A small square loop of wire of side l is plac | vedclass

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Question

(b) Magnetic field produced due to large loop $B = \frac{{{\mu _0}}}{{4\pi }}\frac{{8\sqrt 2 i}}{L}$

Flux linked with smaller loop

$\phi = B({l^

Vedclass · Accepted Answer

${l^2}/L$

Vedclass · Answer

(b) Magnetic field produced due to large loop $B = \frac{{{\mu _0}}}{{4\pi }}\frac{{8\sqrt 2 i}}{L}$

Flux linked with smaller loop

$\phi = B({l^2}) = \frac{{{\mu _0}}}{{4\pi }}\frac{{8\pi i{l^2}}}{L}$

$	herefore \phi = Mi \Rightarrow M = \frac{\phi }{i} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{8\sqrt 2 {l^2}}}{L}$

$\Rightarrow M \propto \frac{{{l^2}}}{L}$

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

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