NCERT Solutions for Class 12 Physics Chapter 5: Magnetism and Matter

These Class 12 Physics Chapter 5 solutions cover Magnetism and Matter with every NCERT exercise reproduced verbatim and solved step by step, with all numericals worked out and verified with proper units. Updated for the NCERT 2026–27 session, this guide explains the bar magnet, magnetic dipole, Gauss’s law for magnetism, magnetisation, magnetic intensity, and the dia-, para- and ferromagnetic classification of materials — exactly the way you will be tested in the CBSE board exam.

Class: 12 Subject: Physics Chapter: 5 Name: Magnetism and Matter Exercises: 5.1 – 5.7 Session: 2026–27

On this page

Chapter overview
Key concepts & definitions
Important formulas
NCERT Exercises (5.1–5.7) — solutions
Extra practice questions
MCQs & Assertion–Reason
Common mistakes
Exam tips
FAQs

Chapter 5 Magnetism and Matter – Overview

After learning in Chapter 4 that moving charges and currents create magnetic fields, Chapter 5 studies magnetism as a subject in its own right. A bar magnet behaves like a magnetic dipole whose field lines form continuous closed loops, and it is equivalent to a current-carrying solenoid of the same magnetic moment. Placed in a uniform field, a dipole feels a torque (but no net force) and stores potential energy depending on its orientation. Gauss’s law for magnetism states that the net magnetic flux through any closed surface is zero, because isolated magnetic poles (monopoles) do not exist. Finally, materials are classified by their magnetic susceptibility χ into diamagnetic (small negative χ), paramagnetic (small positive χ) and ferromagnetic (large positive χ) substances.

Key Concepts & Definitions

Magnetic dipole moment (m): for a bar magnet it points from S to N pole; for a current loop of N turns, area A and current I, m = NIA, measured in A m² (or J T⁻¹).

Torque on a dipole: in a uniform field B, τ = m × B, magnitude τ = mB sinθ, where θ is the angle between m and B.

Potential energy: U = −m·B = −mB cosθ; minimum (−mB) at θ = 0° (stable), maximum (+mB) at θ = 180° (unstable).

Gauss’s law for magnetism: the net magnetic flux through any closed surface is zero, Σ B·ΔS = 0 — magnetic monopoles do not exist.

Magnetisation (M): net magnetic moment per unit volume, M = m_net/V, in A m⁻¹.

Magnetic intensity (H): H = B/μ₀ − M, so B = μ₀(H + M), in A m⁻¹.

Susceptibility & permeability: M = χH, μ_r = 1 + χ, μ = μ₀μ_r, and B = μ₀(1 + χ)H = μH.

Important Formulas

Torque on a magnetic dipole: τ = mB sinθ

Potential energy: U = −mB cosθ

Work to rotate a dipole: W = U_f − U_i = mB(cosθ_i − cosθ_f)

Magnetic moment of a solenoid/loop: m = NIA

Axial field of a short bar magnet: B_A = (μ₀/4π) × (2m/r³)

Equatorial field of a short bar magnet: B_E = (μ₀/4π) × (m/r³)

Inside a solenoid: H = nI, B = μ_rμ₀H, where n = turns per unit length. Use μ₀/4π = 10⁻⁷ T m A⁻¹, 1 T = 10⁴ G.

NCERT Solutions – Magnetism and Matter (Exercises)

Questions are reproduced verbatim from the NCERT textbook (Reprint 2026–27). Answers are original and step-by-step; numerical results have been cross-checked against the NCERT answer key.

5.1 A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10⁻² J. What is the magnitude of magnetic moment of the magnet?

SOLUTION Given: B = 0.25 T, θ = 30°, τ = 4.5 × 10⁻² J (i.e. N m). Torque on a dipole: τ = mB sinθ ⇒ m = τ / (B sinθ). m = (4.5 × 10⁻²) / (0.25 × sin 30°) = (4.5 × 10⁻²) / (0.25 × 0.5) = (4.5 × 10⁻²) / 0.125. m = 0.36 J T⁻¹ (= 0.36 A m²).

5.2 A short bar magnet of magnetic moment m = 0.32 J T⁻¹ is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?

SOLUTION Given: m = 0.32 J T⁻¹, B = 0.15 T. Potential energy U = −mB cosθ. (a) Stable equilibrium: m is parallel to B, θ = 0°. U = −mB cos 0° = −(0.32)(0.15) = −4.8 × 10⁻² J (minimum energy → stable). (b) Unstable equilibrium: m is anti-parallel to B, θ = 180°. U = −mB cos 180° = +(0.32)(0.15) = +4.8 × 10⁻² J (maximum energy → unstable).

5.3 A closely wound solenoid of 800 turns and area of cross section 2.5 × 10⁻⁴ m² carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?

SOLUTION A current-carrying solenoid produces a magnetic field with field lines emerging from one end (acting as a north pole) and entering the other end (south pole), exactly like a bar magnet. Its polarity is fixed by the sense of current flow (right-hand rule). Given: N = 800, A = 2.5 × 10⁻⁴ m², I = 3.0 A. Magnetic moment m = NIA = 800 × 3.0 × 2.5 × 10⁻⁴. m = 800 × 3.0 × 2.5 × 10⁻⁴ = 6000 × 10⁻⁴ = 0.60 J T⁻¹ (= 0.60 A m²), directed along the solenoid axis as set by the current.

5.4 If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?

SOLUTION Here the relevant solenoid is the one of Exercise 5.3, whose magnetic moment was found to be m = 0.60 J T⁻¹. Given: B = 0.25 T, θ = 30°. Torque τ = mB sinθ. τ = 0.60 × 0.25 × sin 30° = 0.60 × 0.25 × 0.5 = 0.075. τ = 7.5 × 10⁻² N m (= 7.5 × 10⁻² J).

5.5 A bar magnet of magnetic moment 1.5 J T⁻¹ lies aligned with the direction of a uniform magnetic field of 0.22 T. (a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction? (b) What is the torque on the magnet in cases (i) and (ii)?

SOLUTION Given: m = 1.5 J T⁻¹, B = 0.22 T, initially aligned so θ_i = 0°. Work done by external torque W = mB(cosθ_i − cosθ_f). (a)(i) Normal to the field, θ_f = 90°: W = mB(cos 0° − cos 90°) = 1.5 × 0.22 × (1 − 0) = 0.33 J. (a)(ii) Opposite to the field, θ_f = 180°: W = mB(cos 0° − cos 180°) = 1.5 × 0.22 × (1 − (−1)) = 2mB = 2 × 0.33 = 0.66 J. (b) Torque τ = mB sinθ: (i) at θ = 90°, τ = 1.5 × 0.22 × sin 90° = 0.33 N m, directed so as to turn m back into alignment with B. (ii) at θ = 180°, τ = 1.5 × 0.22 × sin 180° = 0 (unstable equilibrium, so torque is zero).

5.6 A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10⁻⁴ m², carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane. (a) What is the magnetic moment associated with the solenoid? (b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10⁻² T is set up at an angle of 30° with the axis of the solenoid?

SOLUTION Given: N = 2000, A = 1.6 × 10⁻⁴ m², I = 4.0 A, B = 7.5 × 10⁻² T, θ = 30°. (a) m = NIA = 2000 × 4.0 × 1.6 × 10⁻⁴ = 8000 × 1.6 × 10⁻⁴ = 1.28 ⇒ m = 1.28 A m², directed along the axis (sense given by the right-hand screw rule). (b) Force: in a uniform field the net force on a dipole is zero. Torque: τ = mB sinθ = 1.28 × 7.5 × 10⁻² × sin 30° = 1.28 × 0.075 × 0.5 = 0.048. τ = 0.048 N m, in a direction that tends to align the solenoid’s magnetic moment along B.

5.7 A short bar magnet has a magnetic moment of 0.48 J T⁻¹. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.

SOLUTION Given: m = 0.48 J T⁻¹, r = 10 cm = 0.10 m, so r³ = (0.10)³ = 1.0 × 10⁻³ m³; μ₀/4π = 10⁻⁷ T m A⁻¹. (a) On the axis: B_A = (μ₀/4π)(2m/r³) = 10⁻⁷ × (2 × 0.48) / (1.0 × 10⁻³) = 10⁻⁷ × 960 = 9.6 × 10⁻⁵ T = 0.96 G, directed along the S–N axis (i.e. along m, away from the magnet). (b) On the equatorial line: B_E = (μ₀/4π)(m/r³) = 10⁻⁷ × 0.48 / (1.0 × 10⁻³) = 10⁻⁷ × 480 = 4.8 × 10⁻⁵ T = 0.48 G, directed parallel to the N–S direction (opposite to m).

Extra Practice Questions

Short Answer Type Questions

Q1. Why can a magnetic monopole not exist?

ANSWERMagnetic field lines always form continuous closed loops with no beginning or end, so the net flux through any closed surface is zero (Gauss’s law for magnetism). Cutting a magnet only produces smaller dipoles, each with both an N and an S pole; an isolated pole is never obtained, hence monopoles do not exist.

Q2. State the conditions for stable and unstable equilibrium of a magnetic dipole in a uniform field.

ANSWERA dipole is in stable equilibrium when m is parallel to B (θ = 0°, U = −mB, minimum). It is in unstable equilibrium when m is anti-parallel to B (θ = 180°, U = +mB, maximum). In both cases the torque mB sinθ is zero, but only the first restores the dipole when disturbed.

Q3. A bar magnet is cut into two equal pieces (i) transverse to its length and (ii) along its length. How does the magnetic moment change in each case?

ANSWERMagnetic moment m = pole strength × length, while pole strength depends on the cross-section. (i) Cut transverse to the length: length halves, pole strength unchanged, so each piece has moment m/2. (ii) Cut along the length: length unchanged, cross-section (and hence pole strength) halves, so each piece again has moment m/2. Either way two weaker dipoles result.

Q4. Distinguish between the magnetic field B and the magnetic intensity H.

ANSWERH represents the part of the field due to free (external) currents only and is given by H = B/μ₀ − M; it is independent of the medium. B is the total magnetic field including the contribution of the material’s magnetisation, B = μ₀(H + M). Both are measured in A m⁻¹ and T respectively.

Q5. Why does a paramagnetic substance lose its magnetisation rapidly as temperature rises?

ANSWERIn a paramagnet the atomic dipoles align with the external field, but rising temperature increases random thermal agitation that disturbs this alignment. So magnetisation, and hence susceptibility, decreases with temperature (Curie’s law, χ ∝ 1/T).

Long Answer Type Questions

Q1. Compare diamagnetic, paramagnetic and ferromagnetic materials on the basis of susceptibility, relative permeability, behaviour in a non-uniform field and examples.

ANSWER The three classes differ in how their atoms respond to an external field. The comparison is summarised below.

Property	Diamagnetic	Paramagnetic	Ferromagnetic
Susceptibility χ	Small, negative (−1 ≤ χ < 0)	Small, positive (0 < χ < ε)	Large, positive (χ >> 1)
Relative permeability μ_r	0 ≤ μ_r < 1	1 < μ_r < 1 + ε	μ_r >> 1
In non-uniform field	Moves from strong to weak field (repelled)	Moves from weak to strong field (weakly attracted)	Moves from weak to strong field (strongly attracted)
Examples	Bismuth, copper, water, NaCl	Aluminium, sodium, oxygen (STP)	Iron, cobalt, nickel, gadolinium

Diamagnetism arises from induced moments opposing the field (Lenz’s law) and is universal but weak; paramagnetism and ferromagnetism arise from permanent atomic moments, with ferromagnets showing cooperative domain alignment that gives very large, retainable magnetisation.

Q2. Show that a current-carrying solenoid is equivalent to a bar magnet, and derive the expression for its magnetic moment.

ANSWERA solenoid’s field lines emerge from one end and enter the other, forming continuous closed loops just like those of a bar magnet, and a small compass needle deflects identically near both. The far axial field of a solenoid is B = (μ₀/4π)(2m/r³), the same form as the axial field of a bar magnet, confirming equivalence. Treating the solenoid as a stack of N current loops, each loop of area A and current I has moment IA; for N turns the total magnetic moment is m = NIA, directed along the axis. Thus the solenoid behaves as a magnetic dipole of moment NIA.

Q3. State Gauss’s law for magnetism and explain how it differs from Gauss’s law in electrostatics, giving the physical reason.

ANSWERGauss’s law for magnetism states that the net magnetic flux through any closed surface is zero: Σ B·ΔS = 0. In electrostatics, Σ E·ΔS = q/ε₀, which can be non-zero whenever the surface encloses a net charge. The difference reflects a fundamental fact: isolated electric charges (monopoles) exist, but isolated magnetic poles do not. The simplest magnetic element is a dipole or a current loop, so magnetic field lines have no sources or sinks and always form closed loops — every line entering a closed surface also leaves it, making the net flux zero.

MCQs & Assertion–Reason

1. The SI unit of magnetic dipole moment is:

(a) T m² (b) A m² (c) A m⁻¹ (d) Wb

2. A magnetic dipole in a uniform field is in stable equilibrium when the angle between m and B is:

(a) 0° (b) 90° (c) 180° (d) 45°

3. The net magnetic flux through any closed surface is:

(a) q/ε₀ (b) μ₀I (c) zero (d) infinite

4. For a diamagnetic material, the magnetic susceptibility χ is:

(a) large and positive (b) small and positive (c) small and negative (d) zero

5. The torque on a bar magnet of moment m placed at angle θ to a field B is:

(a) mB cosθ (b) mB sinθ (c) mB tanθ (d) mB

6. A short bar magnet’s axial field at distance r relates to its equatorial field at the same r as:

(a) equal (b) axial = 2 × equatorial (c) axial = ½ × equatorial (d) axial = 4 × equatorial

7. The relation between relative permeability and susceptibility is:

(a) μ_r = χ (b) μ_r = 1 + χ (c) μ_r = 1 − χ (d) μ_r = χ/ε₀

8. The potential energy of a magnetic dipole is maximum when θ equals:

(a) 0° (b) 90° (c) 180° (d) 270°

9. A perfect diamagnet (superconductor) that completely expels the field has:

(a) χ = +1, μ_r = 2 (b) χ = −1, μ_r = 0 (c) χ = 0, μ_r = 1 (d) χ >> 1

10. The magnetic moment of a solenoid of N turns, current I and area A is:

(a) NI/A (b) NIA (c) IA/N (d) N/IA

Answer key: 1-(b), 2-(a), 3-(c), 4-(c), 5-(b), 6-(b), 7-(b), 8-(c), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: The net force on a bar magnet placed in a uniform magnetic field is zero.

Reason: The forces on the two poles of the magnet are equal and opposite in a uniform field.

A-R 2. Assertion: Isolated magnetic monopoles do not exist.

Reason: The net magnetic flux through any closed surface is always zero.

A-R 3. Assertion: Diamagnetic materials are repelled by a magnet.

Reason: They develop an induced magnetic moment in the same direction as the applied field.

A-R 4. Assertion: A current-carrying solenoid behaves like a bar magnet.

Reason: Its far axial field has the same form as that of a bar magnet, B = (μ₀/4π)(2m/r³).

A-R 5. Assertion: A ferromagnet turns into a paramagnet above a certain temperature.

Reason: Thermal agitation destroys the domain structure at high temperature.

Answer key: 1-(A), 2-(A), 3-(C), 4-(A), 5-(A).

Common Mistakes to Avoid

Watch out for these

Using cosθ for torque or sinθ for energy — remember τ = mB sinθ and U = −mB cosθ.
Forgetting the negative sign in U = −mB cosθ, which decides stable vs unstable equilibrium.
Confusing axial and equatorial fields — axial is twice the equatorial for the same r, and they point in opposite senses relative to m.
Forgetting μ₀/4π = 10⁻⁷ and dropping the factor of 2 in the axial-field formula.
Assuming a dipole feels a net force in a uniform field — it feels only a torque; force appears only in a non-uniform field.
Mixing up the signs of χ for dia- (negative) and para-/ferromagnetic (positive) materials.
Unit slips: m in A m² or J T⁻¹, B in tesla, and 1 T = 10⁴ G when reporting field in gauss.

Exam tips for full marks

Write the formula, substitute values with units, then box the final answer with the correct unit and (where asked) direction. For torque/energy numericals, always show the angle clearly. Convert distances to metres before cubing for axial/equatorial fields. State results in both tesla and gauss when the question is framed in gauss. Learn the dia/para/ferro comparison table — it is a frequent 3-mark question. Remember that Gauss’s law for magnetism (net flux = 0) and the non-existence of monopoles are favourite 1–2 mark theory questions.

Frequently Asked Questions

What is Class 12 Physics Chapter 5 Magnetism and Matter about?

It studies magnetism in its own right: the bar magnet as a magnetic dipole, torque and potential energy of a dipole in a field, the bar magnet–solenoid equivalence, Gauss’s law for magnetism, magnetisation and magnetic intensity, and the classification of materials as diamagnetic, paramagnetic and ferromagnetic.

How many exercises are there in Chapter 5 and are all solved here?

The NCERT chapter has exercises 5.1 to 5.7. All of them are reproduced verbatim and solved step by step on this page, with every numerical answer verified against the NCERT answer key and stated with correct units.

What is the difference between B and H in this chapter?

H (magnetic intensity) represents the part of the field due to free currents alone, H = B/μ₀ − M, while B is the total field including the material’s magnetisation, B = μ₀(H + M). H is measured in A m⁻¹ and B in tesla.

Are these Class 12 Physics Chapter 5 solutions free?

Yes. All solutions are free and follow the official NCERT Physics textbook for the 2026–27 session.

← Chapter 4: Moving Charges and Magnetism All Class 12 Physics Chapters Chapter 6: Electromagnetic Induction →