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(A) Let the length of the wire be $l$. For a square loop,the side of the square is $a = l/4$.Hence,its magnetic moment is $M_{sq} = i A = i (l/4)^2 =

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$4: \pi$

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(A) Let the length of the wire be $l$. For a square loop,the side of the square is $a = l/4$.Hence,its magnetic moment is $M_{sq} = i A = i (l/4)^2 = i l^2 / 16$.For a circular loop,the circumference is $2 \pi r = l$,so the radius is $r = l / (2 \pi)$.Hence,its magnetic moment is $M_{cir} = i A = i \pi r^2 = i \pi (l / (2 \pi))^2 = i \pi (l^2 / 4 \pi^2) = i l^2 / (4 \pi)$.The ratio of the magnetic moment of the circle to that of the square is:$\frac{M_{cir}}{M_{sq}} = \frac{i l^2 / (4 \pi)}{i l^2 / 16} = \frac{16}{4 \pi} = \frac{4}{\pi}$.

Two wires of the same length are used to form a circle and a square. They carry the same current. The ratio of the magnetic moment of the circle to that of the square is:

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