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(A) Let the pole strength of the wire be $m$. The magnetic moment of the straight wire is $M = m 	imes \ell$.When the wire is bent into a semicircula

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$2M / \pi$

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(A) Let the pole strength of the wire be $m$. The magnetic moment of the straight wire is $M = m 	imes \ell$.When the wire is bent into a semicircular arc,the length of the arc is $\ell = \pi r$,where $r$ is the radius of the semicircle.Thus,the radius is $r = \ell / \pi$.The distance between the two ends of the semicircular wire (the magnetic length) is the diameter,$d = 2r = 2\ell / \pi$.The new magnetic moment $M'$ is given by the product of the pole strength and the magnetic length:$M' = m 	imes (2r) = m 	imes (2\ell / \pi) = (2 / \pi) 	imes (m 	imes \ell)$.Since $M = m 	imes \ell$,we have $M' = (2 / \pi) M$.

$A$ steel wire of length $\ell$ has a magnetic moment $M$. It is then bent into a semicircular arc. The new magnetic moment is

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