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?

Answer 1

Magnetic field strength, B = 0.25 T
Magnetic moment, M = 0.6 T-1
The angle $\theta$, 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:
$\tau$ =$MB \sin$ $\theta$
= 0.6 × 0.25 sin 30$\degree$
= 7.5 × 10-2 J