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(C) The magnetic moment of a straight wire of length $l$ and pole strength $m$ is given by $M = m 	imes l$.When the wire is bent into a semicircular

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$\frac{2 M}{\pi}$

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(C) The magnetic moment of a straight wire of length $l$ and pole strength $m$ is given by $M = m 	imes l$.When the wire is bent into a semicircular arc of radius $R$,the length of the wire corresponds to the circumference of the semicircle: $l = \pi R$.Therefore,the radius of the arc is $R = \frac{l}{\pi}$.The new magnetic moment $M'$ is the product of the pole strength $m$ and the straight-line distance between the two ends (the diameter of the semicircle).The diameter is $d = 2R = \frac{2l}{\pi}$.Thus,$M' = m 	imes d = m 	imes \frac{2l}{\pi}$.Since $M = ml$,we substitute $ml$ with $M$ to get $M' = \frac{2}{\pi} M$.

$A$ steel wire of length $l$ and magnetic moment $M$ is bent into a semicircular arc of radius $R$. The new magnetic moment is

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