Class 12 Physics NCERT Solutions

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Magnetism and Matter Class 12 Physics NCERT Solutions

The complete NCERT exercise solutions for Chapter 5, Magnetism and Matter — 7 questions from 5.1 to 5.7, each worked through step by step in the CBSE marking pattern. Magnetisation and magnetic susceptibility, hysteresis, the magnetic field of a solenoid and a bar magnet, and the angle of dip.

Class:12Subject:PhysicsChapter:5

4 Key Formulas25 Practice MCQs

DWritten byDeep Narayan

UpdatedSeptember 2026

Key Concept Summary

How many questions are in NCERT Class 12 Physics Chapter 5?

Chapter 5 carries 7 exercise questions, numbered 5.1 to 5.7. All of them are solved step by step on this page, along with the chapter's key formulas and exam pointers.

Table of Contents8 sections

Revision & Practice

Chapter Overview

Magnetism and Matter turns the bar magnet into a measurable object: torque τ = m × B on a short magnet, energy U = –m·B, the axial and equatorial fields B = (μ₀/4π)·2M/r³ and (μ₀/4π)·M/r³, and a current loop (or solenoid) acting as a magnet of moment M = NIA. Boards love the two-equilibrium torque/energy comparison and the field-of-a-dipole pair. Every question below is from the NCERT Class 12 textbook (rationalised edition), solved line by line in the board pattern.

Board pattern

A short bar magnet is a magnetic dipole, so use the dipole fields and the dipole energy U = –mB cosθ. Stable equilibrium means minimum energy (m aligned with B, θ = 0); unstable means maximum energy (m anti-parallel to B, θ = 180°). For a solenoid, apply the right-hand grip rule to the current to decide which face is the north pole. The axial field is twice the equatorial field at the same distance.

$Ex 5.1 — Magnetic moment from the torque on a magnet$
$Ex 5.2 — Stable and unstable equilibrium, potential energies$
$Ex 5.3 — Solenoid acting as a bar magnet, its moment$
$Ex 5.4 — Torque on a rotating solenoid in a horizontal field$
$Ex 5.5 — Work to rotate a magnet, torques at 90° and 180°$
$Ex 5.6 — Moment, force and torque on a suspended solenoid$
$Ex 5.7 — Axial and equatorial fields of a short magnet$

NCERT Exercise 5.1 — Magnetic Moment From the Torque on a Magnet

1Exercise question

Step-by-step solution

1τ = m B sinθ.
24.5 × 10⁻² = m × 0.25 × sin30° = m × 0.25 × 0.5 = 0.125 m.
3m = 4.5 × 10⁻²/0.125 = 0.36 J T⁻¹.

Final answer

m = 0.36 J T⁻¹.

NCERT Exercise 5.2 — Stable and Unstable Equilibrium, Potential Energies

1Exercise question

Step-by-step solution

1U = –mB cosθ. U is minimum when cosθ = 1, i.e. θ = 0°. So (a) stable equilibrium: magnetic moment aligned parallel to the field.
2U_min = –mB = –0.32 × 0.15 = –0.048 J.
3U is maximum when cosθ = –1, i.e. θ = 180°. So (b) unstable equilibrium: magnetic moment anti-parallel to the field.
4U_max = +mB = +0.048 J.

Final answer

Stable: m parallel to B, U = –0.048 J. Unstable: m anti-parallel to B, U = +0.048 J.

NCERT Exercise 5.3 — Solenoid Acting as a Bar Magnet, Its Moment

1Exercise question

Step-by-step solution

1The solenoid's field lines emerge from one face and re-enter at the other, exactly like the field of a bar magnet: the current loop set behaves as a magnetic dipole.
2By the right-hand grip rule (fingers along the current, thumb pointing to the north face), the end from which field lines emerge behaves as a north pole.
3Magnetic moment: m = N I A.
4m = 800 × 3.0 × 2.5 × 10⁻⁴ = 0.6 J T⁻¹.

Final answer

The solenoid's field mimics a bar magnet with the grip-rule end as its north pole; m = NIA = 0.6 J T⁻¹.

NCERT Exercise 5.4 — Torque on a Rotating Solenoid in a Horizontal Field

1Exercise question

Step-by-step solution

1Take m = N I A = 800 × 3.0 × 2.5 × 10⁻⁴ = 0.6 J T⁻¹ (the solenoid of Exercise 5.3).
2τ = m B sinθ = 0.6 × 0.25 × sin30°.
3τ = 0.6 × 0.25 × 0.5 = 0.075 N m.

Final answer

τ = 7.5 × 10⁻² N m.

NCERT Exercise 5.5 — Work to Rotate a Magnet, Torques at 90° and 180°

1Exercise question

Step-by-step solution

1Initially aligned: θ_i = 0°, U_i = –mB = –1.5 × 0.22 = –0.33 J.
2(a)(i) At θ = 90°: U = 0, so work = ΔU = 0 – (–0.33) = 0.33 J.
3(a)(ii) At θ = 180°: U = +mB = +0.33 J, so work = 0.33 – (–0.33) = 0.66 J.
4(b)(i) At θ = 90°: τ = mB sin90° = 1.5 × 0.22 = 0.33 N m (tending to rotate the moment back toward the field).
5(b)(ii) At θ = 180°: τ = mB sin180° = 0 N m.

Final answer

(a) Work: 0.33 J to reach 90°, 0.66 J to reach 180°. (b) Torque: 0.33 N m at 90°, zero at 180°.

NCERT Exercise 5.6 — Moment, Force and Torque on a Suspended Solenoid

1Exercise question

Step-by-step solution

1(a) m = N I A = 2000 × 4.0 × 1.6 × 10⁻⁴ = 1.28 J T⁻¹.
2(b) In a uniform field there is no net force on a magnetic dipole: F = 0.
3Torque: τ = m B sinθ = 1.28 × 7.5 × 10⁻² × sin30°.
4τ = 1.28 × 7.5 × 10⁻² × 0.5 = 0.048 N m.

Final answer

(a) m = 1.28 J T⁻¹. (b) Force = 0; torque = 4.8 × 10⁻² N m.

NCERT Exercise 5.7 — Axial and Equatorial Fields of a Short Magnet

1Exercise question

Step-by-step solution

1On the axis: B_ax = (μ₀/4π) × 2m/r³ = 10⁻⁷ × (2 × 0.48)/(0.10)³.
2B_ax = 10⁻⁷ × 0.96/10⁻³ = 10⁻⁷ × 960 = 9.6 × 10⁻⁵ T. Direction: along the magnetic moment, i.e. from the south to the north pole (the direction of m).
3On the equator: B_eq = (μ₀/4π) × m/r³ = 10⁻⁷ × 0.48/10⁻³ = 4.8 × 10⁻⁵ T.
4B_eq direction: antiparallel to the magnetic moment, i.e. from the north toward the south pole.

Final answer

(a) Axis: 9.6 × 10⁻⁵ T along m (S → N). (b) Equator: 4.8 × 10⁻⁵ T opposite to m (N → S).

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Relation between B, H and M

B = μ_{0} (H + M)

Magnetic susceptibility

χ = \frac{M}{H}

Permeability of a medium

μ_{r} = 1 + χ, μ = μ_{0} μ_{r}

Dipole moment of a bar magnet

m = pole strength \times magnetic length

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

Diamagnetic materials have χ slightly negative, paramagnetic χ small and positive, and ferromagnetic χ very large — the distinction is what the classification questions turn on.
Torque on a magnetic dipole is μ × H, and it is maximum when the dipole is perpendicular to the field, not when it is aligned with it.

FAQ

Frequently asked questions

How many questions are in NCERT Class 12 Physics Chapter 5 (Magnetism and Matter)?

There are 7 exercise questions in this chapter, numbered 5.1 to 5.7. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Magnetism and Matter Class 12 Physics?

The formulas this chapter's questions actually turn on are: Relation between B, H and M, Magnetic susceptibility, Permeability of a medium, Dipole moment of a bar magnet. They are listed with their expressions in the key formulas section below, and the solved questions show where each one is used.

Is Magnetism and Matter important for JEE Main and NEET?

Moderate — mostly conceptual and worth one or two marks, but the B = μ₀(H + M) decomposition and the definitions of χ and μr appear regularly in both JEE and NEET.

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