Question:
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 degree with the direction of applied field?
Answer:
Magnetic field strength, B = 0.25 T
Magnetic moment, M=
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:
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