Question:
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:
Magnetic field strength, B = 0.25 T
Magnetic moment, M = 0.6 T−1
The angle θ, between the axis of the solenoid and the direction of the applied field is 30°.
Therefore, the torque acting on the solenoid is given as:
$\tau=M B \sin \theta$
$=0.6 \times 0.25 \sin 30^{\circ}$
$=7.5 \times 10^{-2} \mathrm{~J}$