# 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?

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, $\dpi{100} \theta$ =  30°.

We know, the torque acting on the solenoid is:

$\dpi{100} \tau$ = m x B = mBsinθ
= (0.6 $\dpi{80} JT^{-1}$)(0.25 T)(sin 30o

= 0.075 J
= 7.5 x $\dpi{80} 10^{-2}$ J

The magnitude of torque is 7.5 x $\dpi{80} 10^{-2}$ J.

Exams
Articles
Questions