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A magnetic needle suspended parallel to a magnetic field requires $\sqrt{3} \mathrm{~J}$ of work to turn it through $60^{\circ}$. The torque needed to maintain the needle in this position will be
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Verified Answer
The correct answer is:
$3 \mathrm{~J}$
In this case, work done
$$
\begin{aligned}
W & =M B\left(\cos \theta_1-\cos \theta_2\right) \\
& =M B\left(\cos 0^{\circ}-\cos 60^{\circ}\right) \\
& =M B\left(1-\frac{1}{2}\right)=\frac{M B}{2} \\
M B & =2 \sqrt{3} \mathrm{~J} \quad(\because \text { given } W=\sqrt{3} \mathrm{~J}) \\
\tau & =M B \sin 60^{\circ}=(2 \sqrt{3})\left(\frac{\sqrt{3}}{2}\right) \mathrm{J}=3 \mathrm{~J}
\end{aligned}
$$
$$
\begin{aligned}
W & =M B\left(\cos \theta_1-\cos \theta_2\right) \\
& =M B\left(\cos 0^{\circ}-\cos 60^{\circ}\right) \\
& =M B\left(1-\frac{1}{2}\right)=\frac{M B}{2} \\
M B & =2 \sqrt{3} \mathrm{~J} \quad(\because \text { given } W=\sqrt{3} \mathrm{~J}) \\
\tau & =M B \sin 60^{\circ}=(2 \sqrt{3})\left(\frac{\sqrt{3}}{2}\right) \mathrm{J}=3 \mathrm{~J}
\end{aligned}
$$
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