Q106 / 120 · VITEEE 2015 · Physics
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$(\sqrt{2}-1) W$
- Correct answerD$\frac{W}{(\sqrt{2}-1)}$
Solution
Work done by magnet to turn from angle
$\begin{array}{l}
\begin{array}{l}
\theta_{1} \text { to } \theta_{2} \\
\mathrm{~W}=\mathrm{MB}\left(\cos \theta_{1}-\cos \theta_{2}\right) \\
=\mathrm{MB}\left(\cos 0^{\circ}-\cos 45^{\circ}\right)
\end{array} \\
=\mathrm{MB}\left(1-\frac{1}{\sqrt{2}}\right)=\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right) \mathrm{MB}
\end{array}$
Also torque acting on the magnet
$\begin{array}{l}
\tau=\mathrm{MB} \sin 45^{\circ}=\frac{\mathrm{MB}}{\sqrt{2}} \\
\mathrm{~W}=(\sqrt{2}-1) \cdot \tau \Rightarrow \tau=\frac{\mathrm{W}}{(\sqrt{2}-1)}
\end{array}$
Hence, the correct answer is option: 'D'.