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CBSE · Class 12 · Physics

Magnetism and Matter

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
A bar magnet of magnetic moment 2 A m2 makes 150° with a uniform field of 6 × 10−2 T. The torque on it is:
  1. 6 × 10−4 N m
  2. 0.12 N m
  3. 0 N m
  4. 0.06 N m
2.
The magnetic field 20 cm from the centre of a short bar magnet of moment 9 A m2, on its axis, is: (μ0 = 4π × 10−7 T m/A)
  1. 1.125 × 10−4 T
  2. 2.25 × 10−6 T
  3. 2.25 × 10−4 T
  4. 4.5 × 10−3 T
3.
A magnet of moment 2 A m2 makes 90° with a field of 2 × 10−2 T. Its potential energy is:
  1. 0.04 J
  2. −0.02 J
  3. −0.04 J
  4. 0 J
4.
The work needed to turn a magnet of moment 8 A m2 from 0° to 180° with a field of 6 × 10−2 T is:
  1. 0.96 J
  2. 1.92 J
  3. −0.96 J
  4. 0.48 J
5.
The magnetic field lines of a bar magnet:
  1. stop at the surface
  2. form closed loops
  3. start at N and end at S only
  4. cross each other at the poles
6.
A material has susceptibility χ = −6 × 10−4. Its relative permeability and type are:
  1. 0.9994, diamagnetic
  2. −6 × 10−4, ferromagnetic
  3. 0.9994, paramagnetic
  4. 1.0006, diamagnetic
7.
Water is:
  1. non-magnetic in every field
  2. paramagnetic
  3. diamagnetic
  4. ferromagnetic
8.
Assertion (A): A material with susceptibility −6 × 10−5 is paramagnetic.
Reason (R): Diamagnetic materials have a small negative susceptibility.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
A coil of 226 turns and area 26 cm2 carries 1 A. Find its magnetic moment and the largest torque on it in a field of 9 × 10−2 T.
10.
A short bar magnet has moment 8 A m2. Find the field 5 cm from its centre (a) on its axis, (b) on its equatorial line. (μ0 = 4π × 10−7 T m/A)
11.
A magnet of moment 7 A m2 lies in a field of 9 × 10−2 T. Find the work done in turning it from 0° to 120° with the field.
12.
A magnet of moment 7 A m2 is in a field of 9 × 10−2 T. Find its potential energy when it points along the field and when it points against it. Which position is stable?
13.
A solenoid of 629 turns, radius 5 cm, carries 4 A. Find its magnetic moment.
14.
A bar magnet of volume 85 cm3 has magnetic moment 8 A m2. Find its magnetisation.
15.
A core of relative permeability 1800 sits in a magnetising field H = 100 A/m. Find the magnetic field B in the core. (μ0 = 4π × 10−7 T m/A)
16.
A magnet of moment 4 A m2 is in a field of 3 × 10−2 T. Find the torque on it at 30° and at 90° to the field.
Section C — Word problems4 × 4 = 16 marks
17.
A short bar magnet of moment 1 A m2 lies along the x-axis with its centre at the origin. Find the field at (20 cm, 0) and at (0, 20 cm), and the ratio of the two. (μ0 = 4π × 10−7 T m/A)
18.
A magnet of moment 9 A m2 is in a field of 2 × 10−2 T. Find the work done to turn it (a) from along the field to 90°, (b) from along the field to against it. Also find the torque at 90°.
19.
A solenoid of 834 turns and radius 4 cm carries 5 A. It is free to turn in a horizontal field of 8 × 10−2 T. Find its magnetic moment and the torque on it when its axis makes 30° with the field.
20.
An iron core has relative permeability 1100 and is placed in a magnetising field of 800 A/m. Find its susceptibility, its magnetisation and the magnetic field in it. (μ0 = 4π × 10−7 T m/A)
AmbifulWorksheets
Practice that never repeats
CBSE · Class 12 · Physics

Answer key

Magnetism and Matter · AW-12P-MAM-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(d) 0.06 N mτ = mB sin θ
2(c) 2.25 × 10−4 TB = (μ0/4π) × 2m/r3
3(d) 0 JU = −mB cos θ
4(a) 0.96 JW = mB(cos θ1 − cos θ2)
5(b) form closed loops—
6(a) 0.9994, diamagneticμr = 1 + χ
7(c) diamagneticsmall negative χ
8(d) A is false, but R is true.A is false: a small negative susceptibility marks a diamagnetic material. R is true.
90.5876 A m2; 0.052884 N mm = NIA; τmax = mB
10(a) 0.0128 T; (b) 6.4 × 10−3 Taxis: 2m/r3 × 10−7; equator: m/r3 × 10−7
110.945 JW = mB(cos 0° − cos 120°)
12along: −0.63 J (stable); against: 0.63 J (unstable)U = −mB cos θ
136.29π A m2m = NIA = 629 × 4 × π(5 × 10−2)2
149.41 × 104 A/mM = m/V = 8 ÷ (85 × 10−6)
157.2π × 10−2 TB = μ0μrH
160.06 N m; 0.12 N mτ = mB sin θ
172.5 × 10−5 T along the axis; 1.25 × 10−5 T, opposite to the moment; ratio 2 : 1B(axis) = 2m/r3 × 10−7, B(equator) = m/r3 × 10−7
18(a) 0.18 J; (b) 0.36 J; torque 0.18 N mW = mB(cos θ1 − cos θ2)
19m = 6.672π A m2; τ = 2.6688π × 10−1 N mm = NIπr2; τ = mB sin θ
20χ = 1099; M = 8.792 × 105 A/m; B = 3.52π × 10−1 Tχ = μr − 1; M = χH; B = μ0μrH