Question

If a solenoid is having magnetic moment of $0.65$J $T^{-1}$ is free to turn about the vertical direction and has a uniform horizontal magnetic field of $0.25$T applied. What is the magnitude of the torque on the solenoid when its axis makes an angle of ${30}^o$ with the direction of applied field?

• A. $0.075$N m
• B. $0.060$N m
• C. $0.081$N m
• D. $0.091$N m

Answer 1

Answer: (C)

$0.081$N m

Given:

The magnetic moment of the solenoid is $M=0.65J{T}^{-1}$

The magnetic field applied across the solenoid is $B=0.25T$

The angle made by the axis of the solenoid is $\theta ={30}^o$

The torque experienced by the solenoid is given by:

$\therefore \tau =MB\sin \theta$

$=0.65\times 0.25\times \sin {30}^o$

$=0.65\times 0.25\times \frac{1}{2}$

$=0.08125=0.08Nm$