A circular loop and a square loop are formed | vedclass

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(B) Let the length of the wire be $l$.For the circular loop,the circumference is $2 \pi r = l$,so the radius $r = \frac{l}{2 \pi}$.The magnetic dipole

Vedclass · Accepted Answer

$\frac{4}{\pi}$

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(B) Let the length of the wire be $l$.For the circular loop,the circumference is $2 \pi r = l$,so the radius $r = \frac{l}{2 \pi}$.The magnetic dipole moment of the circular loop is $M_{1} = i A_{1} = i \pi r^{2}$.Substituting $r$,we get $M_{1} = i \pi \left(\frac{l}{2 \pi}\right)^{2} = \frac{i l^{2}}{4 \pi}$.For the square loop,the perimeter is $4 a = l$,so the side length $a = \frac{l}{4}$.The magnetic dipole moment of the square loop is $M_{2} = i A_{2} = i a^{2}$.Substituting $a$,we get $M_{2} = i \left(\frac{l}{4}\right)^{2} = \frac{i l^{2}}{16}$.The ratio of the dipole moments is $\frac{M_{1}}{M_{2}} = \frac{i l^{2} / 4 \pi}{i l^{2} / 16} = \frac{16}{4 \pi} = \frac{4}{\pi}$.

$A$ circular loop and a square loop are formed from the same wire and the same current is passed through them. Find the ratio of their dipole moments.

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