If the solenoid in Exercise 5.5 is free to turn

Question:

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?

Solution:

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:

$\tau=M B \sin \theta$

$=0.6 \times 0.25 \sin 30^{\circ}$

$=7.5 \times 10^{-2} \mathrm{~J}$

 

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