Q151 / 160 · WBJEE 2012 · Physics
A magnetic needle is placed in a uniform magnetic field and is aligned with the field. The needle is now rotated by an angle of $60^{\circ}$ and the work done is $W$. The torque on the magnetic needle at this position is
- A$2 \sqrt{3} w$
- Correct answerB$\sqrt{3} w$
- C$\frac{\sqrt{3}}{2} w$
- D$\frac{\sqrt{3}}{4} \mathrm{W}$
Solution
Given, work done $=W$ and $\theta=60^{\circ}$ We know that
Hence,
$$
\begin{array}{l}
W=M B(1-\cos \theta) \\
W=M B\left(1-\cos 60^{\circ}\right) \\
W=\frac{M B}{2} \\
|\tau|=M B \sin 60^{\circ}=\sqrt{3} W
\end{array}
$$
Hence, the correct answer is option: 'B'.