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(B) The magnetic moment $M$ of a current loop is given by the product of the current $I$ and the area $A$ of the loop,$M = I A$.When a wire of length

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$\frac{L^2 I}{4 \pi}$

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(B) The magnetic moment $M$ of a current loop is given by the product of the current $I$ and the area $A$ of the loop,$M = I A$.When a wire of length $L$ is bent into a circle,its circumference is $2 \pi r = L$,which implies the radius $r = \frac{L}{2 \pi}$.The area of this circular loop is $A = \pi r^2 = \pi \left( \frac{L}{2 \pi} ight)^2 = \frac{L^2}{4 \pi}$.Substituting the area into the magnetic moment formula,we get $M = I 	imes \frac{L^2}{4 \pi} = \frac{L^2 I}{4 \pi}$.

$A$ wire of length $L \ m$ carrying a current $I \ A$ is bent in the form of a circle. The magnitude of its magnetic moment is

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