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Solenoid and Bar Magnet

If the soleno...

Question

If the solenoid in Exercise 5.5 is free to turn about the verticaldirection and a uniform horizontal magnetic field of 0.25 T is applied,what is the magnitude of torque on the solenoid when its axis makesan angle of 30° with the direction of applied field?

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Solution

Given: The angle between the axis of solenoid and magnetic field is 30°, magnitude of the magnetic field is 0.25 T and the magnetic moment is 0.6 JT -1 .

The torque on the solenoid is given as,

τ=MBsinθ
By substituting the given values, we get,

τ=0.6×0.25sin30° =7.5× 10 −2 J
Thus, the magnitude of torque acting on the solenoid is 7.5× 10 −2 J.

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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?

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