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Question
If the solenoid 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?
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Solution
Given: 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°.
Formula: τ = MB sin θ
= 0.6 × 0.25 × sin 30°
= 7.5 × 10−2 J
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