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:
   (i) normal to the field direction,
   (ii) 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 magnetic field is given as:
$W = -MB(\cos\theta_2-\cos\theta_1)$
= -1.5 $\times$ 0.22(cos 90°-cos 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 magnetic field is given as:
$W = -MB(\cos\theta_2-\cos\theta_1)$
= -1.5 $\times$ 0.22(cos 180°-cos 0°)
= -0.33(-1-1)
= 0.66 J

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(b) For case (i): θ = θ2 = 90°
∴ Torque, τ =$MB\sin\theta$
= 1.5 $\times$ 0.22 sin 90°
= 0.33 J
For case (ii): θ = θ2 = 180°
∴ Torque, τ = $MB\sin\theta$
= $MB\sin$ 180° = 0 J