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(B) Let $l$ be the length of the metal wire.When the wire is bent into a circular coil of radius $r$,the circumference is $2 \pi r = l$,so $r = \frac{

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$\frac{4}{\pi}$

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(B) Let $l$ be the length of the metal wire.When the wire is bent into a circular coil of radius $r$,the circumference is $2 \pi r = l$,so $r = \frac{l}{2 \pi}$.The area of the circular coil is $A_c = \pi r^2 = \pi \left( \frac{l}{2 \pi} \right)^2 = \frac{l^2}{4 \pi}$.The magnetic dipole moment of the circular coil is $\mu_c = i A_c = i \frac{l^2}{4 \pi}$.When the same wire is bent into a square coil,the perimeter is $4a = l$,so the side length is $a = \frac{l}{4}$.The area of the square coil is $A_s = a^2 = \left( \frac{l}{4} \right)^2 = \frac{l^2}{16}$.The magnetic dipole moment of the square coil is $\mu_s = i A_s = i \frac{l^2}{16}$.The ratio of the magnetic dipole moments is $\frac{\mu_c}{\mu_s} = \frac{i l^2 / 4 \pi}{i l^2 / 16} = \frac{16}{4 \pi} = \frac{4}{\pi}$.

$A$ circular and a square coil are prepared from two identical metal wires and a current is passed through them. The ratio of the magnetic dipole moment associated with the circular coil to that with the square coil is

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