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(D) The length of the conductor $L$ forms the circumference of the circular loop.$L = 2 \pi r$,where $r$ is the radius of the loop.Therefore,$r = \fra

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$\frac{IL^2}{4 \pi}$

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(D) The length of the conductor $L$ forms the circumference of the circular loop.$L = 2 \pi r$,where $r$ is the radius of the loop.Therefore,$r = \frac{L}{2 \pi}$.The area $A$ of the circular loop is given by $A = \pi r^2$.Substituting the value of $r$,we get $A = \pi \left( \frac{L}{2 \pi} ight)^2 = \pi \left( \frac{L^2}{4 \pi^2} ight) = \frac{L^2}{4 \pi}$.The magnetic moment $M$ of a current-carrying loop is given by $M = I 	imes A$.Substituting the value of $A$,we get $M = I \left( \frac{L^2}{4 \pi} ight) = \frac{IL^2}{4 \pi}$.

$A$ current $I$ is flowing in a conductor of length $L$. When it is bent in the form of a circular loop,its magnetic moment will be

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