Question 125 of 153

A small square loop of wire of side $l$ is placed inside a large square loop of wire of side $L$$\left(L={l}^{2}\right)$. The loops are coplanar and their centers coincide. The value of the mutual inductance of the system is $\sqrt{x}\times {10}^{-7}H$, where $x=$______.

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Answer: 128

Flux linkage for inner loop
$\phi ={B}_{center}\times {l}^{2}$
$=4\times \dfrac{{\mu }_{0}i}{4\pi \dfrac{L}{2}}\left(\sin 45^{\circ}+\sin 45^{\circ}\right){l}^{2}$
$\Rightarrow \phi =2\sqrt{2}\dfrac{{\mu }_{0}i}{\pi L}{l}^{2}$
Therefore, mutual inductance will be
$M=\dfrac{\phi }{i} =\dfrac{2\sqrt{2}{\mu }_{0}{l}^{2}}{\pi L} =2\sqrt{2}\dfrac{{\mu }_{0}}{\pi }$
$=2\sqrt{2}\dfrac{4\pi }{\pi }\times {10}^{-7}$
$=8\sqrt{2}\times {10}^{-7}$
$=\sqrt{128}\times {10}^{-7}H$
Hence, $x=128$.