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

Vedclass · Accepted Answer

$\frac{4}{\pi}$

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(D) Let $l$ be the length of the metal wire.When the wire is bent into a circular coil of radius $r$,then 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 for the circular coil is $\mu_c = i A_c = \frac{i l^2}{4\pi}$.When the wire is bent into a square coil,the perimeter is $4a = l$,so the side length $a = \frac{l}{4}$.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 for the square coil is $\mu_s = i A_s = \frac{i l^2}{16}$.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 coil and a square coil are prepared from two identical metal wires,and the same current is passed through them. The ratio of the magnetic dipole moment associated with the circular coil to that of the square coil is:

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