Question

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

• A. $\sqrt{2} W$
• B. $\frac{1}{\sqrt{3} W}$
• C. $(\sqrt{2} - 1) W$
• D. $\frac{W}{(\sqrt{2} - 1) }$

Answer 1

Answer: (D)

Work done by magnet to turn from angle $\theta_1$ to $\theta_2$
$W = MB \left(\cos\theta_{1} - \cos\theta_{2}\right)$
$ = MB \left(\cos0^{\circ} - \cos 45^{\circ}\right) $
$= MB \left(1 - \frac{1}{\sqrt{2}}\right)$
$ = \left(\frac{\sqrt{2} - 1}{\sqrt{2}}\right)MB$
Also torque acting on the magnet
$ \tau= MB \sin45^{\circ} = \frac{MB}{\sqrt{2}} $
SO $ W = \left(\sqrt{2} - 1\right). \tau$ $\therefore = \frac{W}{\left(\sqrt{2} - 1\right) } $