Question

If a solenoid is having magnetic moment of $0.65J{T}^{-1}$ is free to turn about the vertical direction and has a uniform horizontal magnetic field of $0.25T$ applied. What is the magnitude of the torque on the solenoid when its axis makes an angle of ${30}^0$ with the direction of applied field is?

• A. $0.075$ $Nm$
• B. $0.080$ $Nm$
• C. $0.081$ $Nm$
• D. $0.091$ $Nm$

Answer 1

Answer: (C)

$0.081$ $Nm$

Given that, $M=0.65J{T}^{-1}$ , $B=0.25T$ and $\theta ={30}^o$

$\therefore$ Torque on the solenoid $,\tau =MB\sin \theta =0.65\times 0.25\times \sin {30}^o$

$=0.65\times 0.25\times \frac{1}{2}=0.08125=0.081Nm$

So, option $\left(C\right)$ is correct.