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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^{\circ} = \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 distance between the two ends (poles) of the arc is the chord length $d = 2r \sin(	heta/2)$.Substituting $	heta = 60^{\circ}$,we get $d = 2r \sin(30^{\circ}) = 2r(1/2) = r$.Substituting $r = 3L/\pi$,we get $d = 3L/\pi$.The new magnetic moment $M'$ is the product of pole strength $m$ and the chord length $d$:$M' = m \cdot d = m \cdot (3L/\pi) = (3/\pi) \cdot (m \cdot L)$.Since $M = m \cdot L$,we have $M' = 3M/\pi$.

$A$ 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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