, R2 = –15 cm.
Refractive index of air is taken as unity.
We use the lens formula of Eq. (9.22). The sign convention has to
be applied for f, R1 and R 2.
Substituting the values, we have
1  1 1 
= (n − 1)  −
12  10 −15 
This gives n = 1.5.
(iii) For a glass lens in air, n 2 = 1.5, n1 = 1, f = +20 cm. Hence, the lens
formula gives
1 1 1 
= 0.5  − 
20  1
R R 2

EXAMPLE 9.7
For the same glass lens in water, n 2 = 1.5, n1 = 1.33. Therefore,
1.33 1 1 
= (1.5 − 1.33)  −  (9.26)
f  R1 R 2 
Combining these two equations, we find f = + 78.2 cm.

9.5.4 Combination of thin lenses in contact

Consider two lenses A and B of focal length f1 and
f2 placed in contact with each other. Let the object
be placed at a point O beyond the focus of the first
lens A (Fig. 9.19). The first lens produces an image
at I1. Since image I1 is real, it serves as a virtual
object for the second lens B, producing the final
image at I. It must, however, be borne in mind that
formation of image by the first lens is presumed
only to facilitate determination of the position of the
FIGURE 9.19 Image formation by a
final image. In fact, the direction of rays emerging combination of two thin lenses in contact.
from the first lens gets modified in accordance with
the angle at which they strike the second lens. Since the lenses are thin,
we assume the optical centres of the lenses to be coincident. Let this
central point be denoted by P.
For the image formed by the first lens A, we get
1 1 1
− = (9.27)
v1 u f1
For the image formed by the second lens B, we get
1 1 1
− = (9.28)
v v1 f 2
Adding Eqs. (9.27) and (9.28), we get
1 1 1 1
− = + (9.29)
v u f1 f2
If the two lens-system is regarded as equivalent to a single lens of
focal length f, we have 237

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1 1 1
− =
v u f
so that we get
1 1 1
= + (9.30)
f f1 f2
The derivation is valid for any number of thin lenses in contact. If
several thin lenses of focal length f1, f2, f3,... are in contact, the effective
focal length of their combination is given by
1 1 1 1
= + + +… (9.31)
f f1 f2 f3
In terms of power, Eq. (9.31) can be written as
P = P1 + P2 + P3 + … (9.32)
where P is the net power of the lens combination. Note that the sum in
Eq. (9.32) is an algebraic sum of individual powers, so some of the terms
on the right side may be positive (for convex lenses) and some negative
(for concave lenses). Combination of lenses helps to obtain diverging or
converging lenses of desired magnification. It also enhances sharpness
of the image. Since the image formed by the first lens becomes the object
for the second, Eq. (9.25) implies that the total magnification m of the
combination is a product of magnification (m1, m 2, m 3,...) of individual
lenses
m = m1 m 2 m 3 ... (9.33)
Such a system of combination of lenses is commonly used in designing
lenses for cameras, microscopes, telescopes and other optical instruments.

Example 9.8 Find the position of the image formed by the lens
combination given in the Fig. 9.20.

FIGURE 9.20

Solution Image formed by the first lens

1 1 1
− =
EXAMPLE 9.8

v1 u1 f 1

1 1 1
− =
v1 −30 10
or v1 = 15 cm
238

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The image formed by the first lens serves as the object for the second.
This is at a distance of (15 – 5) cm = 10 cm to the right of the second
lens. Though the image is real, it serves as a virtual object for the
second lens, which means that the rays appear to come from it for
the second lens.
1 1 1
− =
v2 10 −10
or v2 = ¥
The virtual image is formed at an infinite distance to the left of the
second lens. This acts as an object for the third lens.
1 1 1
− =
v3 u 3 f3

EXAMPLE 9.8
1 1 1
or = +
v3 ∞ 30
or v3 = 30 cm
The final image is formed 30 cm to the right of the third lens.

9.6 REFRACTION THROUGH A PRISM

Figure 9.21 shows the passage of light through
a triangular prism ABC. The angles of incidence
and refraction at the first face AB are i and r1,
while the angle of incidence (from glass to air) at
the second face AC is r2 and the angle of refraction
or emergence e. The angle between the emergent
ray RS and the direction of the incident ray PQ
is called the angle of deviation, d.
In the quadrilateral AQNR, two of the angles
(at the vertices Q and R) are right angles.
Therefore, the sum of the other angles of the
quadrilateral is 180°. FIGURE 9.21 A ray of light passing through
a triangular glass prism.
ÐA + ÐQNR = 180°
From the triangle QNR,
r1 + r2 + ÐQNR = 180°
Comparing these two equations, we get
r1 + r2 = A (9.34)
The total deviation d is the sum of deviations at the two faces,
d = (i – r1 ) + (e – r2 )
that is,
d = i+ e–A (9.35)
Thus, the angle of deviation depends on the angle of incidence. A plot
between the angle of deviation and angle of incidence is shown in
Fig. 9.22. You can see that, in general, any given value of d, except for
i = e, corresponds to two values i and hence of e. This, in fact, is expected
from the symmetry of i and e in Eq. (9.35), i.e., d remains the same if i 239

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and e are interchanged. Physically, this is related
to the fact that the path of ray in Fig. 9.21 can be
traced back, resulting in the same angle of
deviation. At the minimum deviation Dm, the
refracted ray inside the prism becomes parallel
to its base. We have
d = Dm, i = e which implies r1 = r2.
Equation (9.34) gives
A
2r = A or r = (9.36)
2
In the same way, Eq. (9.35) gives
Dm = 2i – A, or i = (A + Dm)/2 (9.37)
The refractive index of the prism is
FIGURE 9.22 Plot of angle of deviation (d )
n 2 sin [( A + Dm )/2]
versus angle of incidence (i ) for a n 21 = = (9.38)
triangular prism. n1 sin[ A / 2]
The angles A and Dm can be measured experimentally. Equation
(9.38) thus provides a method of determining refractive index of the
material of the prism.
For a small angle prism, i.e., a thin prism, Dm is also very small, and
we get
sin [( A + Dm )/ 2] ( A + Dm ) / 2
n 21 = ≃
sin[ A /2] A /2
Dm = (n 21–1)A
It implies that, thin prisms do not deviate light much.

9.7 OPTICAL INSTRUMENTS

A number of optical devices and instruments have been designed utilising
reflecting and refracting properties of mirrors, lenses and prisms.
Periscope, kaleidoscope, binoculars, telescopes, microscopes are some
examples of optical devices and instruments that are in common use.
Our eye is, of course, one of the most important optical device the nature
has endowed us with. We have already studied about the human eye in
Class X. We now go on to describe the principles of working of the
microscope and the telescope.

9.7.1 The microscope

A simple magnifier or microscope is a converging lens of small focal length
(Fig. 9.23). In order to use such a lens as a microscope, the lens is held
near the object, one focal length away or less, and the eye is positioned
close to the lens on the other side. The idea is to get an erect, magnified
and virtual image of the object at a distance so that it can be viewed
comfortably, i.e., at 25 cm or more. If the object is at a distance f, the
240 image is at infinity. However, if the object is at a distance slightly less

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FIGURE 9.23 A simple microscope; (a) the magnifying lens is located

such that the image is at the near point, (b) the angle subtanded by the
object, is the same as that at the near point, and (c) the object near the
focal point of the lens; the image is far off but closer than infinity.

than the focal length of the lens, the image is virtual and closer than
infinity. Although the closest comfortable distance for viewing the image
is when it is at the near point (distance D @ 25 cm), it causes some strain
on the eye. Therefore, the image formed at infinity is often considered
most suitable for viewing by the relaxed eye. We show both cases, the
first in Fig. 9.23(a), and the second in Fig. 9.23(b) and (c).
The linear magnification m, for the image formed at the near point D,
by a simple microscope can be obtained by using the relation 241

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v 1 1   v
m= = v  –  = 1 – 
u v f   f 
Now according to our sign convention, v is negative, and is equal in
magnitude to D. Thus, the magnification is
 D
m = 1 +  (9.39)
 f
Since D is about 25 cm, to have a magnification of six, one needs a convex
lens of focal length, f = 5 cm.
Note that m = h¢/h where h is the size of the object and h¢ the size of
the image. This is also the ratio of the angle subtended by the image
to that subtended by the object, if placed at D for comfortable viewing.
(Note that this is not the angle actually subtended by the object at the
eye, which is h/u.) What a single-lens simple magnifier achieves is that it
allows the object to be brought closer to the eye than D.
We will now find the magnification when the image is at infinity. In
this case we will have to obtained the angular magnification. Suppose
the object has a height h. The maximum angle it can subtend, and be
clearly visible (without a lens), is when it is at the near point, i.e., a distance
D. The angle subtended is then given by
h
tan θo =   » qo (9.40)
 D
We now find the angle subtended at the eye by the image when the
object is at u. From the relations
h′ v
=m =
h u
we have the angle subtended by the image
h′ h v h
tan θi = = ⋅ = »q . The angle subtended by the object, when it
−v −v u −u
is at u = –f.

h
θi =   (9.41)
f
as is clear from Fig. 9.23(c). The angular magnification is, therefore

θ  D
m = i = (9.42)
 θo  f

This is one less than the magnification when the image is at the near
point, Eq. (9.39), but the viewing is more comfortable and the difference
in magnification is usually small. In subsequent discussions of optical
instruments (microscope and telescope) we shall assume the image to be
242 at infinity.

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 Ray Optics and
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FIGURE 9.24 Ray diagram for the formation of image by a

compound microscope.

http://astro.nineplanets.org/bigeyes.html
The world’s largest optical telescopes
A simple microscope has a limited maximum magnification (£ 9) for
realistic focal lengths. For much larger magnifications, one uses two lenses,
one compounding the effect of the other. This is known as a compound
microscope. A schematic diagram of a compound microscope is shown
in Fig. 9.24. The lens nearest the object, called the objective, forms a
real, inverted, magnified image of the object. This serves as the object for
the second lens, the eyepiece, which functions essentially like a simple
microscope or magnifier, produces the final image, which is enlarged
and virtual. The first inverted image is thus near (at or within) the focal
plane of the eyepiece, at a distance appropriate for final image formation
at infinity, or a little closer for image formation at the near point. Clearly,
the final image is inverted with respect to the original object.
We now obtain the magnification due to a compound microscope.
The ray diagram of Fig. 9.24 shows that the (linear) magnification due to
the objective, namely h¢/h, equals
h′ L
mO = = (9.43)
h fo
where we have used the result
 h  h ′
tan β =   =  
 fo   L 

Here h¢ is the size of the first image, the object size being h and fo
being the focal length of the objective. The first image is formed near the
focal point of the eyepiece. The distance L, i.e., the distance between the
second focal point of the objective and the first focal point of the eyepiece
(focal length fe) is called the tube length of the compound microscope. 243

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As the first inverted image is near the focal point of the eyepiece, we
use the result from the discussion above for the simple microscope to
obtain the (angular) magnification me due to it [Eq. (9.39)], when the
final image is formed at the near point, is

 D 
m e = 1 +  [9.44(a)]
 fe 

When the final image is formed at infinity, the angular magnification

due to the eyepiece [Eq. (9.42)] is
me = (D/fe ) [9.44(b)]
Thus, the total magnification [(according to Eq. (9.33)], when the
image is formed at infinity, is

 L  D 
m = m om e =     (9.45)
 fo  fe 
Clearly, to achieve a large magnification of a small object (hence the
name microscope), the objective and eyepiece should have small focal
lengths. In practice, it is difficult to make the focal length much smaller
than 1 cm. Also large lenses are required to make L large.
For example, with an objective with fo = 1.0 cm, and an eyepiece with
focal length fe = 2.0 cm, and a tube length of 20 cm, the magnification is

 L  D 
m = m om e =    
 fo  fe 

20 25
250
1 2
Various other factors such as illumination of the object, contribute to
the quality and visibility of the image. In modern microscopes, multi-
component lenses are used for both the objective and the eyepiece to
improve image quality by minimising various optical aberrations (defects)
in lenses.

9.7.2 Telescope
The telescope is used to provide angular magnification of distant objects
(Fig. 9.25). It also has an objective and an eyepiece. But here, the objective
has a large focal length and a much larger aperture than the eyepiece.
Light from a distant object enters the objective and a real image is formed
in the tube at its second focal point. The eyepiece magnifies this image
producing a final inverted image. The magnifying power m is the ratio of
the angle b subtended at the eye by the final image to the angle a which
the object subtends at the lens or the eye. Hence
h fo fo
m . (9.46)
244 fe h fe

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 Ray Optics and
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In this case, the length of the telescope tube is fo + fe.
Terrestrial telescopes have, in addition, a pair of inverting lenses to
make the final image erect. Refracting telescopes can be used both for
terrestrial and astronomical observations. For example, consider
a telescope whose objective has a focal length of 100 cm and the eyepiece
a focal length of 1 cm. The magnifying power of this telescope is
m = 100/1 = 100.
Let us consider a pair of stars of actual separation 1¢ (one minute of
arc). The stars appear as though they are separated by an angle of 100 ×
1¢ = 100¢ =1.67°.

FIGURE 9.25 A refracting telescope.

The main considerations with an astronomical telescope are its light

gathering power and its resolution or resolving power. The former clearly
depends on the area of the objective. With larger diameters, fainter objects
can be observed. The resolving power, or the ability to observe two objects
distinctly, which are in very nearly the same direction, also depends on
the diameter of the objective. So, the desirable aim in optical telescopes is
to make them with objective of large diameter. The largest lens objective
in use has a diameter of 40 inch (~1.02 m). It is at the Yerkes Observatory
in Wisconsin, USA. Such big lenses tend to be very heavy and therefore,
difficult to make and support by their edges. Further, it is rather difficult
and expensive to make such large sized lenses which form images that
are free from any kind of chromatic aberration and distortions.
For these reasons, modern telescopes use a concave mirror rather
than a lens for the objective. Telescopes with mirror objectives are called
reflecting telescopes. There is no chromatic aberration in a mirror.
Mechanical support is much less of a problem since a mirror weighs
much less than a lens of equivalent optical quality, and can be supported
over its entire back surface, not just over its rim. One obvious problem
with a reflecting telescope is that the objective mirror focusses light inside 245

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FIGURE 9.26 Schematic diagram of a reflecting telescope (Cassegrain).

the telescope tube. One must have an eyepiece and the observer right
there, obstructing some light (depending on the size of the observer cage).
This is what is done in the very large 200 inch (~5.08 m) diameters, Mt.
Palomar telescope, California. The viewer sits near the focal point of the
mirror, in a small cage. Another solution to the problem is to deflect the
light being focussed by another mirror. One such arrangement using a
convex secondary mirror to focus the incident light, which now passes
through a hole in the objective primary mirror, is shown in Fig. 9.26.
This is known as a Cassegrain telescope, after its inventor. It has the
advantages of a large focal length in a short telescope. The largest telescope
in India is in Kavalur, Tamil Nadu. It is a 2.34 m diameter reflecting
telescope (Cassegrain). It was ground, polished, set up, and is being used
by the Indian Institute of Astrophysics, Bangalore. The largest reflecting
telescopes in the world are the pair of Keck telescopes in Hawaii, USA,
with a reflector of 10 metre in diameter.

SUMMARY

1. Reflection is governed by the equation Ði = Ðr¢ and refraction by the

Snell’s law, sini/sinr = n, where the incident ray, reflected ray, refracted
ray and normal lie in the same plane. Angles of incidence, reflection
and refraction are i, r ¢ and r, respectively.
2. The critical angle of incidence ic for a ray incident from a denser to rarer
medium, is that angle for which the angle of refraction is 90°. For
i > ic, total internal reflection occurs. Multiple internal reflections in
diamond (ic @ 24.4°), totally reflecting prisms and mirage, are some
examples of total internal reflection. Optical fibres consist of glass
fibres coated with a thin layer of material of lower refractive index.
Light incident at an angle at one end comes out at the other, after
multiple internal reflections, even if the fibre is bent.
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3. Cartesian sign convention: Distances measured in the same direction

as the incident light are positive; those measured in the opposite
direction are negative. All distances are measured from the pole/optic
centre of the mirror/lens on the principal axis. The heights measured
upwards above x-axis and normal to the principal axis of the mirror/
lens are taken as positive. The heights measured downwards are taken
as negative.
4. Mirror equation:

1 1 1
+ =
v u f

where u and v are object and image distances, respectively and f is the
focal length of the mirror. f is (approximately) half the radius of
curvature R. f is negative for concave mirror; f is positive for a convex
mirror.

5. For a prism of the angle A, of refractive index n 2 placed in a medium

of refractive index n1,

n 2 sin ( A + D m ) / 2 
n 21 = =
n1 sin ( A / 2)
where Dm is the angle of minimum deviation.

6. For refraction through a spherical interface (from medium 1 to 2 of

refractive index n1 and n 2, respectively)

n 2 n1 n 2 − n1
− =
v u R
Thin lens formula
1 1 1
− =
v u f
Lens maker’s formula

1 (n 2 − n1 )  1 1 
=  −
f n1  R1 R 2 

R1 and R2 are the radii of curvature of the lens surfaces. f is positive

for a converging lens; f is negative for a diverging lens. The power of a
lens P = 1/f.
The SI unit for power of a lens is dioptre (D): 1 D = 1 m–1.
If several thin lenses of focal length f1, f2, f3,.. are in contact, the
effective focal length of their combination, is given by
1 1 1 1
= + + +
f f1 f2 f3 …
The total power of a combination of several lenses is
P = P1 + P2 + P3 + …

7. Dispersion is the splitting of light into its constituent colour.

247

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8. Magnifying power m of a simple microscope is given by m = 1 + (D/f),
where D = 25 cm is the least distance of distinct vision and f is the
focal length of the convex lens. If the image is at infinity, m = D/f. For
a compound microscope, the magnifying power is given by m = me × m0
where me = 1 + (D/fe), is the magnification due to the eyepiece and mo
is the magnification produced by the objective. Approximately,
L D
m= ×
fo fe
where fo and fe are the focal lengths of the objective and eyepiece,
respectively, and L is the distance between their focal points.
9. Magnifying power m of a telescope is the ratio of the angle b subtended
at the eye by the image to the angle a subtended at the eye by the
object.
β f
m = = o
α fe
where f0 and fe are the focal lengths of the objective and eyepiece,
respectively.

POINTS TO PONDER
1. The laws of reflection and refraction are true for all surfaces and
pairs of media at the point of the incidence.
2. The real image of an object placed between f and 2f from a convex lens
can be seen on a screen placed at the image location. If the screen is
removed, is the image still there? This question puzzles many, because
it is difficult to reconcile ourselves with an image suspended in air
without a screen. But the image does exist. Rays from a given point
on the object are converging to an image point in space and diverging
away. The screen simply diffuses these rays, some of which reach our
eye and we see the image. This can be seen by the images formed in
air during a laser show.
3. Image formation needs regular reflection/refraction. In principle, all
rays from a given point should reach the same image point. This is
why you do not see your image by an irregular reflecting object, say
the page of a book.
4. Thick lenses give coloured images due to dispersion. The variety in
colour of objects we see around us is due to the constituent colours
of the light incident on them. A monochromatic light may produce an
entirely different perception about the colours on an object as seen in
white light.
5. For a simple microscope, the angular size of the object equals the
angular size of the image. Yet it offers magnification because we can
keep the small object much closer to the eye than 25 cm and hence
have it subtend a large angle. The image is at 25 cm which we can see.
Without the microscope, you would need to keep the small object at
25 cm which would subtend a very small angle.

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EXERCISES
9.1 A small candle, 2.5 cm in size is placed at 27 cm in front of a concave
mirror of radius of curvature 36 cm. At what distance from the mirror
should a screen be placed in order to obtain a sharp image? Describe
the nature and size of the image. If the candle is moved closer to the
mirror, how would the screen have to be moved?
9.2 A 4.5 cm needle is placed 12 cm away from a convex mirror of focal
length 15 cm. Give the location of the image and the magnification.
Describe what happens as the needle is moved farther from the mirror.
9.3 A tank is filled with water to a height of 12.5 cm. The apparent
depth of a needle lying at the bottom of the tank is measured by a
microscope to be 9.4 cm. What is the refractive index of water? If
water is replaced by a liquid of refractive index 1.63 up to the same
height, by what distance would the microscope have to be moved to
focus on the needle again?
9.4 Figures 9.27(a) and (b) show refraction of a ray in air incident at 60°
with the normal to a glass-air and water-air interface, respectively.
Predict the angle of refraction in glass when the angle of incidence
in water is 45° with the normal to a water-glass interface [Fig. 9.27(c)].

FIGURE 9.27

9.5 A small bulb is placed at the bottom of a tank containing water to a

depth of 80cm. What is the area of the surface of water through
which light from the bulb can emerge out? Refractive index of water
is 1.33. (Consider the bulb to be a point source.)
9.6 A prism is made of glass of unknown refractive index. A parallel
beam of light is incident on a face of the prism. The angle of minimum
deviation is measured to be 40°. What is the refractive index of the
material of the prism? The refracting angle of the prism is 60°. If the
prism is placed in water (refractive index 1.33), predict the new
angle of minimum deviation of a parallel beam of light.
9.7 Double-convex lenses are to be manufactured from a glass of
refractive index 1.55, with both faces of the same radius of
curvature. What is the radius of curvature required if the focal length
is to be 20 cm?
9.8 A beam of light converges at a point P. Now a lens is placed in the
path of the convergent beam 12 cm from P. At what point does the
beam converge if the lens is (a) a convex lens of focal length 20 cm,
and (b) a concave lens of focal length 16 cm?
9.9 An object of size 3.0 cm is placed 14cm in front of a concave lens of
focal length 21 cm. Describe the image produced by the lens. What
happens if the object is moved further away from the lens? 249

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9.10 What is the focal length of a convex lens of focal length 30 cm in
contact with a concave lens of focal length 20 cm? Is the system a
converging or a diverging lens? Ignore thickness of the lenses.
9.11 A compound microscope consists of an objective lens of focal length
2.0 cm and an eyepiece of focal length 6.25 cm separated by a
distance of 15 cm. How far from the objective should an object be
placed in order to obtain the final image at (a) the least distance of
distinct vision (25 cm), and (b) at infinity? What is the magnifying
power of the microscope in each case?
9.12 A person with a normal near point (25 cm) using a compound
microscope with objective of focal length 8.0 mm and an eyepiece of
focal length 2.5 cm can bring an object placed at 9.0 mm from the
objective in sharp focus. What is the separation between the two
lenses? Calculate the magnifying power of the microscope,
9.13 A small telescope has an objective lens of focal length 144 cm and
an eyepiece of focal length 6.0 cm. What is the magnifying power of
the telescope? What is the separation between the objective and
the eyepiece?
9.14 (a) A giant refracting telescope at an observatory has an objective
lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is
used, what is the angular magnification of the telescope?
(b) If this telescope is used to view the moon, what is the diameter
of the image of the moon formed by the objective lens? The
diameter of the moon is 3.48 × 106 m, and the radius of lunar
orbit is 3.8 × 108 m.
9.15 Use the mirror equation to deduce that:
(a) an object placed between f and 2 f of a concave mirror produces
a real image beyond 2 f.
(b) a convex mirror always produces a virtual image independent
of the location of the object.
(c) the virtual image produced by a convex mirror is always
diminished in size and is located between the focus and
the pole.
(d) an object placed between the pole and focus of a concave mirror
produces a virtual and enlarged image.
[Note: This exercise helps you deduce algebraically properties of
images that one obtains from explicit ray diagrams.]
9.16 A small pin fixed on a table top is viewed from above from a distance
of 50 cm. By what distance would the pin appear to be raised if it is
viewed from the same point through a 15 cm thick glass slab held
parallel to the table? Refractive index of glass = 1.5. Does the answer
depend on the location of the slab?
9.17 (a) Figure 9.28 shows a cross-section of a ‘light pipe’ made of a
glass fibre of refractive index 1.68. The outer covering of the
pipe is made of a material of refractive index 1.44. What is the
range of the angles of the incident rays with the axis of the pipe
for which total reflections inside the pipe take place, as shown
in the figure.

250
FIGURE 9.28

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(b) What is the answer if there is no outer covering of the pipe?
9.18 The image of a small electric bulb fixed on the wall of a room is to be
obtained on the opposite wall 3 m away by means of a large convex
lens. What is the maximum possible focal length of the lens required
for the purpose?
9.19 A screen is placed 90 cm from an object. The image of the object on
the screen is formed by a convex lens at two different locations
separated by 20 cm. Determine the focal length of the lens.
9.20 (a) Determine the ‘effective focal length’ of the combination of
the two lenses in Exercise 9.10, if they are placed 8.0cm apart
with their principal axes coincident. Does the answer depend
on which side of the combination a beam of parallel light is
incident? Is the notion of effective focal length of this system
useful at all?
(b) An object 1.5 cm in size is placed on the side of the convex lens
in the arrangement (a) above. The distance between the object
and the convex lens is 40 cm. Determine the magnification
produced by the two-lens system, and the size of the image.
9.21 At what angle should a ray of light be incident on the face of a prism
of refracting angle 60° so that it just suffers total internal reflection
at the other face? The refractive index of the material of the prism is
1.524.
9.22 A card sheet divided into squares each of size 1 mm2 is being viewed
at a distance of 9 cm through a magnifying glass (a converging lens
of focal length 9 cm) held close to the eye.
(a) What is the magnification produced by the lens? How much is
the area of each square in the virtual image?
(b) What is the angular magnification (magnifying power) of the
lens ?
(c) Is the magnification in (a) equal to the magnifying power in (b)?
Explain.
9.23 (a) At what distance should the lens be held from the card sheet in
Exercise 9.22 in order to view the squares distinctly with the
maximum possible magnifying power?
(b) What is the magnification in this case?
(c) Is the magnification equal to the magnifying power in this case?
Explain.
9.24 What should be the distance between the object in Exercise 9.23
and the magnifying glass if the virtual image of each square in
the figure is to have an area of 6.25 mm2. Would you be able to
see the squares distinctly with your eyes very close to the
magnifier?
[Note: Exercises 9.22 to 9.24 will help you clearly understand the
difference between magnification in absolute size and the angular
magnification (or magnifying power) of an instrument.]
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9.25 Answer the following questions:
(a) The angle subtended at the eye by an object is equal to the
angle subtended at the eye by the virtual image produced by a
magnifying glass. In what sense then does a magnifying glass
provide angular magnification?
(b) In viewing through a magnifying glass, one usually positions
one’s eyes very close to the lens. Does angular magnification
change if the eye is moved back?
(c) Magnifying power of a simple microscope is inversely proportional
to the focal length of the lens. What then stops us from using a
convex lens of smaller and smaller focal length and achieving
greater and greater magnifying power?
(d) Why must both the objective and the eyepiece of a compound
microscope have short focal lengths?
(e) When viewing through a compound microscope, our eyes should
be positioned not on the eyepiece but a short distance away
from it for best viewing. Why? How much should be that short
distance between the eye and eyepiece?
9.26 An angular magnification (magnifying power) of 30X is desired using
an objective of focal length 1.25 cm and an eyepiece of focal length
5 cm. How will you set up the compound microscope?
9.27 A small telescope has an objective lens of focal length 140 cm and
an eyepiece of focal length 5.0 cm. What is the magnifying power of
the telescope for viewing distant objects when
(a) the telescope is in normal adjustment (i.e., when the final image
is at infinity)?
(b) the final image is formed at the least distance of distinct vision
(25 cm)?
9.28 (a) For the telescope described in Exercise 9.27 (a), what is the
separation between the objective lens and the eyepiece?
(b) If this telescope is used to view a 100 m tall tower 3 km away,
what is the height of the image of the tower formed by the objective
lens?
(c) What is the height of the final image of the tower if it is formed at
25 cm?
9.29 A Cassegrain telescope uses two mirrors as shown in Fig. 9.26. Such
a telescope is built with the mirrors 20 mm apart. If the radius of
curvature of the large mirror is 220 mm and the small mirror is
140 mm, where will the final image of an object at infinity be?
9.30 Light incident normally on a plane mirror attached to a galvanometer
coil retraces backwards as shown in Fig. 9.29. A current in the coil
produces a deflection of 3.5o of the mirror. What is the displacement
of the reflected spot of light on a screen placed 1.5 m away?

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FIGURE 9.29

9.31 Figure 9.30 shows an equiconvex lens (of refractive index 1.50) in
contact with a liquid layer on top of a plane mirror. A small needle
with its tip on the principal axis is moved along the axis until its
inverted image is found at the position of the needle. The distance of
the needle from the lens is measured to be 45.0 cm. The liquid is
removed and the experiment is repeated. The new distance is
measured to be 30.0 cm. What is the refractive index of the liquid?

FIGURE 9.30

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Notes

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Chapter Ten

WAVE OPTICS

10.1 INTRODUCTION
In 1637 Descartes gave the corpuscular model of light and derived Snell’s
law. It explained the laws of reflection and refraction of light at an interface.
The corpuscular model predicted that if the ray of light (on refraction)
bends towards the normal then the speed of light would be greater in the
second medium. This corpuscular model of light was further developed
by Isaac Newton in his famous book entitled OPTICKS and because of
the tremendous popularity of this book, the corpuscular model is very
often attributed to Newton.
In 1678, the Dutch physicist Christiaan Huygens put forward the
wave theory of light – it is this wave model of light that we will discuss in
this chapter. As we will see, the wave model could satisfactorily explain
the phenomena of reflection and refraction; however, it predicted that on
refraction if the wave bends towards the normal then the speed of light
would be less in the second medium. This is in contradiction to the
prediction made by using the corpuscular model of light. It was much
later confirmed by experiments where it was shown that the speed of
light in water is less than the speed in air confirming the prediction of the
wave model; Foucault carried out this experiment in 1850.
The wave theory was not readily accepted primarily because of
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and it was felt that a wave would always require a medium to propagate
from one point to the other. However, when Thomas Young performed
his famous interference experiment in 1801, it was firmly established
that light is indeed a wave phenomenon. The wavelength of visible
light was measured and found to be extremely small; for example, the
wavelength of yellow light is about 0.6 mm. Because of the smallness
of the wavelength of visible light (in comparison to the dimensions of
typical mirrors and lenses), light can be assumed to approximately
travel in straight lines. This is the field of geometrical optics, which we
had discussed in the previous chapter. Indeed, the branch of optics in
which one completely neglects the finiteness of the wavelength is called
geometrical optics and a ray is defined as the path of energy
propagation in the limit of wavelength tending to zero.
After the interference experiment of Young in 1801, for the next 40
years or so, many experiments were carried out involving the
interference and diffraction of lightwaves; these experiments could only
be satisfactorily explained by assuming a wave model of light. Thus,
around the middle of the nineteenth century, the wave theory seemed
to be very well established. The only major difficulty was that since it
was thought that a wave required a medium for its propagation, how
could light waves propagate through vacuum. This was explained
when Maxwell put forward his famous electromagnetic theory of light.
Maxwell had developed a set of equations describing the laws of
electricity and magnetism and using these equations he derived what
is known as the wave equation from which he predicted the existence
of electromagnetic waves*. From the wave equation, Maxwell could
calculate the speed of electromagnetic waves in free space and he found
that the theoretical value was very close to the measured value of speed
o f l i g h t . F r o m t h i s , h e p r o p o u n d e d t h a t light must be an
electromagnetic wave. Thus, according to Maxwell, light waves are
associated with changing electric and magnetic fields; changing electric
field produces a time and space varying magnetic field and a changing
magnetic field produces a time and space varying electric field. The
changing electric and magnetic fields result in the propagation of
electromagnetic waves (or light waves) even in vacuum.
In this chapter we will first discuss the original formulation of the
Huygens principle and derive the laws of reflection and refraction. In
Sections 10.4 and 10.5, we will discuss the phenomenon of interference
which is based on the principle of superposition. In Section 10.6 we
will discuss the phenomenon of diffraction which is based on Huygens-
Fresnel principle. Finally in Section 10.7 we will discuss the
phenomenon of polarisation which is based on the fact that the light
waves are transverse electromagnetic waves.

* Maxwell had predicted the existence of electromagnetic waves around 1855; it

was much later (around 1890) that Heinrich Hertz produced radiowaves in the
laboratory. J.C. Bose and G. Marconi made practical applications of the Hertzian
256 waves

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10.2 HUYGENS PRINCIPLE

We would first define a wavefront: when we drop a small stone on a
calm pool of water, waves spread out from the point of impact. Every
point on the surface starts oscillating with time. At any instant, a
photograph of the surface would show circular rings on which the
disturbance is maximum. Clearly, all points on such a circle are
oscillating in phase because they are at the same distance from the
source. Such a locus of points, which oscillate in phase is called a
wavefront; thus a wavefront is defined as a surface of constant
phase. The speed with which the wavefront moves outwards from the FIGURE 10.1 (a) A
diverging spherical
source is called the speed of the wave. The energy of the wave travels
wave emanating from
in a direction perpendicular to the wavefront. a point source. The
If we have a point source emitting waves uniformly in all directions, wavefronts are
then the locus of points which have the same amplitude and vibrate in spherical.
the same phase are spheres and we have what is known as a spherical
wave as shown in Fig. 10.1(a). At a large distance from the source, a
small portion of the sphere can be considered as a plane and we have
what is known as a plane wave [Fig. 10.1(b)].
Now, if we know the shape of the wavefront at t = 0, then Huygens
principle allows us to determine the shape of the wavefront at a later
time t. Thus, Huygens principle is essentially a geometrical construction,
which given the shape of the wafefront at any time allows us to determine
the shape of the wavefront at a later time. Let us consider a diverging FIGURE 10.1 (b) At a
wave and let F1F2 represent a portion of the spherical wavefront at t = 0 large distance from
(Fig. 10.2). Now, according to Huygens principle, each point of the the source, a small
wavefront is the source of a secondary disturbance and the wavelets portion of the
emanating from these points spread out in all directions with the speed spherical wave can
of the wave. These wavelets emanating from the wavefront are usually be approximated by a
plane wave.
referred to as secondary wavelets and if we draw a common tangent
to all these spheres, we obtain the new position of the wavefront at a
later time.

FIGURE 10.2 F1F2 represents the spherical wavefront (with O as

centre) at t = 0. The envelope of the secondary wavelets
emanating from F1F2 produces the forward moving wavefront G1G2.
The backwave D1D2 does not exist.
257

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Thus, if we wish to determine the shape of the wavefront at t = t, we
draw spheres of radius vt from each point on the spherical wavefront
where v represents the speed of the waves in the medium. If we now draw
a common tangent to all these spheres, we obtain the new position of the
wavefront at t = t. The new wavefront shown as G1G2 in Fig. 10.2 is again
spherical with point O as the centre.
The above model has one shortcoming: we also have a backwave which
is shown as D1D2 in Fig. 10.2. Huygens argued that the amplitude of the
secondary wavelets is maximum in the forward direction and zero in the
backward direction; by making this adhoc assumption, Huygens could
explain the absence of the backwave. However, this adhoc assumption is
not satisfactory and the absence of the backwave is really justified from
more rigorous wave theory.
In a similar manner, we can use Huygens principle to determine the
shape of the wavefront for a plane wave propagating through a medium
FIGURE 10.3 (Fig. 10.3).
Huygens geometrical
construction for a 10.3 REFRACTION AND REFLECTION OF
plane wave
propagating to the
PLANE WAVES USING HUYGENS PRINCIPLE
right. F1 F2 is the
plane wavefront at
10.3.1 Refraction of a plane wave
t = 0 and G1G2 is the We will now use Huygens principle to derive the laws of refraction. Let PP¢
wavefront at a later represent the surface separating medium 1 and medium 2, as shown in
time t. The lines A1A2, Fig. 10.4. Let v1 and v2 represent the speed of light in medium 1 and
B1B2 … etc., are medium 2, respectively. We assume a plane wavefront AB propagating in
normal to both F1F2
the direction A¢A incident on the interface at an angle i as shown in the
and G1G2 and
represent rays.
figure. Let t be the time taken by the wavefront to travel the distance BC.
Thus,
BC = v1 t

FIGURE 10.4 A plane wave AB is incident at an angle i on the surface

PP¢ separating medium 1 and medium 2. The plane wave undergoes
refraction and CE represents the refracted wavefront. The figure
corresponds to v2 < v1 so that the refracted waves bends towards the
258 normal.

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 Wave Optics
In order to determine the shape of the refracted
wavefront, we draw a sphere of radius v2t from the point
A in the second medium (the speed of the wave in the
second medium is v2). Let CE represent a tangent plane
drawn from the point C on to the sphere. Then, AE = v2 t
and CE would represent the refracted wavefront. If we
now consider the triangles ABC and AEC, we readily
obtain
BC v1τ
sin i = = (10.1)
AC AC
and
AE v 2τ
sin r = = (10.2)
AC AC Christiaan Huygens

CHRISTIAAN HUYGENS (1629 – 1695)

where i and r are the angles of incidence and refraction, (1629 – 1695) Dutch
respectively. Thus we obtain physicist, astronomer,
mathematician and the
sin i v
= 1 (10.3)
founder of the wave
sin r v2 theory of light. His book,
From the above equation, we get the important result T reatise on light, makes
fascinating reading even
that if r < i (i.e., if the ray bends toward the normal), the
today. He brilliantly
speed of the light wave in the second medium (v2) will be explained the double
less then the speed of the light wave in the first medium refraction shown by the
(v1). This prediction is opposite to the prediction from mineral calcite in this
the corpuscular model of light and as later experiments work in addition to
showed, the prediction of the wave theory is correct. Now, reflection and refraction.
if c represents the speed of light in vacuum, then, He was the first to
analyse circular and
c
n1 = (10.4)
simple harmonic motion
v1 and designed and built
and improved clocks and
telescopes. He discovered
c the true geometry of
n2 = (10.5) Saturn’s rings.
v2
are known as the refractive indices of medium 1 and
medium 2, respectively. In terms of the refractive indices, Eq. (10.3) can
be written as
n1 sin i = n2 sin r (10.6)
This is the Snell’s law of refraction. Further, if l1 and l 2 denote the
wavelengths of light in medium 1 and medium 2, respectively and if the
distance BC is equal to l 1 then the distance AE will be equal to l 2 ( because
if the crest from B has reached C in time t, then the crest from A should
have also reached E in time t ); thus,
λ1 BC v
= = 1
λ2 AE v2
or
v1 v2
= (10.7) 259
λ1 λ2

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The above equation implies that when a wave gets refracted into a
denser medium (v1 > v2) the wavelength and the speed of propagation
decrease but the frequency n (= v/l) remains the same.

10.3.2 Refraction at a rarer medium

We now consider refraction of a plane wave at a rarer medium, i.e.,
v2 > v1. Proceeding in an exactly similar manner we can construct a
refracted wavefront as shown in Fig. 10.5. The angle of refraction
Demonstration of interference, diffraction, refraction, resonance and Doppler effect

will now be greater than angle of incidence; however, we will still have
n1 sin i = n2 sin r . We define an angle ic by the following equation
n2
sin i c = (10.8)
n1
Thus, if i = ic then sin r = 1 and r = 90°. Obviously, for i > ic, there can
not be any refracted wave. The angle ic is known as the critical angle and
for all angles of incidence greater than the critical angle, we will not have
any refracted wave and the wave will undergo what is known as total
internal reflection. The phenomenon of total internal reflection and its
applications was discussed in Section 9.4.
http://www.falstad.com/ripple/

FIGURE 10.5 Refraction of a plane wave incident on a

rarer medium for which v2 > v1. The plane wave bends
away from the normal.

10.3.3 Reflection of a plane wave by a plane surface

We next consider a plane wave AB incident at an angle i on a reflecting
surface MN. If v represents the speed of the wave in the medium and if t
represents the time taken by the wavefront to advance from the point B
to C then the distance
BC = vt
In order to construct the reflected wavefront we draw a sphere of
radius vt from the point A as shown in Fig. 10.6. Let CE represent the
tangent plane drawn from the point C to this sphere. Obviously
260 AE = BC = vt

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FIGURE 10.6 Reflection of a plane wave AB by the reflecting surface MN.

AB and CE represent incident and reflected wavefronts.

If we now consider the triangles EAC and BAC we will find that they
are congruent and therefore, the angles i and r (as shown in Fig. 10.6)
would be equal. This is the law of reflection.
Once we have the laws of reflection and refraction, the behaviour of
prisms, lenses, and mirrors can be understood. These phenomena were
discussed in detail in Chapter 9 on the basis of rectilinear propagation of
light. Here we just describe the behaviour of the wavefronts as they
undergo reflection or refraction. In Fig. 10.7(a) we consider a plane wave
passing through a thin prism. Clearly, since the speed of light waves is
less in glass, the lower portion of the incoming wavefront (which travels
through the greatest thickness of glass) will get delayed resulting in a tilt
in the emerging wavefront as shown in the figure. In Fig. 10.7(b) we
consider a plane wave incident on a thin convex lens; the central part of
the incident plane wave traverses the thickest portion of the lens and is
delayed the most. The emerging wavefront has a depression at the centre
and therefore the wavefront becomes spherical and converges to the point
F which is known as the focus. In Fig. 10.7(c) a plane wave is incident on
a concave mirror and on reflection we have a spherical wave converging
to the focal point F. In a similar manner, we can understand refraction
and reflection by concave lenses and convex mirrors.
From the above discussion it follows that the total time taken from a
point on the object to the corresponding point on the image is the same
measured along any ray. For example, when a convex lens focusses light
to form a real image, although the ray going through the centre traverses
a shorter path, but because of the slower speed in glass, the time taken
is the same as for rays travelling near the edge of the lens.

FIGURE 10.7 Refraction of a plane wave by (a) a thin prism, (b) a convex lens.
261
(c) Reflection of a plane wave by a concave mirror.

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Example 10.1
(a) When monochromatic light is incident on a surface separating
two media, the reflected and refracted light both have the same
frequency as the incident frequency. Explain why?
(b) When light travels from a rarer to a denser medium, the speed
decreases. Does the reduction in speed imply a reduction in the
energy carried by the light wave?
(c) In the wave picture of light, intensity of light is determined by the
square of the amplitude of the wave. What determines the intensity
of light in the photon picture of light.
Solution
(a) Reflection and refraction arise through interaction of incident light
with the atomic constituents of matter. Atoms may be viewed as
oscillators, which take up the frequency of the external agency
(light) causing forced oscillations. The frequency of light emitted by
a charged oscillator equals its frequency of oscillation. Thus, the
frequency of scattered light equals the frequency of incident light.
EXAMPLE 10.1

(b) No. Energy carried by a wave depends on the amplitude of the

wave, not on the speed of wave propagation.
(c) For a given frequency, intensity of light in the photon picture is
determined by the number of photons crossing an unit area per
unit time.

10.4 COHERENT AND INCOHERENT

ADDITION OF WAVES
In this section we will discuss the
interference pattern produced by the
superposition of two waves. You may recall
that we had discussed the superposition
principle in Chapter 14 of your Class XI
textbook. Indeed the entire field of
interference is based on the superposition
(a) (b) principle according to which at a particular
FIGURE 10.8 (a) Two needles oscillating in point in the medium, the resultant
phase in water represent two coherent sources. displacement produced by a number of
(b) The pattern of displacement of water waves is the vector sum of the displace-
molecules at an instant on the surface of water ments produced by each of the waves.
showing nodal N (no displacement) and Consider two needles S1 and S2 moving
antinodal A (maximum displacement) lines. periodically up and down in an identical
fashion in a trough of water [Fig. 10.8(a)]. They produce two water waves,
and at a particular point, the phase difference between the displacements
produced by each of the waves does not change with time; when this
happens the two sources are said to be coherent. Figure 10.8(b) shows
the position of crests (solid circles) and troughs (dashed circles) at a given
instant of time. Consider a point P for which
262
S1 P = S2 P

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 Wave Optics
Since the distances S1 P and S2 P are equal, waves from S1 and S2 will
take the same time to travel to the point P and waves that emanate from
S1 and S2 in phase will also arrive, at the point P, in phase.
Thus, if the displacement produced by the source S1 at the point P is
given by
y1 = a cos wt
then, the displacement produced by the source S2 (at the point P) will
also be given by
y2 = a cos wt
Thus, the resultant of displacement at P would be given by
y = y1 + y2 = 2 a cos wt
Since the intensity is proportional to the square of the amplitude, the
resultant intensity will be given by
I = 4 I0
where I0 represents the intensity produced by each one of the individual
sources; I0 is proportional to a2. In fact at any point on the perpendicular
bisector of S1S2, the intensity will be 4I0. The two sources are said to
FIGURE 10.9
interfere constructively and we have what is referred to as constructive
(a) Constructive
interference. We next consider a point Q [Fig. 10.9(a)]
interference at a
for which point Q for which the
S2Q –S1Q = 2l path difference is 2l.
(b) Destructive
The waves emanating from S1 will arrive exactly two cycles earlier interference at a
than the waves from S2 and will again be in phase [Fig. 10.9(a)]. Thus, if point R for which the
the displacement produced by S1 is given by path difference is
2.5 l .
y1 = a cos wt
then the displacement produced by S2 will be given by
y2 = a cos (wt – 4p) = a cos wt
where we have used the fact that a path difference of 2l corresponds to a
phase difference of 4p. The two displacements are once again in phase
and the intensity will again be 4 I0 giving rise to constructive interference.
In the above analysis we have assumed that the distances S1Q and S2Q
are much greater than d (which represents the distance between S1 and
S2) so that although S1Q and S2Q are not equal, the amplitudes of the
displacement produced by each wave are very nearly the same.
We next consider a point R [Fig. 10.9(b)] for which
S2R – S1R = –2.5l
The waves emanating from S1 will arrive exactly two and a half cycles
later than the waves from S2 [Fig. 10.10(b)]. Thus if the displacement
FIGURE 10.10 Locus
produced by S1 is given by of points for which
y1 = a cos wt S1P – S2P is equal to
zero, ±l, ± 2l, ± 3l .
then the displacement produced by S2 will be given by
y2 = a cos (wt + 5p) = – a cos wt
263

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 Physics
where we have used the fact that a path difference of 2.5l corresponds to
a phase difference of 5p. The two displacements are now out of phase
and the two displacements will cancel out to give zero intensity. This is
referred to as destructive interference.
To summarise: If we have two coherent sources S1 and S2 vibrating
in phase, then for an arbitrary point P whenever the path difference,
S1P ~ S2P = nl (n = 0, 1, 2, 3,...) (10.9)
we will have constructive interference and the resultant intensity will be
4I0; the sign ~ between S1P and S2 P represents the difference between
S1P and S2 P. On the other hand, if the point P is such that the path
difference,
http://phet.colorado.edu/en/simulation/legacy/wave-interference

1
S1P ~ S2P = (n+ ) l (n = 0, 1, 2, 3, ...) (10.10)
2
we will have destructive interference and the resultant intensity will be
zero. Now, for any other arbitrary point G (Fig. 10.10) let the phase
difference between the two displacements be f. Thus, if the displacement
Ripple Tank experiments on wave interference

produced by S1 is given by
y1 = a cos wt
then, the displacement produced by S2 would be
y2 = a cos (wt + f )
and the resultant displacement will be given by
y = y1 + y2
= a [cos wt + cos (wt +f )]
= 2 a cos (f/2) cos (wt + f/2)

The amplitude of the resultant displacement is 2a cos (f/2) and

therefore the intensity at that point will be
I = 4 I0 cos2 (f/2) (10.11)
If f = 0, ± 2 p, ± 4 p,… which corresponds to the condition given by
Eq. (10.9) we will have constructive interference leading to maximum
intensity. On the other hand, if f = ± p, ± 3p, ± 5p … [which corresponds to
the condition given by Eq. (10.10)] we will have destructive interference
leading to zero intensity.
Now if the two sources are coherent (i.e., if the two needles are going
up and down regularly) then the phase difference f at any point will not
change with time and we will have a stable interference pattern; i.e., the
positions of maxima and minima will not change with time. However, if
the two needles do not maintain a constant phase difference, then the
interference pattern will also change with time and, if the phase difference
changes very rapidly with time, the positions of maxima and minima will
also vary rapidly with time and we will see a “time-averaged” intensity
distribution. When this happens, we will observe an average intensity
that will be given by
I = 2 I0 (10.12)
264 at all points.

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 Wave Optics
When the phase difference between the two vibrating sources changes
rapidly with time, we say that the two sources are incoherent and when
this happens the intensities just add up. This is indeed what happens
when two separate light sources illuminate a wall.

10.5 INTERFERENCE OF LIGHT WAVES AND YOUNG’S

EXPERIMENT
We will now discuss interference using light waves. If
we use two sodium lamps illuminating two pinholes
(Fig. 10.11) we will not observe any interference fringes.
This is because of the fact that the light wave emitted
from an ordinary source (like a sodium lamp) undergoes
abrupt phase changes in times of the order of 10–10
seconds. Thus the light waves coming out from two
independent sources of light will not have any fixed
phase relationship and would be incoherent, when this FIGURE 10.11 If two sodium
happens, as discussed in the previous section, the lamps illuminate two pinholes
intensities on the screen will add up. S1 and S2, the intensities will add
The British physicist Thomas Young used an up and no interference fringes will
ingenious technique to “lock” the phases of the waves be observed on the screen.
emanating from S1 and S2. He made two pinholes S1
and S2 (very close to each other) on an opaque screen [Fig. 10.12(a)].
These were illuminated by another pinholes that was in turn, lit by a
bright source. Light waves spread out from S and fall on both S1 and S2.
S1 and S2 then behave like two coherent sources because light waves
coming out from S1 and S2 are derived from the same original source
and any abrupt phase change in S will manifest in exactly similar phase
changes in the light coming out from S1 and S2. Thus, the two sources S1
and S2 will be locked in phase; i.e., they will be coherent like the two
vibrating needle in our water wave example [Fig. 10.8(a)].
The spherical waves emanating from S 1 and S 2 will produce
interference fringes on the screen GG¢, as shown in Fig. 10.12(b). The
positions of maximum and minimum intensities can be calculated by
using the analysis given in Section 10.4.

(a) (b)

FIGURE 10.12 Young’s arrangement to produce interference pattern. 265

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 Physics
We will have constructive interference resulting in a bright
xd
region when = nl. That is,
D
n λD
x = xn = ; n = 0, ± 1, ± 2, ... (10.13)
d
On the other hand, we will have destructive
xd 1
interference resulting in a dark region when = (n+ ) l
THOMAS YOUNG (1773 – 1829)

D 2
that is
1 D
x = xn = (n+ ) ; n  0,  1,  2 (10.14)
Thomas Young
2 d
Thus dark and bright bands appear on the screen,
(1773 – 1829) English
as shown in Fig. 10.13. Such bands are called fringes.
physicist, physician and
Egyptologist. Young worked Equations (10.13) and (10.14) show that dark and
on a wide variety of bright fringes are equally spaced.
scientific problems, ranging
from the structure of the eye
and the mechanism of
vision to the decipherment
of the Rosetta stone. He
revived the wave theory of
light and recognised that
interference phenomena
provide proof of the wave
properties of light.

FIGURE 10.13 Computer generated fringe pattern produced by two point

source S1 and S2 on the screen GG¢ (Fig. 10.12); correspond to d = 0.025
mm, D = 5 cm and l = 5 × 10–5 cm.) (Adopted from OPTICS by A. Ghatak,
Tata McGraw Hill Publishing Co. Ltd., New Delhi, 2000.)

10.6 DIFFRACTION
If we look clearly at the shadow cast by an opaque object, close to the
region of geometrical shadow, there are alternate dark and bright regions
just like in interference. This happens due to the phenomenon of
diffraction. Diffraction is a general characteristic exhibited by all types of
waves, be it sound waves, light waves, water waves or matter waves. Since
the wavelength of light is much smaller than the dimensions of most
266 obstacles; we do not encounter diffraction effects of light in everyday

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 Wave Optics
observations. However, the finite resolution of our eye or of optical
instruments such as telescopes or microscopes is limited due to the
phenomenon of diffraction. Indeed the colours that you see when a CD is
viewed is due to diffraction effects. We will now discuss the phenomenon
of diffraction.

10.6.1 The single slit

In the discussion of Young’s experiment, we stated that a single narrow
slit acts as a new source from which light spreads out. Even before Young,
early experimenters – including Newton – had noticed that light spreads
out from narrow holes and slits. It seems to turn around corners and
enter regions where we would expect a shadow. These effects, known as
diffraction, can only be properly understood using wave ideas. After all,
you are hardly surprised to hear sound
waves from someone talking around a corner!
When the double slit in Young’s
experiment is replaced by a single narrow
slit (illuminated by a monochromatic
source), a broad pattern with a central bright
region is seen. On both sides, there are
alternate dark and bright regions, the
intensity becoming weaker away from the
centre (Fig. 10.15). To understand this, go
to Fig. 10.14, which shows a parallel beam
of light falling normally on a single slit LN of
width a. The diffracted light goes on to meet FIGURE 10.14 The geometry of path
a screen. The midpoint of the slit is M. differences for diffraction by a single slit.
A straight line through M perpendicular
to the slit plane meets the screen at C. We want the
intensity at any point P on the screen. As before, straight
lines joining P to the different points L,M,N, etc., can be
treated as parallel, making an angle q with the
normal MC.
The basic idea is to divide the slit into much smaller
parts, and add their contributions at P with the proper
phase differences. We are treating different parts of the
wavefront at the slit as secondary sources. Because the
incoming wavefront is parallel to the plane of the slit, these
sources are in phase.
It is observed that the intensity has a central
maximum at q = 0 and other secondary maxima at q l
(n+1/2) l/a, which go on becoming weaker and weaker
with increasing n. The minima (zero intensity) are at q l
nl/a, n = ±1, ±2, ±3, .... FIGURE 10.15 Intensity
The photograph and intensity pattern corresponding distribution and photograph of
to it is shown in Fig. 10.15. fringes due to diffraction
There has been prolonged discussion about at single slit.
difference between intereference and diffraction among 267

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scientists since the discovery of these phenomena. In this context, it is
interesting to note what Richard Feynman* has said in his famous
Feynman Lectures on Physics:
No one has ever been able to define the difference between
interference and diffraction satisfactorily. It is just a question
of usage, and there is no specific, important physical difference
between them. The best we can do is, roughly speaking, is to
say that when there are only a few sources, say two interfering
sources, then the result is usually called interference, but if there
is a large number of them, it seems that the word diffraction is
more often used.
In the double-slit experiment, we must note that the pattern on the
screen is actually a superposition of single-slit diffraction from each slit
or hole, and the double-slit interference pattern.

10.6.2 Seeing the single slit diffraction pattern

It is surprisingly easy to see the single-slit diffraction pattern for oneself.
The equipment needed can be found in most homes –– two razor blades
and one clear glass electric bulb preferably with a straight filament. One
has to hold the two blades so that the edges are parallel and have a
narrow slit in between. This is easily done with the thumb and forefingers
(Fig. 10.16).
Keep the slit parallel to the filament, right in front of the eye. Use
spectacles if you normally do. With slight adjustment of the width of
the slit and the parallelism of the edges, the pattern should be seen
with its bright and dark bands. Since the position of all the bands
(except the central one) depends on wavelength, they will show some
colours. Using a filter for red or blue will make the fringes clearer.
With both filters available, the wider fringes for red compared to blue
FIGURE 10.16
can be seen.
Holding two blades to
form a single slit. A In this experiment, the filament plays the role of the first slit S in
bulb filament viewed Fig. 10.15. The lens of the eye focuses the pattern on the screen (the
through this shows retina of the eye).
clear diffraction With some effort, one can cut a double slit in an aluminium foil with
bands. a blade. The bulb filament can be viewed as before to repeat Young’s
experiment. In daytime, there is another suitable bright source subtending
a small angle at the eye. This is the reflection of the Sun in any shiny
convex surface (e.g., a cycle bell). Do not try direct sunlight – it can damage
the eye and will not give fringes anyway as the Sun subtends an angle
of (1/2)°.
In interference and diffraction, light energy is redistributed. If it
reduces in one region, producing a dark fringe, it increases in another
region, producing a bright fringe. There is no gain or loss of energy,
which is consistent with the principle of conservation of energy.

* Richand Feynman was one of the recipients of the 1965 Nobel Prize in Physics
268 for his fundamental work in quantum electrodynamics.

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10.7 POLARISATION
Consider holding a long string that is held horizontally, the other end of
which is assumed to be fixed. If we move the end of the string up and
down in a periodic manner, we will generate a wave propagating in the
+x direction (Fig. 10.17). Such a wave could be described by the following
equation

FIGURE 10.17 (a) The curves represent the displacement of a string at

t = 0 and at t = Dt, respectively when a sinusoidal wave is propagating
in the +x-direction. (b) The curve represents the time variation
of the displacement at x = 0 when a sinusoidal wave is propagating
in the +x-direction. At x = Dx, the time variation of the
displacement will be slightly displaced to the right.

y (x,t ) = a sin (kx – wt) (10.15)

where a and w (= 2pn ) represent the amplitude and the angular frequency
of the wave, respectively; further,

2π
λ= (10.16)
k
represents the wavelength associated with the wave. We had discussed
propagation of such waves in Chapter 14 of Class XI textbook. Since the
displacement (which is along the y direction) is at right angles to the
direction of propagation of the wave, we have what is known as a
transverse wave. Also, since the displacement is in the y direction, it is
often referred to as a y-polarised wave. Since each point on the string
moves on a straight line, the wave is also referred to as a linearly polarised 269

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wave. Further, the string always remains confined to the x-y plane and
therefore it is also referred to as a plane polarised wave.
In a similar manner we can consider the vibration of the string in the
x-z plane generating a z-polarised wave whose displacement will be given
by
z (x,t ) = a sin (kx – wt ) (10.17)
It should be mentioned that the linearly polarised waves [described
by Eqs. (10.15) and (10.17)] are all transverse waves; i.e., the
displacement of each point of the string is always at right angles to the
direction of propagation of the wave. Finally, if the plane of vibration of
the string is changed randomly in very short intervals of time, then we
have what is known as an unpolarised wave. Thus, for an unpolarised
wave the displacement will be randomly changing with time though it
will always be perpendicular to the direction of propagation.
Light waves are transverse in nature; i.e., the electric field associated
with a propagating light wave is always at right angles to the direction of
propagation of the wave. This can be easily demonstrated using a simple
polaroid. You must have seen thin plastic like sheets, which are called
polaroids. A polaroid consists of long chain molecules aligned in a
particular direction. The electric vectors (associated with the propagating
light wave) along the direction of the aligned molecules get absorbed.
Thus, if an unpolarised light wave is incident on such a polaroid then
the light wave will get linearly polarised with the electric vector oscillating
along a direction perpendicular to the aligned molecules; this direction
is known as the pass-axis of the polaroid.
Thus, if the light from an ordinary source (like a sodium lamp) passes
through a polaroid sheet P1, it is observed that its intensity is reduced by
half. Rotating P1 has no effect on the transmitted beam and transmitted
intensity remains constant. Now, let an identical piece of polaroid P2 be
placed before P1. As expected, the light from the lamp is reduced in
intensity on passing through P2 alone. But now rotating P1 has a dramatic
effect on the light coming from P2. In one position, the intensity transmitted
by P2 followed by P1 is nearly zero. When turned by 90° from this position,
P1 transmits nearly the full intensity emerging from P2 (Fig. 10.18).
The experiment at figure 10.18 can be easily understood by assuming
that light passing through the polaroid P2 gets polarised along the pass-
axis of P2. If the pass-axis of P2 makes an angle q with the pass-axis of
P1, then when the polarised beam passes through the polaroid P2, the
component E cos q (along the pass-axis of P2) will pass through P2.
Thus, as we rotate the polaroid P1 (or P2), the intensity will vary as:
I = I0 cos2q (10.18)
where I0 is the intensity of the polarized light after passing through
P1. This is known as Malus’ law. The above discussion shows that the

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FIGURE 10.18 (a) Passage of light through two polaroids P2 and P1. The
transmitted fraction falls from 1 to 0 as the angle between them varies
from 0° to 90°. Notice that the light seen through a single polaroid
P1 does not vary with angle. (b) Behaviour of the electric vector
when light passes through two polaroids. The transmitted
polarisation is the component parallel to the polaroid axis.
The double arrows show the oscillations of the electric vector.

intensity coming out of a single polaroid is half of the incident intensity.

By putting a second polaroid, the intensity can be further controlled
from 50% to zero of the incident intensity by adjusting the angle between
the pass-axes of two polaroids.
Polaroids can be used to control the intensity, in sunglasses,
windowpanes, etc. Polaroids are also used in photographic cameras and
3D movie cameras.

Example 10.2 Discuss the intensity of transmitted light when a

polaroid sheet is rotated between two crossed polaroids?
Solution Let I0 be the intensity of polarised light after passing through
the first polariser P1. Then the intensity of light after passing through
second polariser P2 will be
I = I 0 cos2θ ,

where q is the angle between pass axes of P1 and P2. Since P1 and P3
are crossed the angle between the pass axes of P2 and P3 will be
(p/2 – q ). Hence the intensity of light emerging from P3 will be
EXAMPLE 10.2

π 
I = I 0cos 2θ cos2  – θ 
2 
= I0 cos2q sin2q =(I0/4) sin22q
Therefore, the transmitted intensity will be maximum when q = p/4.

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SUMMARY

1. Huygens’ principle tells us that each point on a wavefront is a source

of secondary waves, which add up to give the wavefront at a later time.
2. Huygens’ construction tells us that the new wavefront is the forward
envelope of the secondary waves. When the speed of light is
independent of direction, the secondary waves are spherical. The rays
are then perpendicular to both the wavefronts and the time of travel
is the same measured along any ray. This principle leads to the well
known laws of reflection and refraction.
3. The principle of superposition of waves applies whenever two or more
sources of light illuminate the same point. When we consider the
intensity of light due to these sources at the given point, there is an
interference term in addition to the sum of the individual intensities.
But this term is important only if it has a non-zero average, which
occurs only if the sources have the same frequency and a stable phase
difference.
4. Young’s double slit of separation d gives equally spaced interference
fringes.
5. A single slit of width a gives a diffraction pattern with a central
λ 2λ
maximum. The intensity falls to zero at angles of ± , ± , etc.,
a a
with successively weaker secondary maxima in between.
6. Natural light, e.g., from the sun is unpolarised. This means the electric
vector takes all possible directions in the transverse plane, rapidly
and randomly, during a measurement. A polaroid transmits only one
component (parallel to a special axis). The resulting light is called
linearly polarised or plane polarised. When this kind of light is viewed
through a second polaroid whose axis turns through 2p, two maxima
and minima of intensity are seen.

POINTS TO PONDER
1. Waves from a point source spread out in all directions, while light was
seen to travel along narrow rays. It required the insight and experiment
of Huygens, Young and Fresnel to understand how a wave theory could
explain all aspects of the behaviour of light.
2. The crucial new feature of waves is interference of amplitudes from different
sources which can be both constructive and destructive, as shown in
Young’s experiment.
3. Diffraction phenomena define the limits of ray optics. The limit of the
ability of microscopes and telescopes to distinguish very close objects is
set by the wavelength of light.
4. Most interference and diffraction effects exist even for longitudinal waves
like sound in air. But polarisation phenomena are special to transverse
waves like light waves.

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EXERCISES
10.1 Monochromatic light of wavelength 589 nm is incident from air on a
water surface. What are the wavelength, frequency and speed of
(a) reflected, and (b) refracted light? Refractive index of water is
1.33.
10.2 What is the shape of the wavefront in each of the following cases:
(a) Light diverging from a point source.
(b) Light emerging out of a convex lens when a point source is placed
at its focus.
(c) The portion of the wavefront of light from a distant star intercepted
by the Earth.
10.3 (a) The refractive index of glass is 1.5. What is the speed of light in
glass? (Speed of light in vacuum is 3.0 × 108 m s–1)
(b) Is the speed of light in glass independent of the colour of light? If
not, which of the two colours red and violet travels slower in a
glass prism?
10.4 In a Young’s double-slit experiment, the slits are separated by
0.28 mm and the screen is placed 1.4 m away. The distance between
the central bright fringe and the fourth bright fringe is measured
to be 1.2 cm. Determine the wavelength of light used in the
experiment.
10.5 In Young’s double-slit experiment using monochromatic light of
wavelength l, the intensity of light at a point on the screen where
path difference is l, is K units. What is the intensity of light at a
point where path difference is l/3 ?
10.6 A beam of light consisting of two wavelengths, 650 nm and 520 nm,
is used to obtain interference fringes in a Young’s double-slit
experiment.
(a) Find the distance of the third bright fringe on the screen from
the central maximum for wavelength 650 nm.
(b) What is the least distance from the central maximum where the
bright fringes due to both the wavelengths coincide?

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Chapter Eleven

DUAL NATURE OF
RADIATION AND
MATTER

11.1 INTRODUCTION
The Maxwell’s equations of electromagnetism and Hertz experiments on
the generation and detection of electromagnetic waves in 1887 strongly
established the wave nature of light. Towards the same period at the end
of 19th century, experimental investigations on conduction of electricity
(electric discharge) through gases at low pressure in a discharge tube led
to many historic discoveries. The discovery of X-rays by Roentgen in 1895,
and of electron by J. J. Thomson in 1897, were important milestones in
the understanding of atomic structure. It was found that at sufficiently
low pressure of about 0.001 mm of mercury column, a discharge took
place between the two electrodes on applying the electric field to the gas
in the discharge tube. A fluorescent glow appeared on the glass opposite
to cathode. The colour of glow of the glass depended on the type of glass,
it being yellowish-green for soda glass. The cause of this fluorescence
was attributed to the radiation which appeared to be coming from the
cathode. These cathode rays were discovered, in 1870, by William
Crookes who later, in 1879, suggested that these rays consisted of streams
of fast moving negatively charged particles. The British physicist
J. J. Thomson (1856 -1940) confirmed this hypothesis. By applying
mutually perpendicular electric and magnetic fields across the discharge
274 tube, J. J. Thomson was the first to determine experimentally the speed

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 Dual Nature of Radiation
and Matter
and the specific charge [charge to mass ratio (e/m )] of the cathode ray
particles. They were found to travel with speeds ranging from about 0.1
to 0.2 times the speed of light (3 ×108 m/s). The presently accepted value
of e/m is 1.76 × 1011 C/kg. Further, the value of e/m was found to be
independent of the nature of the material/metal used as the cathode
(emitter), or the gas introduced in the discharge tube. This observation
suggested the universality of the cathode ray particles.
Around the same time, in 1887, it was found that certain metals, when
irradiated by ultraviolet light, emitted negatively charged particles having
small speeds. Also, certain metals when heated to a high temperature were
found to emit negatively charged particles. The value of e/m of these particles
was found to be the same as that for cathode ray particles. These
observations thus established that all these particles, although produced
under different conditions, were identical in nature. J. J. Thomson, in 1897,
named these particles as electrons, and suggested that they were
fundamental, universal constituents of matter. For his epoch-making
discovery of electron, through his theoretical and experimental
investigations on conduction of electricity by gasses, he was awarded the
Nobel Prize in Physics in 1906. In 1913, the American physicist R. A.
Millikan (1868 -1953) performed the pioneering oil-drop experiment for
the precise measurement of the charge on an electron. He found that the
charge on an oil-droplet was always an integral multiple of an elementary
charge, 1.602 × 10 –19 C. Millikan’s experiment established that electric
charge is quantised. From the values of charge (e ) and specific charge
(e/m ), the mass (m) of the electron could be determined.

11.2 ELECTRON EMISSION

We know that metals have free electrons (negatively charged particles) that
are responsible for their conductivity. However, the free electrons cannot
normally escape out of the metal surface. If an electron attempts to come
out of the metal, the metal surface acquires a positive charge and pulls the
electron back to the metal. The free electron is thus held inside the metal
surface by the attractive forces of the ions. Consequently, the electron can
come out of the metal surface only if it has got sufficient energy to overcome
the attractive pull. A certain minimum amount of energy is required to be
given to an electron to pull it out from the surface of the metal. This
minimum energy required by an electron to escape from the metal surface
is called the work function of the metal. It is generally denoted by f0 and
measured in eV (electron volt). One electron volt is the energy gained by an
electron when it has been accelerated by a potential difference of 1 volt, so
that 1 eV = 1.602 ×10 –19 J.
This unit of energy is commonly used in atomic and nuclear physics.
The work function (f0 ) depends on the properties of the metal and the
nature of its surface.
The minimum energy required for the electron emission from the metal
surface can be supplied to the free electrons by any one of the following
physical processes:
(i) Thermionic emission: By suitably heating, sufficient thermal energy
can be imparted to the free electrons to enable them to come out of the
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(ii) Field emission : By applying a very strong electric field (of the order of
108 V m–1) to a metal, electrons can be pulled out of the metal, as in a
spark plug.
(iii) Photoelectric emission : When light of suitable frequency illuminates
a metal surface, electrons are emitted from the metal surface. These
photo(light)-generated electrons are called photoelectrons.

11.3 PHOTOELECTRIC EFFECT

11.3.1 Hertz’s observations
The phenomenon of photoelectric emission was discovered in 1887 by
Heinrich Hertz (1857-1894), during his electromagnetic wave experiments.
In his experimental investigation on the production of electromagnetic
waves by means of a spark discharge, Hertz observed that high voltage
sparks across the detector loop were enhanced when the emitter plate
was illuminated by ultraviolet light from an arc lamp.
Light shining on the metal surface somehow facilitated the escape of
free, charged particles which we now know as electrons. When light falls
on a metal surface, some electrons near the surface absorb enough energy
from the incident radiation to overcome the attraction of the positive ions
in the material of the surface. After gaining sufficient energy from the
incident light, the electrons escape from the surface of the metal into the
surrounding space.

11.3.2 Hallwachs’ and Lenard’s observations

Wilhelm Hallwachs and Philipp Lenard investigated the phenomenon of
photoelectric emission in detail during 1886-1902.
Lenard (1862-1947) observed that when ultraviolet radiations were
allowed to fall on the emitter plate of an evacuated glass tube enclosing
two electrodes (metal plates), current flows in the circuit (Fig. 11.1). As
soon as the ultraviolet radiations were stopped, the current flow also
stopped. These observations indicate that when ultraviolet radiations fall
on the emitter plate C, electrons are ejected from it which are attracted
towards the positive, collector plate A by the electric field. The electrons
flow through the evacuated glass tube, resulting in the current flow. Thus,
light falling on the surface of the emitter causes current in the external
circuit. Hallwachs and Lenard studied how this photo current varied with
collector plate potential, and with frequency and intensity of incident light.
Hallwachs, in 1888, undertook the study further and connected a
negatively charged zinc plate to an electroscope. He observed that the
zinc plate lost its charge when it was illuminated by ultraviolet light.
Further, the uncharged zinc plate became positively charged when it was
irradiated by ultraviolet light. Positive charge on a positively charged
zinc plate was found to be further enhanced when it was illuminated by
ultraviolet light. From these observations he concluded that negatively
charged particles were emitted from the zinc plate under the action of
ultraviolet light.
After the discovery of the electron in 1897, it became evident that the
incident light causes electrons to be emitted from the emitter plate. Due
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 Dual Nature of Radiation
and Matter
to negative charge, the emitted electrons are pushed towards the collector
plate by the electric field. Hallwachs and Lenard also observed that when
ultraviolet light fell on the emitter plate, no electrons were emitted at all
when the frequency of the incident light was smaller than a certain
minimum value, called the threshold frequency. This minimum frequency
depends on the nature of the material of the emitter plate.
It was found that certain metals like zinc, cadmium, magnesium, etc.,
responded only to ultraviolet light, having short wavelength, to cause
electron emission from the surface. However, some alkali metals such as
lithium, sodium, potassium, caesium and rubidium were sensitive
even to visible light. All these photosensitive substances emit electrons
when they are illuminated by light. After the discovery of electrons, these
electrons were termed as photoelectrons. The phenomenon is called
photoelectric effect.

11.4 EXPERIMENTAL STUDY OF PHOTOELECTRIC

EFFECT
Figure 11.1 depicts a schematic view of the arrangement used for the
experimental study of the photoelectric effect. It consists of an evacuated
glass/quartz tube having a thin photosensitive plate C and another metal
plate A. Monochromatic light from the source S of sufficiently short
wavelength passes through the window W and falls on the photosensitive
plate C (emitter). A transparent quartz window is sealed on to the glass
tube, which permits ultraviolet radiation to pass through it and irradiate
the photosensitive plate C. The electrons are emitted by the plate C and
are collected by the plate A (collector), by the electric field created by the
battery. The battery maintains the potential difference between the plates
C and A, that can be varied. The polarity of the plates C and A can be
reversed by a commutator. Thus, the plate A can be maintained at a desired
positive or negative potential with respect to emitter C.
When the collector plate A is positive with respect to the
emitter plate C, the electrons are attracted to it. The
emission of electrons causes flow of electric current in
the circuit. The potential difference between the emitter
and collector plates is measured by a voltmeter (V)
whereas the resulting photo current flowing in the circuit
is measured by a microammeter (mA). The photoelectric
current can be increased or decreased by varying the
potential of collector plate A with respect to the emitter
plate C. The intensity and frequency of the incident light
can be varied, as can the potential difference V between
the emitter C and the collector A.
We can use the experimental arrangement of Fig.
11.1 to study the variation of photocurrent with (a)
intensity of radiation, (b) frequency of incident radiation,
FIGURE 11.1 Experimental
(c) the potential difference between the plates A and C, arrangement for study of
and (d) the nature of the material of plate C. Light of photoelectric effect.
different frequencies can be used by putting appropriate
coloured filter or coloured glass in the path of light falling 277

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on the emitter C. The intensity of light is varied by changing
the distance of the light source from the emitter.

11.4.1 Effect of intensity of light on photocurrent

The collector A is maintained at a positive potential with
respect to emitter C so that electrons ejected from C are
attracted towards collector A. Keeping the frequency of the
incident radiation and the potential fixed, the intensity of
light is varied and the resulting photoelectric current is
measured each time. It is found that the photocurrent
increases linearly with intensity of incident light as shown
graphically in Fig. 11.2. The photocurrent is directly
proportional to the number of photoelectrons emitted per
FIGURE 11.2 Variation of
Photoelectric current with second. This implies that the number of photoelectrons
intensity of light. emitted per second is directly proportional to the intensity
of incident radiation.

11.4.2 Effect of potential on photoelectric current

We first keep the plate A at some positive potential with respect to the
plate C and illuminate the plate C with light of fixed frequency n and fixed
intensity I1. We next vary the positive potential of plate A gradually and
measure the resulting photocurrent each time. It is found that the
photoelectric current increases with increase in positive (accelerating)
potential. At some stage, for a certain positive potential of plate A, all the
emitted electrons are collected by the plate A and the photoelectric current
becomes maximum or saturates. If we increase the accelerating potential
of plate A further, the photocurrent does not increase. This maximum
value of the photoelectric current is called saturation current. Saturation
current corresponds to the case when all the photoelectrons emitted by
the emitter plate C reach the collector plate A.
We now apply a negative (retarding)
potential to the plate A with respect to the
plate C and make it increasingly negative
gradually. When the polarity is reversed,
the electrons are repelled and only the
sufficiently energetic electrons are able to
reach the collector A. The photocurrent
is found to decrease rapidly until it drops
to zero at a certain sharply defined,
critical value of the negative potential V0
on the plate A. For a particular frequency
of incident radiation, the minimum
negative (retarding) potential V0 given to
the plate A for which the photocurrent
stops or becomes zero is called the cut-
off or stopping potential.
The interpretation of the observation
FIGURE 11.3 Variation of photocurrent with in terms of photoelectrons is
collector plate potential for different straightforward. All the photoelectrons
278 intensity of incident radiation. emitted from the metal do not have the

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 Dual Nature of Radiation
and Matter
same energy. Photoelectric current is zero when the stopping potential is
sufficient to repel even the most energetic photoelectrons, with the
maximum kinetic energy (Kmax), so that
Kmax = e V0 (11.1)
We can now repeat this experiment with incident radiation of the same
frequency but of higher intensity I2 and I3 (I3 > I2 > I1). We note that the
saturation currents are now found to be at higher values. This shows
that more electrons are being emitted per second, proportional to the
intensity of incident radiation. But the stopping potential remains the
same as that for the incident radiation of intensity I1, as shown graphically
in Fig. 11.3. Thus, for a given frequency of the incident radiation, the
stopping potential is independent of its intensity. In other words, the
maximum kinetic energy of photoelectrons depends on the light source
and the emitter plate material, but is independent of intensity of incident
radiation.

11.4.3 Effect of frequency of incident radiation on

stopping potential
We now study the relation between the frequency
n of the incident radiation and the stopping
potential V0 . We suitably adjust the same
intensity of light radiation at various frequencies
and study the variation of photocurrent with
collector plate potential. The resulting variation
is shown in Fig. 11.4. We obtain different values
of stopping potential but the same value of the
saturation current for incident radiation of
different frequencies. The energy of the emitted
electrons depends on the frequency of the
incident radiations. The stopping potential is FIGURE 11.4 Variation of photoelectric current
more negative for higher frequencies of incident with collector plate potential for different
radiation. Note from Fig. 11.4 that the stopping frequencies of incident radiation.
potentials are in the order V03 > V02 > V01 if the
frequencies are in the order n3 > n2 > n1 . This
implies that greater the frequency of incident
light, greater is the maximum kinetic energy of
the photoelectrons. Consequently, we need
greater retarding potential to stop them
completely. If we plot a graph between the
frequency of incident radiation and the
corresponding stopping potential for different
metals we get a straight line, as shown in Fig. 11.5.
The graph shows that
(i) the stopping potential V0 varies linearly with
the frequency of incident radiation for a given
photosensitive material. FIGURE 11.5 Variation of stopping potential V0
(ii) there exists a certain minimum cut-off with frequency n of incident radiation for a
frequency n0 for which the stopping potential given photosensitive material.
is zero.
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These observations have two implications:
(i) The maximum kinetic energy of the photoelectrons varies linearly
with the frequency of incident radiation, but is independent of its
intensity.
(ii) For a frequency n of incident radiation, lower than the cut-off
frequency n0, no photoelectric emission is possible even if the
intensity is large.
This minimum, cut-off frequency n0, is called the threshold frequency.
It is different for different metals.
Different photosensitive materials respond differently to light. Selenium
is more sensitive than zinc or copper. The same photosensitive substance
gives different response to light of different wavelengths. For example,
ultraviolet light gives rise to photoelectric effect in copper while green or
red light does not.
Note that in all the above experiments, it is found that, if frequency of
the incident radiation exceeds the threshold frequency, the photoelectric
emission starts instantaneously without any apparent time lag, even if
the incident radiation is very dim. It is now known that emission starts in
a time of the order of 10 – 9 s or less.
We now summarise the experimental features and observations
described in this section.
(i) For a given photosensitive material and frequency of incident radiation
(above the threshold frequency), the photoelectric current is directly
proportional to the intensity of incident light (Fig. 11.2).
(ii) For a given photosensitive material and frequency of incident radiation,
saturation current is found to be proportional to the intensity of
incident radiation whereas the stopping potential is independent of
its intensity (Fig. 11.3).
(iii) For a given photosensitive material, there exists a certain minimum
cut-off frequency of the incident radiation, called the threshold
frequency, below which no emission of photoelectrons takes place,
no matter how intense the incident light is. Above the threshold
frequency, the stopping potential or equivalently the maximum kinetic
energy of the emitted photoelectrons increases linearly with the
frequency of the incident radiation, but is independent of its intensity
(Fig. 11.5).
(iv) The photoelectric emission is an instantaneous process without any
apparent time lag (~10 – 9s or less), even when the incident radiation is
made exceedingly dim.
11.5 PHOTOELECTRIC EFFECT AND WAVE THEORY
OF LIGHT
The wave nature of light was well established by the end of the nineteenth
century. The phenomena of interference, diffraction and polarisation were
explained in a natural and satisfactory way by the wave picture of light.
According to this picture, light is an electromagnetic wave consisting of
electric and magnetic fields with continuous distribution of energy over
280 the region of space over which the wave is extended. Let us now see if this

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 Dual Nature of Radiation
and Matter
wave picture of light can explain the observations on photoelectric
emission given in the previous section.
According to the wave picture of light, the free electrons at the surface
of the metal (over which the beam of radiation falls) absorb the radiant
energy continuously. The greater the intensity of radiation, the greater are
the amplitude of electric and magnetic fields. Consequently, the greater
the intensity, the greater should be the energy absorbed by each electron.
In this picture, the maximum kinetic energy of the photoelectrons on the
surface is then expected to increase with increase in intensity. Also, no
matter what the frequency of radiation is, a sufficiently intense beam of
radiation (over sufficient time) should be able to impart enough energy to
the electrons, so that they exceed the minimum energy needed to escape
from the metal surface . A threshold frequency, therefore, should not exist.
These expectations of the wave theory directly contradict observations (i),
(ii) and (iii) given at the end of sub-section 11.4.3.
Further, we should note that in the wave picture, the absorption of
energy by electron takes place continuously over the entire
wavefront of the radiation. Since a large number of electrons absorb energy,
the energy absorbed per electron per unit time turns out to be small.
Explicit calculations estimate that it can take hours or more for a single
electron to pick up sufficient energy to overcome the work function and
come out of the metal. This conclusion is again in striking contrast to
observation (iv) that the photoelectric emission is instantaneous. In short,
the wave picture is unable to explain the most basic features of
photoelectric emission.

11.6 EINSTEIN’S PHOTOELECTRIC EQUATION: ENERGY

QUANTUM OF RADIATION
In 1905, Albert Einstein (1879 -1955) proposed a radically new picture
of electromagnetic radiation to explain photoelectric effect. In this picture,
photoelectric emission does not take place by continuous absorption of
energy from radiation. Radiation energy is built up of discrete units – the
so called quanta of energy of radiation. Each quantum of radiant energy
has energy h n, where h is Planck’s constant and n the frequency of light.
In photoelectric effect, an electron absorbs a quantum of energy (hn ) of
radiation. If this quantum of energy absorbed exceeds the minimum
energy needed for the electron to escape from the metal surface (work
function f0), the electron is emitted with maximum kinetic energy
Kmax = hn – f0 (11.2)
More tightly bound electrons will emerge with kinetic energies less
than the maximum value. Note that the intensity of light of a given
frequency is determined by the number of photons incident per second.
Increasing the intensity will increase the number of emitted electrons per
second. However, the maximum kinetic energy of the emitted
photoelectrons is determined by the energy of each photon.
Equation (11.2) is known as Einstein’s photoelectric equation. We
now see how this equation accounts in a simple and elegant manner all
the observations on photoelectric effect given at the end of sub-section 281
11.4.3.

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 Physics
· According to Eq. (11.2), Kmax depends linearly on n,
and is independent of intensity of radiation, in
agreement with observation. This has happened
because in Einstein’s picture, photoelectric effect arises
from the absorption of a single quantum of radiation
by a single electron. The intensity of radiation (that is
proportional to the number of energy quanta per unit
area per unit time) is irrelevant to this basic process.
· Since Kmax must be non-negative, Eq. (11.2 ) implies
that photoelectric emission is possible only if
h n > f0
or n > n0 , where
φ0
Albert Einstein (1879 – n0 = (11.3)
1955) Einstein, one of the h
greatest physicists of all Equation (11.3) shows that the greater the work
time, was born in Ulm, function f0, the higher the minimum or threshold
Germany. In 1905, he
published three path-
frequency n0 needed to emit photoelectrons. Thus,
breaking papers. In the there exists a threshold frequency n0 (= f0/h) for the
first paper, he introduced metal surface, below which no photoelectric emission
the notion of light quanta is possible, no matter how intense the incident
(now called photons) and radiation may be or how long it falls on the surface.
used it to explain the
features of photoelectric · In this picture, intensity of radiation as noted above,
effect. In the second paper, is proportional to the number of energy quanta per
he developed a theory of unit area per unit time. The greater the number of
Brownian motion, energy quanta available, the greater is the number of
confirmed experimentally a electrons absorbing the energy quanta and greater,
few years later and provided
a convincing evidence of therefore, is the number of electrons coming out of
the atomic picture of matter. the metal (for n > n0 ). This explains why, for n > n0 ,
The third paper gave birth photoelectric current is proportional to intensity.
to the special theory of · In Einstein’s picture, the basic elementary process
ALBERT EINSTEIN (1879 – 1955)

relativity. In 1916, he involved in photoelectric effect is the absorption of a

published the general
theory of relativity. Some of light quantum by an electron. This process is
Einstein’s most significant instantaneous. Thus, whatever may be the intensity
later contributions are: the i.e., the number of quanta of radiation per unit area
notion of stimulated per unit time, photoelectric emission is instantaneous.
emission introduced in an
Low intensity does not mean delay in emission, since
alternative derivation of
Planck’s blackbody the basic elementary process is the same. Intensity
radiation law, static model only determines how many electrons are able to
of the universe which participate in the elementary process (absorption of a
started modern cosmology, light quantum by a single electron) and, therefore, the
quantum statistics of a gas photoelectric current.
of massive bosons, and a
critical analysis of the Using Eq. (11.1), the photoelectric equation, Eq. (11.2),
foundations of quantum can be written as
mechanics. In 1921, he was e V0 = h n – f 0; for ν ≥ ν 0
awarded the Nobel Prize in
physics for his contribution
h φ
to theoretical physics and or V0 = ν− 0 (11.4)
the photoelectric effect. e e
This is an important result. It predicts that the V0
282 versus n curve is a straight line with slope = (h/e),

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 Dual Nature of Radiation
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independent of the nature of the material. During 1906-1916, Millikan
performed a series of experiments on photoelectric effect, aimed at
disproving Einstein’s photoelectric equation. He measured the slope of
the straight line obtained for sodium, similar to that shown in Fig. 11.5.
Using the known value of e, he determined the value of Planck’s constant
h. This value was close to the value of Planck’s contant (= 6.626 × 10–
34
J s) determined in an entirely different context. In this way, in 1916,
Millikan proved the validity of Einstein’s photoelectric equation, instead
of disproving it.
The successful explanation of photoelectric effect using the hypothesis
of light quanta and the experimental determination of values of h and f0,
in agreement with values obtained from other experiments, led to the
acceptance of Einstein’s picture of photoelectric effect. Millikan verified
photoelectric equation with great precision, for a number of alkali metals
over a wide range of radiation frequencies.

11.7 PARTICLE NATURE OF LIGHT: THE PHOTON

Photoelectric effect thus gave evidence to the strange fact that light in
interaction with matter behaved as if it was made of quanta or packets of
energy, each of energy h n.
Is the light quantum of energy to be associated with a particle? Einstein
arrived at the important result, that the light quantum can also be
associated with momentum (h n/c ). A definite value of energy as well as
momentum is a strong sign that the light quantum can be associated
with a particle. This particle was later named photon. The particle-like
behaviour of light was further confirmed, in 1924, by the experiment of
A.H. Compton (1892-1962) on scattering of X-rays from electrons. In
1921, Einstein was awarded the Nobel Prize in Physics for his contribution
to theoretical physics and the photoelectric effect. In 1923, Millikan was
awarded the Nobel Prize in physics for his work on the elementary
charge of electricity and on the photoelectric effect.
We can summarise the photon picture of electromagnetic radiation
as follows:
(i) In interaction of radiation with matter, radiation behaves as if it is
made up of particles called photons.
(ii) Each photon has energy E (=hn ) and momentum p (= h n/c), and
speed c, the speed of light.
(iii) All photons of light of a particular frequency n, or wavelength l, have
the same energy E (=hn = hc/l) and momentum p (= hn/c= h/l),
whatever the intensity of radiation may be. By increasing the intensity
of light of given wavelength, there is only an increase in the number of
photons per second crossing a given area, with each photon having
the same energy. Thus, photon energy is independent of intensity of
radiation.
(iv) Photons are electrically neutral and are not deflected by electric and
magnetic fields.
(v) In a photon-particle collision (such as photon-electron collision), the
total energy and total momentum are conserved. However, the number
of photons may not be conserved in a collision. The photon may be
absorbed or a new photon may be created. 283

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 Physics
Example 11.1 Monochromatic light of frequency 6.0 ´1014 Hz is
produced by a laser. The power emitted is 2.0 ´10–3 W. (a) What is the
energy of a photon in the light beam? (b) How many photons per second,
on an average, are emitted by the source?
Solution
(a) Each photon has an energy
E = h n = ( 6.63 ´10–34 J s) (6.0 ´1014 Hz)
= 3.98 ´ 10–19 J
(b) If N is the number of photons emitted by the source per second,
EXAMPLE 11.1

the power P transmitted in the beam equals N times the energy

per photon E, so that P = N E. Then
P 2.0 × 10 −3 W
N= =
E 3.98 × 10 −19 J
= 5.0 ´1015 photons per second.

Example 11.2 The work function of caesium is 2.14 eV. Find (a) the
threshold frequency for caesium, and (b) the wavelength of the incident
light if the photocurrent is brought to zero by a stopping potential of
0.60 V.
Solution
(a) For the cut-off or threshold frequency, the energy h n0 of the incident
radiation must be equal to work function f 0, so that

n0 = φ0 = 2.14 eV
h 6.63 × 10 −34 J s

2.14 × 1.6 × 10−19 J

= = 5.16 × 1014 Hz
6.63 × 10 −34 J s
Thus, for frequencies less than this threshold frequency, no
photoelectrons are ejected.
(b) Photocurrent reduces to zero, when maximum kinetic energy of
the emitted photoelectrons equals the potential energy e V0 by the
retarding potential V0. Einstein’s Photoelectric equation is
hc
eV0 = hn – f 0 = – f0
λ
or, l = hc/(eV0 + f0 )
(6.63 × 10 −34 J s) × (3 × 108 m/s)
=
(0.60 eV + 2.14 eV)
EXAMPLE 11.2

19.89 × 10 −26 J m
=
(2.74 eV)
19.89 × 10 −26 J m
λ= = 454 nm
2.74 × 1.6 × 10 −19 J

11.8 WAVE NATURE OF MATTER

The dual (wave-particle) nature of light (electromagnetic radiation, in
general) comes out clearly from what we have learnt in this and the
preceding chapters. The wave nature of light shows up in the phenomena
284
of interference, diffraction and polarisation. On the other hand, in

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 Dual Nature of Radiation
and Matter
photoelectric effect and Compton effect which involve
energy and momentum transfer, radiation behaves as if it
is made up of a bunch of particles – the photons. Whether
a particle or wave description is best suited for

LOUIS VICTOR DE BROGLIE (1892 – 1987)

understanding an experiment depends on the nature of
the experiment. For example, in the familiar phenomenon
of seeing an object by our eye, both descriptions are
important. The gathering and focussing mechanism of
light by the eye-lens is well described in the wave picture.
But its absorption by the rods and cones (of the retina)
requires the photon picture of light.
A natural question arises: If radiation has a dual (wave-
particle) nature, might not the particles of nature (the
electrons, protons, etc.) also exhibit wave-like character?
In 1924, the French physicist Louis Victor de Broglie Louis Victor de Broglie
(pronounced as de Broy) (1892-1987) put forward the (1892 – 1987) French
physicist who put forth
bold hypothesis that moving particles of matter should
revolutionary idea of wave
display wave-like properties under suitable conditions. nature of matter. This idea
He reasoned that nature was symmetrical and that the was developed by Erwin
two basic physical entities – matter and energy, must have Schródinger into a full-
symmetrical character. If radiation shows dual aspects, fledged theory of quantum
so should matter. De Broglie proposed that the wave mechanics commonly
length l associated with a particle of momentum p is known as wave mechanics.
given as In 1929, he was awarded the
Nobel Prize in Physics for his
h h discovery of the wave nature
l = = (11.5)
p mv of electrons.
where m is the mass of the particle and v its speed.
Equation (11.5) is known as the de Broglie relation and
the wavelength l of the matter wave is called de Broglie wavelength. The
dual aspect of matter is evident in the de Broglie relation. On the left hand
side of Eq. (11.5), l is the attribute of a wave while on the right hand side
the momentum p is a typical attribute of a particle. Planck’s constant h
relates the two attributes.
Equation (11.5) for a material particle is basically a hypothesis whose
validity can be tested only by experiment. However, it is interesting to see
that it is satisfied also by a photon. For a photon, as we have seen,
p = hn /c (11.6)
Therefore,
h c
= =λ (11.7)
p ν
That is, the de Broglie wavelength of a photon given by Eq. (11.5) equals
the wavelength of electromagnetic radiation of which the photon is a
quantum of energy and momentum.
Clearly, from Eq. (11.5 ), l is smaller for a heavier particle ( large m ) or
more energetic particle (large v). For example, the de Broglie wavelength
of a ball of mass 0.12 kg moving with a speed of 20 m s–1 is easily
calculated:
285

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 Physics
p = m v = 0.12 kg × 20 m s–1 = 2.40 kg m s–1

h 6.63 × 10−34 J s
l= = = 2.76 × 10–34 m
p 2.40 kg m s −1
This wavelength is so small that it is beyond any measurement. This
is the reason why macroscopic objects in our daily life do not show wave-
like properties. On the other hand, in the sub-atomic domain, the wave
character of particles is significant and measurable.

Example 11.3 What is the de Broglie wavelength associated with (a) an

electron moving with a speed of 5.4´106 m/s, and (b) a ball of mass 150 g
travelling at 30.0 m/s?
Solution
(a) For the electron:
Mass m = 9.11´10–31 kg, speed v = 5.4´106 m/s. Then, momentum
p = m v = 9.11´10–31 (kg) ´ 5.4 ´ 106 (m/s)
p = 4.92 ´ 10–24 kg m/s
de Broglie wavelength, l = h/p
6.63 × 10 –34 J s
=
4.92 × 10 –24 kg m/s
l = 0.135 nm
(b) For the ball:
Mass m ’ = 0.150 kg, speed v ’ = 30.0 m/s.
Then momentum p ’ = m’ v ’ = 0.150 (kg) ´ 30.0 (m/s)
p ’= 4.50 kg m/s
de Broglie wavelength l’ = h/p’.
6.63 × 10 –34 J s
EXAMPLE 11.3

=
4.50 × kg m/s
l ’ = 1.47 ´10–34 m
The de Broglie wavelength of electron is comparable with X-ray
wavelengths. However, for the ball it is about 10–19 times the size of
the proton, quite beyond experimental measurement.

SUMMARY

1. The minimum energy needed by an electron to come out from a metal

surface is called the work function of the metal. Energy (greater than
the work function (fo) required for electron emission from the metal
surface can be supplied by suitably heating or applying strong electric
field or irradiating it by light of suitable frequency.
2. Photoelectric effect is the phenomenon of emission of electrons by metals
when illuminated by light of suitable frequency. Certain metals respond
to ultraviolet light while others are sensitive even to the visible light.
Photoelectric effect involves conversion of light energy into electrical
energy. It follows the law of conservation of energy. The photoelectric
emission is an instantaneous process and possesses certain special
features.
286

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 Dual Nature of Radiation
and Matter

3. Photoelectric current depends on (i) the intensity of incident light, (ii)

the potential difference applied between the two electrodes, and (iii)
the nature of the emitter material.
4. The stopping potential (Vo) depends on (i) the frequency of incident
light, and (ii) the nature of the emitter material. For a given frequency
of incident light, it is independent of its intensity. The stopping potential
is directly related to the maximum kinetic energy of electrons emitted:
e V0 = (1/2) m v 2max = Kmax.
5. Below a certain frequency (threshold frequency) n 0 , characteristic of
the metal, no photoelectric emission takes place, no matter how large
the intensity may be.
6. The classical wave theory could not explain the main features of
photoelectric effect. Its picture of continuous absorption of energy
from radiation could not explain the independence of K max on
intensity, the existence of n o and the instantaneous nature of the
process. Einstein explained these features on the basis of photon
picture of light. According to this, light is composed of discrete
packets of energy called quanta or photons. Each photon carries an
energy E (= h n ) and momentum p (= h/l), which depend on the
frequency (n ) of incident light and not on its intensity. Photoelectric
emission from the metal surface occurs due to absorption of a photon
by an electron.
7. Einstein’s photoelectric equation is in accordance with the energy
conservation law as applied to the photon absorption by an electron in
the metal. The maximum kinetic energy (1/2)m v 2max is equal to
the photon energy (hn ) minus the work function f0 (= hn0 ) of the
target metal:
1
m v 2max = V0 e = hn – f0 = h (n – n0 )
2
This photoelectric equation explains all the features of the photoelectric
effect. Millikan’s first precise measurements confirmed the Einstein’s
photoelectric equation and obtained an accurate value of Planck’s
constant h . This led to the acceptance of particle or photon description
(nature) of electromagnetic radiation, introduced by Einstein.
8. Radiation has dual nature: wave and particle. The nature of experiment
determines whether a wave or particle description is best suited for
understanding the experimental result. Reasoning that radiation and
matter should be symmetrical in nature, Louis Victor de Broglie
attributed a wave-like character to matter (material particles). The waves
associated with the moving material particles are called matter waves
or de Broglie waves.
9. The de Broglie wavelength (l) associated with a moving particle is
related to its momentum p as: l = h/p. The dualism of matter is
inherent in the de Broglie relation which contains a wave concept
( l ) and a particle concept (p). The de Broglie wavelength is
independent of the charge and nature of the material particle. It is
significantly measurable (of the order of the atomic-planes spacing
in crystals) only in case of sub-atomic particles like electrons,
protons, etc. (due to smallness of their masses and hence, momenta).
However, it is indeed very small, quite beyond measurement, in case
of macroscopic objects, commonly encountered in everyday life.
287

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Physical Symbol Dimensions Unit Remarks
Quantity

Planck’s h [ML2 T –1] Js E = hn

constant
Stopping V0 [ML2 T –3A–1] V e V0= Kmax
potential
Work f0 [ML2 T –2] J; eV Kmax = E –f0
function
Threshold n0 [T –1] Hz n0 = f0 /h
frequency
de Broglie l [L] m  = h/p
wavelength

POINTS TO PONDER

1. Free electrons in a metal are free in the sense that they move inside the
metal in a constant potential (This is only an approximation). They are
not free to move out of the metal. They need additional energy to get
out of the metal.
2. Free electrons in a metal do not all have the same energy. Like molecules
in a gas jar, the electrons have a certain energy distribution at a given
temperature. This distribution is different from the usual Maxwell’s
distribution that you have learnt in the study of kinetic theory of gases.
You will learn about it in later courses, but the difference has to do
with the fact that electrons obey Pauli’s exclusion principle.
3. Because of the energy distribution of free electrons in a metal, the energy
required by an electron to come out of the metal is different for different
electrons. Electrons with higher energy require less additional energy to
come out of the metal than those with lower energies. Work function is
the least energy required by an electron to come out of the metal.
4. Observations on photoelectric effect imply that in the event of matter-
light interaction, absorption of energy takes place in discrete units of hn.
This is not quite the same as saying that light consists of particles,
each of energy hn.
5. Observations on the stopping potential (its independence of intensity
and dependence on frequency) are the crucial discriminator between
the wave-picture and photon-picture of photoelectric effect.
h
6. The wavelength of a matter wave given by λ = has physical
p
significance; its phase velocity vp has no physical significance. However,
the group velocity of the matter wave is physically meaningful and
equals the velocity of the particle.

EXERCISES
11.1 Find the
(a) maximum frequency, and
288 (b) minimum wavelength of X-rays produced by 30 kV electrons.

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 Dual Nature of Radiation
and Matter
11.2 The work function of caesium metal is 2.14 eV. When light of
frequency 6 ×1014Hz is incident on the metal surface, photoemission
of electrons occurs. What is the
(a) maximum kinetic energy of the emitted electrons,
(b) Stopping potential, and
(c) maximum speed of the emitted photoelectrons?
11.3 The photoelectric cut-off voltage in a certain experiment is 1.5 V.
What is the maximum kinetic energy of photoelectrons emitted?
11.4 Monochromatic light of wavelength 632.8 nm is produced by a
helium-neon laser. The power emitted is 9.42 mW.
(a) Find the energy and momentum of each photon in the light beam,
(b) How many photons per second, on the average, arrive at a target
irradiated by this beam? (Assume the beam to have uniform
cross-section which is less than the target area ), and
(c) How fast does a hydrogen atom have to travel in order to have
the same momentum as that of the photon?
11.5 In an experiment on photoelectric effect, the slope of the cut-off voltage
versus frequency of incident light is found to be 4.12 × 10–15 V s. Calculate
the value of Planck’s constant.
11.6 The threshold frequency for a certain metal is 3.3 × 1014 Hz. If light
of frequency 8.2 × 1014 Hz is incident on the metal, predict the cut-
off voltage for the photoelectric emission.
11.7 The work function for a certain metal is 4.2 eV. Will this metal give
hotoelectric emission for incident radiation of wavelength 330 nm?
11.8 Light of frequency 7.21 × 1014 Hz is incident on a metal surface.
Electrons with a maximum speed of 6.0 × 105 m/s are ejected from
the surface. What is the threshold frequency for photoemission of
electrons?
11.9 Light of wavelength 488 nm is produced by an argon laser which is
used in the photoelectric effect. When light from this spectral line is
incident on the emitter, the stopping (cut-of f) potential of
photoelectrons is 0.38 V. Find the work function of the material from
which the emitter is made.
11.10 What is the de Broglie wavelength of
(a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,
(b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and
(c) a dust particle of mass 1.0 × 10–9 kg drifting with a speed of 2.2
m/s ?
11.11 Show that the wavelength of electromagnetic radiation is equal to
the de Broglie wavelength of its quantum (photon).

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Chapter Twelve

ATOMS

12.1 INTRODUCTION
By the nineteenth century, enough evidence had accumulated in favour of
atomic hypothesis of matter. In 1897, the experiments on electric discharge
through gases carried out by the English physicist J. J. Thomson (1856 –
1940) revealed that atoms of different elements contain negatively charged
constituents (electrons) that are identical for all atoms. However, atoms on a
whole are electrically neutral. Therefore, an atom must also contain some
positive charge to neutralise the negative charge of the electrons. But what
is the arrangement of the positive charge and the electrons inside the atom?
In other words, what is the structure of an atom?
The first model of atom was proposed by J. J. Thomson in 1898.
According to this model, the positive charge of the atom is uniformly
distributed throughout the volume of the atom and the negatively charged
electrons are embedded in it like seeds in a watermelon. This model was
picturesquely called plum pudding model of the atom. However
subsequent studies on atoms, as described in this chapter, showed that
the distribution of the electrons and positive charges are very different
from that proposed in this model.
We know that condensed matter (solids and liquids) and dense gases at
all temperatures emit electromagnetic radiation in which a continuous
distribution of several wavelengths is present, though with different
290 intensities. This radiation is considered to be due to oscillations of atoms

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 Atoms
and molecules, governed by the interaction of each atom or
molecule with its neighbours. In contrast, light emitted from
rarefied gases heated in a flame, or excited electrically in a
glow tube such as the familiar neon sign or mercury vapour
light has only certain discrete wavelengths. The spectrum
appears as a series of bright lines. In such gases, the
average spacing between atoms is large. Hence, the
radiation emitted can be considered due to individual atoms
rather than because of interactions between atoms or
molecules.
In the early nineteenth century it was also established
that each element is associated with a characteristic
spectrum of radiation, for example, hydrogen always gives
a set of lines with fixed relative position between the lines. Ernst Rutherford (1871 –

ERNST RUTHERFORD (1871 – 1937)

This fact suggested an intimate relationship between the 1937) New Zealand born,
internal structure of an atom and the spectrum of British physicist who did
radiation emitted by it. In 1885, Johann Jakob Balmer pioneering work on
(1825 – 1898) obtained a simple empirical formula which radioactive radiation. He
gave the wavelengths of a group of lines emitted by atomic discovered alpha-rays and
hydrogen. Since hydrogen is simplest of the elements beta-rays. Along with
known, we shall consider its spectrum in detail in this Federick Soddy, he created
chapter. the modern theory of
Ernst Rutherford (1871–1937), a former research radioactivity. He studied
student of J. J. Thomson, was engaged in experiments on the ‘emanation’ of thorium
a-particles emitted by some radioactive elements. In 1906, and discovered a new noble
he proposed a classic experiment of scattering of these gas, an isotope of radon,
a-particles by atoms to investigate the atomic structure. now known as thoron. By
This experiment was later performed around 1911 by Hans scattering alpha-rays from
Geiger (1882–1945) and Ernst Marsden (1889–1970, who the metal foils, he
was 20 year-old student and had not yet earned his discovered the atomic
nucleus and proposed the
bachelor’s degree). The details are discussed in Section
plenatery model of the
12.2. The explanation of the results led to the birth of
atom. He also estimated the
Rutherford’s planetary model of atom (also called the approximate size of the
nuclear model of the atom). According to this the entire nucleus.
positive charge and most of the mass of the atom is
concentrated in a small volume called the nucleus with electrons revolving
around the nucleus just as planets revolve around the sun.
Rutherford’s nuclear model was a major step towards how we see
the atom today. However, it could not explain why atoms emit light of
only discrete wavelengths. How could an atom as simple as hydrogen,
consisting of a single electron and a single proton, emit a complex
spectrum of specific wavelengths? In the classical picture of an atom, the
electron revolves round the nucleus much like the way a planet revolves
round the sun. However, we shall see that there are some serious
difficulties in accepting such a model.

12.2 ALPHA-PARTICLE SCATTERING AND

RUTHERFORD’S NUCLEAR MODEL OF ATOM
At the suggestion of Ernst Rutherford, in 1911, H. Geiger and E. Marsden
performed some experiments. In one of their experiments, as shown in 291

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 Physics
Fig. 12.1, they directed a beam of
5.5 MeV a-particles emitted from a
214
83 Bi
radioactive source at a thin metal
foil made of gold. Figure 12.2 shows a
schematic diagram of this experiment.
Alpha-particles emitted by a 214 83 Bi
radioactive source were collimated into
a narrow beam by their passage
through lead bricks. The beam was
allowed to fall on a thin foil of gold of
thickness 2.1 × 10–7 m. The scattered
alpha-particles were observed through
a rotatable detector consisting of zinc
sulphide screen and a microscope. The
scattered alpha-particles on striking
the screen produced brief light flashes
or scintillations. These flashes may be
viewed through a microscope and the
FIGURE 12.1 Geiger -Marsden scattering experiment. distribution of the number of scattered
The entire apparatus is placed in a vacuum chamber particles may be studied as a function
(not shown in this figure).
of angle of scattering.

FIGURE 12.2 Schematic arrangement of the Geiger-Marsden experiment.

A typical graph of the total number of a-particles scattered at different

angles, in a given interval of time, is shown in Fig. 12.3. The dots in this
figure represent the data points and the solid curve is the theoretical
prediction based on the assumption that the target atom has a small,
dense, positively charged nucleus. Many of the a-particles pass through
the foil. It means that they do not suffer any collisions. Only about 0.14%
of the incident a-particles scatter by more than 1°; and about 1 in 8000
deflect by more than 90°. Rutherford argued that, to deflect the a-particle
292 backwards, it must experience a large repulsive force. This force could

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 Atoms
be provided if the greater part of the
mass of the atom and its positive charge
were concentrated tightly at its centre.
Then the incoming a-particle could get
very close to the positive charge without
penetrating it, and such a close
encounter would result in a large
deflection. This agreement supported
the hypothesis of the nuclear atom. This
is why Rutherford is credited with the
discovery of the nucleus.
In Rutherford’s nuclear model of
the atom, the entire positive charge and
most of the mass of the atom are
concentrated in the nucleus with the
electrons some distance away. The
electrons would be moving in orbits
about the nucleus just as the planets FIGURE 12.3 Experimental data points (shown by
dots) on scattering of a-particles by a thin foil at
do around the sun. Rutherford’s
different angles obtained by Geiger and Marsden
experiments suggested the size of using the setup shown in Figs. 12.1 and
the nucleus to be about 10–15 m to 12.2. Rutherford’s nuclear model predicts the solid
10–14 m. From kinetic theory, the size curve which is seen to be in good agreement with
of an atom was known to be 10–10 m, experiment.
about 10,000 to 100,000 times larger
than the size of the nucleus (see Chapter 10, Section 10.6 in Class XI
Physics textbook). Thus, the electrons would seem to be at a distance
from the nucleus of about 10,000 to 100,000 times the size of the nucleus
itself. Thus, most of an atom is empty space. With the atom being largely
empty space, it is easy to see why most a-particles go right through a
thin metal foil. However, when a-particle happens to come near a nucleus,
the intense electric field there scatters it through a large angle. The atomic
electrons, being so light, do not appreciably affect the a-particles.
The scattering data shown in Fig. 12.3 can be analysed by employing
Rutherford’s nuclear model of the atom. As the gold foil is very thin, it
can be assumed that a-particles will suffer not more than one scattering
during their passage through it. Therefore, computation of the trajectory
of an alpha-particle scattered by a single nucleus is enough. Alpha-
particles are nuclei of helium atoms and, therefore, carry two units, 2e,
of positive charge and have the mass of the helium atom. The charge of
the gold nucleus is Ze, where Z is the atomic number of the atom; for
gold Z = 79. Since the nucleus of gold is about 50 times heavier than an
a-particle, it is reasonable to assume that it remains stationary
throughout the scattering process. Under these assumptions, the
trajectory of an alpha-particle can be computed employing Newton’s
second law of motion and the Coulomb’s law for electrostatic
force of repulsion between the alpha-particle and the positively
charged nucleus. 293

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 Physics
The magnitude of this force is
1 (2e )( Ze )
F= (12.1)
4πε 0 r2
where r is the distance between the a-particle and the nucleus. The force
is directed along the line joining the a-particle and the nucleus. The
magnitude and direction of the force on an a-particle continuously
changes as it approaches the nucleus and recedes away from it.

12.2.1 Alpha-particle trajectory

The trajectory traced by an a-particle depends on the impact parameter,
b of collision. The impact parameter is the perpendicular distance of the
initial velocity vector of the a-particle from the centre of the nucleus (Fig.
12.4). A given beam of a-particles has a
distribution of impact parameters b, so that
the beam is scattered in various directions
with different probabilities (Fig. 12.4). (In
a beam, all particles have nearly same
kinetic energy.) It is seen that an a-particle
close to the nucleus (small impact
parameter) suffers large scattering. In case
of head-on collision, the impact parameter
is minimum and the a-particle rebounds
back (q @ p). For a large impact parameter,
the a-particle goes nearly undeviated and
has a small deflection (q @ 0).
FIGURE 12.4 Trajectory of a-particles in the The fact that only a small fraction of the
coulomb field of a target nucleus. The impact
number of incident particles rebound back
parameter, b and scattering angle q
are also depicted.
indicates that the number of a-particles
undergoing head on collision is small. This,
in turn, implies that the mass and positive charge of the atom is
concentrated in a small volume. Rutherford scattering therefore, is a
powerful way to determine an upper limit to the size of the nucleus.

Example 12.1 In the Rutherford’s nuclear model of the atom, the

nucleus (radius about 10–15 m) is analogous to the sun about which
the electron move in orbit (radius » 10–10 m) like the earth orbits
around the sun. If the dimensions of the solar system had the same
proportions as those of the atom, would the earth be closer to or
farther away from the sun than actually it is ? The radius of earth’s
orbit is about 1.5 ´ 1011 m. The radius of sun is taken as 7 ´ 108 m.
Solution The ratio of the radius of electron’s orbit to the radius of
nucleus is (10–10 m)/(10–15 m) = 105, that is, the radius of the electron’s
orbit is 105 times larger than the radius of nucleus. If the radius of
the earth’s orbit around the sun were 105 times larger than the radius
EXAMPLE 12.1

of the sun, the radius of the earth’s orbit would be 105 ´ 7 ´ 108 m =
7 ´ 1013 m. This is more than 100 times greater than the actual
orbital radius of earth. Thus, the earth would be much farther away
from the sun.
It implies that an atom contains a much greater fraction of empty
294 space than our solar system does.

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 Atoms

Example 12.2 In a Geiger-Marsden experiment, what is the distance

of closest approach to the nucleus of a 7.7 MeV a-particle before it
comes momentarily to rest and reverses its direction?
Solution The key idea here is that throughout the scattering process,
the total mechanical energy of the system consisting of an a-particle
and a gold nucleus is conserved. The system’s initial mechanical
energy is Ei, before the particle and nucleus interact, and it is equal
to its mechanical energy Ef when the a-particle momentarily stops.
The initial energy E i is just the kinetic energy K of the incoming
a- particle. The final energy Ef is just the electric potential energy U
of the system. The potential energy U can be calculated from
Eq. (12.1).
Let d be the centre-to-centre distance between the a-particle and
the gold nucleus when the a-particle is at its stopping point. Then
we can write the conservation of energy Ei = Ef as
1 (2e )( Ze ) 2 Ze 2
K = =
4 πε 0 d 4 πε 0d
Thus the distance of closest approach d is given by
2 Ze 2
d=
4 πε 0 K
The maximum kinetic energy found in a-particles of natural origin is
7.7 MeV or 1.2 × 10–12 J. Since 1/4pe0 = 9.0 × 109 N m2/C2. Therefore
with e = 1.6 × 10–19 C, we have,
(2)(9.0 × 109 Nm 2 / C 2 )(1.6 × 10 –19 C )2 Z
d =
1.2 × 10 −12 J
–16
= 3.84 × 10 Z m
The atomic number of foil material gold is Z = 79, so that
d (Au) = 3.0 × 10–14 m = 30 fm. (1 fm (i.e. fermi) = 10–15 m.)
The radius of gold nucleus is, therefore, less than 3.0 × 10–14 m. This
EXAMPLE 12.2

is not in very good agreement with the observed result as the actual
radius of gold nucleus is 6 fm. The cause of discrepancy is that the
distance of closest approach is considerably larger than the sum of
the radii of the gold nucleus and the a-particle. Thus, the a-particle
reverses its motion without ever actually touching the gold nucleus.

12.2.2 Electron orbits

The Rutherford nuclear model of the atom which involves classical
concepts, pictures the atom as an electrically neutral sphere consisting
of a very small, massive and positively charged nucleus at the centre
surrounded by the revolving electrons in their respective dynamically
stable orbits. The electrostatic force of attraction, Fe between the revolving
electrons and the nucleus provides the requisite centripetal force (Fc ) to
keep them in their orbits. Thus, for a dynamically stable orbit in a
hydrogen atom
Fe = Fc
1 e 2 mv 2
= (12.2) 295
4 πε 0 r 2 r

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 Physics
Thus the relation between the orbit radius and the electron
velocity is
e2
r = (12.3)
4 πε 0mv 2
The kinetic energy (K ) and electrostatic potential energy (U ) of the electron
in hydrogen atom are
1 e2 e2
K= mv 2 = and U = −
2 8πε 0r 4 πε 0r
(The negative sign in U signifies that the electrostatic force is in the –r
direction.) Thus the total energy E of the electron in a hydrogen atom is
e2 e2
E = K +U = −
8πε 0r 4 πε 0r

e2
=− (12.4)
8πε 0r
The total energy of the electron is negative. This implies the fact that
the electron is bound to the nucleus. If E were positive, an electron will
not follow a closed orbit around the nucleus.

Example 12.3 It is found experimentally that 13.6 eV energy is

required to separate a hydrogen atom into a proton and an electron.
Compute the orbital radius and the velocity of the electron in a
hydrogen atom.
Solution Total energy of the electron in hydrogen atom is –13.6 eV =
–13.6 × 1.6 × 10–19 J = –2.2 ×10–18 J. Thus from Eq. (12.4), we have
e2
E=− = −2.2 × 10−18 J
8πε 0r
This gives the orbital radius
e2 (9 × 109 N m 2/C2 )(1.6 × 10−19 C)2
r =− =−
8πε 0 E (2)(–2.2 × 10 −18 J)
EXAMPLE 12.3

= 5.3 × 10–11 m.
The velocity of the revolving electron can be computed from Eq. (12.3)
with m = 9.1 × 10–31 kg,
e
v= = 2.2 × 106 m/s.
4 πε0mr

12.3 ATOMIC SPECTRA

As mentioned in Section 12.1, each element has a characteristic spectrum
of radiation, which it emits. When an atomic gas or vapour is excited at
low pressure, usually by passing an electric current through it, the emitted
radiation has a spectrum which contains certain specific wavelengths
296 only. A spectrum of this kind is termed as emission line spectrum and it

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 Atoms
consists of bright lines on a
dark background. The
spectrum emitted by atomic
hydrogen is shown in
Fig. 12.5. Study of emission
line spectra of a material can
therefore serve as a type of
“fingerprint” for identification
of the gas. When white light
passes through a gas and we
analyse the transmitted light
using a spectrometer we find
some dark lines in the FIGURE 12.5 Emission lines in the spectrum of hydrogen.
spectrum. These dark lines
correspond precisely to those wavelengths which were found in the
emission line spectrum of the gas. This is called the absorption spectrum
of the material of the gas.

12.4 BOHR MODEL OF THE HYDROGEN

ATOM
The model of the atom proposed by Rutherford assumes
that the atom, consisting of a central nucleus and
revolving electron is stable much like sun-planet system
which the model imitates. However, there are some

NIELS HENRIK DAVID BOHR (1885 – 1962)

fundamental differences between the two situations.
While the planetary system is held by gravitational
force, the nucleus-electron system being charged
objects, interact by Coulomb’s Law of force. We know
that an object which moves in a circle is being
constantly accelerated – the acceleration being
centripetal in nature. According to classical
electromagnetic theory, an accelerating charged particle
emits radiation in the form of electromagnetic waves.
The energy of an accelerating electron should therefore, Niels Henrik David Bohr
continuously decrease. The electron would spiral (1885 – 1962) Danish
inward and eventually fall into the nucleus (Fig. 12.6). physicist who explained the
Thus, such an atom can not be stable. Further, spectrum of hydrogen atom
according to the classical electromagnetic theory, the based on quantum ideas.
frequency of the electromagnetic waves emitted by the He gave a theory of nuclear
fission based on the liquid-
revolving electrons is equal to the frequency of
drop model of nucleus.
revolution. As the electrons spiral inwards, their angular
Bohr contributed to the
velocities and hence their frequencies would change clarification of conceptual
continuously, and so will the frequency of the light problems in quantum
emitted. Thus, they would emit a continuous spectrum, mechanics, in particular by
in contradiction to the line spectrum actually observed. proposing the comple-
Clearly Rutherford model tells only a part of the story mentary principle.
implying that the classical ideas are not sufficient to
explain the atomic structure. 297

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 Physics

FIGURE 12.6 An accelerated atomic electron must spiral into the

nucleus as it loses energy.

Example 12.4 According to the classical electromagnetic theory,

calculate the initial frequency of the light emitted by the electron
revolving around a proton in hydrogen atom.
Solution From Example 12.3 we know that velocity of electron moving
around a proton in hydrogen atom in an orbit of radius 5.3 × 10–11 m
is 2.2 × 10–6 m/s. Thus, the frequency of the electron moving around
the proton is
v 2.2 × 106 m s −1
ν= =
(
2πr 2π 5.3 × 10 −11 m )
EXAMPLE 12.4

15
» 6.6 × 10 Hz.
According to the classical electromagnetic theory we know that the
frequency of the electromagnetic waves emitted by the revolving
electrons is equal to the frequency of its revolution around the nucleus.
Thus the initial frequency of the light emitted is 6.6 × 1015 Hz.

It was Niels Bohr (1885 – 1962) who made certain modifications in

this model by adding the ideas of the newly developing quantum
hypothesis. Niels Bohr studied in Rutherford’s laboratory for several
months in 1912 and he was convinced about the validity of Rutherford
nuclear model. Faced with the dilemma as discussed above, Bohr, in
1913, concluded that in spite of the success of electromagnetic theory in
explaining large-scale phenomena, it could not be applied to the processes
at the atomic scale. It became clear that a fairly radical departure from
the established principles of classical mechanics and electromagnetism
would be needed to understand the structure of atoms and the relation
of atomic structure to atomic spectra. Bohr combined classical and early
quantum concepts and gave his theory in the form of three postulates.
These are :
(i) Bohr’s first postulate was that an electron in an atom could revolve
in certain stable orbits without the emission of radiant energy,
contrary to the predictions of electromagnetic theory. According to
298 this postulate, each atom has certain definite stable states in which it

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 Atoms
can exist, and each possible state has definite total energy. These are
called the stationary states of the atom.
(ii) Bohr’s second postulate defines these stable orbits. This postulate
states that the electron revolves around the nucleus only in those
orbits for which the angular momentum is some integral multiple of
h/2p where h is the Planck’s constant (= 6.6 ´ 10–34 J s). Thus the
angular momentum (L) of the orbiting electron is quantised. That is
L = nh/2p (12.5)
(iii) Bohr’s third postulate incorporated into atomic theory the early
quantum concepts that had been developed by Planck and Einstein.
It states that an electron might make a transition from one of its
specified non-radiating orbits to another of lower energy. When it
does so, a photon is emitted having energy equal to the energy
difference between the initial and final states. The frequency of the
emitted photon is then given by
hn = Ei – Ef (12.6)
where Ei and Ef are the energies of the initial and final states and Ei > Ef .
For a hydrogen atom, Eq. (12.4) gives the expression to determine
the energies of different energy states. But then this equation requires
the radius r of the electron orbit. To calculate r, Bohr’s second postulate
about the angular momentum of the electron–the quantisation
condition – is used.
The radius of nth possible orbit thus found is

4 πε 0
2
n2 h
rn = (12.7)
m 2π e2
The total energy of the electron in the stationary states of the hydrogen
atom can be obtained by substituting the value of orbital radius in
Eq. (12.4) as
2π
2
e2 m e2
En = −
8 πε 0 n2 h 4 πε 0

me 4
or En = − (12.8)
8n 2 ε 02h 2
Substituting values, Eq. (12.8) yields
2.18 × 10 −18
En = − J (12.9)
n2
Atomic energies are often expressed in electron volts (eV) rather than
joules. Since 1 eV = 1.6 ´ 10–19 J, Eq. (12.9) can be rewritten as
13.6
En = − eV (12.10)
n2
The negative sign of the total energy of an electron moving in an orbit
means that the electron is bound with the nucleus. Energy will thus be
required to remove the electron from the hydrogen atom to a distance
infinitely far away from its nucleus (or proton in hydrogen atom). 299

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 Physics
12.4.1 Energy levels
The energy of an atom is the least (largest negative value)
when its electron is revolving in an orbit closest to the
nucleus i.e., the one for which n = 1. For n = 2, 3, ... the
absolute value of the energy E is smaller, hence the energy
is progressively larger in the outer orbits. The lowest state
of the atom, called the ground state, is that of the lowest
energy, with the electron revolving in the orbit of smallest
radius, the Bohr radius, a 0. The energy of this state (n = 1),
E1 is –13.6 eV. Therefore, the minimum energy required to
free the electron from the ground state of the hydrogen atom
is 13.6 eV. It is called the ionisation energy of the hydrogen
atom. This prediction of the Bohr’s model is in excellent
agreement with the experimental value of ionisation energy.
At room temperature, most of the hydrogen atoms are
in ground state. When a hydrogen atom receives energy
by processes such as electron collisions, the atom may
acquire sufficient energy to raise the electron to higher
energy states. The atom is then said to be in an excited
state. From Eq. (12.10), for n = 2; the energy E2 is
–3.40 eV. It means that the energy required to excite an
electron in hydrogen atom to its first excited state, is an
FIGURE 12.7 The energy level
diagram for the hydrogen atom.
energy equal to E2 – E1 = –3.40 eV – (–13.6) eV = 10.2 eV.
The electron in a hydrogen atom Similarly, E3 = –1.51 eV and E3 – E1 = 12.09 eV, or to excite
at room temperature spends the hydrogen atom from its ground state (n = 1) to second
most of its time in the ground excited state (n = 3), 12.09 eV energy is required, and so
state. To ionise a hydrogen on. From these excited states the electron can then fall back
atom an electron from the to a state of lower energy, emitting a photon in the process.
ground state, 13.6 eV of energy Thus, as the excitation of hydrogen atom increases (that is
must be supplied. (The horizontal as n increases) the value of minimum energy required to
lines specify the presence of free the electron from the excited atom decreases.
allowed energy states.) The energy level diagram* for the stationary states of a
hydrogen atom, computed from Eq. (12.10), is given in
Fig. 12.7. The principal quantum number n labels the stationary
states in the ascending order of energy. In this diagram, the highest
energy state corresponds to n = ¥ in Eq, (12.10) and has an energy
of 0 eV. This is the energy of the atom when the electron is
completely removed (r = ¥) from the nucleus and is at rest. Observe how
the energies of the excited states come closer and closer together as
n increases.

12.5 THE LINE SPECTRA OF THE HYDROGEN ATOM

According to the third postulate of Bohr’s model, when an atom makes a
transition from the higher energy state with quantum number ni to the
lower energy state with quantum number nf (nf < ni ), the difference of
energy is carried away by a photon of frequency nif such that

* An electron can have any total energy above E = 0 eV. In such situations the
300 electron is free. Thus there is a continuum of energy states above E = 0 eV, as
shown in Fig. 12.7.

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 Atoms
hvif = Eni – Enf (12.11)
Since both nf and ni are integers, this immediately shows that in
transitions between different atomic levels, light is radiated in various
discrete frequencies.
The various lines in the atomic spectra are produced when electrons
jump from higher energy state to a lower energy state and photons are
emitted. These spectral lines are called emission lines. But when an atom
absorbs a photon that has precisely the same energy needed by the
electron in a lower energy state to make transitions to a higher energy
state, the process is called absorption. Thus if photons with a continuous
range of frequencies pass through a rarefied gas and then are analysed
with a spectrometer, a series of dark spectral absorption lines appear in
the continuous spectrum. The dark lines indicate the frequencies that
have been absorbed by the atoms of the gas.
The explanation of the hydrogen atom spectrum provided by Bohr’s
model was a brilliant achievement, which greatly stimulated progress
towards the modern quantum theory. In 1922, Bohr was awarded Nobel
Prize in Physics.

12.6 DE BROGLIE’S EXPLANATION OF BOHR’S

SECOND POSTULATE OF QUANTISATION
Of all the postulates, Bohr made in his model of the atom,
perhaps the most puzzling is his second postulate. It states
that the angular momentum of the electron orbiting around
the nucleus is quantised (that is, Ln = nh/2p; n = 1, 2, 3 …).
Why should the angular momentum have only those values
that are integral multiples of h/2p? The French physicist Louis
de Broglie explained this puzzle in 1923, ten years after Bohr
proposed his model.
We studied, in Chapter 11, about the de Broglie’s
hypothesis that material particles, such as electrons, also
have a wave nature. C. J. Davisson and L. H. Germer later
experimentally verified the wave nature of electrons in 1927.
Louis de Broglie argued that the electron in its circular orbit,
as proposed by Bohr, must be seen as a particle wave. In
analogy to waves travelling on a string, particle waves too
can lead to standing waves under resonant conditions. From FIGURE 12.8 A standing wave
Chapter 14 of Class XI Physics textbook, we know that when is shown on a circular orbit
a string is plucked, a vast number of wavelengths are excited. where four de Broglie
However only those wavelengths survive which have nodes wavelengths fit into the
circumference of the orbit.
at the ends and form the standing wave in the string. It means
that in a string, standing waves are formed when the total distance
travelled by a wave down the string and back is one wavelength, two
wavelengths, or any integral number of wavelengths. Waves with other
wavelengths interfere with themselves upon reflection and their
amplitudes quickly drop to zero. For an electron moving in nth circular
orbit of radius rn, the total distance is the circumference of the orbit,
2prn. Thus 301

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 Physics
2p rn = nl, n = 1, 2, 3... (12.12)
Figure 12.8 illustrates a standing particle wave on a circular orbit
for n = 4, i.e., 2prn = 4l, where l is the de Broglie wavelength of the electron
moving in nth orbit. From Chapter 11, we have l = h/p, where p is the
magnitude of the electron’s momentum. If the speed of the electron is
much less than the speed of light, the momentum is mvn. Thus, l = h/
mvn. From Eq. (12.12), we have
2p rn = n h/mvn or m vn rn = nh/2p
This is the quantum condition proposed by Bohr for the angular
momentum of the electron [Eq. (12.15)]. In Section 12.5, we saw that
this equation is the basis of explaining the discrete orbits and energy
levels in hydrogen atom. Thus de Broglie hypothesis provided an
explanation for Bohr’s second postulate for the quantisation of angular
momentum of the orbiting electron. The quantised electron orbits and
energy states are due to the wave nature of the electron and only resonant
standing waves can persist.
Bohr’s model, involving classical trajectory picture (planet-like electron
orbiting the nucleus), correctly predicts the gross features of the
hydrogenic atoms*, in particular, the frequencies of the radiation emitted
or selectively absorbed. This model however has many limitations.
Some are:
(i) The Bohr model is applicable to hydrogenic atoms. It cannot be
extended even to mere two electron atoms such as helium. The analysis
of atoms with more than one electron was attempted on the lines of
Bohr’s model for hydrogenic atoms but did not meet with any success.
Difficulty lies in the fact that each electron interacts not only with the
positively charged nucleus but also with all other electrons.
The formulation of Bohr model involves electrical force between
positively charged nucleus and electron. It does not include the
electrical forces between electrons which necessarily appear in
multi-electron atoms.
(ii) While the Bohr’s model correctly predicts the frequencies of the light
emitted by hydrogenic atoms, the model is unable to explain the
relative intensities of the frequencies in the spectrum. In emission
spectrum of hydrogen, some of the visible frequencies have weak
intensity, others strong. Why? Experimental observations depict that
some transitions are more favoured than others. Bohr’s model is
unable to account for the intensity variations.
Bohr’s model presents an elegant picture of an atom and cannot be
generalised to complex atoms. For complex atoms we have to use a new
and radical theory based on Quantum Mechanics, which provides a more
complete picture of the atomic structure.

* Hydrogenic atoms are the atoms consisting of a nucleus with positive charge
+Ze and a single electron, where Z is the proton number. Examples are hydrogen
atom, singly ionised helium, doubly ionised lithium, and so forth. In these
302 atoms more complex electron-electron interactions are nonexistent.

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 Atoms

SUMMARY

1. Atom, as a whole, is electrically neutral and therefore contains equal

amount of positive and negative charges.
2. In Thomson’s model, an atom is a spherical cloud of positive charges
with electrons embedded in it.
3. In Rutherford’s model, most of the mass of the atom and all its positive
charge are concentrated in a tiny nucleus (typically one by ten thousand
the size of an atom), and the electrons revolve around it.
4. Rutherford nuclear model has two main difficulties in explaining the
structure of atom: (a) It predicts that atoms are unstable because the
accelerated electrons revolving around the nucleus must spiral into
the nucleus. This contradicts the stability of matter. (b) It cannot
explain the characteristic line spectra of atoms of different elements.
5. Atoms of most of the elements are stable and emit characteristic
spectrum. The spectrum consists of a set of isolated parallel lines
termed as line spectrum. It provides useful information about the
atomic structure.
6. To explain the line spectra emitted by atoms, as well as the stability
of atoms, Niel’s Bohr proposed a model for hydrogenic (single elctron)
atoms. He introduced three postulates and laid the foundations of
quantum mechanics:
(a) In a hydrogen atom, an electron revolves in certain stable orbits
(called stationary orbits) without the emission of radiant energy.
(b) The stationary orbits are those for which the angular momentum
is some integral multiple of h/2p. (Bohr’s quantisation condition.)
That is L = nh/2p, where n is an integer called the principal
quantum number.
(c) The third postulate states that an electron might make a transition
from one of its specified non-radiating orbits to another of lower
energy. When it does so, a photon is emitted having energy equal
to the energy difference between the initial and final states. The
frequency (n) of the emitted photon is then given by
hn = Ei – Ef
An atom absorbs radiation of the same frequency the atom emits,
in which case the electron is transferred to an orbit with a higher
value of n.
Ei + hn = Ef
7. As a result of the quantisation condition of angular momentum, the
electron orbits the nucleus at only specific radii. For a hydrogen atom
it is given by
 n 2   h  4 πε 0
2

rn =    
 m   2π  e 2
The total energy is also quantised:
me 4
En = −
8n 2 ε02h 2
= –13.6 eV/n2
The n = 1 state is called ground state. In hydrogen atom the ground
state energy is –13.6 eV. Higher values of n correspond to excited
states (n > 1). Atoms are excited to these higher states by collisions
with other atoms or electrons or by absorption of a photon of right
frequency. 303

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 Physics
8. de Broglie’s hypothesis that electrons have a wavelength l = h/mv gave
an explanation for Bohr’s quantised orbits by bringing in the wave-
particle duality. The orbits correspond to circular standing waves in
which the circumference of the orbit equals a whole number of
wavelengths.
9. Bohr’s model is applicable only to hydrogenic (single electron) atoms.
It cannot be extended to even two electron atoms such as helium.
This model is also unable to explain for the relative intensities of the
frequencies emitted even by hydrogenic atoms.

POINTS TO PONDER

1. Both the Thomson’s as well as the Rutherford’s models constitute an

unstable system. Thomson’s model is unstable electrostatically, while
Rutherford’s model is unstable because of electromagnetic radiation
of orbiting electrons.
2. What made Bohr quantise angular momentum (second postulate) and
not some other quantity? Note, h has dimensions of angular
momentum, and for circular orbits, angular momentum is a very
relevant quantity. The second postulate is then so natural!
3. The orbital picture in Bohr’s model of the hydrogen atom was
inconsistent with the uncertainty principle. It was replaced by modern
quantum mechanics in which Bohr’s orbits are regions where the
electron may be found with large probability.
4. Unlike the situation in the solar system, where planet-planet
gravitational forces are very small as compared to the gravitational
force of the sun on each planet (because the mass of the sun is so
much greater than the mass of any of the planets), the electron-electron
electric force interaction is comparable in magnitude to the electron-
nucleus electrical force, because the charges and distances are of the
same order of magnitude. This is the reason why the Bohr’s model
with its planet-like electron is not applicable to many electron atoms.
5. Bohr laid the foundation of the quantum theory by postulating specific
orbits in which electrons do not radiate. Bohr’s model include only
one quantum number n. The new theory called quantum mechanics
supportes Bohr’s postulate. However in quantum mechanics (more
generally accepted), a given energy level may not correspond to just
one quantum state. For example, a state is characterised by four
quantum numbers (n, l, m, and s), but for a pure Coulomb potential
(as in hydrogen atom) the energy depends only on n.
6. In Bohr model, contrary to ordinary classical expectation, the frequency
of revolution of an electron in its orbit is not connected to the frequency
of spectral line. The later is the difference between two orbital energies
divided by h. For transitions between large quantum numbers (n to n
– 1, n very large), however, the two coincide as expected.
7. Bohr’s semiclassical model based on some aspects of classical physics
and some aspects of modern physics also does not provide a true picture
of the simplest hydrogenic atoms. The true picture is quantum
mechanical affair which differs from Bohr model in a number of
fundamental ways. But then if the Bohr model is not strictly correct,
why do we bother about it? The reasons which make Bohr’s model
still useful are:
304

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 Atoms

(i) The model is based on just three postulates but accounts for almost
all the general features of the hydrogen spectrum.
(ii) The model incorporates many of the concepts we have learnt in
classical physics.
(iii) The model demonstrates how a theoretical physicist occasionally
must quite literally ignore certain problems of approach in hopes
of being able to make some predictions. If the predictions of the
theory or model agree with experiment, a theoretician then must
somehow hope to explain away or rationalise the problems that
were ignored along the way.

EXERCISES
12.1 Choose the correct alternative from the clues given at the end of
the each statement:
(a) The size of the atom in Thomson’s model is .......... the atomic
size in Rutherford’s model. (much greater than/no different
from/much less than.)
(b) In the ground state of .......... electrons are in stable equilibrium,
while in .......... electrons always experience a net force.
(Thomson’s model/ Rutherford’s model.)
(c) A classical atom based on .......... is doomed to collapse.
(Thomson’s model/ Rutherford’s model.)
(d) An atom has a nearly continuous mass distribution in a ..........
but has a highly non-uniform mass distribution in ..........
(Thomson’s model/ Rutherford’s model.)
(e) The positively charged part of the atom possesses most of the
mass in .......... (Rutherford’s model/both the models.)
12.2 Suppose you are given a chance to repeat the alpha-particle
scattering experiment using a thin sheet of solid hydrogen in place
of the gold foil. (Hydrogen is a solid at temperatures below 14 K.)
What results do you expect?
12.3 A difference of 2.3 eV separates two energy levels in an atom. What
is the frequency of radiation emitted when the atom make a
transition from the upper level to the lower level?
12.4 The ground state energy of hydrogen atom is –13.6 eV. What are the
kinetic and potential energies of the electron in this state?
12.5 A hydrogen atom initially in the ground level absorbs a photon,
which excites it to the n = 4 level. Determine the wavelength and
frequency of photon.
12.6 (a) Using the Bohr’s model calculate the speed of the electron in a
hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital
period in each of these levels.
12.7 The radius of the innermost electron orbit of a hydrogen atom is
5.3×10–11 m. What are the radii of the n = 2 and n =3 orbits?
12.8 A 12.5 eV electron beam is used to bombard gaseous hydrogen at
room temperature. What series of wavelengths will be emitted?
12.9 In accordance with the Bohr’s model, find the quantum number
that characterises the earth’s revolution around the sun in an orbit
of radius 1.5 × 1011 m with orbital speed 3 × 104 m/s. (Mass of earth 305
= 6.0 × 1024 kg.)

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Chapter Thirteen

NUCLEI

13.1 INTRODUCTION
In the previous chapter, we have learnt that in every atom, the positive
charge and mass are densely concentrated at the centre of the atom
forming its nucleus. The overall dimensions of a nucleus are much smaller
than those of an atom. Experiments on scattering of a -particles
demonstrated that the radius of a nucleus was smaller than the radius
of an atom by a factor of about 104. This means the volume of a nucleus
is about 10–12 times the volume of the atom. In other words, an atom is
almost empty. If an atom is enlarged to the size of a classroom, the nucleus
would be of the size of pinhead. Nevertheless, the nucleus contains most
(more than 99.9%) of the mass of an atom.
Does the nucleus have a structure, just as the atom does? If so, what
are the constituents of the nucleus? How are these held together? In this
chapter, we shall look for answers to such questions. We shall discuss
various properties of nuclei such as their size, mass and stability, and
also associated nuclear phenomena such as radioactivity, fission and fusion.

13.2 ATOMIC MASSES AND COMPOSITION OF NUCLEUS

The mass of an atom is very small, compared to a kilogram; for example,
the mass of a carbon atom, 12C, is 1.992647 × 10–26 kg. Kilogram is not
306 a very convenient unit to measure such small quantities. Therefore, a

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 Nuclei
different mass unit is used for expressing atomic masses. This unit is the
atomic mass unit (u), defined as 1/12th of the mass of the carbon (12C)
atom. According to this definition
12
mass of one C atom
1u =
12

1.992647 ×10−26 kg
=
12
= 1.660539 × 10 −27 kg (13.1)
The atomic masses of various elements expressed in atomic mass
unit (u) are close to being integral multiples of the mass of a hydrogen
atom. There are, however, many striking exceptions to this rule. For
example, the atomic mass of chlorine atom is 35.46 u.
Accurate measurement of atomic masses is carried out with a mass
spectrometer, The measurement of atomic masses reveals the existence
of different types of atoms of the same element, which exhibit the same
chemical properties, but differ in mass. Such atomic species of the same
element differing in mass are called isotopes. (In Greek, isotope means
the same place, i.e. they occur in the same place in the periodic table of
elements.) It was found that practically every element consists of a mixture
of several isotopes. The relative abundance of different isotopes differs
from element to element. Chlorine, for example, has two isotopes having
masses 34.98 u and 36.98 u, which are nearly integral multiples of the
mass of a hydrogen atom. The relative abundances of these isotopes are
75.4 and 24.6 per cent, respectively. Thus, the average mass of a chlorine
atom is obtained by the weighted average of the masses of the two
isotopes, which works out to be
75.4 × 34.98 + 24.6 × 36.98
=
100
= 35.47 u
which agrees with the atomic mass of chlorine.
Even the lightest element, hydrogen has three isotopes having masses
1.0078 u, 2.0141 u, and 3.0160 u. The nucleus of the lightest atom of
hydrogen, which has a relative abundance of 99.985%, is called the
proton. The mass of a proton is
m p = 1.00727 u = 1.67262 × 10 −27 kg (13.2)
This is equal to the mass of the hydrogen atom (= 1.00783u), minus
the mass of a single electron (me = 0.00055 u). The other two isotopes of
hydrogen are called deuterium and tritium. Tritium nuclei, being
unstable, do not occur naturally and are produced artificially in
laboratories.
The positive charge in the nucleus is that of the protons. A proton
carries one unit of fundamental charge and is stable. It was earlier thought
that the nucleus may contain electrons, but this was ruled out later using
arguments based on quantum theory. All the electrons of an atom are
outside the nucleus. We know that the number of these electrons outside
the nucleus of the atom is Z, the atomic number. The total charge of the 307

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atomic electrons is thus (–Ze), and since the atom is neutral, the charge
of the nucleus is (+Ze). The number of protons in the nucleus of the atom
is, therefore, exactly Z, the atomic number.
Discovery of Neutron
Since the nuclei of deuterium and tritium are isotopes of hydrogen, they
must contain only one proton each. But the masses of the nuclei of
hydrogen, deuterium and tritium are in the ratio of 1:2:3. Therefore, the
nuclei of deuterium and tritium must contain, in addition to a proton,
some neutral matter. The amount of neutral matter present in the nuclei
of these isotopes, expressed in units of mass of a proton, is approximately
equal to one and two, respectively. This fact indicates that the nuclei of
atoms contain, in addition to protons, neutral matter in multiples of a
basic unit. This hypothesis was verified in 1932 by James Chadwick
who observed emission of neutral radiation when beryllium nuclei were
bombarded with alpha-particles (a-particles are helium nuclei, to be
discussed in a later section). It was found that this neutral radiation
could knock out protons from light nuclei such as those of helium, carbon
and nitrogen. The only neutral radiation known at that time was photons
(electromagnetic radiation). Application of the principles of conservation
of energy and momentum showed that if the neutral radiation consisted
of photons, the energy of photons would have to be much higher than is
available from the bombardment of beryllium nuclei with a-particles.
The clue to this puzzle, which Chadwick satisfactorily solved, was to
assume that the neutral radiation consists of a new type of neutral
particles called neutrons. From conservation of energy and momentum,
he was able to determine the mass of new particle ‘as very nearly the
same as mass of proton’.
The mass of a neutron is now known to a high degree of accuracy. It is
m n = 1.00866 u = 1.6749×10–27 kg (13.3)
Chadwick was awarded the 1935 Nobel Prize in Physics for his
discovery of the neutron.
A free neutron, unlike a free proton, is unstable. It decays into a
proton, an electron and a antineutrino (another elementary particle), and
has a mean life of about 1000s. It is, however, stable inside the nucleus.
The composition of a nucleus can now be described using the following
terms and symbols:
Z - atomic number = number of protons [13.4(a)]
N - neutron number = number of neutrons [13.4(b)]
A - mass number = Z + N
= total number of protons and neutrons [13.4(c)]
One also uses the term nucleon for a proton or a neutron. Thus the
number of nucleons in an atom is its mass number A.
Nuclear species or nuclides are shown by the notation ZA X where X is
the chemical symbol of the species. For example, the nucleus of gold is
denoted by 197
79 Au
. It contains 197 nucleons, of which 79 are protons
308 and the rest118 are neutrons.

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 Nuclei
The composition of isotopes of an element can now be readily
explained. The nuclei of isotopes of a given element contain the same
number of protons, but differ from each other in their number of neutrons.
Deuterium, 12 H , which is an isotope of hydrogen, contains one proton
and one neutron. Its other isotope tritium, 13 H , contains one proton and
two neutrons. The element gold has 32 isotopes, ranging from A =173 to
A = 204. We have already mentioned that chemical properties of elements
depend on their electronic structure. As the atoms of isotopes have
identical electronic structure they have identical chemical behaviour and
are placed in the same location in the periodic table.
All nuclides with same mass number A are called isobars. For
example, the nuclides 13 H and 32 He are isobars. Nuclides with same
neutron number N but different atomic number Z, for example 198 80 Hg
and 197
79 Au , are called isotones.

13.3 SIZE OF THE NUCLEUS

As we have seen in Chapter 12, Rutherford was the pioneer who
postulated and established the existence of the atomic nucleus. At
Rutherford’s suggestion, Geiger and Marsden performed their classic
experiment: on the scattering of a-particles from thin gold foils. Their
experiments revealed that the distance of closest approach to a gold
nucleus of an a-particle of kinetic energy 5.5 MeV is about 4.0 × 10–14 m.
The scattering of a-particle by the gold sheet could be understood by
Rutherford by assuming that the coulomb repulsive force was solely
responsible for scattering. Since the positive charge is confined to the
nucleus, the actual size of the nucleus has to be less than 4.0 × 10–14 m.
If we use a-particles of higher energies than 5.5 MeV, the distance of
closest approach to the gold nucleus will be smaller and at some point
the scattering will begin to be affected by the short range nuclear forces,
and differ from Rutherford’s calculations. Rutherford’s calculations are
based on pure coulomb repulsion between the positive charges of the a-
particle and the gold nucleus. From the distance at which deviations set
in, nuclear sizes can be inferred.
By performing scattering experiments in which fast electrons, instead
of a-particles, are projectiles that bombard targets made up of various
elements, the sizes of nuclei of various elements have been accurately
measured.
It has been found that a nucleus of mass number A has a radius
R = R 0 A1/3 (13.5)
–15 –15
where R 0 = 1.2 × 10 m (=1.2 fm; 1 fm = 10 m). This means the volume
of the nucleus, which is proportional to R 3 is proportional to A. Thus the
density of nucleus is a constant, independent of A, for all nuclei. Different
nuclei are like a drop of liquid of constant density. The density of nuclear
matter is approximately 2.3 × 1017 kg m–3. This density is very large
compared to ordinary matter, say water, which is 103 kg m–3. This is
understandable, as we have already seen that most of the atom is empty.
Ordinary matter consisting of atoms has a large amount of empty space. 309

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Example 13.1 Given the mass of iron nucleus as 55.85u and A=56,
find the nuclear density?
Solution
mFe = 55.85, u = 9.27 × 10–26 kg
EXAMPLE 13.1

−26
mass 9.27 × 10 1
Nuclear density = = ×
volume (4 π /3)(1.2 × 10 −15 )3 56
= 2.29 × 1017 kg m–3
The density of matter in neutron stars (an astrophysical object) is
comparable to this density. This shows that matter in these objects
has been compressed to such an extent that they resemble a big nucleus.

13.4 MASS-ENERGY AND NUCLEAR BINDING ENERGY

13.4.1 Mass – Energy
Einstein showed from his theory of special relativity that it is necessary
to treat mass as another form of energy. Before the advent of this theory
of special relativity it was presumed that mass and energy were conserved
separately in a reaction. However, Einstein showed that mass is another
form of energy and one can convert mass-energy into other forms of
energy, say kinetic energy and vice-versa.
Einstein gave the famous mass-energy equivalence relation
E = mc 2 (13.6)
Here the energy equivalent of mass m is related by the above equation
and c is the velocity of light in vacuum and is approximately equal to
3×108 m s–1.

Example 13.2 Calculate the energy equivalent of 1 g of substance.

EXAMPLE 13.2

Solution
Energy, E = 10–3 × ( 3 × 108)2 J
E = 10–3 × 9 × 1016 = 9 × 1013 J
Thus, if one gram of matter is converted to energy, there is a release
of enormous amount of energy.

Experimental verification of the Einstein’s mass-energy relation has

been achieved in the study of nuclear reactions amongst nucleons, nuclei,
electrons and other more recently discovered particles. In a reaction the
conservation law of energy states that the initial energy and the final
energy are equal provided the energy associated with mass is also
included. This concept is important in understanding nuclear masses
and the interaction of nuclei with one another. They form the subject
matter of the next few sections.

13.4.2 Nuclear binding energy

In Section 13.2 we have seen that the nucleus is made up of neutrons
and protons. Therefore it may be expected that the mass of the nucleus
310 is equal to the total mass of its individual protons and neutrons. However,

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 Nuclei
the nuclear mass M is found to be always less than this. For example, let
us consider 168 O ; a nucleus which has 8 neutrons and 8 protons. We
have
Mass of 8 neutrons = 8 × 1.00866 u
Mass of 8 protons = 8 × 1.00727 u
Mass of 8 electrons = 8 × 0.00055 u
Therefore the expected mass of 168 O nucleus
= 8 × 2.01593 u = 16.12744 u.
The atomic mass of 168 O found from mass spectroscopy experiments
is seen to be 15.99493 u. Substracting the mass of 8 electrons (8 × 0.00055 u)
from this, we get the experimental mass of 168 O nucleus to be 15.99053 u.
Thus, we find that the mass of the 168 O nucleus is less than the total
mass of its constituents by 0.13691u. The difference in mass of a nucleus
and its constituents, DM, is called the mass defect, and is given by
∆M = [ Zm p + ( A − Z ) m n ] − M (13.7)
What is the meaning of the mass defect? It is here that Einstein’s
equivalence of mass and energy plays a role. Since the mass of the oxygen
nucleus is less that the sum of the masses of its constituents (8 protons
and 8 neutrons, in the unbound state), the equivalent energy of the oxygen
nucleus is less than that of the sum of the equivalent energies of its
constituents. If one wants to break the oxygen nucleus into 8 protons
and 8 neutrons, this extra energy DM c2, has to supplied. This energy
required Eb is related to the mass defect by
Eb = D M c 2 (13.8)

Example 13.3 Find the energy equivalent of one atomic mass unit,
first in Joules and then in MeV. Using this, express the mass defect
of 168 O in MeV/c 2.
Solution
1u = 1.6605 × 10–27 kg
To convert it into energy units, we multiply it by c 2 and find that
energy equivalent = 1.6605 × 10–27 × (2.9979 × 108)2 kg m2/s2
= 1.4924 × 10–10 J
1.4924 ×10 −10
= eV
1.602 × 10 −19
= 0.9315 × 109 eV
= 931.5 MeV
EXAMPLE 13.3

or, 1u = 931.5 MeV/c 2

For 168 O , DM = 0.13691 u = 0.13691×931.5 MeV/c 2
= 127.5 MeV/c 2
16
The energy needed to separate 8 O into its constituents is thus
2
127.5 MeV/c .

If a certain number of neutrons and protons are brought together to

form a nucleus of a certain charge and mass, an energy Eb will be released 311

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in the process. The energy Eb is called the binding energy of the nucleus.
If we separate a nucleus into its nucleons, we would have to supply a
total energy equal to Eb, to those particles. Although we cannot tear
apart a nucleus in this way, the nuclear binding energy is still a convenient
measure of how well a nucleus is held together. A more useful measure
of the binding between the constituents of the nucleus is the binding
energy per nucleon, Ebn, which is the ratio of the binding energy Eb of a
nucleus to the number of the nucleons, A, in that nucleus:
Ebn = Eb / A (13.9)
We can think of binding energy per nucleon as the average energy
per nucleon needed to separate a nucleus into its individual nucleons.
Figure 13.1 is a plot of the
binding energy per nucleon Ebn
versus the mass number A for a
large number of nuclei. We notice
the following main features of
the plot:
(i) the binding energy per
nucleon, Ebn, is practically
constant, i.e. practically
independent of the atomic
number for nuclei of middle
mass number ( 30 < A < 170).
The curve has a maximum of
about 8.75 MeV for A = 56
and has a value of 7.6 MeV
FIGURE 13.1 The binding energy per nucleon for A = 238.
as a function of mass number.
(ii) Ebn is lower for both light
nuclei (A<30) and heavy
nuclei (A>170).
We can draw some conclusions from these two observations:
(i) The force is attractive and sufficiently strong to produce a binding
energy of a few MeV per nucleon.
(ii) The constancy of the binding energy in the range 30 < A < 170 is a
consequence of the fact that the nuclear force is short-ranged. Consider
a particular nucleon inside a sufficiently large nucleus. It will be under
the influence of only some of its neighbours, which come within the
range of the nuclear force. If any other nucleon is at a distance more
than the range of the nuclear force from the particular nucleon it will
have no influence on the binding energy of the nucleon under
consideration. If a nucleon can have a maximum of p neighbours
within the range of nuclear force, its binding energy would be
proportional to p. Let the binding energy of the nucleus be pk, where
k is a constant having the dimensions of energy. If we increase A by
adding nucleons they will not change the binding energy of a nucleon
inside. Since most of the nucleons in a large nucleus reside inside it
and not on the surface, the change in binding energy per nucleon
would be small. The binding energy per nucleon is a constant and is
312 approximately equal to pk. The property that a given nucleon

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 Nuclei
influences only nucleons close to it is also referred to as saturation
property of the nuclear force.
(iii) A very heavy nucleus, say A = 240, has lower binding energy per
nucleon compared to that of a nucleus with A = 120. Thus if a
nucleus A = 240 breaks into two A = 120 nuclei, nucleons get more
tightly bound. This implies energy would be released in the process.
It has very important implications for energy production through
fission, to be discussed later in Section 13.7.1.
(iv) Consider two very light nuclei (A ≤ 10) joining to form a heavier
nucleus. The binding energy per nucleon of the fused heavier nuclei
is more than the binding energy per nucleon of the lighter nuclei.
This means that the final system is more tightly bound than the initial
system. Again energy would be released in such a process of
fusion. This is the energy source of sun, to be discussed later in
Section 13.7.2.

13.5 NUCLEAR FORCE

The force that determines the motion of atomic electrons is the familiar
Coulomb force. In Section 13.4, we have seen that for average mass
nuclei the binding energy per nucleon is approximately 8 MeV, which is
much larger than the binding energy in atoms. Therefore, to bind a
nucleus together there must be a strong attractive force of a totally
different kind. It must be strong enough to overcome the repulsion
between the (positively charged) protons and to bind both protons and
neutrons into the tiny nuclear volume. We have already seen
that the constancy of binding energy per nucleon can be
understood in terms of its short-range. Many features of the
nuclear binding force are summarised below. These are
obtained from a variety of experiments carried out during 1930
to 1950.
(i) The nuclear force is much stronger than the Coulomb force
acting between charges or the gravitational forces between
masses. The nuclear binding force has to dominate over
the Coulomb repulsive force between protons inside the
nucleus. This happens only because the nuclear force is
much stronger than the coulomb force. The gravitational
force is much weaker than even Coulomb force.
FIGURE 13.2 Potential energy
(ii) The nuclear force between two nucleons falls rapidly to
of a pair of nucleons as a
zero as their distance is more than a few femtometres. This function of their separation.
leads to saturation of forces in a medium or a large-sized For a separation greater
nucleus, which is the reason for the constancy of the than r0, the force is attractive
binding energy per nucleon. and for separations less
A rough plot of the potential energy between two nucleons than r0, the force is
as a function of distance is shown in the Fig. 13.2. The strongly repulsive.
potential energy is a minimum at a distance r0 of about
0.8 fm. This means that the force is attractive for distances larger
than 0.8 fm and repulsive if they are separated by distances less
than 0.8 fm. 313

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 Physics
(iii) The nuclear force between neutron-neutron, proton-neutron and
proton-proton is approximately the same. The nuclear force does not
depend on the electric charge.
Unlike Coulomb’s law or the Newton’s law of gravitation there is no
simple mathematical form of the nuclear force.

13.6 RADIOACTIVITY
A. H. Becquerel discovered radioactivity in 1896 purely by accident. While
studying the fluorescence and phosphorescence of compounds irradiated
with visible light, Becquerel observed an interesting phenomenon. After
illuminating some pieces of uranium-potassium sulphate with visible
light, he wrapped them in black paper and separated the package from a
photographic plate by a piece of silver. When, after several hours of
exposure, the photographic plate was developed, it showed blackening
due to something that must have been emitted by the compound and
was able to penetrate both black paper and the silver.
Experiments performed subsequently showed that radioactivity was
a nuclear phenomenon in which an unstable nucleus undergoes a decay.
This is referred to as radioactive decay. Three types of radioactive decay
occur in nature :
(i) a-decay in which a helium nucleus 42 He is emitted;
(ii) b-decay in which electrons or positrons (particles with the same mass
as electrons, but with a charge exactly opposite to that of electron)
are emitted;
(iii) g-decay in which high energy (hundreds of keV or more) photons are
emitted.
Each of these decay will be considered in subsequent sub-sections.

13.7 NUCLEAR ENERGY

The curve of binding energy per nucleon Ebn, given in Fig. 13.1, has
a long flat middle region between A = 30 and A = 170. In this region
the binding energy per nucleon is nearly constant (8.0 MeV). For
the lighter nuclei region, A < 30, and for the heavier nuclei region,
A > 170, the binding energy per nucleon is less than 8.0 MeV, as we
have noted earlier. Now, the greater the binding energy, the less is the
total mass of a bound system, such as a nucleus. Consequently, if nuclei
with less total binding energy transform to nuclei with greater binding
energy, there will be a net energy release. This is what happens when a
heavy nucleus decays into two or more intermediate mass fragments
(fission) or when light nuclei fuse into a havier nucleus (fusion.)
Exothermic chemical reactions underlie conventional energy sources
such as coal or petroleum. Here the energies involved are in the range of
electron volts. On the other hand, in a nuclear reaction, the energy release
is of the order of MeV. Thus for the same quantity of matter, nuclear
sources produce a million times more energy than a chemical source.
Fission of 1 kg of uranium, for example, generates 1014 J of energy;
314 compare it with burning of 1 kg of coal that gives 107 J.

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 Nuclei
13.7.1 Fission
New possibilities emerge when we go beyond natural radioactive decays
and study nuclear reactions by bombarding nuclei with other nuclear
particles such as proton, neutron, a-particle, etc.
A most important neutron-induced nuclear reaction is fission. An
235
example of fission is when a uranium isotope U bombarded with a
92
neutron breaks into two intermediate mass nuclear fragments
1
0n + 235
92 U → 92 U → 56 Ba + 36 Kr + 3 0 n
236 144 89 1
(13.10)
The same reaction can produce other pairs of intermediate mass
fragments
n + 235
1
0 92 U → 92 U → 51 Sb +
236 133 99
41 Nb + 4 10 n (13.11)
Or, as another example,
1
0n + 235
92 U → 54 Xe + 38 Sr + 2 0 n
140 94 1
(13.12)
The fragment products are radioactive nuclei; they emit b particles in
succession to achieve stable end products.
The energy released (the Q value ) in the fission reaction of nuclei like
uranium is of the order of 200 MeV per fissioning nucleus. This is
estimated as follows:
Let us take a nucleus with A = 240 breaking into two fragments each
of A = 120. Then
Ebn for A = 240 nucleus is about 7.6 MeV,
Ebn for the two A = 120 fragment nuclei is about 8.5 MeV.
\ Gain in binding energy for nucleon is about 0.9 MeV.
Hence the total gain in binding energy is 240×0.9 or 216 MeV.
The disintegration energy in fission events first appears as the kinetic
energy of the fragments and neutrons. Eventually it is transferred to the
surrounding matter appearing as heat. The source of energy in nuclear
reactors, which produce electricity, is nuclear fission. The enormous
energy released in an atom bomb comes from uncontrolled nuclear
fission.

13.7.2 Nuclear fusion – energy generation in stars

When two light nuclei fuse to form a larger nucleus, energy is released,
since the larger nucleus is more tightly bound, as seen from the binding
energy curve in Fig.13.1. Some examples of such energy liberating nuclear
fusion reactions are :
H + 11 H → 21 H + e + n + 0.42 MeV
1 +
1
[13.13(a)]

2
1 H + 21 H → 23 He + n + 3.27 MeV [13.13(b)]

2
1 H + 21 H → 31 H + 11 H + 4.03 MeV [13.13(c)]
In the first reaction, two protons combine to form a deuteron and
a positron with a release of 0.42 MeV energy. In reaction [13.13(b)], two 315

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deuterons combine to form the light isotope of helium. In reaction
(13.13c), two deuterons combine to form a triton and a proton. For
fusion to take place, the two nuclei must come close enough so that
attractive short-range nuclear force is able to affect them. However,
since they are both positively charged particles, they experience coulomb
repulsion. They, therefore, must have enough energy to overcome this
coulomb barrier. The height of the barrier depends on the charges and
radii of the two interacting nuclei. It can be shown, for example, that
the barrier height for two protons is ~ 400 keV, and is higher for nuclei
with higher charges. We can estimate the temperature at which two
protons in a proton gas would (averagely) have enough energy to
overcome the coulomb barrier:
(3/2)k T = K ≃ 400 keV, which gives T ~ 3 × 109 K.
When fusion is achieved by raising the temperature of the system so
that particles have enough kinetic energy to overcome the coulomb
repulsive behaviour, it is called thermonuclear fusion.
Thermonuclear fusion is the source of energy output in the interior
of stars. The interior of the sun has a temperature of 1.5×107 K, which
is considerably less than the estimated temperature required for fusion
of particles of average energy. Clearly, fusion in the sun involves protons
whose energies are much above the average energy.
The fusion reaction in the sun is a multi-step process in which the
hydrogen is burned into helium. Thus, the fuel in the sun is the hydrogen
in its core. The proton-proton (p, p) cycle by which this occurs is
represented by the following sets of reactions:
H + 11 H → 21 H + e + n + 0.42 MeV
1 +
1
(i)
e + + e – ® g + g + 1.02 MeV (ii)
2
1 H + 11 H → 32 He + g + 5.49 MeV (iii)

3
2 He + 32 He → 42 He + 11 H + 11 H + 12.86 MeV (iv) (13.14)
For the fourth reaction to occur, the first three reactions must occur
twice, in which case two light helium nuclei unite to form ordinary helium
nucleus. If we consider the combination 2(i) + 2(ii) + 2(iii) +(iv), the net
effect is
4 11 H + 2e − → 42 He + 2ν + 6γ + 26.7 MeV
or (4 11 H + 4e − ) → ( 42 He + 2e − ) + 2ν + 6γ + 26.7 MeV (13.15)
4
Thus, four hydrogen atoms combine to form an 2 He atom with a
release of 26.7 MeV of energy.
Helium is not the only element that can be synthesized in the interior of
a star. As the hydrogen in the core gets depleted and becomes helium, the
core starts to cool. The star begins to collapse under its own gravity which
increases the temperature of the core. If this temperature increases to about
108 K, fusion takes place again, this time of helium nuclei into carbon.
This kind of process can generate through fusion higher and higher mass
number elements. But elements more massive than those near the peak of
316 the binding energy curve in Fig. 13.1 cannot be so produced.

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 Nuclei
The age of the sun is about 5×109 y and it is estimated that there is
enough hydrogen in the sun to keep it going for another 5 billion years.
After that, the hydrogen burning will stop and the sun will begin to cool
and will start to collapse under gravity, which will raise the core
temperature. The outer envelope of the sun will expand, turning it into
the so called red giant.

13.7.3 Controlled thermonuclear fusion

The natural thermonuclear fusion process in a star is replicated in a
thermonuclear fusion device. In controlled fusion reactors, the aim is to
generate steady power by heating the nuclear fuel to a temperature in the
range of 108 K. At these temperatures, the fuel is a mixture of positive
ions and electrons (plasma). The challenge is to confine this plasma, since
no container can stand such a high temperature. Several countries
around the world including India are developing techniques in this
connection. If successful, fusion reactors will hopefully supply almost
unlimited power to humanity.

Example 13.4 Answer the following questions:

(a) Are the equations of nuclear reactions (such as those given in
Section 13.7) ‘balanced’ in the sense a chemical equation (e.g.,
2H2 + O2® 2 H2O) is? If not, in what sense are they balanced on
both sides?
(b) If both the number of protons and the number of neutrons are
conserved in each nuclear reaction, in what way is mass converted
into energy (or vice-versa) in a nuclear reaction?
(c) A general impression exists that mass-energy interconversion
takes place only in nuclear reaction and never in chemical
reaction. This is strictly speaking, incorrect. Explain.
Solution
(a) A chemical equation is balanced in the sense that the number of
atoms of each element is the same on both sides of the equation.
A chemical reaction merely alters the original combinations of
atoms. In a nuclear reaction, elements may be transmuted. Thus,
the number of atoms of each element is not necessarily conserved
in a nuclear reaction. However, the number of protons and the
number of neutrons are both separately conserved in a nuclear
reaction. [Actually, even this is not strictly true in the realm of
very high energies – what is strictly conserved is the total charge
and total ‘baryon number’. We need not pursue this matter here.]
In nuclear reactions (e.g., Eq. 13.10), the number of protons and
the number of neutrons are the same on the two sides of the equation.
(b) We know that the binding energy of a nucleus gives a negative
contribution to the mass of the nucleus (mass defect). Now, since
EXAMPLE 13.4

proton number and neutron number are conserved in a nuclear

reaction, the total rest mass of neutrons and protons is the same
on either side of a reaction. But the total binding energy of nuclei
on the left side need not be the same as that on the right hand
side. The difference in these binding energies appears as energy
released or absorbed in a nuclear reaction. Since binding energy 317

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contributes to mass, we say that the difference in the total mass
of nuclei on the two sides get converted into energy or vice-versa.
It is in these sense that a nuclear reaction is an example of mass-
energy interconversion.
(c) From the point of view of mass-energy interconversion, a chemical
reaction is similar to a nuclear reaction in principle. The energy
released or absorbed in a chemical reaction can be traced to the
difference in chemical (not nuclear) binding energies of atoms
and molecules on the two sides of a reaction. Since, strictly
speaking, chemical binding energy also gives a negative
contribution (mass defect) to the total mass of an atom or molecule,
we can equally well say that the difference in the total mass of
EXAMPLE 13.4

atoms or molecules, on the two sides of the chemical reaction

gets converted into energy or vice-versa. However, the mass
defects involved in a chemical reaction are almost a million times
smaller than those in a nuclear reaction.This is the reason for
the general impression, (which is incorrect ) that mass-energy
interconversion does not take place in a chemical reaction.

SUMMARY

1. An atom has a nucleus. The nucleus is positively charged. The radius

of the nucleus is smaller than the radius of an atom by a factor of
104. More than 99.9% mass of the atom is concentrated in the nucleus.
2. On the atomic scale, mass is measured in atomic mass units (u). By
definition, 1 atomic mass unit (1u) is 1/12th mass of one atom of 12C;
1u = 1.660563 × 10–27 kg.
3. A nucleus contains a neutral particle called neutron. Its mass is almost
the same as that of proton
4. The atomic number Z is the number of protons in the atomic nucleus
of an element. The mass number A is the total number of protons and
neutrons in the atomic nucleus; A = Z+N; Here N denotes the number
of neutrons in the nucleus.
A
A nuclear species or a nuclide is represented as Z X , where X is the
chemical symbol of the species.
Nuclides with the same atomic number Z, but different neutron number
N are called isotopes. Nuclides with the same A are isobars and those
with the same N are isotones.
Most elements are mixtures of two or more isotopes. The atomic mass
of an element is a weighted average of the masses of its isotopes and
calculated in accordance to the relative abundances of the isotopes.
5. A nucleus can be considered to be spherical in shape and assigned a
radius. Electron scattering experiments allow determination of the
nuclear radius; it is found that radii of nuclei fit the formula
R = R0 A1/3,
where R0 = a constant = 1.2 fm. This implies that the nuclear density
is independent of A. It is of the order of 1017 kg/m3.
6. Neutrons and protons are bound in a nucleus by the short-range strong
nuclear force. The nuclear force does not distinguish between neutron
318 and proton.

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 Nuclei

7. The nuclear mass M is always less than the total mass, Sm, of its
constituents. The difference in mass of a nucleus and its constituents
is called the mass defect,
DM = (Z mp + ( A – Z )mn ) – M
Using Einstein’s mass energy relation, we express this mass difference
in terms of energy as
DEb = DM c2
The energy DEb represents the binding energy of the nucleus. In the
mass number range A = 30 to 170, the binding energy per nucleon is
nearly constant, about 8 MeV/nucleon.
8. Energies associated with nuclear processes are about a million times
larger than chemical process.
9. The Q-value of a nuclear process is
Q = final kinetic energy – initial kinetic energy.
Due to conservation of mass-energy, this is also,
Q = (sum of initial masses – sum of final masses)c2
10. Radioactivity is the phenomenon in which nuclei of a given species
transform by giving out a or b or g rays; a-rays are helium nuclei;
b-rays are electrons. g-rays are electromagnetic radiation of wavelengths
shorter than X-rays.
11. Energy is released when less tightly bound nuclei are transmuted into
235
more tightly bound nuclei. In fission, a heavy nucleus like 92 U breaks
into two smaller fragments, e.g., 235
92 U+ n →
1
0
133
51 Sb + 99
41 Nb + 4 10 n
12. In fusion, lighter nuclei combine to form a larger nucleus. Fusion of
hydrogen nuclei into helium nuclei is the source of energy of all stars
including our sun.

Physical Quantity Symbol Dimensions Units Remarks

Atomic mass unit [M] u Unit of mass for

expressing atomic or
nuclear masses. One
atomic mass unit equals
1/12th of the mass of 12C
atom.
Disintegration or l [T –1] s–1
decay constant
Half-life T1/2 [T] s Time taken for the decay
of one-half of the initial
number of nuclei present
in a radioactive sample.
Mean life t [T] s Time at which number of
nuclei has been reduced to
e–1 of its initial value
Activity of a radio- R [ T–1] Bq Measure of the activity
active sample of a radioactive source.

319

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 Physics
POINTS TO PONDER
1. The density of nuclear matter is independent of the size of the nucleus.
The mass density of the atom does not follow this rule.
2. The radius of a nucleus determined by electron scattering is found to
be slightly different from that determined by alpha-particle scattering.
This is because electron scattering senses the charge distribution of
the nucleus, whereas alpha and similar particles sense the nuclear
matter.
3. After Einstein showed the equivalence of mass and energy, E = mc 2,
we cannot any longer speak of separate laws of conservation of mass
and conservation of energy, but we have to speak of a unified law of
conservation of mass and energy. The most convincing evidence that
this principle operates in nature comes from nuclear physics. It is
central to our understanding of nuclear energy and harnessing it as a
source of power. Using the principle, Q of a nuclear process (decay or
reaction) can be expressed also in terms of initial and final masses.
4. The nature of the binding energy (per nucleon) curve shows that
exothermic nuclear reactions are possible, when two light nuclei fuse
or when a heavy nucleus undergoes fission into nuclei with intermediate
mass.
5. For fusion, the light nuclei must have sufficient initial energy to
overcome the coulomb potential barrier. That is why fusion requires
very high temperatures.
6. Although the binding energy (per nucleon) curve is smooth and slowly
varying, it shows peaks at nuclides like 4He, 16O etc. This is considered
as evidence of atom-like shell structure in nuclei.
7. Electrons and positron are a particle-antiparticle pair. They are
identical in mass; their charges are equal in magnitude and opposite.
( It is found that when an electron and a positron come together, they
annihilate each other giving energy in the form of gamma-ray photons.)
8. Radioactivity is an indication of the instability of nuclei. Stability
requires the ratio of neutron to proton to be around 1:1 for light
nuclei. This ratio increases to about 3:2 for heavy nuclei. (More
neutrons are required to overcome the effect of repulsion among the
protons.) Nuclei which are away from the stability ratio, i.e., nuclei
which have an excess of neutrons or protons are unstable. In fact,
only about 10% of knon isotopes (of all elements), are stable. Others
have been either artificially produced in the laboratory by bombarding
a, p, d, n or other particles on targets of stable nuclear species or
identified in astronomical observations of matter in the universe.

320

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 Nuclei

EXERCISES
You may find the following data useful in solving the exercises:
e = 1.6×10–19C N = 6.023×1023 per mole
1/(4pe0) = 9 × 109 N m2/C2 k = 1.381×10–23J K–1

1 MeV = 1.6×10–13J 1 u = 931.5 MeV/c2

1 year = 3.154×107 s

mH = 1.007825 u mn = 1.008665 u
4
m( He ) = 4.002603 u
2
me = 0.000548 u
13.1 Obtain the binding energy (in MeV) of a nitrogen nucleus ( 14
7 )
N ,
(
given m 147 N =14.00307 u)
56 209
13.2 Obtain the binding energy of the nuclei 26 Fe and 83 Bi in units of
MeV from the following data:
m ( 56
26 Fe
) = 55.934939 u m ( 209
83 Bi
) = 208.980388 u
13.3 A given coin has a mass of 3.0 g. Calculate the nuclear energy that
would be required to separate all the neutrons and protons from
each other. For simplicity assume that the coin is entirely made of
63
29 Cu atoms (of mass 62.92960 u).

13.4 Obtain approximately the ratio of the nuclear radii of the gold isotope
197
79 Au
and the silver isotope 107
47 Ag
.
13.5 The Q value of a nuclear reaction A + b ® C + d is defined by
Q = [ mA + mb – mC – md]c2
where the masses refer to the respective nuclei. Determine from the
given data the Q-value of the following reactions and state whether
the reactions are exothermic or endothermic.
(i) 11 H+13H →12 H+12 H

(ii) 12
6 C+126 C →10
20
Ne+24 He
Atomic masses are given to be
m ( 12 H ) = 2.014102 u
m ( 13 H ) = 3.016049 u
m ( 126 C ) = 12.000000 u
20
m ( 10 Ne ) = 19.992439 u
13.6 Suppose, we think of fission of a 56 26 Fe
nucleus into two equal
28
fragments, 13 Al . Is the fission energetically possible? Argue by
working out Q of the process. Given m ( 56 26 Fe
) = 55.93494 u and
28
m ( 13 Al ) = 27.98191 u.

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13.7 The fission properties of 239
94 Pu
are very similar to those of 235
92 U
. The
average energy released per fission is 180 MeV. How much energy,
in MeV, is released if all the atoms in 1 kg of pure 239 94 Pu undergo
fission?
13.8 How long can an electric lamp of 100W be kept glowing by fusion of
2.0 kg of deuterium? Take the fusion reaction as
2
1 H+ 21 H → 32 He +n +3.27 MeV
13.9 Calculate the height of the potential barrier for a head on collision
of two deuterons. (Hint: The height of the potential barrier is given
by the Coulomb repulsion between the two deuterons when they
just touch each other. Assume that they can be taken as hard
spheres of radius 2.0 fm.)
13.10 From the relation R = R0 A1/3, where R 0 is a constant and A is the
mass number of a nucleus, show that the nuclear matter density is
nearly constant (i.e. independent of A).

322

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Chapter Fourteen

SEMICONDUCTOR
ELECTRONICS:
MATERIALS, DEVICES
AND SIMPLE CIRCUITS
14.1 INTRODUCTION
Devices in which a controlled flow of electrons can be obtained are the
basic building blocks of all the electronic circuits. Before the discovery of
transistor in 1948, such devices were mostly vacuum tubes (also called
valves) like the vacuum diode which has two electrodes, viz., anode (often
called plate) and cathode; triode which has three electrodes – cathode,
plate and grid; tetrode and pentode (respectively with 4 and 5 electrodes).
In a vacuum tube, the electrons are supplied by a heated cathode and
the controlled flow of these electrons in vacuum is obtained by varying
the voltage between its different electrodes. Vacuum is required in the
inter-electrode space; otherwise the moving electrons may lose their
energy on collision with the air molecules in their path. In these devices
the electrons can flow only from the cathode to the anode (i.e., only in one
direction). Therefore, such devices are generally referred to as valves.
These vacuum tube devices are bulky, consume high power, operate
generally at high voltages (~100 V) and have limited life and low reliability.
The seed of the development of modern solid-state semiconductor
electronics goes back to 1930’s when it was realised that some solid-
state semiconductors and their junctions offer the possibility of controlling
the number and the direction of flow of charge carriers through them.
Simple excitations like light, heat or small applied voltage can change
the number of mobile charges in a semiconductor. Note that the supply

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and flow of charge carriers in the semiconductor devices are within the
solid itself, while in the earlier vacuum tubes/valves, the mobile electrons
were obtained from a heated cathode and they were made to flow in an
evacuated space or vacuum. No external heating or large evacuated space
is required by the semiconductor devices. They are small in size, consume
low power, operate at low voltages and have long life and high reliability.
Even the Cathode Ray Tubes (CRT) used in television and computer
monitors which work on the principle of vacuum tubes are being replaced
by Liquid Crystal Display (LCD) monitors with supporting solid state
electronics. Much before the full implications of the semiconductor devices
was formally understood, a naturally occurring crystal of galena (Lead
sulphide, PbS) with a metal point contact attached to it was used as
detector of radio waves.
In the following sections, we will introduce the basic concepts of
semiconductor physics and discuss some semiconductor devices like
junction diodes (a 2-electrode device) and bipolar junction transistor (a
3-electrode device). A few circuits illustrating their applications will also
be described.

14.2 CLASSIFICATION OF METALS, CONDUCTORS AND

SEMICONDUCTORS
On the basis of conductivity
On the basis of the relative values of electrical conductivity (s ) or resistivity
( r = 1/s ), the solids are broadly classified as:
(i) Metals: They possess very low resistivity (or high conductivity).
r ~ 10–2 – 10–8 W m
s ~ 102 – 108 S m–1
(ii) Semiconductors: They have resistivity or conductivity intermediate
to metals and insulators.
r ~ 10–5 – 106 W m
s ~ 105 – 10–6 S m–1
(iii)Insulators: They have high resistivity (or low conductivity).
r ~ 1011 – 1019 W m
s ~ 10–11 – 10–19 S m–1
The values of r and s given above are indicative of magnitude and
could well go outside the ranges as well. Relative values of the resistivity
are not the only criteria for distinguishing metals, insulators and
semiconductors from each other. There are some other differences, which
will become clear as we go along in this chapter.
Our interest in this chapter is in the study of semiconductors which
could be:
(i) Elemental semiconductors: Si and Ge
(ii) Compound semiconductors: Examples are:
· Inorganic: CdS, GaAs, CdSe, InP, etc.
· Organic: anthracene, doped pthalocyanines, etc.
· Organic polymers: polypyrrole, polyaniline, polythiophene, etc.
Most of the currently available semiconductor devices are based on
324 elemental semiconductors Si or Ge and compound inorganic
semiconductors. However, after 1990, a few semiconductor devices using

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 Semiconductor Electronics:
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Simple Circuits
organic semiconductors and semiconducting polymers have been
developed signalling the birth of a futuristic technology of polymer-
electronics and molecular-electronics. In this chapter, we will restrict
ourselves to the study of inorganic semiconductors, particularly
elemental semiconductors Si and Ge. The general concepts introduced
here for discussing the elemental semiconductors, by-and-large, apply
to most of the compound semiconductors as well.
On the basis of energy bands
According to the Bohr atomic model, in an isolated atom the energy of
any of its electrons is decided by the orbit in which it revolves. But when
the atoms come together to form a solid they are close to each other. So
the outer orbits of electrons from neighbouring atoms would come very
close or could even overlap. This would make the nature of electron motion
in a solid very different from that in an isolated atom.
Inside the crystal each electron has a unique position and no two
electrons see exactly the same pattern of surrounding charges. Because
of this, each electron will have a different energy level. These different
energy levels with continuous energy variation form what are called
energy bands. The energy band which includes the energy levels of the
valence electrons is called the valence band. The energy band above the
valence band is called the conduction band. With no external energy, all
the valence electrons will reside in the valence band. If the lowest level in
the conduction band happens to be lower than the highest level of the
valence band, the electrons from the valence band can easily move into
the conduction band. Normally the conduction band is empty. But when
it overlaps on the valence band electrons can move freely into it. This is
the case with metallic conductors.
If there is some gap between the conduction band and the valence
band, electrons in the valence band all remain bound and no free electrons
are available in the conduction band. This makes the material an
insulator. But some of the electrons from the valence band may gain
external energy to cross the gap between the conduction band and the
valence band. Then these electrons will move into the conduction band.
At the same time they will create vacant energy levels in the valence band
where other valence electrons can move. Thus the process creates the
possibility of conduction due to electrons in conduction band as well as
due to vacancies in the valence band.
Let us consider what happens in the case of Si or Ge crystal containing
N atoms. For Si, the outermost orbit is the third orbit (n = 3), while for Ge
it is the fourth orbit (n = 4). The number of electrons in the outermost
orbit is 4 (2s and 2p electrons). Hence, the total number of outer electrons
in the crystal is 4N. The maximum possible number of electrons in the
outer orbit is 8 (2s + 6p electrons). So, for the 4N valence electrons there
are 8N available energy states. These 8N discrete energy levels can either
form a continuous band or they may be grouped in different bands
depending upon the distance between the atoms in the crystal (see box
on Band Theory of Solids).
At the distance between the atoms in the crystal lattices of Si and Ge,
the energy band of these 8N states is split apart into two which are
separated by an energy gap Eg (Fig. 14.1). The lower band which is
completely occupied by the 4N valence electrons at temperature of absolute 325
zero is the valence band. The other band consisting of 4N energy states,
called the conduction band, is completely empty at absolute zero.
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 Physics
The lowest energy level in the
conduction band is shown as EC and
highest energy level in the valence band
is shown as EV . Above EC and below EV
there are a large number of closely spaced
energy levels, as shown in Fig. 14.1.
The gap between the top of the valence
band and bottom of the conduction band
is called the energy band gap (Energy gap
Eg ). It may be large, small, or zero,
depending upon the material. These
different situations, are depicted in Fig.
14.2 and discussed below:
Case I: This refers to a situation, as
shown in Fig. 14.2(a). One can have a
metal either when the conduction band
FIGURE 14.1 The energy band positions in a is partially filled and the balanced band
semiconductor at 0 K. The upper band, called the is partially empty or when the conduction
conduction band, consists of infinitely large number and valance bands overlap. When there
of closely spaced energy states. The lower band,
is overlap electrons from valence band can
called the valence band, consists of closely spaced
easily move into the conduction band.
completely filled energy states.
This situation makes a large number of
electrons available for electrical conduction. When the valence band is
partially empty, electrons from its lower level can move to higher level
making conduction possible. Therefore, the resistance of such materials
is low or the conductivity is high.

FIGURE 14.2 Difference between energy bands of (a) metals,

326 (b) insulators and (c) semiconductors.

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 Semiconductor Electronics:
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Simple Circuits
Case II: In this case, as shown in Fig. 14.2(b), a large band gap Eg exists
(Eg > 3 eV). There are no electrons in the conduction band, and therefore
no electrical conduction is possible. Note that the energy gap is so large
that electrons cannot be excited from the valence band to the conduction
band by thermal excitation. This is the case of insulators.
Case III: This situation is shown in Fig. 14.2(c). Here a finite but small
band gap (Eg < 3 eV) exists. Because of the small band gap, at room
temperature some electrons from valence band can acquire enough
energy to cross the energy gap and enter the conduction band. These
electrons (though small in numbers) can move in the conduction band.
Hence, the resistance of semiconductors is not as high as that of the
insulators.
In this section we have made a broad classification of metals,
conductors and semiconductors. In the section which follows you will
learn the conduction process in semiconductors.

14.3 INTRINSIC SEMICONDUCTOR

We shall take the most common case of Ge and Si whose
lattice structure is shown in Fig. 14.3. These structures
are called the diamond-like structures. Each atom is
surrounded by four nearest neighbours. We know that
Si and Ge have four valence electrons. In its crystalline
structure, every Si or Ge atom tends to share one of its
four valence electrons with each of its four nearest
neighbour atoms, and also to take share of one electron
from each such neighbour. These shared electron pairs
are referred to as forming a covalent bond or simply a
valence bond. The two shared electrons can be assumed
to shuttle back-and-forth between the associated atoms
holding them together strongly. Figure 14.4 schematically
shows the 2-dimensional representation of Si or Ge FIGURE 14.3 Three-dimensional dia-
structure shown in Fig. 14.3 which overemphasises the mond-like crystal structure for Carbon,
covalent bond. It shows an idealised picture in which no Silicon or Germanium with
bonds are broken (all bonds are intact). Such a situation respective lattice spacing a equal
arises at low temperatures. As the temperature increases, to 3.56, 5.43 and 5.66 Å.
more thermal energy becomes available to these electrons and some of
these electrons may break–away (becoming free electrons contributing to
conduction). The thermal energy effectively ionises only a few atoms in the
crystalline lattice and creates a vacancy in the bond as shown in Fig. 14.5(a).
The neighbourhood, from which the free electron (with charge –q) has come
out leaves a vacancy with an effective charge (+q). This vacancy with the
effective positive electronic charge is called a hole. The hole behaves as an
apparent free particle with effective positive charge.
In intrinsic semiconductors, the number of free electrons, ne is equal to
the number of holes, nh. That is
ne = nh = ni (14.1)
where ni is called intrinsic carrier concentration.
Semiconductors posses the unique property in which, apart from
electrons, the holes also move. Suppose there is a hole at site 1 as shown 327

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 Physics
in Fig. 14.5(a). The movement of holes can be
visualised as shown in Fig. 14.5(b). An electron
from the covalent bond at site 2 may jump to
the vacant site 1 (hole). Thus, after such a jump,
the hole is at site 2 and the site 1 has now an
electron. Therefore, apparently, the hole has
moved from site 1 to site 2. Note that the electron
originally set free [Fig. 14.5(a)] is not involved
in this process of hole motion. The free electron
moves completely independently as conduction
electron and gives rise to an electron current, Ie
under an applied electric field. Remember that
the motion of hole is only a convenient way of
FIGURE 14.4 Schematic two-dimensional describing the actual motion of bound electrons,
representation of Si or Ge structure showing
whenever there is an empty bond anywhere in
covalent bonds at low temperature
(all bonds intact). +4 symbol
the crystal. Under the action of an electric field,
indicates inner cores of Si or Ge. these holes move towards negative potential
giving the hole current, Ih. The total current, I is
thus the sum of the electron current Ie and the
hole current Ih:
I = Ie + Ih (14.2)
It may be noted that apart from the process of generation of conduction
electrons and holes, a simultaneous process of recombination occurs in
which the electrons recombine with the holes. At equilibrium, the rate of
generation is equal to the rate of recombination of charge carriers. The
recombination occurs due to an electron colliding with a hole.

(a) (b)
FIGURE 14.5 (a) Schematic model of generation of hole at site 1 and conduction electron
due to thermal energy at moderate temperatures. (b) Simplified representation of
possible thermal motion of a hole. The electron from the lower left hand covalent bond
(site 2) goes to the earlier hole site1, leaving a hole at its site indicating an
328 apparent movement of the hole from site 1 to site 2.

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 Semiconductor Electronics:
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Simple Circuits
An intrinsic semiconductor
will behave like an insulator at
T = 0 K as shown in Fig. 14.6(a).
It is the thermal energy at
higher temperatures (T > 0K),
which excites some electrons
from the valence band to the
conduction band. These
thermally excited electrons at
T > 0 K, partially occupy the
conduction band. Therefore,
the energy-band diagram of an
intrinsic semiconductor will be
as shown in Fig. 14.6(b). Here, FIGURE 14.6 (a) An intrinsic semiconductor at T = 0 K
some electrons are shown in behaves like insulator. (b) At T > 0 K, four thermally generated
the conduction band. These electron-hole pairs. The filled circles ( ) represent electrons
have come from the valence and empty circles ( ) represent holes.
band leaving equal number of
holes there.

Example 14.1 C, Si and Ge have same lattice structure. Why is C

insulator while Si and Ge intrinsic semiconductors?
Solution The 4 bonding electrons of C, Si or Ge lie, respectively, in

EXAMPLE 14.1
the second, third and fourth orbit. Hence, energy required to take
out an electron from these atoms (i.e., ionisation energy Eg ) will be
least for Ge, followed by Si and highest for C. Hence, number of free
electrons for conduction in Ge and Si are significant but negligibly
small for C.

14.4 EXTRINSIC SEMICONDUCTOR

The conductivity of an intrinsic semiconductor depends on its
temperature, but at room temperature its conductivity is very low. As
such, no important electronic devices can be developed using these
semiconductors. Hence there is a necessity of improving their
conductivity. This can be done by making use of impurities.
When a small amount, say, a few parts per million (ppm), of a suitable
impurity is added to the pure semiconductor, the conductivity of the
semiconductor is increased manifold. Such materials are known as
extrinsic semiconductors or impurity semiconductors. The deliberate
addition of a desirable impurity is called doping and the impurity atoms
are called dopants. Such a material is also called a doped semiconductor.
The dopant has to be such that it does not distort the original pure
semiconductor lattice. It occupies only a very few of the original
semiconductor atom sites in the crystal. A necessary condition to attain
this is that the sizes of the dopant and the semiconductor atoms should
be nearly the same.
There are two types of dopants used in doping the tetravalent Si
or Ge:
(i) Pentavalent (valency 5); like Arsenic (As), Antimony (Sb), Phosphorous
(P), etc. 329

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 Physics
(ii) Trivalent (valency 3); like Indium (In),
Boron (B), Aluminium (Al), etc.
We shall now discuss how the doping
changes the number of charge carriers (and
hence the conductivity) of semiconductors.
Si or Ge belongs to the fourth group in the
Periodic table and, therefore, we choose the
dopant element from nearby fifth or third
group, expecting and taking care that the
size of the dopant atom is nearly the same as
that of Si or Ge. Interestingly, the pentavalent
and trivalent dopants in Si or Ge give two
entirely different types of semiconductors as
discussed below.
(i) n-type semiconductor
Suppose we dope Si or Ge with a pentavalent
element as shown in Fig. 14.7. When an atom
of +5 valency element occupies the position
of an atom in the crystal lattice of Si, four of
its electrons bond with the four silicon
neighbours while the fifth remains very
weakly bound to its parent atom. This is
because the four electrons participating in
FIGURE 14.7 (a) Pentavalent donor atom (As, Sb, bonding are seen as part of the effective core
P, etc.) doped for tetravalent Si or Ge giving n- of the atom by the fifth electron. As a result
type semiconductor, and (b) Commonly used the ionisation energy required to set this
schematic representation of n-type material
electron free is very small and even at room
which shows only the fixed cores of the
substituent donors with one additional effective
temperature it will be free to move in the
positive charge and its associated extra electron. lattice of the semiconductor. For example, the
energy required is ~ 0.01 eV for germanium,
and 0.05 eV for silicon, to separate this
electron from its atom. This is in contrast to the energy required to jump
the forbidden band (about 0.72 eV for germanium and about 1.1 eV for
silicon) at room temperature in the intrinsic semiconductor. Thus, the
pentavalent dopant is donating one extra electron for conduction and
hence is known as donor impurity. The number of electrons made
available for conduction by dopant atoms depends strongly upon the
doping level and is independent of any increase in ambient temperature.
On the other hand, the number of free electrons (with an equal number
of holes) generated by Si atoms, increases weakly with temperature.
In a doped semiconductor the total number of conduction electrons
ne is due to the electrons contributed by donors and those generated
intrinsically, while the total number of holes nh is only due to the holes
from the intrinsic source. But the rate of recombination of holes would
increase due to the increase in the number of electrons. As a result, the
number of holes would get reduced further.
Thus, with proper level of doping the number of conduction electrons
330 can be made much larger than the number of holes. Hence in an extrinsic

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 Semiconductor Electronics:
Materials, Devices and
Simple Circuits
semiconductor doped with pentavalent impurity, electrons
become the majority carriers and holes the minority carriers.
These semiconductors are, therefore, known as n-type
semiconductors. For n-type semiconductors, we have,
ne >> nh (14.3)
(ii) p-type semiconductor
This is obtained when Si or Ge is doped with a trivalent impurity
like Al, B, In, etc. The dopant has one valence electron less than
Si or Ge and, therefore, this atom can form covalent bonds with
neighbouring three Si atoms but does not have any electron to
offer to the fourth Si atom. So the bond between the fourth
neighbour and the trivalent atom has a vacancy or hole as
shown in Fig. 14.8. Since the neighbouring Si atom in the lattice
wants an electron in place of a hole, an electron in the outer
orbit of an atom in the neighbourhood may jump to fill this
vacancy, leaving a vacancy or hole at its own site. Thus the hole
is available for conduction. Note that the trivalent foreign atom
becomes effectively negatively charged when it shares fourth
electron with neighbouring Si atom. Therefore, the dopant atom
of p-type material can be treated as core of one negative charge
along with its associated hole as shown in Fig. 14.8(b). It is
obvious that one acceptor atom gives one hole. These holes are
in addition to the intrinsically generated holes while the source
of conduction electrons is only intrinsic generation. Thus, for
such a material, the holes are the majority carriers and electrons
are minority carriers. Therefore, extrinsic semiconductors doped FIGURE 14.8 (a) Trivalent
with trivalent impurity are called p-type semiconductors. For acceptor atom (In, Al, B etc.)
doped in tetravalent Si or Ge
p-type semiconductors, the recombination process will reduce
lattice giving p-type semicon-
the number (ni )of intrinsically generated electrons to ne.
ductor. (b) Commonly used
We have, for p-type semiconductors schematic representation of
nh >> ne (14.4) p-type material which shows
Note that the crystal maintains an overall charge neutrality only the fixed core of the
substituent acceptor with
as the charge of additional charge carriers is just equal and
one effective additional
opposite to that of the ionised cores in the lattice. negative charge and its
In extrinsic semiconductors, because of the abundance of associated hole.
majority current carriers, the minority carriers produced
thermally have more chance of meeting majority carriers and
thus getting destroyed. Hence, the dopant, by adding a large number of
current carriers of one type, which become the majority carriers, indirectly
helps to reduce the intrinsic concentration of minority carriers.
The semiconductor’s energy band structure is affected by doping. In
the case of extrinsic semiconductors, additional energy states due to donor
impurities (ED ) and acceptor impurities (EA ) also exist. In the energy band
diagram of n-type Si semiconductor, the donor energy level ED is slightly
below the bottom EC of the conduction band and electrons from this level
move into the conduction band with very small supply of energy. At room
temperature, most of the donor atoms get ionised but very few (~1012)
atoms of Si get ionised. So the conduction band will have most electrons
coming from the donor impurities, as shown in Fig. 14.9(a). Similarly, 331

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for p-type semiconductor, the acceptor energy level EA is slightly above
the top EV of the valence band as shown in Fig. 14.9(b). With very small
supply of energy an electron from the valence band can jump to the level
EA and ionise the acceptor negatively. (Alternately, we can also say that
with very small supply of energy the hole from level EA sinks down into
the valence band. Electrons rise up and holes fall down when they gain
external energy.) At room temperature, most of the acceptor atoms get
ionised leaving holes in the valence band. Thus at room temperature the
density of holes in the valence band is predominantly due to impurity in
the extrinsic semiconductor. The electron and hole concentration in a
semiconductor in thermal equilibrium is given by
nenh = ni2 (14.5)
Though the above description is grossly approximate and
hypothetical, it helps in understanding the difference between metals,
insulators and semiconductors (extrinsic and intrinsic) in a simple
manner. The difference in the resistivity of C, Si and Ge depends upon
the energy gap between their conduction and valence bands. For C
(diamond), Si and Ge, the energy gaps are 5.4 eV, 1.1 eV and 0.7 eV,
respectively. Sn also is a group IV element but it is a metal because the
energy gap in its case is 0 eV.

FIGURE 14.9 Energy bands of (a) n-type semiconductor at T > 0K, (b) p-type
semiconductor at T > 0K.

Example 14.2 Suppose a pure Si crystal has 5 × 1028 atoms m–3. It is

doped by 1 ppm concentration of pentavalent As. Calculate the
number of electrons and holes. Given that ni =1.5 × 1016 m–3.
Solution Note that thermally generated electrons (ni ~1016 m–3 ) are
EXAMPLE 14.2

negligibly small as compared to those produced by doping.

Therefore, ne » ND.
Since nenh = ni2, The number of holes
nh = (2.25 × 1032 )/(5 ×1022 )

332 ~ 4.5 × 109 m–3

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14.5 p-n JUNCTION

A p-n junction is the basic building block of many semiconductor devices
like diodes, transistor, etc. A clear understanding of the junction behaviour
is important to analyse the working of other semiconductor devices.
We will now try to understand how a junction is formed and how the
junction behaves under the influence of external applied voltage (also
called bias).

http://hyperphysics.phy-astr.gsu.edu/hbase/solids/pnjun.html
Formation and working of p-n junction diode
14.5.1 p-n junction formation
Consider a thin p-type silicon (p-Si) semiconductor wafer. By adding
precisely a small quantity of pentavelent impurity, part of the p-Si wafer
can be converted into n-Si. There are several processes by which a
semiconductor can be formed. The wafer now contains p-region and
n-region and a metallurgical junction between p-, and n- region.
Two important processes occur during the formation of a p-n junction:
diffusion and drift. We know that in an n-type semiconductor, the
concentration of electrons (number of electrons per unit volume) is more
compared to the concentration of holes. Similarly, in a p-type
semiconductor, the concentration of holes is more than the concentration
of electrons. During the formation of p-n junction, and due to the
concentration gradient across p-, and n- sides, holes diffuse from p-side
to n-side (p ® n) and electrons diffuse from n-side to p-side (n ® p). This
motion of charge carries gives rise to diffusion current across the junction.
When an electron diffuses from n ® p, it leaves behind an ionised
donor on n-side. This ionised donor (positive charge) is immobile as it is
bonded to the surrounding atoms. As the electrons continue to diffuse
from n ® p, a layer of positive charge (or positive space-charge region) on
n-side of the junction is developed.
Similarly, when a hole diffuses from p ® n due to the concentration
gradient, it leaves behind an ionised acceptor (negative charge) which is
immobile. As the holes continue to diffuse, a layer of negative charge (or
negative space-charge region) on the p-side of the junction is developed.
This space-charge region on either side of the junction together is known
as depletion region as the electrons and holes taking
part in the initial movement across the junction depleted
the region of its free charges (Fig. 14.10). The thickness
of depletion region is of the order of one-tenth of a
micrometre. Due to the positive space-charge region on
n-side of the junction and negative space charge region
on p-side of the junction, an electric field directed from
positive charge towards negative charge develops. Due
to this field, an electron on p-side of the junction moves
to n-side and a hole on n-side of the junction moves to p- FIGURE 14.10 p-n junction
side. The motion of charge carriers due to the electric field formation process.
is called drift. Thus a drift current, which is opposite in
direction to the diffusion current (Fig. 14.10) starts. 333

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Initially, diffusion current is large and drift current is small.
As the diffusion process continues, the space-charge regions
on either side of the junction extend, thus increasing the electric
field strength and hence drift current. This process continues
until the diffusion current equals the drift current. Thus a p-n
junction is formed. In a p-n junction under equilibrium there
is no net current.
The loss of electrons from the n-region and the gain of
electron by the p-region causes a difference of potential across
the junction of the two regions. The polarity of this potential is
such as to oppose further flow of carriers so that a condition of
equilibrium exists. Figure 14.11 shows the p-n junction at
equilibrium and the potential across the junction. The
n-material has lost electrons, and p material has acquired
electrons. The n material is thus positive relative to the p
FIGURE 14.11 (a) Diode under material. Since this potential tends to prevent the movement of
equilibrium (V = 0), (b) Barrier electron from the n region into the p region, it is often called a
potential under no bias. barrier potential.

Example 14.3 Can we take one slab of p-type semiconductor and

EXAMPLE 14.3

physically join it to another n-type semiconductor to get p-n junction?

Solution No! Any slab, howsoever flat, will have roughness much
larger than the inter-atomic crystal spacing (~2 to 3 Å) and hence
continuous contact at the atomic level will not be possible. The junction
will behave as a discontinuity for the flowing charge carriers.

14.6 SEMICONDUCTOR DIODE

A semiconductor diode [Fig. 14.12(a)] is basically a p-n
junction with metallic contacts provided at the ends for
the application of an external voltage. It is a two terminal
device. A p-n junction diode is symbolically represented
as shown in Fig. 14.12(b).
The direction of arrow indicates the conventional
direction of current (when the diode is under forward
p n bias). The equilibrium barrier potential can be altered
by applying an external voltage V across the diode. The
FIGURE 14.12 (a) Semiconductor diode, situation of p-n junction diode under equilibrium
(b) Symbol for p-n junction diode. (without bias) is shown in Fig. 14.11(a) and (b).

14.6.1 p-n junction diode under forward bias

When an external voltage V is applied across a semiconductor diode such
that p-side is connected to the positive terminal of the battery and n-side
to the negative terminal [Fig. 14.13(a)], it is said to be forward biased.
The applied voltage mostly drops across the depletion region and the
voltage drop across the p-side and n-side of the junction is negligible.
(This is because the resistance of the depletion region – a region where
there are no charges – is very high compared to the resistance of n-side
334 and p-side.) The direction of the applied voltage (V ) is opposite to the

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built-in potential V0. As a result, the depletion layer width
decreases and the barrier height is reduced [Fig. 14.13(b)]. The
effective barrier height under forward bias is (V0 – V ).
If the applied voltage is small, the barrier potential will be
reduced only slightly below the equilibrium value, and only a
small number of carriers in the material—those that happen to
be in the uppermost energy levels—will possess enough energy
to cross the junction. So the current will be small. If we increase
the applied voltage significantly, the barrier height will be reduced
and more number of carriers will have the required energy. Thus
the current increases.
Due to the applied voltage, electrons from n-side cross the
depletion region and reach p-side (where they are minority
carries). Similarly, holes from p-side cross the junction and reach
the n-side (where they are minority carries). This process under
forward bias is known as minority carrier injection. At the
junction boundary, on each side, the minority carrier FIGURE 14.13 (a) p-n
junction diode under forward
concentration increases significantly compared to the locations
bias, (b) Barrier potential
far from the junction.
(1) without battery, (2) Low
Due to this concentration gradient, the injected electrons on
battery voltage, and (3) High
p-side diffuse from the junction edge of p-side to the other end voltage battery.
of p-side. Likewise, the injected holes on n-side diffuse from the
junction edge of n-side to the other end of n-side
(Fig. 14.14). This motion of charged carriers on either side
gives rise to current. The total diode forward current is sum
of hole diffusion current and conventional current due to
electron diffusion. The magnitude of this current is usually
in mA.

14.6.2 p-n junction diode under reverse bias

When an external voltage (V ) is applied across the diode such FIGURE 14.14 Forward bias
that n-side is positive and p-side is negative, it is said to be minority carrier injection.
reverse biased [Fig.14.15(a)]. The applied voltage mostly
drops across the depletion region. The direction of applied voltage is same
as the direction of barrier potential. As a result, the barrier height increases
and the depletion region widens due to the change in the electric field.
The effective barrier height under reverse bias is (V0 + V ), [Fig. 14.15(b)].
This suppresses the flow of electrons from n ® p and holes from p ® n.
Thus, diffusion current, decreases enormously compared to the diode
under forward bias.
The electric field direction of the junction is such that if electrons on
p-side or holes on n-side in their random motion come close to the
junction, they will be swept to its majority zone. This drift of carriers
gives rise to current. The drift current is of the order of a few mA. This is
quite low because it is due to the motion of carriers from their minority
side to their majority side across the junction. The drift current is also
there under forward bias but it is negligible (mA) when compared with
current due to injected carriers which is usually in mA.
The diode reverse current is not very much dependent on the applied
voltage. Even a small voltage is sufficient to sweep the minority carriers
from one side of the junction to the other side of the junction. The current 335

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is not limited by the magnitude of the applied voltage but is
limited due to the concentration of the minority carrier on either
side of the junction.
The current under reverse bias is essentially voltage
independent upto a critical reverse bias voltage, known as
breakdown voltage (Vbr ). When V = Vbr, the diode reverse current
increases sharply. Even a slight increase in the bias voltage causes
large change in the current. If the reverse current is not limited by
an external circuit below the rated value (specified by the
manufacturer) the p-n junction will get destroyed. Once it exceeds
the rated value, the diode gets destroyed due to overheating. This
can happen even for the diode under forward bias, if the forward
current exceeds the rated value.
The circuit arrangement for studying the V-I characteristics
of a diode, (i.e., the variation of current as a function of applied
FIGURE 14.15 (a) Diode
voltage) are shown in Fig. 14.16(a) and (b). The battery is connected
under reverse bias,
(b) Barrier potential under to the diode through a potentiometer (or reheostat) so that the
reverse bias. applied voltage to the diode can be changed. For different values
of voltages, the value of the current is noted. A graph between V
and I is obtained as in Fig. 14.16(c). Note that in forward bias
measurement, we use a milliammeter since the expected current is large
(as explained in the earlier section) while a micrometer is used in reverse
bias to measure the current. You can see in Fig. 14.16(c) that in forward

FIGURE 14.16 Experimental circuit arrangement for studying V-I characteristics of

a p-n junction diode (a) in forward bias, (b) in reverse bias. (c) Typical V-I
336 characteristics of a silicon diode.

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bias, the current first increases very slowly, almost negligibly, till the
voltage across the diode crosses a certain value. After the characteristic
voltage, the diode current increases significantly (exponentially), even for
a very small increase in the diode bias voltage. This voltage is called the
threshold voltage or cut-in voltage (~0.2V for germanium diode and
~0.7 V for silicon diode).
For the diode in reverse bias, the current is very small (~mA) and almost
remains constant with change in bias. It is called reverse saturation
current. However, for special cases, at very high reverse bias (break down
voltage), the current suddenly increases. This special action of the diode
is discussed later in Section 14.8. The general purpose diode are not
used beyond the reverse saturation current region.
The above discussion shows that the p-n junction diode primerly
allows the flow of current only in one direction (forward bias). The forward
bias resistance is low as compared to the reverse bias resistance. This
property is used for rectification of ac voltages as discussed in the next
section. For diodes, we define a quantity called dynamic resistance as
the ratio of small change in voltage DV to a small change in current DI:
∆V
rd = (14.6)
∆I

Example 14.4 The V-I characteristic of a silicon diode is shown in

the Fig. 14.17. Calculate the resistance of the diode at (a) ID = 15 mA
and (b) VD = –10 V.

FIGURE 14.17

Solution Considering the diode characteristics as a straight line

between I = 10 mA to I = 20 mA passing through the origin, we can
calculate the resistance using Ohm’s law.
EXAMPLE 14.4

(a) From the curve, at I = 20 mA, V = 0.8 V; I = 10 mA, V = 0.7 V

rf b = DV/DI = 0.1V/10 mA = 10 W
(b) From the curve at V = –10 V, I = –1 mA,
Therefore,
rrb = 10 V/1mA= 1.0 × 107 W
337

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14.7 APPLICATION OF JUNCTION DIODE AS A RECTIFIER
From the V-I characteristic of a junction diode we see that it allows current
to pass only when it is forward biased. So if an alternating voltage is
applied across a diode the current flows only in that part of the cycle
when the diode is forward biased. This property
is used to rectify alternating voltages and the
circuit used for this purpose is called a rectifier.
If an alternating voltage is applied across a
diode in series with a load, a pulsating voltage will
appear across the load only during the half cycles
of the ac input during which the diode is forward
biased. Such rectifier circuit, as shown in
Fig. 14.18, is called a half-wave rectifier. The
secondary of a transformer supplies the desired
ac voltage across terminals A and B. When the
voltage at A is positive, the diode is forward biased
and it conducts. When A is negative, the diode is
reverse-biased and it does not conduct. The reverse
saturation current of a diode is negligible and can
be considered equal to zero for practical purposes.
(The reverse breakdown voltage of the diode must
be sufficiently higher than the peak ac voltage at
the secondary of the transformer to protect the
diode from reverse breakdown.)
Therefore, in the positive half-cycle of ac there
FIGURE 14.18 (a) Half-wave rectifier is a current through the load resistor RL and we
circuit, (b) Input ac voltage and output
get an output voltage, as shown in Fig. 14.18(b),
voltage waveforms from the rectifier circuit.
whereas there is no current in the negative half-
cycle. In the next positive half-cycle, again we get
the output voltage. Thus, the output voltage, though still varying, is
restricted to only one direction and is said to be rectified. Since the
rectified output of this circuit is only for half of the input ac wave it is
called as half-wave rectifier.
The circuit using two diodes, shown in Fig. 14.19(a), gives output
rectified voltage corresponding to both the positive as well as negative
half of the ac cycle. Hence, it is known as full-wave rectifier. Here the
p-side of the two diodes are connected to the ends of the secondary of the
transformer. The n-side of the diodes are connected together and the
output is taken between this common point of diodes and the midpoint
of the secondary of the transformer. So for a full-wave rectifier the
secondary of the transformer is provided with a centre tapping and so it
is called centre-tap transformer. As can be seen from Fig.14.19(c) the
voltage rectified by each diode is only half the total secondary voltage.
Each diode rectifies only for half the cycle, but the two do so for alternate
cycles. Thus, the output between their common terminals and the centre-
tap of the transformer becomes a full-wave rectifier output. (Note that
there is another circuit of full wave rectifier which does not need a centre-
338 tap transformer but needs four diodes.) Suppose the input voltage to A

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with respect to the centre tap at any instant
is positive. It is clear that, at that instant,
voltage at B being out of phase will be
negative as shown in Fig.14.19(b). So, diode
D1 gets forward biased and conducts (while
D2 being reverse biased is not conducting).
Hence, during this positive half cycle we get
an output current (and a output voltage
across the load resistor RL) as shown in
Fig.14.19(c). In the course of the ac cycle
when the voltage at A becomes negative with
respect to centre tap, the voltage at B would
be positive. In this part of the cycle diode
D1 would not conduct but diode D2 would,
giving an output current and output
voltage (across RL ) during the negative half
cycle of the input ac. Thus, we get output
voltage during both the positive as well as
the negative half of the cycle. Obviously,
this is a more efficient circuit for getting
rectified voltage or current than the half-
wave rectifier.
The rectified voltage is in the form of
pulses of the shape of half sinusoids.
Though it is unidirectional it does not have
a steady value. To get steady dc output
from the pulsating voltage normally a
capacitor is connected across the output
terminals (parallel to the load RL). One can
also use an inductor in series with RL for
the same purpose. Since these additional FIGURE 14.19 (a) A Full-wave rectifier
circuits appear to filter out the ac ripple circuit; (b) Input wave forms given to the
diode D1 at A and to the diode D2 at B;
and give a pure dc voltage, so they are
(c) Output waveform across the
called filters. load RL connected in the full-wave
Now we shall discuss the role of rectifier circuit.
capacitor in filtering. When the voltage
across the capacitor is rising, it gets
charged. If there is no external load, it remains charged to the peak voltage
of the rectified output. When there is a load, it gets discharged through
the load and the voltage across it begins to fall. In the next half-cycle of
rectified output it again gets charged to the peak value (Fig. 14.20). The
rate of fall of the voltage across the capacitor depends inversely upon the
product of capacitance C and the effective resistance RL used in the circuit
and is called the time constant. To make the time constant large value of
C should be large. So capacitor input filters use large capacitors. The
output voltage obtained by using capacitor input filter is nearer to the
peak voltage of the rectified voltage. This type of filter is most widely
used in power supplies. 339

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FIGURE 14.20 (a) A full-wave rectifier with capacitor filter, (b) Input and output
voltage of rectifier in (a).

SUMMARY

1. Semiconductors are the basic materials used in the present solid state
electronic devices like diode, transistor, ICs, etc.
2. Lattice structure and the atomic structure of constituent elements
decide whether a particular material will be insulator, metal or
semiconductor.
3. Metals have low resistivity (10–2 to 10–8 W m), insulators have very high
resistivity (>108 W m–1), while semiconductors have intermediate values
of resistivity.
4. Semiconductors are elemental (Si, Ge) as well as compound (GaAs,
CdS, etc.).
5. Pure semiconductors are called ‘intrinsic semiconductors’. The presence
of charge carriers (electrons and holes) is an ‘intrinsic’ property of the
material and these are obtained as a result of thermal excitation. The
number of electrons (ne ) is equal to the number of holes (nh ) in intrinsic
conductors. Holes are essentially electron vacancies with an effective
positive charge.
6. The number of charge carriers can be changed by ‘doping’ of a suitable
impurity in pure semiconductors. Such semiconductors are known as
extrinsic semiconductors. These are of two types (n-type and p-type).
7. In n-type semiconductors, ne >> nh while in p-type semiconductors nh >> ne.
8. n-type semiconducting Si or Ge is obtained by doping with pentavalent
atoms (donors) like As, Sb, P, etc., while p-type Si or Ge can be obtained
by doping with trivalent atom (acceptors) like B, Al, In etc.
9. nenh = ni2 in all cases. Further, the material possesses an overall charge
neutrality.
10. There are two distinct band of energies (called valence band and
conduction band) in which the electrons in a material lie. Valence
band energies are low as compared to conduction band energies. All
energy levels in the valence band are filled while energy levels in the
conduction band may be fully empty or partially filled. The electrons in
the conduction band are free to move in a solid and are responsible for
the conductivity. The extent of conductivity depends upon the energy
gap (Eg ) between the top of valence band (EV ) and the bottom of the
conduction band EC. The electrons from valence band can be excited by
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heat, light or electrical energy to the conduction band and thus, produce
a change in the current flowing in a semiconductor.
11. For insulators Eg > 3 eV, for semiconductors Eg is 0.2 eV to 3 eV, while
for metals Eg » 0.
12. p-n junction is the ‘key’ to all semiconductor devices. When such a
junction is made, a ‘depletion layer’ is formed consisting of immobile
ion-cores devoid of their electrons or holes. This is responsible for a
junction potential barrier.
13. By changing the external applied voltage, junction barriers can be
changed. In forward bias (n-side is connected to negative terminal of the
battery and p-side is connected to the positive), the barrier is decreased
while the barrier increases in reverse bias. Hence, forward bias current
is more (mA) while it is very small (mA) in a p-n junction diode.
14. Diodes can be used for rectifying an ac voltage (restricting the ac voltage
to one direction). With the help of a capacitor or a suitable filter, a dc
voltage can be obtained.

POINTS TO PONDER
1. The energy bands (EC or EV ) in the semiconductors are space delocalised
which means that these are not located in any specific place inside the
solid. The energies are the overall averages. When you see a picture in
which EC or EV are drawn as straight lines, then they should be
respectively taken simply as the bottom of conduction band energy levels
and top of valence band energy levels.
2. In elemental semiconductors (Si or Ge), the n-type or p-type
semiconductors are obtained by introducing ‘dopants’ as defects. In
compound semiconductors, the change in relative stoichiometric ratio
can also change the type of semiconductor. For example, in ideal GaAs
the ratio of Ga:As is 1:1 but in Ga-rich or As-rich GaAs it could
respectively be Ga1.1 As0.9 or Ga0.9 As1.1. In general, the presence of
defects control the properties of semiconductors in many ways.

EXERCISES
14.1 In an n-type silicon, which of the following statement is true:
(a) Electrons are majority carriers and trivalent atoms are the
dopants.
(b) Electrons are minority carriers and pentavalent atoms are the
dopants.
(c) Holes are minority carriers and pentavalent atoms are the
dopants.
(d) Holes are majority carriers and trivalent atoms are the dopants.
14.2 Which of the statements given in Exercise 14.1 is true for p-type
semiconductos.
14.3 Carbon, silicon and germanium have four valence electrons each.
These are characterised by valence and conduction bands separated
341

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by energy band gap respectively equal to (Eg)C, (Eg)Si and (Eg)Ge. Which
of the following statements is true?
(a) (Eg)Si < (Eg)Ge < (Eg)C
(b) (Eg)C < (Eg)Ge > (Eg)Si
(c) (Eg)C > (Eg)Si > (Eg)Ge
(d) (Eg)C = (Eg)Si = (Eg)Ge
14.4 In an unbiased p-n junction, holes diffuse from the p-region to
n-region because
(a) free electrons in the n-region attract them.
(b) they move across the junction by the potential difference.
(c) hole concentration in p-region is more as compared to n-region.
(d) All the above.
14.5 When a forward bias is applied to a p-n junction, it
(a) raises the potential barrier.
(b) reduces the majority carrier current to zero.
(c) lowers the potential barrier.
(d) None of the above.
14.6 In half-wave rectification, what is the output frequency if the input
frequency is 50 Hz. What is the output frequency of a full-wave rectifier
for the same input frequency.

342

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APPENDICES

APPENDIX A 1
THE GREEK ALPHABET

APPENDIX A 2
COMMON SI PREFIXES AND SYMBOLS FOR MULTIPLES AND SUB-MULTIPLES

344

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Answers
APPENDIX A 3
SOME IMPORTANT CONSTANTS

OTHER USEFUL CONSTANTS

345

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ANSWERS

CHAPTER 9

9.1 v = –54 cm. The image is real, inverted and magnified. The size of the
image is 5.0 cm. As u ® f, v ® ¥; for u < f, image is virtual.
9.2 v = 6.7 cm. Magnification = 5/9, i.e., the size of the image is 2.5 cm. As
u ® ¥; v ® f (but never beyond) while m ® 0.
9.3 1.33; 1.7 cm
9.4 nga = 1.51; n wa = 1.32; ngw = 1.144; which gives sin r = 0.6181 i.e.,
r ~ 38°.
9.5 r = 0.8 × tan ic and sin i c = 1/1.33 ≅ 0.75 , where r is the radius (in m)
of the largest circle from which light comes out and ic is the critical
angle for water-air interface, Area = 2.6 m2
9.6 n ≅ 1.53 and Dm for prism in water ≅ 10°
9.7 R = 22 cm
9.8 Here the object is virtual and the image is real. u = +12 cm (object on
right; virtual)
(a) f = +20 cm. Image is real and at 7.5 cm from the lens on its right
side.
(b) f = –16 cm. Image is real and at 48cm from the lens on its right side.
9.9 v = 8.4 cm, image is erect and virtual. It is diminished to a size
1.8 cm. As u ® ¥, v ® f (but never beyond f while m ® 0).
Note that when the object is placed at the focus of the concave lens
(21 cm), the image is located at 10.5 cm (not at infinity as one might
wrongly think).
9.10 A diverging lens of focal length 60 cm
9.11 (a) ve = –25 cm and fe = 6.25 cm give ue = –5 cm; vO = (15 – 5) cm =
10 cm,
fO = u O = – 2.5 cm; Magnifying power = 20
(b) u O = – 2.59 cm.
Magnifying power = 13.5.
9.12 Angular magnification of the eye-piece for image at 25 cm
25 25
  1  11; | u e |=
cm = 2.27cm ; v = 7.2 cm
2.5 11 O

Separation = 9.47 cm; Magnifying power = 88

9.13 24; 150 cm
9.14 (a) Angular magnification = 1500
346 (b) Diameter of the image = 13.7 cm.

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9.15 Apply mirror equation and the condition:
(a) f < 0 (concave mirror); u < 0 (object on left)
(b) f > 0; u < 0
(c) f > 0 (convex mirror) and u < 0
(d) f < 0 (concave mirror); f < u < 0
to deduce the desired result.
9.16 The pin appears raised by 5.0 cm. It can be seen with an explicit ray
diagram that the answer is independent of the location of the slab
(for small angles of incidence).
9.17 (a) sin ic¢ = 1.44/1.68 which gives ic¢ = 59°. Total internal reflection
takes place when i > 59° or when r < r max = 31°. Now,
(sin i /sin r ) = 1.68 , which gives i ~ 60°. Thus, all
max max max
incident rays of angles in the range 0 < i < 60° will suffer total
internal reflections in the pipe. (If the length of the pipe is
finite, which it is in practice, there will be a lower limit on i
determined by the ratio of the diameter to the length of the
pipe.)
(b) If there is no outer coating, ic¢ = sin –1(1/1.68) = 36.5°. Now,
i = 90° will have r = 36.5° and i ¢ = 53.5° which is greater than
ic¢. Thus, all incident rays (in the range 53.5° < i < 90°) will
suffer total internal reflections.
9.18 For fixed distance s between object and screen, the lens equation
does not give a real solution for u or v if f is greater than s/4.
Therefore, fmax = 0.75 m.
9.19 21.4 cm
9.20 (a) (i) Let a parallel beam be the incident from the left on the convex
lens first.
f1 = 30 cm and u1 = –  , give v1 = + 30 cm. This image becomes
a virtual object for the second lens.
f 2 = –20 cm, u 2 = + (30 – 8) cm = + 22 cm which gives,
v2 = – 220 cm. The parallel incident beam appears to diverge
from a point 216 cm from the centre of the two-lens system.
(ii) Let the parallel beam be incident from the left on the concave
lens first: f1 = – 20 cm, u1 = – ¥, give v1 = – 20 cm. This image
becomes a real object for the second lens: f2 = + 30 cm, u2 =
– (20 + 8) cm = – 28 cm which gives, v2 = – 420 cm. The parallel
incident beam appears to diverge from a point 416 cm on the
left of the centre of the two-lens system.
Clearly, the answer depends on which side of the lens system
the parallel beam is incident. Further we do not have a simple lens
equation true for all u (and v) in terms of a definite constant of the
system (the constant being determined by f1 and f2, and the separation
between the lenses). The notion of effective focal length, therefore,
does not seem to be meaningful for this system.
(b) u1 = – 40 cm, f1 = 30 cm, gives v1= 120 cm.
Magnitude of magnification due to the first (convex) lens is 3.
u 2 = + (120 – 8) cm = +112 cm (object virtual);
112 × 20
f2 = – 20 cm which gives v2 = − cm
92
Magnitude of magnification due to the second (concave) 347

Rationalised 2023-24
 Physics
lens = 20/92.
Net magnitude of magnification = 0.652
Size of the image = 0.98 cm
9.21 If the refracted ray in the prism is incident on the second face at the
critical angle ic, the angle of refraction r at the first face is (60°–ic).
Now, ic = sin–1 (1/1.524) ~ 41°
Therefore, r = 19°
sin i = 0.4962; i ~ 30°
1 1 1
9.22 (a) + =
v 9 10
i.e., v = – 90 cm,
Magnitude of magnification = 90/9 = 10.
Each square in the virtual image has an area 10 × 10 × 1 mm2
= 100 mm2 = 1 cm2
(b) Magnifying power = 25/9 = 2.8
(c) No, magnification of an image by a lens and angular magnification
(or magnifying power) of an optical instrument are two separate
things. The latter is the ratio of the angular size of the object
(which is equal to the angular size of the image even if the image
is magnified) to the angular size of the object if placed at the near
point (25 cm). Thus, magnification magnitude is |(v/u )| and
magnifying power is (25/ |u|). Only when the image is located at
the near point |v| = 25 cm, are the two quantities equal.
9.23 (a) Maximum magnifying power is obtained when the image is at
the near point (25 cm)
u = – 7.14 cm.
(b) Magnitude of magnification = (25/ |u|) = 3.5.
(c) Magnifying power = 3.5
Yes, the magnifying power (when the image is produced at 25 cm)
is equal to the magnitude of magnification.
9.24 Magnification = (6.25 / 1) = 2.5
v = +2.5u
1 1 1
  
2.5u u 10
i.e.,u = – 6 cm
|v| = 15 cm
The virtual image is closer than the normal near point (25 cm) and
cannot be seen by the eye distinctly.
9.25 (a) Even though the absolute image size is bigger than the object
size, the angular size of the image is equal to the angular size of
the object. The magnifier helps in the following way: without it
object would be placed no closer than 25 cm; with it the object
can be placed much closer. The closer object has larger angular
size than the same object at 25 cm. It is in this sense that angular
magnification is achieved.
(b) Yes, it decreases a little because the angle subtended at the eye
is then slightly less than the angle subtended at the lens. The
348

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 Answers
effect is negligible if the image is at a very large distance away.
[Note: When the eye is separated from the lens, the angles
subtended at the eye by the first object and its image are not
equal.]
(c) First, grinding lens of very small focal length is not easy. More
important, if you decrease focal length, aberrations ( both spherical
and chromatic ) become more pronounced. So, in practice, you
cannot get a magnifying power of more than 3 or so with a simple
convex lens. However, using an aberration corrected lens system,
one can increase this limit by a factor of 10 or so.
(d) Angular magnification of eye-piece is [( 25/fe ) + 1] ( fe in cm) which
increases if fe is smaller. Further, magnification of the objective
vO 1
is given by =
|u O | (|u O |/ f O ) − 1

which is large when |u O | is slightly greater than fO. The micro-

scope is used for viewing very close object. So |u O | is small, and
so is fO.
(e) The image of the objective in the eye-piece is known as ‘eye-ring’.
All the rays from the object refracted by objective go through the
eye-ring. Therefore, it is an ideal position for our eyes for viewing.
If we place our eyes too close to the eye-piece, we shall not collect
much of the light and also reduce our field of view. If we position
our eyes on the eye-ring and the area of the pupil of our eye is
greater or equal to the area of the eye-ring, our eyes will collect
all the light refracted by the objective. The precise location of
the eye-ring naturally depends on the separation between the
objective and the eye-piece. When you view through a microscope
by placing your eyes on one end,the ideal distance between the
eyes and eye-piece is usually built-in the design of the
instrument.
9.26 Assume microscope in normal use i.e., image at 25 cm. Angular
magnification of the eye-piece
25
= 1  6
5
Magnification of the objective
30
= 5
6
1 1 1
− =
5u O u O 1.25

which gives uO= –1.5 cm; v0= 7.5 cm. |u e| (25/6) cm = 4.17 cm. The
separation between the objective and the eye-piece should be (7.5 +
4.17) cm = 11.67 cm. Further the object should be placed 1.5 cm from
the objective to obtain the desired magnification.
9.27 (a) m = ( fO/fe ) = 28

fO  fO 
(b) m = 1 + 25  = 33.6
fe
349

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 Physics
9.28 (a) fO + fe = 145 cm
(b) Angle subtended by the tower = (100/3000) = (1/30) rad.
Angle subtended by the image produced by the objective
h h
= =
f O 140
Equating the two, h = 4.7 cm.
(c) Magnification (magnitude) of the eye-piece = 6. Height of the
final image (magnitude) = 28 cm.
9.29 The image formed by the larger (concave) mirror acts as virtual object
for the smaller (convex) mirror. Parallel rays coming from the object
at infinity will focus at a distance of 110 mm from the larger mirror.
The distance of virtual object for the smaller mirror = (110 – 20) =
90 mm. The focal length of smaller mirror is 70 mm. Using the mirror
formula, image is formed at 315 mm from the smaller mirror.
9.30 The reflected rays get deflected by twice the angle of rotation of the
mirror. Therefore, d/1.5 = tan 7°. Hence d = 18.4 cm.
9.31 n = 1.33

CHAPTER 10
10.1 (a) Reflected light: (wavelength, frequency, speed same as incident
light)
l = 589 nm, n = 5.09 ´ 1014 Hz, c = 3.00 ´ 108 m s–1
(b) Refracted light: (frequency same as the incident frequency)
n = 5.09 ´ 1014Hz
v = (c/n) = 2.26 × 108 m s–1, l = (v/n ) = 444 nm
10.2 (a) Spherical
(b) Plane
(c) Plane (a small area on the surface of a large sphere is nearly
planar).
10.3 (a) 2.0 × 108 m s–1
(b) No. The refractive index, and hence the speed of light in a
medium, depends on wavelength. [When no particular
wavelength or colour of light is specified, we may take the given
refractive index to refer to yellow colour.] Now we know violet
colour deviates more than red in a glass prism, i.e. nv > nr.
Therefore, the violet component of white light travels slower than
the red component.

–2 –3
1.2 10  0.28 10
10.4  m = 600 nm
4 1.4

10.5 K/4
10.6 (a) 1.17 mm (b) 1.56 mm
10.7 0.15°
350 10.8 tan–1(1.5) ~ 56.3o

Rationalised 2023-24
 Answers
10.9 5000 Å, 6 × 1014 Hz; 45°
10.10 40 m

CHAPTER 11
11.1 (a) 7.24 × 1018 Hz (b) 0.041 nm
11.2 (a) 0.34 eV = 0.54 × 10–19J (b) 0.34 V (c) 344 km/s
11.3 1.5 eV = 2.4 × 10–19 J
11.4 (a) 3.14 × 10–19J, 1.05 × 10–27 kg m/s (b) 3 × 1016 photons/s
(c) 0.63 m/s
11.5 6.59 × 10–34 J s
11.6 2.0 V
11.7 No, because n < no
11.8 4.73 × 1014 Hz
11.9 2.16 eV = 3.46 × 10–19J
11.10 (a) 1.7 × 10–35 m (b) 1.1 × 10–32 m (c) 3.0 × 10–23 m
11.11 l = h/p = h/(hn/c) = c/n

CHAPTER 12
12.1 (a) No different from
(b) Thomson’s model; Rutherford’s model
(c) Rutherford’s model
(d) Thomson’s model; Rutherford’s model
(e) Both the models
12.2 The nucleus of a hydrogen atom is a proton. The mass of it is
1.67 × 10–27 kg, whereas the mass of an incident a-particle is
6.64 × 10–27 kg. Because the scattering particle is more massive than
the target nuclei (proton), the a-particle won’t bounce back in even
in a head-on collision. It is similar to a football colliding with a tenis
ball at rest. Thus, there would be no large-angle scattering.
12.3 5.6 ´ 1014 Hz
12.4 13.6 eV; –27.2 eV
12.5 9.7 × 10 – 8 m; 3.1 × 1015 Hz.
12.6 (a) 2.18 × 106 m/s; 1.09 × 106 m/s; 7.27 × 105 m/s
(b) 1.52 × 10–16 s; 1.22 × 10–15 s; 4.11 × 10–15 s.
12.7 2.12´10–10 m; 4.77 ´ 10–10 m
12.8 Lyman series: 103 nm and 122 nm; Balmer series: 656 nm.
12.9 2.6 × 1074

CHAPTER 13
13.1 104.7 MeV
13.2 8.79 MeV, 7.84 MeV
13.3 1.584 × 1025 MeV or 2.535×1012J
13.4 1.23 351

Rationalised 2023-24
 Physics
13.5 (i) Q = – 4.03 MeV; endothermic
(ii) Q = 4.62 MeV; exothermic
13.6 Q= m ( 56
26 )
Fe – 2m ( 28
13 )
Al = 26.90 MeV; not possible.
26
13.7 4.536 × 10 MeV
13.8 About 4.9 × 104 y
13.9 360 KeV

CHAPTER 14
14.1 (c)
14.2 (d)
14.3 (c)
14.4 (c)
14.5 (c)
14.6 50 Hz for half-wave, 100 Hz for full-wave

352

Rationalised 2023-24
 Bibligraphy

BIBLIOGRAPHY

TEXTBOOKS

For additional reading on the topics covered in this book, you may like to consult one or more of the following
books. Some of these books however are more advanced and contain many more topics than this book.

1 Ordinary Level Physics, A.F. Abbott, Arnold-Heinemann (1984).

2 Advanced Level Physics, M. Nelkon and P. Parker, 6th Edition, Arnold-Heinemann (1987).
3 Advanced Physics, Tom Duncan, John Murray (2000).
4 Fundamentals of Physics, David Halliday, Robert Resnick and Jearl Walker, 7th Edition
John Wily (2004).
5 University Physics (Sears and Zemansky’s), H.D. Young and R.A. Freedman, 11th Edition,
Addison—Wesley (2004).
6 Problems in Elementary Physics, B. Bukhovtsa, V. Krivchenkov, G. Myakishev and
V. Shalnov, MIR Publishers, (1971).
7 Lectures on Physics (3 volumes), R.P. Feynman, Addision – Wesley (1965).
8 Berkeley Physics Course (5 volumes) McGraw Hill (1965).
a. Vol. 1 – Mechanics: (Kittel, Knight and Ruderman)
b. Vol. 2 – Electricity and Magnetism (E.M. Purcell)
c. Vol. 3 – Waves and Oscillations (Frank S. Crawford)
d. Vol. 4 – Quantum Physics (Wichmann)
e. Vol. 5 – Statistical Physics (F. Reif )
9 Fundamental University Physics, M. Alonso and E. J. Finn, Addison – Wesley (1967).
10 College Physics, R.L. Weber, K.V. Manning, M.W. White and G.A. Weygand, Tata McGraw
Hill (1977).
11 Physics: Foundations and Frontiers, G. Gamow and J.M. Cleveland, Tata McGraw Hill
(1978).
12 Physics for the Inquiring Mind, E.M. Rogers, Princeton University Press (1960).
13 PSSC Physics Course, DC Heath and Co. (1965) Indian Edition, NCERT (1967).
14 Physics Advanced Level, Jim Breithampt, Stanley Thornes Publishers (2000).
15 Physics, Patrick Fullick, Heinemann (2000).
16 Conceptual Physics, Paul G. Hewitt, Addision—Wesley (1998).
17 College Physics, Raymond A. Serway and Jerry S. Faughn, Harcourt Brace and Co. (1999).
18 University Physics, Harris Benson, John Wiley (1996).
19 University Physics, William P. Crummet and Arthur B. Western, Wm.C. Brown (1994).
20 General Physics, Morton M. Sternheim and Joseph W. Kane, John Wiley (1988).
21 Physics, Hans C. Ohanian, W.W. Norton (1989).

Rationalised 2023-24
 Physics
22 Advanced Physics, Keith Gibbs, Cambridge University Press (1996).
23 Understanding Basic Mechanics, F. Reif, John Wiley (1995).
24 College Physics, Jerry D. Wilson and Anthony J. Buffa, Prentice Hall (1997).
25 Senior Physics, Part – I, I.K. Kikoin and A.K. Kikoin, MIR Publishers (1987).
26 Senior Physics, Part – II, B. Bekhovtsev, MIR Publishers (1988).
27 Understanding Physics, K. Cummings, Patrick J. Cooney, Priscilla W. Laws and Edward F.
Redish, John Wiley (2005).
28 Essentials of Physics, John D. Cutnell and Kenneth W. Johnson, John Wiley (2005).

GENERAL BOOKS

For instructive and entertaining general reading on science, you may like to read some of the following books.
Remember however, that many of these books are written at a level far beyond the level of the present book.

1 Mr. Tompkins in paperback, G. Gamow, Cambridge University Press (1967).

2 The Universe and Dr. Einstein, C. Barnett, Time Inc. New York (1962).
3 Thirty years that Shook Physics, G. Gamow, Double Day, New York (1966).
4 Surely You’re Joking, Mr. Feynman, R.P. Feynman, Bantam books (1986).
5 One, Two, Three… Infinity, G. Gamow, Viking Inc. (1961).
6 The Meaning of Relativity, A. Einstein, (Indian Edition) Oxford and IBH Pub. Co. (1965).
7 Atomic Theory and the Description of Nature, Niels Bohr, Cambridge (1934).
8 The Physical Principles of Quantum Theory, W. Heisenberg, University of Chicago Press
(1930).
9 The Physics—Astronomy Frontier, F. Hoyle and J.V. Narlikar, W.H. Freeman (1980).
10 The Flying Circus of Physics with Answer, J. Walker, John Wiley and Sons (1977).
11 Physics for Everyone (series), L.D. Landau and A.I. Kitaigorodski, MIR Publisher (1978).
Book 1: Physical Bodies
Book 2: Molecules
Book 3: Electrons
Book 4: Photons and Nuclei.
12 Physics can be Fun, Y. Perelman, MIR Publishers (1986).
13 Power of Ten, Philip Morrison and Eames, W.H. Freeman (1985).
14 Physics in your Kitchen Lab., I.K. Kikoin, MIR Publishers (1985).
15 How Things Work: The Physics of Everyday Life, Louis A. Bloomfield, John Wiley (2005).
16 Physics Matters: An Introduction to Conceptual Physics, James Trefil and Robert M.
Hazen, John Wiley (2004).

354

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 ANSWERS

CHAPTER 1

1.1 6 × 10–3 N (repulsive)

1.2 (a) 12 cm
(b) 0.2 N (attractive)
1.3 2.4 × 1039. This is the ratio of electric force to the gravitational force
(at the same distance) between an electron and a proton.
1.5 Charge is not created or destroyed. It is merely transferred from one
body to another.
1.6 Zero N
1.8 (a) 5.4 × 106 N C–1 along OB
(b) 8.1 × 10–3 N along OA
1.9 Total charge is zero. Dipole moment = 7.5 × 10–8 C m along z-axis.
1.10 10–4 N m
1.11 (a) 2 × 1012, from wool to polythene.
(b) Yes, but of a negligible amount ( = 2 × 10–18 kg in the example).
1.12 (a) 1.5 × 10–2 N
(b) 0.24 N
1.13 Charges 1 and 2 are negative, charge 3 is positive. Particle 3 has
the highest charge to mass ratio.
1.14 (a) 30Nm2/C, (b) 15 Nm2/C
1.15 Zero. The number of lines entering the cube is the same as the
number of lines leaving the cube.
1.16 (a) 0.07 mC
(b) No, only that the net charge inside is zero.
1.17 2.2 × 105 N m2/C
1.18 1.9 × 105 N m2/C
1.19 (a) –103 N m2/C; because the charge enclosed is the same in the
two cases.
(b) –8.8 nC
1.20 – 6.67 nC
1.21 (a) 1.45 × 10–3 C
(b) 1.6 × 108 Nm2/C
1.22 10 mC/m
1.23 (a) Zero, (b) Zero, (c) 1.9 N/C

Rationalised 2023-24
 Physics
CHAPTER 2

2.1 10 cm, 40 cm away from the positive charge on the side of the
negative charge.
2.2 2.7 × 106 V
2.3 (a) The plane normal to AB and passing through its mid-point has
zero potential everywhere.
(b) Normal to the plane in the direction AB.
2.4 (a) Zero
(b) 105 N C–1
(c) 4.4 × 104 N C–1
2.5 96 pF
2.6 (a) 3 pF
(b) 40 V
2.7 (a) 9 pF
(b) 2 × 10–10 C, 3 × 10–10 C, 4 × 10–10 C
2.8 18 pF, 1.8 × 10–9 C
2.9 (a) V = 100 V, C = 108 pF, Q = 1.08 × 10–8 C
(b) Q = 1.8 × 10–9 C, C = 108 pF, V = 16.6 V
2.10 1.5 × 10–8 J
2.11 6 × 10–6 J

CHAPTER 3

3.1 30 A
3.2 17 W, 8.5 V
3.3 1027 °C
3.4 2.0 ´ 10–7 W m
3.5 0.0039 °C–1
3.6 867 °C
3.7 Current in branch AB = (4/17) A,
in BC = (6/17) A, in CD = (–4/17) A,
in AD = (6/17) A, in BD. = (–2/17) A, total current = (10/17) A.
3.8 11.5 V; the series resistor limits the current drawn from the external
source. In its absence, the current will be dangerously high.
3.9 2.7 ´ 104 s (7.5 h)

CHAPTER 4

4.1 p × 10–4 T ≃ 3.1 × 10–4 T

4.2 3.5 × 10–5 T
4.3 4 × 10–6 T, vertical up

216 4.4 1.2 × 10–5 T, towards south

Rationalised 2023-24
 Answers
4.5 0.6 N m–1
4.6 8.1 × 10–2 N; direction of force given by Fleming’s left-hand rule
4.7 2 × 10–5 N; attractive force normal to A towards B
4.8 8p × 10–3 T ≃ 2.5 × 10–2 T
4.9 0.96 N m
4.10 (a) 1.4, (b) 1
4.11 4.2 cm
4.12 18 MHz
4.13 (a) 3.1 Nm, (b) No, the answer is unchanged because the formula
t = N I A × B is true for a planar loop of any shape.

CHAPTER 5

5.1 0.36 J T –1
5.2 (a) m parallel to B; U = –mB = – 4.8 × 10–2 J: stable.
(b) m anti-parallel to B; U = +mB = +4.8 × 10–2 J; unstable.
5.3 0.60 J T –1 along the axis of the solenoid determined by the sense of
flow of the current.
5.4 7.5 ×10–2 J
5.5 (a) (i) 0.33 J (ii) 0.66 J
(b) (i) Torque of magnitude 0.33 J in a direction that tends to align
the magnitude moment vector along B. (ii) Zero.
5.6 (a) 1.28 A m2 along the axis in the direction related to the sense of
current via the right-handed screw rule.
(b) Force is zero in uniform field; torque = 0.048 Nm in a direction
that tends to align the axis of the solenoid (i.e., its magnetic
moment vector) along B.
5.7 (a) 0.96 g along S-N direction.
(b) 0.48 G along N-S direction.

CHAPTER 6

6.1 (a) Along qrpq

(b) Along prq, along yzx
(c) Along yzx
(d) Along zyx
(e) Along xry
(f ) No induced current since field lines lie in the plane of the loop.
6.2 (a) Along adcd (flux through the surface increases during shape
change, so induced current produces opposing flux).
(b) Along a¢d¢c¢b¢ (flux decreases during the process)
6.3 7.5 × 10–6 V
6.4 (1) 2.4 × 10–4 V, lasting 2 s 217

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 Physics
(2) 0.6 × 10–4 V, lasting 8 s
6.5 100 V
6.6 (a) 1.5 × 10–3 V, (b) West to East, (c) Eastern end.
6.7 4H
6.8 30 Wb

CHAPTER 7

7.1 (a) 2.20 A

(b) 484 W
300
7.2 (a)  212.1 V
2

(b) 10 2  14.1 A

7.3 15.9 A
7.4 2.49 A
7.5 Zero in each case.
7.6 125 s–1; 25
7.7 1.1 ´ 103 s–1
7.8 0.6 J, same at later times.
7.9 2,000 W

1 1 1
7.10 ν= , i.e., C =
2π LC 4 π 2ν 2 L
For L = 200 mH, n = 1200 kHz, C = 87.9 pF.
For L = 200 mH, n = 800 kHz, C = 197.8 pF.
The variable capacitor should have a range of about 88 pF to 198 pF.
7.11 (a) 50 rad s–1
(b) 40 W, 8.1 A
(c) VLrms  1437.5 V, VCrms  1437.5 V , V Rrms  230 V

 1 
V LCrms = I rms  ω 0 L −  =0
 ω0 C 

CHAPTER 8

8.1 (a) C = ε0 A / d = 8.00 pF

dQ dV
=C
dt dt

dV 0.15
= = 1.87 × 10 9 V s –1
218 dt 80.1 × 10 –12

Rationalised 2023-24
 Answers
d
(b) i d = ε0 ΦΕ. . Now across the capacitor FE = EA, ignoring end
dt
corrections.

dΦΕ
Therefore, i d = ε0 A
dt

Q dE i
Now, E = . Therefore, = , which implies id = i = 0.15 A.
εA
0
dt ε0 A

(c) Yes, provided by ‘current’ we mean the sum of conduction and

displacement currents.
8.2 (a) Irms = Vrms wC = 6.9mA
(b) Yes. The derivation in Exercise 8.1(b) is true even if i is oscillating
in time.

µ0 r
(c) The formula B = id
2π R 2

goes through even if id (and therefore B ) oscillates in time. The

formula shows they oscillate in phase. Since id = i, we have
µ r
B0 = 0 2 i 0 , where B and i are the amplitudes of the oscillating
2π R 0 0

magnetic field and current, respectively. i0= 2I rms = 9.76 mA. For
r = 3 cm, R = 6 cm, B0 = 1.63 × 10–11 T.
8.3 The speed in vacuum is the same for all: c = 3 ´ 108 m s–1.
8.4 E and B in x-y plane and are mutually perpendicular, 10 m.
8.5 Wavelength band: 40 m – 25 m.
8.6 109 Hz
8.7 153 N/C
8.8 (a) 400 nT, 3.14 ´ 108 rad/s, 1.05 rad/m, 6.00 m.
(b) E = { (120 N/C) sin[(1.05 rad/m)]x – (3.14 ´ 108 rad/s)t]} ĵ
B = { (400 nT) sin[(1.05 rad/m)]x – (3.14 ´ 108 rad/s)t ]} k̂
8.9 Photon energy (for l = 1 m)

6.63 × 10−34 × 3 × 108

= eV = 1.24 × 10 −6 eV
1.6 × 10 −19
Photon energy for other wavelengths in the figure for electromagnetic
spectrum can be obtained by multiplying approximate powers of
ten. Energy of a photon that a source produces indicates the spacings
of the relevant energy levels of the source. For example, l = 10–12 m
corresponds to photon energy = 1.24 ´ 106 eV = 1.24 MeV. This
indicates that nuclear energy levels (transition between which causes
g-ray emission) are typically spaced by 1 MeV or so. Similarly, a
visible wavelength l = 5 ´ 10–7 m, corresponds to photon energy
= 2.5 eV. This implies that energy levels (transition between which
219
gives visible radiation) are typically spaced by a few eV.

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 Physics
8.10 (a) l = (c/n) = 1.5 ´ 10–2 m
(b) B0 = (E0/c) = 1.6 ´ 10–7 T
(c) Energy density in E field: uE = (1/2)e0 E 2
Energy density in B field: uB = (1/2m0)B 2
1
Using E = cB, and c = , uE = uB
µ0 ε 0

220

Rationalised 2023-24