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

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$(3M / \pi)$

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(D) Let the original length of the wire be $L$ and its pole strength be $m$. The original magnetic moment is $M = m \cdot L$.When the wire is bent into an arc of a circle of radius $r$ subtending an angle $	heta = 60^o = \pi/3$ radians at the center,the length of the arc is $L = r	heta$.Thus,$r = L / 	heta = L / (\pi/3) = 3L / \pi$.The new magnetic moment $M'$ is the product of the pole strength $m$ and the straight-line distance (chord length) between the two ends of the arc.The chord length $d = 2r \sin(	heta/2)$.Substituting $	heta = 60^o$,we get $d = 2r \sin(30^o) = 2r(1/2) = r$.Substituting $r = 3L / \pi$,we get $d = 3L / \pi$.Therefore,the new magnetic moment $M' = m \cdot d = m \cdot (3L / \pi) = (3 / \pi) \cdot (mL) = 3M / \pi$.

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

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