Question

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

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

Answer 1

The correct option is B

$\sqrt{3}W$

$W=MB(1-cos{60}^{\circ })=\frac{MB}{2}$

$\Rightarrow MB=2W$

Again,

$\tau =MBsin{60}^{\circ }$

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