A circular loop and a square loop are formed | vedclass

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(D) Let the length of the wire be $L$. For the circular loop, the circumference is $2 \pi r = L$, so $r = \frac{L}{2 \pi}$. The area is $A_1 = \pi r^2

Vedclass · Accepted Answer

$\frac{4}{\pi}$

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(D) Let the length of the wire be $L$. For the circular loop, the circumference is $2 \pi r = L$, so $r = \frac{L}{2 \pi}$. The area is $A_1 = \pi r^2 = \pi (\frac{L}{2 \pi})^2 = \frac{L^2}{4 \pi}$. For the square loop, the perimeter is $4a = L$, so $a = \frac{L}{4}$. The area is $A_2 = a^2 = (\frac{L}{4})^2 = \frac{L^2}{16}$. The magnetic dipole moment is $M = iA$. Since the current $i$ is the same, the ratio of the dipole moments is $\frac{M_1}{M_2} = \frac{i A_1}{i A_2} = \frac{A_1}{A_2}$. Substituting the areas: $\frac{M_1}{M_2} = \frac{L^2 / 4 \pi}{L^2 / 16} = \frac{16}{4 \pi} = \frac{4}{\pi}$.

$A$ circular loop and a square loop are formed from two wires of same length and cross-section. The same current is passed through them. The ratio of their dipole moments is

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