Consider a conducting wire of length L bent i | vedclass

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(B) $1$. For the circular loop: The circumference is $2 \pi R = L$, so $R = \frac{L}{2 \pi}$.$2$. For the square loop: The perimeter is $4s = a$, wher

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$\mu_{0} \frac{\pi a^{2}}{16 L}$

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(B) $1$. For the circular loop: The circumference is $2 \pi R = L$, so $R = \frac{L}{2 \pi}$.$2$. For the square loop: The perimeter is $4s = a$, where $s$ is the side length of the square. Thus, $s = \frac{a}{4}$.$3$. The magnetic field $B$ at the centre of the circular loop due to a current $I$ flowing through it is $B = \frac{\mu_{0} I}{2R}$.$4$. Since $a \ll R$, the magnetic field $B$ is approximately uniform over the area of the square loop. The magnetic flux $\phi$ linked with the square loop is $\phi = B 	imes 	ext{Area of square} = B 	imes s^2$.$5$. Substituting the values: $\phi = \left( \frac{\mu_{0} I}{2R} ight) 	imes s^2 = \left( \frac{\mu_{0} I}{2(L / 2 \pi)} ight) 	imes \left( \frac{a}{4} ight)^2 = \left( \frac{\mu_{0} I \pi}{L} ight) 	imes \frac{a^2}{16} = \frac{\mu_{0} \pi a^2}{16 L} I$.$6$. The mutual inductance $M$ is given by $M = \frac{\phi}{I} = \frac{\mu_{0} \pi a^2}{16 L}$.

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