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A magnetic needle lying parallel to a magnetic field is turned through $60^{\circ}$. The work done on it is $W$. The torque required to maintain the magnetic needle in the position mentioned above is
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Verified Answer
The correct answer is:
$\sqrt{3} W$
Work done
$\begin{aligned} W & =M B\left(1-\cos 60^{\circ}\right) \\ & =\frac{M B}{2}\end{aligned}$
The torque required to maintain the magnetic needle
$\begin{aligned} \tau & =M B \sin \theta \\ & =M B \sin 60^{\circ} \\ & =M B \frac{\sqrt{3}}{2} \\ \tau & =\sqrt{3} \mathrm{~W}\end{aligned}$
$\begin{aligned} W & =M B\left(1-\cos 60^{\circ}\right) \\ & =\frac{M B}{2}\end{aligned}$
The torque required to maintain the magnetic needle
$\begin{aligned} \tau & =M B \sin \theta \\ & =M B \sin 60^{\circ} \\ & =M B \frac{\sqrt{3}}{2} \\ \tau & =\sqrt{3} \mathrm{~W}\end{aligned}$
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