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0% found this document useful (0 votes)
54 views6 pages

B PH 5 S Cbse 1

Thank you

Uploaded by

s30090674
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Class XII (CBSE) Physics

Chapter 5: Magnetism and Matter [PH-5]


Important Questions & Answers 2024-25

Note:
Section-1: Objective type question carries 1 mark each
Section-2: Very short answer type question carries 2 marks each
Section-3: Short answer type question carries 3 marks each
Section-4: Long answer type question carries 5 marks each

Section-1: Objective type question


Q.1. Which of the following is another term for magnetization?
a) Magnetic neutrality
b) Magnetic polarization
c) Magnetic power
d) Magnetic moment
Answer: b

Q.2. The material in which the magnetic susceptibility is independent on temperature-


a) Diamagnetic
b) Paramagnetic
c) Ferromagnetic
d) Ferrite.
Answer: a

Q.3. The magnetic field lines of force inside a bar magnet:


a) From S pole to N pole of the magnet
b) Do not exist
c) From N pole to S pole of the magnet
d) Area of the cross-section of the magnet
Answer: b

Q.4. The SI unit of magnetic flux is


a) Dyne
b) Tesla
c) Weber
d) Ohm
Answer: c

Q.5. The angle of dip at the magnetic equator is-


a) 45 degree
b) 0 degree
c) 90 degree
d) 60 degree
Answer: b
Q.6. The finest material for the core of a transformer is:
a) hard steel
b) stainless steel
c) mild steel
d) soft iron.
Answer: d

Q.7. Which among the following materials display higher magnetic susceptibility?
a) Ferromagnetic material
b) Paramagnetic material
c) Diamagnetic material
d) None of these options
Answer: (a) Ferromagnetic material

Q.8. Fill in the blanks :


1. At ........................... temperature, a ferro-magnetic material passes over to para-
magnetics.
2. Two magnetic lines ........................... cross each other.
3. Angle of dip is ........................... at the poles.
4. Surface integral of ........................... field over a closed surface is zero.
Answer:
1. curie
2. never
3. 90°
4. magnetic

Section-2: Very Short Answer type Questions


Q. 1. Where on the surface of Earth is the vertical component of Earth’s magnetic field
zero?
Answer:
At the Magnetic equator the vertical component of Earth’s magnetic field is zero.

Q. 2. An iron bar magnet will not retain is heated to 1000°C and then cooled in a
magnetic field free space. Will it retain magnetism? Why?
Answer:
Curie temperature of iron is 770°C When it is heated to a very high temperature magnetism
of iron is lost and does not retain its magnetism further.

Q. 3. What are permanent magnets? Give one example.


Answer:
Permanent magnets are materials that retain their ferromagnetic properties for an extended
period of time at normal temperature.
Example: Steel

Q. 4. What is meant by “magnetic susceptibility” and how does it relate to “relative


magnetic permeability”?
Answer:
It describes how easily a substance might become magnetized. It is described as the
proportion of the magnetization’s intensity to the magnetic field. The necessary relationship
is µr = 1 + χm.

Q. 5. What type of magnetic material is used in making permanent magnets?


Answer:
Materials having high coercivity are used in making permanent magnets. Steel is better
choice than soft iron.

Section-3: Short Answer type Questions


Q.1. Compare to a bar magnet and a current – carrying solenoid.
Answer:
Comparison of a bar magnet and a solenoid :
Bar magnet:

 It attracts magnetic substances.


 When it is suspended freely it rests in the direction of N – S.
 It has two poles.
 Like poles of magnet repel and unlike poles attract.
Solenoid:

 It also attracts magnetic substances.


 It also rests in N – S direction if suspended freely.
 It has also two poles.
 Like poles of solenoid also repel and unlike poles attract.

Q. 2. Write the difference between the magnetic properties of soft iron and steel.
Answer:
Soft Iron Steel
High Magnetic Permeability Less Magnetic Permeability
Small area of hysteresis loop Large area of hysteresis loop
Coercivity is low Coercivity is high

Q. 3. A vector needs three quantities for its specification. Name the three independent
quantities conventionally used to specify the earth’s magnetic field.
Answer:
The three independent conventional quantities used for determining the earth’s magnetic field
are: (i) Magnetic declination, (ii) Angle of dip (iii) The horizontal component of the earth’s
magnetic field.

Q. 4. Make a list of two properties of a material used to make permanent magnets.


Answer:
The following are two features of a material used to make permanent magnets:
a. High retentivity, resulting in a powerful magnetic field.
b. High coercivity, so that high magnetic fields, temperature changes, or small
mechanical damage do not remove its magnetism.

Q. 5. When kept in an external magnetic field, how is the behavior of a diamagnetic


substance different from that of a paramagnetic?
Answer:
a. A diamagnetic specimen would gravitate toward the weaker part of the field, whereas
a paramagnetic specimen would gravitate toward the stronger part.
b. A magnet repels a diamagnetic specimen, while a paramagnetic specimen travels
towards the magnet.
c. The diamagnetic is positioned perpendicular to the field, whereas the paramagnetic is
aligned along the field.

Section-4: Long Answer type Questions

Q. 1. State and prove Gauss's law of magnetism. Describe its significance.


b) Write down the four most important properties of magnetic field lines caused by a
bar magnet.
Answer:
a) According to Gauss's Law of Magnetism, the total flux of the magnetic field through any
closed surface is always zero.

i.e.

Proof: Consider a magnetic field produced by a bar magnet as shown in the figure. Let us
consider a closed Gaussian surface enclosing the N-pole of this magnet is shown by dotted
circle. We find that number of magnetic lines entering the Gaussian surface is equal to
number of magnetic lines leaving it i.e. total normal flux for whole Gaussian surface is zero.

i.e. Φm =
This theorem implies that isolated magnetic poles do not exist.
or
Magnetic poles always exist in pairs.

This law implies the absence of magnetic monopoles or magnetic field lines forms closed
loops.

b) The following are the four properties of magnetic field lines:


i. Magnetic field lines always form closed loops that are continuous.
ii. The direction of the net magnetic field at a given position is represented by the
tangent to the magnetic field line at that point.
iii. The magnitude of the magnetic field increases as the number of field lines crossing
per unit area increases.
iv. Magnetic field lines do not cross.

Q. 2. State and illustrate Curie law in magnetism.


Answer:
Curie law states that susceptibility of a material varies inversely with its temperature in
kelvin.
According to Curie’s law, the intensity of magnetisation I of a para and ferromagnetic
material is
(i) directly proportional to the magnetic induction,
i.e. I ∝ B
(ii) inversely proportional to the temperature.
1
i.e. I ∝
𝑇

𝐵
⸫ I∝
𝑇

Since B ∝ H
𝐻
⸫ I∝
𝑇
𝐻
or I=C
𝑇

𝐼 𝐶
or =
𝐻 𝑇

𝐶
or χm=
𝑇

(𝜒𝑚 = HI is magnetic susceptibility)

I
or 𝜒𝑚 =
H

This is Curie's law.


The graph between I and H/T is shown in Fig. and it shows that as the external magnetic field
is increased and T is decreased, the atomic dipoles get aligned more and more in the direction
of applied magnetic field and after sometime saturation is reached.
Q. 3. A particle of mass m and charge q moving with a uniform speed normal to a
uniform magnetic field B describes a circular path of radius & Derive expressions for
(1) Radius of the circular path (2) time period of revolution (3) Kinetic energy of the
particle
Answer:
particle of mass (m) and change (q) moving with velocity normal to describes a circular path
if

𝒎𝒗𝟐
=qBvsinθ
𝒓

𝒎𝒗𝟐
⇒ =qBv(∵θ=90°)
𝒓
𝒎𝒗
⇒ r= …… (1)
𝑩𝒒
This is the required radius of the circular path.
Now, since
𝑪𝒊𝒓𝒄𝒖𝒎𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝒐𝒇 𝒄𝒊𝒓𝒄𝒍𝒆
Time period of Revolution during circular path = ;
𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚

𝟐𝝅𝒓
T=
𝒗
𝟐𝝅𝒓.𝒎
⇒T= ……. (from 1)
𝑩𝒒𝒓
𝟐𝝅𝒎
⇒T= …… (2)
𝑩𝒒
This is the required time period.
Now,
𝟏
Kinetic energy K.E= 𝒎𝒗𝟐
𝟐

𝟏 𝑩𝒒𝒓 𝟐
⇒K.E= 𝟐 𝒎 ( )
𝒎

𝑩𝟐 𝒒𝟐 𝒓𝟐
⇒K.E=
𝟐𝒎

, which is the required kinetic energy.

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