A small square loop of wire of side $l$ is placed inside a large square loop of wire of side $L(L>>L)$. The loops are coplanar and their centers coincide. The mutual inductance of the system is proportional to

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A small square loop of wire of side $l$ is placed inside a large square loop of wire of side $L(L>>L)$. The loops are coplanar and their centers coincide. The mutual inductance of the system is proportional to

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kritika · Answer 1 · 2021-12-24T04:29:06+0000

Correct option (B) $\mathrm{l}^{2} / \mathrm{L}$
Explanation:
The magnetic flux that links the larger loop with the smaller loop of side $l(l<L)$ is
$$
\phi_{12}=\mathrm{B}^{2}=\frac{2 \sqrt{2} \mu_{0} H^{2}}{\pi L}
$$
$\therefore$ Mutual inductance $M_{12}$
$$
\begin{gathered}
=\frac{\phi_{12}}{I}=\frac{2 \sqrt{2} \mu_{6}}{\pi}\left(\frac{l^{2}}{L}\right) \\
\text { i.e. } M_{\text {ए }} \propto \frac{l^{2}}{\mathrm{~L}}
\end{gathered}
$$

A small square loop of wire of side \(l\) is placed inside a large square loop of wire of side \(L(L>>L)\). The loops are coplanar and their centers coincide. The mutual inductance of the system is proportional to

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