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An iron bar of length $L$ has magnetic moment $M$. It is bent at the middle of its length such that the two arms make an angle $60^{\circ}$ with each other. The magnetic moment of this new magnet is :
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The correct answer is:
$\frac{M}{2}$
$\begin{aligned} & \Delta I=2 \frac{I}{2} \sin 30^{\circ} \\ & =\frac{1}{2} \\ & M^{\prime}=m / / 2 \\ & =M / 2\end{aligned}$
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