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(C) The magnetic field $B$ at the center of a large square loop of side $L$ carrying current $i$ is given by:$B = 4 	imes \frac{\mu_0 i}{4 \pi (L/2)}

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$128$

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(C) The magnetic field $B$ at the center of a large square loop of side $L$ carrying current $i$ is given by:$B = 4 	imes \frac{\mu_0 i}{4 \pi (L/2)} (\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 i}{\pi L/2} 	imes \sqrt{2} = \frac{2\sqrt{2} \mu_0 i}{\pi L}$.Since $L \gg \ell$, we assume the magnetic field is uniform over the area of the small loop.The magnetic flux $\phi$ through the small loop of area $\ell^2$ is:$\phi = B 	imes \ell^2 = \frac{2\sqrt{2} \mu_0 i}{\pi L} \ell^2$.Given $L = \ell^2$, we substitute this into the expression:$\phi = \frac{2\sqrt{2} \mu_0 i}{\pi \ell^2} \ell^2 = \frac{2\sqrt{2} \mu_0 i}{\pi}$.The mutual inductance $M$ is defined as $M = \phi / i$:$M = \frac{2\sqrt{2} \mu_0}{\pi} = \frac{2\sqrt{2} (4\pi 	imes 10^{-7})}{\pi} = 8\sqrt{2} 	imes 10^{-7} 	ext{ H}$.$M = \sqrt{64 	imes 2} 	imes 10^{-7} 	ext{ H} = \sqrt{128} 	imes 10^{-7} 	ext{ H}$.Thus, $x = 128$.

$A$ small square loop of wire of side $\ell$ is placed inside a large square loop of wire of side $L$ $(L \gg \ell)$. The loops are coplanar and their centers coincide. The value of the mutual inductance of the system is $\sqrt{x} \times 10^{-7} \text{ H}$, where $x = \dots$

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