If the solenoid in the above question is free to turn about the vertical direction, and a uniform horizontal magnetic field of `0*25T` is applied, what is the magnitude of the torque on the solenoid when its axis makes an angle of `30^@` with the direction of the applied field?

Explanation

Magnetic field strength, B = 0.25 T , Magnetic moment, `M=0.6T^(-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=MBsin theta , =0.6xx0.25sin30^(@), =7.5xx10^(-2)J`