Question

A magnetic needle lying parallel to a magnetic field requires $W$ units of work to turn it through ${60}^{\circ }$. The torque required to maintain the needle in this position will be

• A. $\sqrt{3}W$
• B. $W$
• C. $\frac{\sqrt{3}}{2}W$
• D. $2W$

Answer 1

The correct option is A $\sqrt{3}W$

Work done to rotate the needle = potential energy of needle = $W$
If $M$ = magnetic dipole moment of the needle and $B$ = magnetic field then :
$W=MB(\cos {\theta }_1-\cos {\theta }_2)=MB(\cos 0^{\circ }-\cos {60}^{\circ })$
$=MB(1-\frac{1}{2})=\frac{MB}{2}$
$\tau =MB\sin \theta =MB\sin {60}^{\circ }=MB\frac{\sqrt{3}}{2}$
$\therefore \tau =\left(\frac{MB}{2}\right)\sqrt{3}\Rightarrow \tau =\sqrt{3}W$