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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
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Verified Answer
The correct answer is:
$\sqrt{3} w$
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,
$$
\begin{array}{l}
W=M B(1-\cos \theta) \\
W=M B\left(1-\cos 60^{\circ}\right) \\
W=\frac{M B}{2} \\
|\tau|=M B \sin 60^{\circ}=\sqrt{3} W
\end{array}
$$
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