What is Magnetism?

Magnetism is the force of attraction or repulsion

of a magnetic material due to the arrangement of
its atoms, particularly its electrons.

The Domain Theory aims to explain why

metals get magnetised. You need to think
of the magnetic elements having little
molecular magnets inside them. These
are randomly orientated in an unmagnetized
piece of metal but point in a particular direction
in a magnetised piece.
Magnetic Fields

The region where the magnetic forces

act is called the “magnetic field”
Field Lines Around a Magnetic Sphere
 Defining Magnetic Field Direction
Magnetic Field vectors as written as B

Magnitude of the B-vector is proportional to the force acting on the moving

charge, magnitude of the moving charge, the magnitude of its velocity, and
the angle between v and the B-field.
Unit is the Tesla or the Gauss (1 T = 104 G).

F=q (v×B)

F= qvBsinθ
 Magnetic Field Lines
• The direction of the magnetic field at any point is tangent to the
magnetic field line at that point.
N

S
N

S
A vector area is the area under consideration,
and whose direction is perpendicular to the
surface area
 Electromagnetic Induction
•The production of an electromotive force (i.e., voltage) across an electrical
conductor in a changing magnetic field.

•Faraday demonstrated that when the magnetic flux linking a conductor changes, an emf is
induced in the conductor and hence current will flow
in it. This phenomenon is called electromagnetic induction

•Lenz's law describes the direction of the induced field.

Electromagnetic induction has found many applications, including electrical components such
as inductors and transformers, and devices such as electric motors and generators.
 Faraday's law of induction and Lenz's law

The induced electromotive force(emf) in any closed circuit is equal to the rate of
change of the magnetic flux enclosed by the circuit:

For N turns of wire,

N

ɛ is the EMF and ΦB is the magnetic flux

The direction of the electromotive force is given by Lenz's law which states that an induced current
will flow in the direction that will oppose the change which produced it. So –ve sign is used.
When charged particles are stationary they have
no magnetic field around them, but when they
move they have!

F=Bqvsinθ
All magnetic phenomena result
from forces between electric
charges in motion.
Not steady magnetic field but change in magnetic field
Mag lines

When switch is closed, current will flow and that creates magnetic field
Magnetic lines
force
Loop 2,around solenoid

Loop 1

Whenever, current flows through loop1,mf produced on solenoid,

But no current is seen in loop2
It is a change in the magnetic field flux(varying field) that results in an induced electromotive
force (or voltage) and hence current is observed passing through loop2
• When the switch is closed, a magnetic field is produced in the coil on the top part of the iron ring and
transmitted (or guided) to the coil on the bottom part of the ring. And galvanometer detects a current in
one direction in the coil on the bottom

•Each time the switch is opened, the galvanometer detects a current in the opposite direction

•Closing and opening the switch induces the current. It is the change in magnetic field that creates
the current.
 Lenz’s Law
Statement: Direction of induced current is such that it opposes the cause which produces it

An aluminium pipe is NOT magnetic but when a magnet is dropped through it , travels
slower than it does through a glass tube, Why ?

Because an electric current is induced in the aluminium pipe that produces a field that
opposes the field of the magnet - the repulsion therefore acts to oppose its movement and it
falls slowly...
 Needle deflection of galvanometer caused by a
magnetic field force acting upon a current-carrying
wire.
•A magnet and a coil of wire can be used to produce an electric current. A voltage is
produced when a magnet moves into a coil of wire. This process is called electromagnetic
induction.
•The direction of the induced voltage is reversed when the magnet is moved out of the coil again.
It can also be reversed if the other pole of the magnet is moved into the coil.
Conservation of energy
Magnet is moving as a result current is getting induced and when the magnet is
not moving there is no induced current in the coil.
There is a mechanical energy involved with the movement of the as magnet as a
result mechanical energy is converted into electrical energy which is produced
by the induced current.
Magnetic energy used in moving magnet is getting transformed into electrical
energy which is produced by the induced current.
This shows that Lenz’s law supports conservation of energy where one form of
energy is getting transformed into another form.
Work done in moving the magnet is dissipated by Joule heating produced by the
induced current.
•Electromagnetic induction is the process by which a current can be induced to flow
due to a changing magnetic field.

•ɛ= dΦ Faraday's law

dt
•Lenz's law is typically incorporated into Faraday's law with a minus sign,
•It states that the direction is always such that it will oppose the change in flux which
produced it.
•ɛ= - dΦ
dt

For N turns, •ɛ= -N dΦ

dt
An emf induced by motion relative to a magnetic field is called a motional emf

If conductor of length L moving at right angle to

uniform B with velocity v then
Φ=B× area swept
= B × L x, where, x is small distance through which
conductor moves in time t

•ɛ= dΦ =BL dx =BLv

dt dt

Also, ɛ=BLvsinθ
Induced emf in a coil rotating in a magnetic field

Let us consider a rectangular coil of N turns kept in a

uniform magnetic field B figure (a). The coil rotates in
anti-clockwise direction with an angular velocity ω about
an axis, perpendicular to the field. At time t= 0, the plane
of the coil is perpendicular to the field and the flux linked
with the coil has its maximum value Φ = BA (where A is
the area of the coil).
Induced emf in a coil rotating in a magnetic field
In a time t seconds, the coil is rotated through an angle θ (= ωt) in anti-clockwise direction. In this position, the flux
linked is
Flux linked to coil Φ=N(B.A)
Φ=NBAcosθ, where θ=wt
Due to rotation, flux keep changing

ɛ= -dΦ =NBAwsinwt
dt
ɛ is maximum when sinwt=1
For which ɛo =NBA

ɛ= ɛo sinwt

Induced current I= ɛ = ɛo sinwt = Io sinwt

R R
The thumb indicates the direction of the motion,
the fore finger indicates the direction of the magnetic field and
the middle finger represents the direction of the induced current.
Generator: Device that converts mechanical energy into electrical energy
in form of alternative emf or alternating current.
Based on the output obtained, we classify generators as:
AC Generators Bidirectional
DC Generators unidirectional
The main work of this device is to convert mechanical energy into electrical energy and this
electrical energy is what we call as alternating current
AC generator works on the principle of
”Electromagnetic Induction”.

whenever amount of magnetic flux linked with a coil changes, an e.m.f. is induced
in the coil. The direction of current induced is given by Fleming’s right hand rule
Construction of The A.C. Generator
Construction of The A.C. Generator It’s main parts are:

(1) THE COIL (ARMATURE) :

A rectangular coil ABCD consist of a large number of turns of copper
bound over a soft iron core is called armature. The soft iron core is used to
increase the magnetic flux. Axis of rotation is perpendicular to the magnetic
field lines

(2) MAGNETIC FIELD : It is usually a permanent sponge magnet having

concave poles. The armature is rotated of a magnet so that axis of the armature
is perpendicular to magnetic field lines.
 3) SLIP RINGS : Slip rings are the magnetic rings which are
connected in the terminal of the armature. These rings are
rotated with the coil and these are used to draw the current from
the generator.

4) BRUSHES : The brushes B1 & B2 just touch the slip rings.

They are not rotating with the coil. Fixed carbon brushes
electrically connect the armature to the external circuit
throughout the rotation lead to the output of load resistance

5)Load : Brushes are connected to a load resistance R in the

external circuit. Here load means the device through which
the generated current is to be passed.
•let the armature ABCD is initially in the
horizontal position in the magnetic field
• Now let it be rotated in the clockwise direction,
the side AB of the coil moves in the upward
direction and side CD moves down.

•According to Fleming's right hand rule, the

current induced flows from A to B in the arm
AB and C to D in the arm CD as shown in the
fig (a).

Thus, the induced current flows in the external

circuit through brush B1 to B2.
•After half the rotation of the armature, the arm
AB moves down and the arm CD moves up.
•Now the direction of induced current in the
armature is DCBA while in the external circuit,
current flows from B2 and B1 as shown in fig (b).

•Thus, in one complete rotation of the armature, in

half the rotation the current induced flows in the
direction ABCD and in the external circuit from
B1 to B2; in next half rotation of the armature, the
induced current flows in the direction DCBA and
in the external circuit from B2 to B1
Thus in one rotation of the coil the current changes
its direction twice alternately
 Expression for Instaneous e.m.f. produced

Let position of the coil at any time t and θ be angle. If 𝜔 is uniform

angular speed of the coil. Then θ =ωt, B be the strength of magnetic
field N be the number of turns in the coil and A area of the coil then
magnetic flux with the coil in this position is given by :

Φ = NBA Cosθ = NBA Cos ωt.

Differentiate w.r.t. time : The generated EMF depends on the number of armature
𝑑∅ =NBA 𝑑 cos(𝜔𝑡) coil turns, magnetic field strength and the speed of the
dt dt rotating field.
𝑑∅/𝑑𝑡 = NBA[-sin(𝜔𝑡)𝜔]
𝑑∅/𝑑𝑡 = -NBA 𝜔 sin(𝜔𝑡)
ε= - 𝑑∅/𝑑𝑡 =-[-NBA 𝜔 sin(𝜔𝑡)] =NBA𝜔 sin(𝜔𝑡)
Maximum value of e.m.f. say ɛo ,
so ɛ= ɛo sin(𝜔𝑡)
 Result :- Hence, maximum value of instantaneous e.m.f. in a AC generator is given
by ɛ= ɛo sin(𝜔𝑡)

•When the coil is perpendicular to field, induced emf

is the greatest

•When the coil is parallel to field, induced emf

is zero

•If the coil is rotated at constant speed,

then the induced emf will vary sinusoidally
with the same frequency as the rotation
Simple loop generator
with slip ring
Applications of A.C. Generator
1. Aircraft auxiliary power generation, wind generators, high speed gas turbine
generators.
2. Hybrid electric vehicle (HEV) drive systems, automotive starter generators.
3. An ac generator, or 'alternator', is used to produce ac voltages for transmission via
the grid system or, locally, as portable generators.
4. All of our household appliances runs on ac current. Ex: Refrigerator, washing
machines, oven, lights, fan etc.
5. The main advantage of AC is ease of power distribution. It is more efficient to use
high voltage to distribute power, but it is not safe to have high voltage at home. It is
easy to step up (and step down) AC voltage using a transformer.
Advantages of AC Generators Following are a few advantages of AC generators over
DC generators:

•AC generators can be easily stepped up and stepped down through

transformers.
•Transmission link size might be thinner because of the step-up
feature
•Losses are relatively lesser than DC machine
•Size of the AC generators are relatively smaller than DC generators

Some disadvantages of AC generators

•Generating large scale AC power using AC generators can be extremely hazardous.

•The flow of electricity through generator and transformer coils produces resistive heat. This
heat can damage the insulation and cause a fire.
 MCQs

1.When a coil rotated in magnetic field the induced current in it

a) Remains same b)Becomes zero
c) Becomes maximum d) continuously changes

Answer: d

2 The direction of induced emf or current is determined by

a)Fleming’s left hand rule b)Fleming’s right hand rule
c)Faraday’s law of electromagnetic induction d) None

Answer: a
•When a current-carrying conductor is placed under a magnetic field, a force acts on the
conductor. The direction of this force can be identified using Fleming’s Left Hand Rule.
•Fleming’s Left-Hand Rule is mainly applicable to electric motors
•Likewise, if a moving conductor is brought under a magnetic field, electric
current will be induced in that conductor. The direction of the induced current can
be found using Fleming’s Right Hand Rule.
•and Fleming’s Right-Hand Rule is mainly
applicable to electric generators.

Note: these rules do not determine the magnitude, instead show only the
direction of the three parameters (magnetic field, current, force) when the
direction of the other two parameters is known.
Simple loop generator with split ring
Q.The magnetic flux passing perpendicular to the plane of coil is given by φ = 4t2+5t+2
where φ is in weber and t is in second. Calculate the magnitude of instantaneous emf
induced in the coil when t = 2 sec.

Given,
magnetic flux, φ = 4t2+5t+2
time, t = 2 sec
induced emf, ε = ?

From Faraday's law, E=−dϕ/dt

E=−d(4t2+5t+2)/dt
E=−(8t+5)
At t = 2 sec, the induced emf is ,
E = −(8×2+5)
∴ Es=−21V
Thus, the magnitude of induced emf is 21V
Q. A straight conductor of length 25 cm is moving perpendicular to its length with a
uniform speed of 10 m/s making an angle of 450 with a uniform magnetic field of 10 T.
Calculate the emf induced across its length.
Solution:
Given,
length, l = 25 cm = 0.25 m
speed, v = 10 m/s
magnetic field, B = 10 T
angle between the velocity vector of motion of conductor and magnetic field, θ = 450
emf induced, E= ?

Now, the emf induced in a straight current carrying conductor is,

E=Blvsinθ
=10×0.25×10×sin45∘=17.67V
Thus, the emf induced across the length is 17.67 V .
Q. A coil of 100 turns, each of area 2×10-3 m2 has a resistance of 12Ω. It lies in a
horizontal plane in a vertical magnetic flux density of 3×10-3 Wbm-2. What charge
circulates through the coil if its ends are short-circuited and the coil is rotated through
1800 about a diametrical axis?
Solution:
Given,
number of turns of coil, N = 100 turns
area of each coil, A = 2×10-3 m2
resistance of coil, R = 12Ω
magnetic flux density, B = 3×10-3 Wbm-2
charge circulated through the coil, Δq = ?

We know, the emf induced in the coil can be written as,

We know, the emf induced in the coil can be written as,
E=IR
Δϕ/Δt=(Δq/Δt)R
Δq=Δϕ/R
The change in flux is,Δϕ=NBAcos0∘−NBAcos180∘=NBA+NBA=2NBA=2×100×3×10−3 ×2×10−3
Δϕ=1.2×10−3 wb
Now, the charge circulated through the coil is,

Now, the charge circulated through the coil is,Δq=Δϕ/R=1.2×10−3 / 12

∴Δq=10−4 C
Thus, the charge circulated through the coil is 10-4 C.
Q.Find the emf induced in a straight conductor of length 25 cm, on the armature of a dynamo
and 12 cm from the axis, when the conductor is moving in a uniform radial magnetic field of 0.5
T. The armature is rotating at 1000 revolutions per minute.
Solution:
Given,
length of conductor, l = 25 cm = 0.25 m
distance from the axis of rotation, r = 12 cm = 0.12 m
magnetic field, B = 0.5 T
rate of rotation, f = 1000 rev / min = 503 rev / s
emf induced, E= ?

Now,E=Blv=Blrω since, v=rw

=Blr×2πf ω= 2πf
=2πBlrf
=2π×0.5×0.25×0.12×503
∴E=1.57V
Thus, the emf induced in the conductor is 1.57 V .
Q. A rectangular coil of 100 turns has dimensions 15×10 cm. It is rotated at the rate of
300 revolutions per minute in a uniform magnetic field of flux density 0.6 T. Calculate
the maximum emf induced in it.
Solution:Given,
number of turns, N = 100
rate of rotation, f = 300 rev / min = 5 rev / s
magnetic field , B = 0.6 T
maximum emf induced ,E = ?
area, A = 150 cm2 = 150×10-4 m2
The emf induced in the coil is given by,
E=NBAωsinωt
For the maximum emf, sinωt=1.
So,E=NBAω
=100×0.6×150×10−4 ×2πf
=100×0.6×150×10−4×2π×5
∴ E=28.27V
Thus, the maximum emf induced in the coil is 28.27 V .
Q. When a wheel with metal spokes 1.2 m long is rotated in a magnetic field of flux density
5×10-5 T normal to the plane of wheel, an emf of 10-2 V is induced between the rim and axle.
Find the rate of rotation of the wheel.
Solution:Given,
length of metal spokes, r = 1.2 m
magnetic field, B = 5×10-5 T
emf induced, ε = 10-2 V
rate of rotation , f = ?
It is the case of motional emf. The emf induced is thus, E=Blv.
Here, the velocity is an average velocity (axle is at rest and rim is moving with maximum
velocity).So,v=(0+ωr)/2=ωr/2
Now,E=Blv=Brωr/2=Bωr2 /2
=B×2πfr2 /2
E=πBr2f
Or, 10−2 =π×5×10−5×(1.2)2×f
Or, f=10−2 / π×5×10−5×(1.2)2
∴f=44.2rev/s
Thus, the rate of rotation of the wheel is 44.2 rev/s .
 An inductor typically consists of an insulated wire wound into a coil. So, it
Inductor:
is also known as coil or choke or reactor.

The electrical symbol for an inductor is L.

L
Defination
An inductor, also called a coil, choke, or reactor, is an electrical component that store
magnetic energy in the form of a magnetic field generated by the current in the coil.
 Inductance is specified by the behavior of a coil of
Inductance wire in resisting any change of electric current
through the coil.

•Self Inductance

•Mutual Inductance
 Self Inductance

The ability of an inductor to induce emf (voltage ) in itself when the current changes is called
its self inductance or simply inductance. It is denoted by the letter L .

The unit of inductance is Henry (H).

Self-induced e.m.f is the e.m.f which is produced in the

coil due to the change of its own flux linked with it. If
the current of the coil is changed, then the flux linked
with its own turns will also change which will produce
an e.m.f that is called self-induced e.m.f.
If the current in the coil is increasing, the self-
induced emf produced in the coil will oppose the
rise of current, that means the direction of the
induced emf is opposite to the applied voltage.

If the current in the coil is decreasing, the emf

induced in the coil is in such a direction as to
oppose the fall of current; this means that the
direction of the self-induced emf is same as that of
the applied voltage.
.
This property of the coil only opposes the changing current (alternating current) and does not
affect the steady current that is (direct current) when flows through it.

Coils can store electrical energy in a form of

magnetic energy using the property that an electric
current flowing through a coil produces a magnetic
field, which in turn produces an electric current.
Expression for self inductance(L):

The magnetic flux (Φ) linked with a coil is directly

proportional to current(I) flowing through it.

ΦαI
Φ = LI
Where, L is proportionality constant known as coefficient of self induction
or self inductance.

From above, L=Φ/I, mathematically, self inductance is ratio of total magnetic flux
Linked to the current flowing in the coil.
 According to Faraday’s law of induction, induced emf due to change in flux is

E= -(dΦ/dt)
= -L(dI/ dt)

If dI/ dt= 1 AS-1 then, E= -L

Thus, Self inductance is also defined as (EMF) across a coil to unit rate of change
of current through the coil.

NOTE:Unit: Henry= Weber per Ampere (WbA-1 ) = Volt per Ampere second(VA-1S)

One Henry: It is defined as induced emf of 1 volt when the current changes at the rate of
1 ampere per second in the circuit.
 One Henry

We have, L = E/(dI/dt)

Where E is the induced emf in volts

dI/dt is the rate of change of current in amperes per second.

If dI/dt = 1 A/s and E = 1v, then L = 1H

Hence a coil has an inductance of one henry if an emf of one volt is induced in it when current
through it changes at the rate of 1 A/s.
NOTE:
•The value of the inductance will be high if the magnetic flux (Φ) is stronger for the given
value of current.

•The value of the Inductance also depends upon the material of the core and the number of
turns in the coil or solenoid.

•The higher will be the value of the inductance in Henry, the rate of change of current will be
lower.

1 Henry is also equal to 1 Weber/ampere

Q. A long solenoid of 1000 turns and cross-sectional area 2× 10-3 m2 carries a current of 2A
and produce a flux density 52 × 10-3 T in the middle of the coil. Assuming the value of flux
density at all sections of the solenoid, calculate its self inductance.

self inductance(L)= ?
L= Φ/I L= NBA/I
= 0.052 H
Φ=? Φ= NBA
Q. A long solenoid with 15 turns per cm has a small loop of area 2 cm2 placed inside, normal
to the axis of the solenoid. If the current carried by the solenoid changes steadily from
2A to 4A in 0.1 second, what is the induced voltage in the loop, while the current is changing?

Number of turns(n)= 15 turns/cm= 1500/m

Area(A)= 2 cm2 = 2 × 10-4 m2
Change in current (dI)= I2 – I1 = 4-2= 2 A
Time(dt)= 0.1 sec
Induced emf (E)=?

E= dΦ/ dt
= A µ0 n (dI/dt ) Φ= BA, for solenoid B=µ0 n I
= 7.5 × 10-6 V µ0 = 4π × 10-7 Hm-1
Energy Stored In An Inductor
When a current passes through an inductor an emf is induced in it. This back emf opposes
the flow of current through the inductor. So in order to establish a current in the inductor
work has to be done against this emf by the voltage source

Let us consider a time interval dt.

During this period, work done dW is given by
dW = Pdt ;Power(P)= work (dw)/time (dt)
= IEdt [ as P= IV ]
= I (L dI /dt) x dt [E= L dI /dt ]
= LI dI 1
To find the total work done, the above expression must be integrated.

W = ∫dW

= 0∫ILIdI Using equation (1)

= L 0∫IIdI

= ½ LI2
Therefore energy stored (U) in an inductor is given by the equation,
W =U= ½ LI2

U= ½ LI2
Q. A plane circular coil has 200 turns and its radius is 0.10m. It is connected to a battery. After
switching on the circuit a current of 2A is set up in the coil. Calculate the energy stored in the
Coil. (µ0 = 4π × 10-7 Hm-1 )

U= ½ LI2 I= 2A
L= ?
L= Φ/ I
Given,
number of turns, N = 200
radius of circular coil, r = 0.10 m
current, I = 2 A
energy stored in the coil , U = ?
The energy stored in the coil is given by, U=1/2 LI2
At first, we need to calculate the self inductance of the coil, L=ϕ/I, where ϕ= NBA, I = 2A
B= μ0N I/ 2r
So, L=NA×μ0NI
(I×2r)
=μ0N2A/2r
=μ0N2×πr2 /2r
=4π×10−7 ×2002×π×0.10/ 2
=7.89×10−3 H
So, U=12×7.89×10−3×(2)2
∴U=0.016J
Thus, the energy stored in the coil is 0.016 J .
Mutual Inductance (M):
Mutual Induction is the phenomenon of inducing emf in the secondary coil due to change in current in
the primary coil and hence the change in magnetic flux in the secondary coil
This effect is used in transformers.

The ability of varying current in one coil to induce voltage in a nearby coil is known as the mutual
inductance. It is denoted by the letter M or LM and its unit is Henry (H).
Two coils namely coil A and coil B are placed Secondary coil
nearer to each other. When the switch S is closed,
and the current flows in the coil, it sets up the flux
φ in the coil A and emf is induced in the coil and if
the value of the current is changed by varying the
value of the resistance (R), the flux linking with
the coil B also changes because of this changing
current.
Primary coil
Mutually induced e.m.f occurs in between two coils.
Let, A & B are two coils which are placed close to each
other. If coil A(primary coil) is joined to a battery and a
variable resistance R and coil B is connected to a
sensitive voltmeter G. When the switch S is closed , a
current will flow through the coil A and produce a
magnetic field in which partly links with the coil B. As
current through A is changed, the flux linked with B is
also changed. According to Faraday’s law, induced e.m.f
is produced in the coil B and This e.m.f is known as
mutually induced e.m.f.
In the above example, there is no movement of any
conductor, the flux variation being brought about by
variation in current strength Only. Such an e.m.f induced
in one coil by influence of the other coil is called
mutually induced e.m.f
 The phenomenon of inducing emf in the secondary
Mutual Induction: coil due to change in current in the primary coil is
called mutual induction.

Let us consider two coils; primary coil(A) and secondary coil(B) . Ep

When current in coil A changes, the magnetic flux linked with both
coil changes and hence emf is induced in coil B.
If Ip is current passing through primary coil(A) and ϕs be the flux linked
with the secondary, then A
Φ s α Ip B
Φs = M Ip (1)
Where, M is proportionality constant known as
Mutual inductance or coefficient of mutual induction Es

Mathematically, M = Φs / Ip, mutual induction is defined as ratio of magnetic flux linked with
Secondary to the current in the primary coil.
 If Φs = M Ip , using Faraday’s law of electromagnetic induction, induced emf in
secondary coil is given as
Es = -dΦs
dt
Es = - M dIp
dt

If dIp/dt =1 AS-1, Es = - M

Thus, mutual inductance is defined as the induced emf in the secondary coil when the
current in the primary coil changes at the rate of 1 ampere per second.
.
Mutual inductance of two concentric plane coils
Let us consider two concentric circles with radii ‘r1’ and ‘r2’ such that r2 < r1 as shown in figure.

Let us consider that a current I1 flows through the outer coil(1) having N1
number of turns. Due to this current, there is a magnetic field at the centre of
r2
the coil given by, r1
B1=μ0 N I
1 1

2r1 (1)

The magnetic flux linked across the inner coil(2) is

given by,
Φ2 = N2 B1 A2 Where, A2 is area of inner coil(2)
Φ2 = N2 μ0 N1 I1 A2
2r1 And N2 be its total turns
(2)
 Φ2 = μ0 N1 N2 I1 A2 A2 is area of inner coil(2) of radius r2
So , A2 = π r2
2r1 2

(2)
We know that magnetic flux in the inner coil also can be written as,
(3)
ϕ2 = M21I1 r2 r1
where ‘M21’ is mutual induction between the two coils.

Equating (2) and (3), we get

Where, A2 = π r22
μ0 N1 N2 I1 A2 = M21I1 M21 = μ0 N1 N2 A2 is required equation for mutual
2r1 2r1 induction between two coils.

For mutual inductance , magnetic field is considered as uniform such that

M21= M12
QTwo plane coils having number of turns 1000 and 2000, and radii
5 cm and 10 cm respectively are placed co-axially in the same
plane. Calculate their mutual inductance.
Solution:
Given,
number of turns in first coil, N1 = 1000
number of turns in second coil, N2 = 2000
radius of first coil, r1 = 5 cm = 0.05 m
radius of second coil, r2 = 10 cm = 0.10 m
mutual inductance, M = ?
Let the current I is passed through the inner coil. A magnetic field
B1=μ0N1I / 2r1 is produced inside the inner coil whereas the field outside it is zero.

The flux linked with each turn of inner coil is, ϕ2 =B1A2=(μ0N1I /2r1 )×πr22
The flux through all turns of inner coil is,ϕ=N2B1A2
=N2×(μ0N1I / 2r1 )×πr22 Or,

So, the mutual inductance is, M=ϕ/I=μ0N1N2πr 22 /2r1 M21 = μ0 N1 N2 A2

=4π×10− 7×1000×2000×π×(0.10)2 /2×(0.05) 2r1
∴L=0.79H
Thus, the mutual inductance is 0.79 H .
M= μ0N1N2πr 22
2r1
Q. Mutual inductance M between two concentric coils of radii 1 m and 2 m is

A. μ0π/2 B. μ0π/4
C. μ0π/8 D. μ0π/10

Answer B
 Mutual induction between two solenoids

Due to current I1 through solenoid of radius r1,

flux linked with second solenoid
Φ2 = M21I1 .....(i)
The flux linked with each turn of inner solenoid coil is
Φ2 = N2 B1A2, .....(ii)
where A2 is the cross-sectional area of
inner solenoid.
B1 =μ0 n1 I1 ; n1= N1 /L
.
M21I1 = N2 B1A2
M21I1 = N2 μ0 (N1 /L) I1 A2
M21 = μ0 N1 N2 A2 = M12
L
 Transformers
• Consider two coils side by side:
– Primary Coil: Connected to a voltage
source
– Secondary Coil: Connected to a
galvanometer

The magnetic field building around the primary extends

to the secondary
Changes in magnetic field intensity induce voltage in
the secondary
Placing a core within the coils will intensify the
magnetic field
The secondary will intercept more of the field
change
 Transformer:

Transformer is an electrical device that step up

(increase) or step down (decrease) AC voltage
using the principle of electromagnetic induction -
mutual induction.

.
Symbol of transformer
. Transformer:

Transformer is a device which

converts lower alternating voltage
(at higher current ) into higher
alternating voltage (at lower current) P
and vice-versa. S Load

Notice that the two coil windings are not electrically connected but are
only linked magnetically.
 Transformer:

A changing current in the primary coil induces an

e.m.f in the secondary. Since the e.m.f generated
depends on the number of turns, the voltage
induced in the secondary can be changed - stepped
up or down - by altering the turn's ratio.

The transformer is basically a voltage control device

that is used widely in the distribution and transmission
of alternating current power.
 Let,
Np = Number of turns in primary coil
Ns = Number of turns in secondary coil
Ep Es
If dΦ/dt be the rate at which the flux linked with
secondary changes, then using Faraday’s law,
Emf induced (Es) in the secondary is,

Es = - Ns dΦ/dt 1

Also to change in current in primary coil, emf(Ep) is induced in primary coil due to self
induction which is given as Ep = - Np dΦ/dt 2
 Dividing equation 1 by 2, we get

Es = Ns
This is required transformer equation
Ep Np

3
For an ideal transformer,
Input power = Output power Combining 3 and 4, we get
Ep Ip = Es Is
Es = Ns = Ip
Es = Ip 4 Ep Np Is
Ep Is

or
Vs = Ns = Ip
Vp Np Is
Transformer:
.

NP NS

NOTE: P S Load
ES / EP = NS / NP = K

(where K is called Transformation Ratio or

Turns Ratio)
 Step - up Transformer: Step - down Transformer:

P S Load P S
Load

NS > NP i.e. K > 1 NS < NP i.e. K < 1

ES < EP & IS > IP
ES > EP & IS < IP
 Types
Step-Up Transformer
When a transformer is used to “increase” the voltage on its secondary
winding with respect to the primary, it is called a Step-up transformer.

Step-down transformer.
When it is used to “decrease” the voltage on the
secondary winding with respect to the primary
it is called a Step-down transformer.

The difference in voltage between the primary

and the secondary windings is achieved by
changing the number of coil turns in the
primary winding ( NP ) compared to the
number of coil turns on the secondary winding
( NS ).
 Transformers
• Voltage is stepped up:
–When the number of coils (turns) in the secondary >
number of turns in primary
• Voltage is stepped down:
–When the number of coils (turns) in the secondary <
number of turns in primary
• Since voltages can be stepped up and down so easily,
electrical power is primarily in the form of AC
Efficiency of transformer (η)
It is defined as the ratio of output power to input power.

Since output power is developed in secondary coil,

Output power(Pout) = Is Es

And input power is supplied from primary coil,

Input power(Pin )= Ip Ep

For an ideal transformer, η = 100 ٪

So, efficiency(η ) = Is Es × 100 ٪
Ip Ep
Q. A transformer has 500 turns in the primary coil and 100 turns in the secondary coil.
What is the output voltage if the input voltage is 4000 volts? If the transformer is assumed
to have an efficiency of 100%, what primary current is required to draw 2000 watts from
the secondary?
 Q. A transformer has 500 turns in the primary coil and 100 turns in the secondary coil.
What is the output voltage if the input voltage is 4000 volts? If the transformer is assumed to
have an efficiency of 100%, what primary current is required to draw 2000 watts from the
secondary?
Solution:
Given,
Number of turns in the primary coil, Np = 500
Number of turns in the secondary coil, Ns = 100
Input voltage, εp = 4000 v
Efficiency of transformer, η = 100%
output power, Po = 2000 W Now,
Primary current Ip = ? For the transformer, Es / Ep = Ns / Np
Es / 4000 = 100/500
∴E s=800V
.
Since the transformer has 100 % efficiency, the input power and output power
is same.

Pi=Po
EpIp=Po
4000×Ip=Po
Ip=2000/4000
∴Ip=0.5A
Thus, the primary current is 0.5 A .
Q. A step down transformer transforms a supply line voltage 220 volts into 100 volts.
Primary coil has 500 turns. The efficiency and power transmitted by the transformer are
80 % and 80KW
Find a) the number of turns in the secondary coil b) power supplied

Given, For (a)

No. of turns in primary coil(Np )= 500 Np /Ns = Ep /Es
Ns /Np = Es /Ep
No. of turns in secondary coil(Ns )= ? Ns =(100/220) × 500
Input voltage (Ep )= 220V = 227
Output voltage(Es )= 100V For (b) Power supplied(Pin )=?
Efficiency (η )=80 % η= Po / Pin
Power transmitted(P0 )= 80 KW= 80000W 80% =80 × 103 / Pin

Pin = 105 watt

Applications of a transformer

• To step up voltage and current.

• To Step down voltage and current
• To prevent DC – transformers can pass only Alternating
Currents so they totally prevent DC from passing to the next
circuit
Why Transformers Can’t step Up Or Step Down A DC Voltage or Current?
A transformer cannot step up or step down a DC voltage. It is not recommendable to connect a DC
supply to a transformer because if a DC rated voltage is applied to the coil (primary) of a transformer, the
flux produced in the transformer will not change in its magnitude but rather remain the same and as a
result EMF will not be induced in the secondary coil except at the moment of switching on, So the
transformer may start to smock and burn because;
In case of DC supply, Frequency is zero. When you apply voltage across a pure inductive circuit, then
according to
XL= 2 π f L
Where:
•XL = Inductive Reactance
•L = Inductance
•f = Frequency
if we put frequency = 0, then the overall XL (inductive reactance) would be zero as well.
Now come to the current, I = V / R (and in case of inductive circuit, I = V / XL) …. basic Ohm’s Law
If we put Inductive reactance as 0, then the current would be infinite (Short circuit)…
So, If we apply DC voltage to a pure inductive circuit, The circuit may start to smoke and burn
 Transformer
Transformer only step-up or step-down the
value of AC voltage and current, why?

On using AC, there will be change in

current in the primary coil and hence the
change in magnetic flux in the secondary
coil to induce emf

So, Transformer won’t operate on DC

 Eddy or Focault currents

Eddy current is a localized electric current

induced in a conductor by a varying magnetic field
According to Lenz’s law, a conducting loop when subjected
to varying magnetic field gets an emf induced into it causing
flow of current in a direction opposing the change causing it.

Currents are generated in a nearby conductor whenever a

changing magnetic field is present. Because they swirl like
eddies in a river, they are called "eddy currents".
 Applications of eddy currents While they cause undesirable high heat losses in the material such as transformer core,
eddy currents find applications in various industrial processes like induction heating,
metallurgy, welding, braking etc

1. Eddy currents are used in electric brake. (examples;Braking of trains, braking of a roller coaster
electric saw or drill for its emergency shut-off)
Due to eddy current losses, Kinetic energy converted into heat giving such numerous applications.
2. Eddy currents are used for Induction Heating using high frequency electromagnet.
3. Eddy current heating used for tissue heating or Hyperthermia Cancer Treatment.
4. Metal Detectors: It detects present of metals inside rocks, soils etc. with the help of eddy current
induction in the metal if present.
5. It is used in car speedometer
 Energy losses in transformer

Even though transformers are very efficient machines, they do result in small energy losses due
to four main causes:
1. The resistance of windings(copper loss) – Heat is produced due to the resistance of the
copper windings of Primary and Secondary coils when current flows through them.
This can be avoided by using thick wires of considerably low resistance for winding

2. Leakage of flux (Flux Loss): In actual transformer coupling

between Primary and Secondary coil is not
perfect. So, a certain amount of magnetic flux
is wasted.
Linking can be maximised by winding the coils
over one another.
3. Iron Losses:
a) Eddy Currents Losses:
When a changing magnetic flux is linked with the iron core, eddy currents are set up which in turn produce
heat and energy is wasted.
Eddy currents are reduced by using laminated core instead of a solid iron block because in laminated core
the eddy currents are confined with in the lamination and they do not get added up to produce larger
current. In other words their paths are broken instead of continuous ones.
b) Hysteresis Loss:
When alternating current is passed, the iron core is
magnetised and demagnetised repeatedly over the
cycles and some energy is being lost in the process.

Solid Core Laminated Core

This can be minimised by using suitable material with thin hysteresis loop.
4. Losses due to vibration of core: Some electrical energy is lost in the form of mechanical
energy due to vibration of the core and humming noise. It can be minimised by the use
of varnish.