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(C) Let the length of each wire be $l$.For the square,the side length is $a = l/4$. The area is $A_{square} = a^2 = (l/4)^2 = l^2/16$.For the circle,t

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$\pi : 4$

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(C) Let the length of each wire be $l$.For the square,the side length is $a = l/4$. The area is $A_{square} = a^2 = (l/4)^2 = l^2/16$.For the circle,the circumference is $2\pi r = l$,so the radius is $r = l/(2\pi)$. The area is $A_{circle} = \pi r^2 = \pi (l/(2\pi))^2 = l^2/(4\pi)$.The magnetic moment is given by $M = iA$. Since the current $i$ is the same for both,$\frac{M_{square}}{M_{circle}} = \frac{A_{square}}{A_{circle}} = \frac{l^2/16}{l^2/(4\pi)} = \frac{4\pi}{16} = \frac{\pi}{4}$.Thus,the ratio is $\pi : 4$.

Two wires of the same length are shaped into a square and a circle. If they carry the same current,the ratio of their magnetic moments is:

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