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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 fi

5.6. 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?

Answers (1)

best_answer

Given,

Magnetic field strength, B = 0.25 T

Magnetic moment, m = 0.6 JT−1

The angle between the axis of the solenoid and the direction of the applied field, \theta =  30°.

We know, the torque acting on the solenoid is:

\tau = m x B = mBsinθ 
= (0.6 JT^{-1})(0.25 T)(sin 30o) 

= 0.075 J
= 7.5 x 10^{-2} J

The magnitude of torque is 7.5 x 10^{-2} J.

Posted by

HARSH KANKARIA

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