Q. 64.8( 13 Votes )

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 JT−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:



τ = MBsinθ
= 0.6 JT-1 × 0.25 T × sin 30o

= 0.6 JT-1 × 0.25 T × (1/2)

(because sin 30 = 1/2)

= 0.075 J
= 7.5 × 10-2 J


The magnitude of torque is 7.5 × 10-2 J.

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