Discussion Forum

A magnetic needle lying parallel to the magnetic field requires W units of work to turn it through an angle of 45°. The torque required to maintain the needle in this position will be

Answer:

Explanation:

 Work done by magnet to turn from angle  $\theta_{1}$ to $\theta_{2}$

 $W=MB(\cos\theta_{1}-\cos\theta_{2})$

$=MB(\cos0^{0}-\cos 45^{0})$

$W=MB(1-\frac{1}{\sqrt{2}})=(\frac{\sqrt{2}-1}{\sqrt{2}})MB$

 Also torque acting on the magnet

 $\tau=MB\sin 45^{0}=\frac{MB}{\sqrt{2}}$

$W=(\sqrt{2}-1).\tau\Rightarrow\tau=\frac{W}{(\sqrt{2}-1)}$