Question

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.

1. What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment:
   1. normal to the field direction,
   2. opposite to the field direction?
2. What is the torque on the magnet in cases (i) and (ii)?

Answer 1

(a) Magnetic moment, M = 1.5 J T⁻¹

Magnetic field strength, B = 0.22 T

(i) Initial angle between the axis and the magnetic field, θ₁ = 0°

Final angle between the axis and the magnetic field, θ₂ = 90°

The work required to make the magnetic moment normal to the direction of the magnetic field is given as:

W = −MB (cos θ₂ − cos θ₁)

= −1.5 × 0.22 (cos 90° − 0°)

= −0.33 (0 − 1)

= 0.33 J

(ii) Initial angle between the axis and the magnetic field, θ₁ = 0°

Final angle between the axis and the magnetic field, θ₂ = 180°

The work required to make the magnetic moment opposite to the direction of the magnetic field is given as:

W = −MB (cos θ₂ − cos θ₁)

= −1.5 × 0.22 (cos 180° − 0°)

= −0.33 (−1 − 1)

= 0.66 J

(b) For case (i), θ = θ₂ = 90°

∴ Torque, τ = MB sin θ

= 1.5 × 0.22 × sin 90°

= 0.33 J

For case (ii), θ = θ₂ = 180°

∴ Torque, τ = MB sin θ

= MB sin 180°

= 0 J