Question

The magnetic needle lying parallel to the magnetic field requires $W$ units of work to rotate it through $60^{\circ}$. The torque needed to maintain the needle in this position is :

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

Answer 1

Answer: (B)

$\sqrt{3}\,\,W$

Answer 2

The instaneous moment of the deflecting couple or torque acting on the needle is $\tau=$ force $\times$ perpendicular distance $=$ work done when axis of needle makes an angle $\theta$ with the magnetic field, then for magnetic moment $M$ and magnetic field $B$, we have $\tau=M B \sin \theta \ldots(1)$ $W=M B \cos \theta \ldots(2)$ Dividing E (1) by E (2), we get $\frac{\tau}{W}=\frac{M B \sin \theta}{M B \cos \theta}$ Given, $\theta=60^{\circ}$ $\tau=W \frac{\sin 60^{\circ}}{\cos 60^{\circ}}$ $W=\sqrt{3}$