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5.6. 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?
Answers (1)
Given,
Magnetic field strength, B = 0.25 T
Magnetic moment, m = 0.6 JT−1
The angle between the axis of the solenoid and the direction of the applied field, = 30°.
We know, the torque acting on the solenoid is:
= m x B = mBsinθ
= (0.6 )(0.25 T)(sin 30o)
= 0.075 J
= 7.5 x J
The magnitude of torque is 7.5 x J.
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