Question

A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through ${60}^0$. The torque needed to maintain the needle in this position will be:

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

Answer 1

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

Work done to rotate the needle from $0^{\circ }$ to ${60}^{\circ }=MB(\cos 0^{\circ }-\cos {60}^{\circ })=W=\frac{1}{2}MB$........(1)

Torque required to keep needle at ${60}^{\circ }$ from the magnetic field $\tau =MB\sin {60}^{\circ }=\frac{\sqrt{3}}{2}MB$........(2)

From equation 1 and 2

$\tau =\frac{\sqrt{3}}{2}\times 2W$

$\tau =\sqrt{3}W$

Option A