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(C) For a square loop of side $l$,the perimeter is $P = 4l$ and the area is $A_s = l^2$. The magnetic dipole moment is $m = I A_s = I l^2$.When reshap

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$\frac{4m}{\pi}$

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(C) For a square loop of side $l$,the perimeter is $P = 4l$ and the area is $A_s = l^2$. The magnetic dipole moment is $m = I A_s = I l^2$.When reshaped into a circular loop of radius $r$,the perimeter remains the same: $2\pi r = 4l$,which gives $r = \frac{2l}{\pi}$.The area of the circular loop is $A_c = \pi r^2 = \pi \left(\frac{2l}{\pi}\right)^2 = \frac{4l^2}{\pi}$.The new magnetic dipole moment $m'$ is $m' = I A_c = I \left(\frac{4l^2}{\pi}\right)$.Since $m = I l^2$,we substitute this into the expression for $m'$:$m' = \frac{4}{\pi} (I l^2) = \frac{4m}{\pi}$.

$A$ square loop is carrying a steady current $I$ and the magnitude of its magnetic dipole moment is $m$. If this square loop is changed to a circular loop and it carries the same current,the magnitude of the magnetic dipole moment of the circular loop will be

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