Class 12 Physics

Chapter 5 Solutions — Magnetism and Matter

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Solutions

Overview

Step-by-step NCERT solutions for Magnetism and Matter (Chapter 5, CBSE Class 12 Physics) — every question and answer worked out in full, not just the final result. You can also read the Magnetism and Matter textbook chapter.

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What these solutions cover

All 7 questions in Magnetism and Matter are solved in the PDF. Here's what's inside, exercise by exercise:

Exercises

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?
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?
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?
If the solenoid in Exercise 5.3 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?
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)?
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…
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.

Sample

Worked examples

3 questions from Magnetism and Matter, solved step by step.

Exercises

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?

Given: angle θ = 30°, B = 0.25 T, torque τ = 4.5 × 10⁻² N·m

The torque on a magnetic dipole in a uniform field is:

τ = mB sinθ

Solving for m:

m = τ / (B sinθ)

m = (4.5 × 10⁻²) / (0.25 × sin 30°)

m = (4.5 × 10⁻²) / (0.25 × 0.5)

m = (4.5 × 10⁻²) / 0.125

m = 0.36 J T⁻¹

Answer: Magnetic moment m = 0.36 J T⁻¹

Exercises

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?

Given: m = 0.32 J T⁻¹, B = 0.15 T

Magnetic potential energy: U = -m · B = -mB cosθ

(a) Stable equilibrium:

Stable equilibrium occurs when potential energy is minimum, i.e., at θ = 0°, when the magnetic moment m is parallel to (aligned with) the field B.

U_stable = -mB cos 0° = -mB = -(0.32)(0.15) = -4.8 × 10⁻² J

(b) Unstable equilibrium:

Unstable equilibrium occurs when potential energy is maximum, i.e., at θ = 180°, when m is anti-parallel to B.

U_unstable = -mB cos 180° = +mB = +(0.32)(0.15) = +4.8 × 10⁻² J

Answer: (a) Stable: m parallel to B (θ = 0°), U = -4.8 × 10⁻² J; (b) Unstable: m anti-parallel to B (θ = 180°), U = +4.8 × 10⁻² J

Exercises

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?

Given: N = 800 turns, A = 2.5 × 10⁻⁴ m², I = 3.0 A

A current-carrying solenoid acts like a bar magnet in the following sense:

Its magnetic field pattern — with field lines emerging from one face (north pole) and entering the other (south pole) — is identical to that of a bar magnet. At large distances, the axial field of both obeys the same formula B = (µ0/4π)(2m/r³). Cutting the solenoid in half yields two smaller solenoids, just as cutting a bar magnet yields two smaller magnets. The magnetic moment of the solenoid equals that of an equivalent bar magnet.

The magnetic moment of a current loop (or solenoid) is:

m = NIA

m = 800 × 3.0 × 2.5 × 10⁻⁴

m = 0.60 J T⁻¹

Answer: The solenoid acts like a bar magnet because their field patterns and far-field formulas are identical. Magnetic moment m = 0.60 J T⁻¹

All 7 questions in Magnetism and Matter are solved step by step in the Solutions PDF above.

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Chapter 5 Solutions — Magnetism and Matter

Overview

What these solutions cover

Exercises

Worked examples

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?

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?

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