A magnetic needle lying parallel to a magnetic field requires $W$ units of work to turn it through $

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\

Professor PC Thomas Classes · Answer

(a) : $W=-M B\left(\cos 	heta_{2}-\cos 	heta_{1}ight)$\ $=-M B\left(\cos 60^{\circ}-\cos 0^{\circ}ight)=\frac{M B}{2}$ $$\begin{equation*} 	herefore \quad M B=2 W 	ag{i} \end{equation*}$$ Torque $=M B \sin 60^{\circ}=(2 W) \sin 60^{\circ}$ $=\frac{2 W 	imes \sqrt{3}}{2}=\sqrt{3} W$\