A small square loop of wire of side l is placed inside a large square loop of wire of side L L = l 2 . The…

Flux linkage for inner loop

$\varphi =B_{\mathrm{center}}\times l^2$

$=4\times \frac{{\mu }_{0}i}{4\mathrm{\pi }{\displaystyle \frac{L}{2}}}\left(\sin 45°+\sin 45°\right)l^2$

$\Rightarrow \varphi =2\sqrt{2}\frac{{\mu }_{0}i}{\pi L}{l}^2$

Therefore, mutual inductance will be

$$\begin{gathered} M=\frac{\varphi }{i} \\ =\frac{2\sqrt{2}{\mu }_0{l}^2}{\pi L} \\ =2\sqrt{2}\frac{{\mu }_0}{\mathrm{\pi }} \end{gathered}$$

$=2\sqrt{2}\frac{4\mathrm{\pi }}{\mathrm{\pi }}\times {10}^{-7}$

$=8\sqrt{2}\times {10}^{-7}$

$=\sqrt{128}\times {10}^{-7} \mathrm{H}$

Hence, $x=128$.