Question

A magnetic needle lying parallel to a magnetic field requires \(W\) units of work to turn it through \(60^{\circ}\). The torque needed to maintain the needle in this position will be
(a) \(\sqrt{3} \mathrm{~W}\)
(b) \(W\)
(c) \((\sqrt{3} / 2) \mathrm{W}\)
(d) \(2 \mathrm{~W}\).

Answer 1

The correct answer is:
$$
\begin{aligned}
\text { (a) : } W &=-M B\left(\cos \theta_{2}-\cos \theta_{1}\right) \\
&=-M B\left(\cos 60^{\circ}-\cos 0\right)=\frac{M B}{2} \\
\therefore \quad & M B=2 W \\
& \text { Torque }=M B \sin 60^{\circ}=(2 W) \sin 60^{\circ} \\
=& \frac{2 W \times \sqrt{3}}{2}=\sqrt{3} W .
\end{aligned}
$$