Question

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

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

Answer 1

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

We know that

The work done will be stored as potential energy change of needle

$W=MB(1-\cos 0^0)=0$

$W=MB(1-\cos {60}^0)$

$W=\frac{MB}{2}$

$2W=MB$

Where M and B are magnetic dipole and magnetic field respectively

Now, the torque

$\tau =MB\sin {60}^0$

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

$\tau =W\sqrt{3}$

Hence, the torque required to maintain the needle in this position is $W\sqrt{3}$