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

The Correct Option is B

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, $\theta=60^{\circ}$ $\tau=W \frac{\sin 60^{\circ}}{\cos 60^{\circ}}$ $W=\sqrt{3}$