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(D) Let the original length of the wire be $L$. The original magnetic moment is $M = m \cdot L$,where $m$ is the pole strength.When the wire is bent i

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$\frac{3M}{\pi}$

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(D) Let the original length of the wire be $L$. The original magnetic moment is $M = m \cdot L$,where $m$ is the pole strength.When the wire is bent into an arc of a circle of radius $r$ subtending an angle $	heta = 60^{\circ} = \frac{\pi}{3}$ radians at the center,the arc length is $L = r	heta$.Thus,$r = \frac{L}{	heta} = \frac{L}{\pi/3} = \frac{3L}{\pi}$.The distance between the two ends of the arc (the chord length) is $d = 2r \sin(\frac{	heta}{2})$.Substituting $	heta = 60^{\circ}$,we get $d = 2r \sin(30^{\circ}) = 2r \cdot \frac{1}{2} = r$.The new magnetic moment $M'$ is given by $M' = m \cdot d = m \cdot r$.Substituting $r = \frac{3L}{\pi}$,we get $M' = m \cdot \frac{3L}{\pi} = \frac{3}{\pi} (m \cdot L)$.Since $M = m \cdot L$,the new magnetic moment is $M' = \frac{3M}{\pi}$.

$A$ magnetized wire of magnetic moment $M$ and length $L$ is bent into an arc of a circle subtending an angle of $60^{\circ}$ at the center. The new magnetic moment is:

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