Question 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?

Answer

Magnetic field strength, B = 0.25 T

Magnetic moment, M= 0.6 T-1

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:

T = MB sinθ
  = 0.6 x 0.25 sin30°
  = 7.5 x 102 J

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