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 ${\theta }_2$
$W=MB\left(\cos {\theta }_1-\cos {\theta }_2\right)$
$=MB\left(\cos 0^{\circ }-\cos 45^{\circ }\right)$
$=MB\left(1-\frac{1}{\sqrt{2}}\right)$
$=\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)MB$
Also torque acting on the magnet
$\tau =MB\sin 45^{\circ }=\frac{MB}{\sqrt{2}}$
SO $W=\left(\sqrt{2}-1\right).\tau$ $\therefore =\frac{W}{\left(\sqrt{2}-1\right)}$