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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° with the direction of applied field?
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?
Updated On: Sep 30, 2023
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Solution and Explanation
Magnetic field strength, B = 0.25 T
Magnetic moment, M = 0.6 T-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:
\(\tau\) =\(MB \sin\) \(\theta\)
= 0.6 × 0.25 sin 30\(\degree\)
= 7.5 × 10-2 J
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