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?
Magnetic field strength, B = 0.25 T
Magnetic moment, M = 0.6 JT−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:
τ = MBsinθ = 0.6 JT-1 x 0.25 T x sin 30o
= 0.6 JT-1 x 0.25 T x 1/2
= 7.5x10-3 J
The magnitude of torque is 7.5 x 10-3 J.
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