Magnetism and Matter Class 12 Physics Numericals 2026-27 | CBSE

CBSE Class 12 Physics · Chapter 5 · 2026–27

Magnetism and Matter Class 12 Physics Numericals

30 core solved numericals covering magnetic dipole moment, torque, magnetisation and quantitative relationships, followed by 5 clearly labelled extension problems.

SEO focus: Magnetism and Matter Class 12 Physics Numericals · Chapter 5 Numericals · CBSE 2026–27

How to use this page: Try every numerical yourself before opening the solution. Write the given data, select the correct relationship, substitute SI units, calculate, and finish with a clearly stated answer. These are original practice problems, not official CBSE board questions.

Core Numericals SQP-Linked Extension Formula Map Common Mistakes Chapter Resources Final Revision

1. Core Chapter 5 Numericals

These problems prioritise the quantitative relationships that naturally arise from the current Chapter 5 concepts. A few classification/interpretation problems are included where the numerical data are used to identify a magnetic response. The current CBSE 2026–27 curriculum describes the bar magnet, magnetic dipole, qualitative field behaviour, torque, magnetic materials, magnetisation and temperature effects.

Numerical 1

Magnetic dipole moment from pole strength

Question: A bar magnet has pole strength 0.50 A m and magnetic length 0.12 m. Find its magnetic dipole moment.

Solution: Use $m=p(2l)$ when 2l is the magnetic length. Hence, $m=0.50\times0.12=0.060\;A\,m^2$.

Final Answer: 0.060 A m²

Numerical 2

Magnetic moment of a current-carrying coil

Question: A circular coil of 20 turns carries 0.50 A current. Its area is 4.0 × 10⁻³ m². Find its magnetic dipole moment.

Solution: For a current-carrying coil, $m=NIA$. Therefore, $m=20\times0.50\times4.0\times10^{-3}=4.0\times10^{-2}\;A\,m^2$.

Final Answer: 4.0 × 10⁻² A m²

Numerical 3

Torque at 90°

Question: A magnetic dipole of moment 0.80 A m² is placed in a uniform magnetic field of 0.25 T at 90°. Find the torque.

Solution: Use $\tau=mB\sin\theta$. With θ = 90°, sin θ = 1, so $\tau=0.80\times0.25=0.20\;N\,m$.

Final Answer: 0.20 N m

Numerical 4

Torque at 30°

Question: A magnetic dipole of moment 1.5 A m² is placed in a 0.40 T field at 30°. Calculate the torque.

Solution: Use $\tau=mB\sin\theta$. Thus $\tau=1.5\times0.40\times0.5=0.30\;N\,m$.

Final Answer: 0.30 N m

Numerical 5

Zero torque

Question: A magnetic dipole of moment 0.60 A m² is placed in a 0.50 T field. At what orientations is the torque zero?

Solution: Torque is $mB\sin\theta$. It is zero when sin θ = 0, i.e. when θ = 0° or 180° (modulo 180°).

Final Answer: 0° or 180°

Numerical 6

Maximum torque

Question: A magnetic dipole of moment 2.0 A m² is placed in a uniform field of 0.30 T. Find the maximum torque.

Solution: Maximum torque occurs at θ = 90°. Therefore, $\tau_{max}=mB=2.0\times0.30=0.60\;N\,m$.

Final Answer: 0.60 N m

Numerical 7

Finding magnetic moment from torque

Question: A dipole experiences a torque of 0.45 N m in a 0.30 T field when it is at 90°. Find its magnetic moment.

Solution: At 90°, $\tau=mB$. Hence $m=\tau/B=0.45/0.30=1.5\;A\,m^2$.

Final Answer: 1.5 A m²

Numerical 8

Finding field from torque

Question: A magnetic dipole of moment 1.2 A m² experiences a torque of 0.36 N m at 90°. Find the field.

Solution: At 90°, $B=\tau/m=0.36/1.2=0.30\;T$.

Final Answer: 0.30 T

Numerical 9

Effect of changing current

Question: A 10-turn coil of area 5 × 10⁻³ m² carries 2 A. If the current is doubled while all other quantities remain unchanged, by what factor does its magnetic moment change?

Solution: Since $m=NIA$, magnetic moment is directly proportional to current. Doubling I doubles m.

Final Answer: 2 times

Numerical 10

Effect of changing turns and area

Question: A coil has N turns, current I and area A. If N is doubled and A is halved, with I unchanged, compare the new magnetic moment with the old one.

Solution: Since $m=NIA$, the new moment is $(2N)I(A/2)=NIA$.

Final Answer: Unchanged

Numerical 11

Torque ratio

Question: The same magnetic dipole is placed in the same field at 30° and 60°. Find the ratio of torques τ₃₀ : τ₆₀.

Solution: Since $\tau\propto\sin\theta$, the ratio is sin30° : sin60° = 1/2 : √3/2 = 1 : √3.

Final Answer: 1 : √3

Numerical 12

Torque after field change

Question: A dipole at 90° experiences 0.50 N m in a field B. If the field becomes 1.5B, find the new torque.

Solution: At fixed m and angle, torque is directly proportional to B. Thus the new torque is 1.5 × 0.50.

Final Answer: 0.75 N m

Numerical 13

Magnetic moment of a rectangular coil

Question: A 25-turn rectangular coil has dimensions 20 cm × 10 cm and carries 0.40 A. Find its magnetic moment.

Solution: Area = 0.20 × 0.10 = 0.020 m². Then $m=NIA=25\times0.40\times0.020=0.20\;A\,m^2$.

Final Answer: 0.20 A m²

Numerical 14

Dipole moment and magnetic length

Question: A bar magnet has pole strength 0.80 A m and magnetic length 15 cm. Find its magnetic dipole moment.

Solution: Here the magnetic length is 0.15 m. Hence $m=p(2l)=0.80\times0.15=0.12\;A\,m^2$.

Final Answer: 0.12 A m²

Numerical 15

Classification using susceptibility sign

Question: A material has a small negative magnetic susceptibility. Identify its magnetic class.

Solution: A negative susceptibility corresponds to diamagnetic behaviour. The response is weakly repulsive in a non-uniform magnetic field.

Final Answer: Diamagnetic

Numerical 16

Relative permeability and material type

Question: A material has relative permeability slightly greater than 1 and a small positive susceptibility. Identify its class.

Solution: For a weak paramagnet, μr is slightly greater than 1 and χ is small and positive.

Final Answer: Paramagnetic

Numerical 17

Ferromagnetic identification

Question: A material shows very strong attraction to an external magnetic field and very large positive magnetic response. Identify the class.

Solution: Strong magnetic response with large positive susceptibility is characteristic of ferromagnetic materials.

Final Answer: Ferromagnetic

Numerical 18

Temperature and ferromagnetism

Question: A ferromagnetic material is heated above its Curie temperature. What happens to its magnetic behaviour?

Solution: Above the Curie temperature, ferromagnetic order is lost and the material becomes paramagnetic.

Final Answer: It becomes paramagnetic

Numerical 19

Magnetisation unit

Question: A specimen has a magnetic dipole moment of 0.24 A m² and volume 3.0 × 10⁻⁴ m³. Find its magnetisation M.

Solution: Magnetisation is magnetic moment per unit volume: $M=m/V$. Thus $M=0.24/(3.0\times10^{-4})=800\;A\,m^{-1}$.

Final Answer: 800 A m⁻¹

Numerical 20

Magnetisation scaling

Question: A sample's magnetic moment is doubled while its volume is unchanged. What happens to its magnetisation?

Solution: Since $M=m/V$, with V fixed, doubling m doubles M.

Final Answer: It doubles

Numerical 21

Volume scaling

Question: A specimen has fixed magnetic moment. Its volume is reduced to half without changing the moment. What happens to M?

Solution: Since $M=m/V$, halving V doubles M.

Final Answer: It doubles

Numerical 22

Dipole orientation

Question: A dipole of moment 0.50 A m² is placed in a 0.20 T field at 180°. Find the torque and state the orientation.

Solution: At 180°, sin180° = 0, so τ = 0. The dipole is antiparallel to the field.

Final Answer: 0 N m; antiparallel

Numerical 23

Torque with a changed angle

Question: A dipole has m = 2 A m² and is in B = 0.5 T. Compare its torque at 90° and 30°.

Solution: τ₉₀ = mB = 1 N m. τ₃₀ = mB/2 = 0.5 N m.

Final Answer: 1 N m and 0.5 N m

Numerical 24

Equivalent-solenoid concept

Question: A coil has 100 turns, current 0.20 A and area 2 × 10⁻³ m². Find the magnetic moment of the equivalent magnetic dipole.

Solution: For a current loop, $m=NIA=100\times0.20\times2\times10^{-3}=0.040\;A\,m^2$.

Final Answer: 0.040 A m²

Numerical 25

Unit conversion practice

Question: A magnetic dipole moment is 250 mA m². Express it in A m².

Solution: Since 250 mA = 0.250 A, the moment is 0.250 A m².

Final Answer: 0.250 A m²

Numerical 26

Area from magnetic moment

Question: A 40-turn coil carrying 0.25 A has magnetic moment 0.50 A m². Find its area.

Solution: From $m=NIA$, $A=m/(NI)=0.50/(40\times0.25)=0.050\;m^2$.

Final Answer: 0.050 m²

Numerical 27

Current from magnetic moment

Question: A 50-turn coil of area 4 × 10⁻³ m² has magnetic moment 0.20 A m². Find the current.

Solution: From $I=m/(NA)=0.20/(50\times4\times10^{-3})=1.0\;A$.

Final Answer: 1.0 A

Numerical 28

Turns from magnetic moment

Question: A coil of area 5 × 10⁻³ m² carries 0.40 A and has magnetic moment 0.10 A m². Find the number of turns.

Solution: From $N=m/(IA)=0.10/(0.40\times5\times10^{-3})=50$.

Final Answer: 50 turns

Numerical 29

Qualitative axial-equatorial comparison

Question: At equal distances from a short magnetic dipole, the field magnitude on the equatorial line is how many times the axial-line magnitude?

Solution: For a short dipole, the equatorial field magnitude is half the axial field magnitude at the same distance.

Final Answer: B_eq = ½ B_axial

Numerical 30

Vector addition of dipole fields

Question: Two magnetic field contributions at a point are perpendicular and have magnitudes 3.0 × 10⁻⁵ T and 4.0 × 10⁻⁵ T. Find the resultant magnitude.

Solution: For perpendicular vectors, $B=\sqrt{B_1^2+B_2^2}=\sqrt{9+16}\times10^{-5}=5.0\times10^{-5}\;T$.

Final Answer: 5.0 × 10⁻⁵ T

2. Official 2026–27 SQP-Linked Extension

Important scope note: These five problems are separated from the core set because some assessment/reference material may use quantitative relationships such as relative permeability/susceptibility and dipole-field scaling even where the curriculum wording emphasises qualitative treatment. Use this section as assessment-aware extension practice, not as a reason to ignore the current curriculum wording. The official CBSE Class XII 2026–27 SQP/MS page lists Physics among the released subjects.

Numerical E1

Relative permeability and susceptibility

Question: A material has relative permeability μr = 1.002. Using the SI relation μr = 1 + χ, find χ.

Solution: χ = μr − 1 = 1.002 − 1 = 0.002.

Final Answer: 0.002

Numerical E2

Diamagnetic susceptibility

Question: A material has χ = −0.0008. Find its relative permeability using μr = 1 + χ.

Solution: μr = 1 − 0.0008 = 0.9992.

Final Answer: 0.9992

Numerical E3

Magnetic field scaling

Question: For a short magnetic dipole, the axial field magnitude varies as 1/r³. If the distance is doubled, by what factor does the field change?

Solution: B ∝ 1/r³. Replacing r by 2r gives B' = B/8.

Final Answer: One-eighth

Numerical E4

Axial vs equatorial field

Question: At the same distance from a short dipole, the axial field is 6 × 10⁻⁵ T. Find the equatorial field magnitude.

Solution: The equatorial magnitude is half the axial magnitude at the same distance.

Final Answer: 3 × 10⁻⁵ T

Numerical E5

Dipole field distance scaling

Question: At a point on the axial line the field is B. At three times the distance, what is the field for a short dipole?

Solution: Because B ∝ 1/r³, increasing r by 3 reduces B by 27.

Final Answer: B/27

3. Formula & Relationship Map

Magnetic dipole moment of a current loop: m = NIA

Torque on a magnetic dipole: τ = mB sin θ

Magnetisation: M = magnetic moment / volume

For a short dipole: B_axial ∝ 1/r³ and B_equatorial ∝ 1/r³

At equal distance: |B_equatorial| = ½ |B_axial|

SI units: m → A m², B → tesla (T), τ → N m, M → A m⁻¹

4. Common Numerical Mistakes

Unit trap
Convert cm to m, mA to A and cm² to m² before substitution.

Angle trap
In τ = mB sin θ, θ is the angle between the magnetic moment vector and the magnetic field.

Factor trap
For a short dipole, remember the axial/equatorial comparison at the same distance.

Moment trap
For a coil, magnetic moment is NIA—not merely IA when there are multiple turns.

Sign/classification trap
Small negative χ indicates diamagnetism; small positive χ indicates paramagnetism.

Temperature trap
Above the Curie temperature, a ferromagnetic substance loses ferromagnetic ordering.

5. Chapter 5 Study Resources

Chapter 5 Complete Notes Chapter 5 Important Questions Chapter 5 MCQs CBSE 2026–27 Physics Curriculum CBSE Class XII 2026–27 SQP & Marking Scheme

6. Final Numerical Revision Checklist

Skill	Can you do it without notes?
Calculate magnetic dipole moment using m = NIA	□ Yes / □ Revise
Calculate torque using τ = mB sin θ	□ Yes / □ Revise
Identify zero and maximum torque conditions	□ Yes / □ Revise
Calculate magnetisation from moment and volume	□ Yes / □ Revise
Compare axial and equatorial dipole fields	□ Yes / □ Revise
Classify materials from magnetic response	□ Yes / □ Revise
Handle SI-unit conversions correctly	□ Yes / □ Revise

Next in the Learn Revise Hub Chapter 5 sequence: Case-Based Questions. The remaining companion pages will be linked here only after their actual live URLs are published and verified.