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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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$\frac{L^{2} i}{4 π}$

$\frac{L^{2}}{4 π}$

$\frac{4 π}{L^{2} i}$

$4 π L^{2} i$

answer is A.

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Detailed Solution

$L = 2 π r, r = \frac{L}{2 π}$
$M = i A$
$M = i π r^{2} = \frac{L^{2} i}{4 π}$

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