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(A) The length of the wire $L$ forms the circumference of the circular loop, so $L = 2 \pi r$, where $r$ is the radius of the loop.From this, the radi

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

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(A) The length of the wire $L$ forms the circumference of the circular loop, so $L = 2 \pi r$, where $r$ is the radius of the loop.From this, the radius is $r = \frac{L}{2 \pi}$.The magnetic moment $M$ of a current-carrying loop is given by $M = iA$, where $A$ is the area of the loop.The area $A = \pi r^2 = \pi \left( \frac{L}{2 \pi} ight)^2 = \pi \left( \frac{L^2}{4 \pi^2} ight) = \frac{L^2}{4 \pi}$.Substituting this into the formula for magnetic moment, we get $M = i 	imes \frac{L^2}{4 \pi} = \frac{i L^2}{4 \pi}$.

$A$ current $i$ is flowing through a wire of length $L$. If it is made into a circular loop of one turn, then its magnetic moment is

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