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 the torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?

Answer –

Magnetic field strength is given by B = 0.25 T

Magnetic moment is by M = 0.6  T^-1      

The angle θ between the direction of the applied field and the axis of the solenoid is 30°.

Therefore, the torque acting on the solenoid is given by the relation –

\[T=MB\sin \theta =0.6\times 0.25\times \sin {{30}^{\circ }}\]

\[T=7.5\times {{10}^{-2}}J\]