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?

Answer 1

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.