A magnetic needle lying parallel to a magnetic field is turned through $60^{\circ}$. The work done on it is…
A magnetic needle lying parallel to a magnetic field is turned through $60^{\circ}$. The work done on it is $W$. The torque required to maintain the magnetic needle in the position mentioned above is
$\sqrt{3} W$
$\frac{\sqrt{3}}{2} W$
$\frac{W}{2}$
$2 W$
Solution
Work done
$\begin{aligned} W & =M B\left(1-\cos 60^{\circ}\right) \\ & =\frac{M B}{2}\end{aligned}$
The torque required to maintain the magnetic needle
$\begin{aligned} \tau & =M B \sin \theta \\ & =M B \sin 60^{\circ} \\ & =M B \frac{\sqrt{3}}{2} \\ \tau & =\sqrt{3} \mathrm{~W}\end{aligned}$