A small square loop of wire of side l is placed inside a large...

Let a current, $I$ pass through the square loop of side, $L$. The magnetic field at the centre $O$,

$B = 4 \times Magneticfieldduetoeach side$

$⇒ B = 4 \times \dfrac{{\mu}_{0}}{4\pi} ⋅ \dfrac{I}{L / 2}\left(sin{45}^{∘} + sin{45}^{∘}\right)$

$⇒ B = \dfrac{2{\mu}_{0}I}{\piL}\left(\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}}\right) = \dfrac{2\sqrt{2}{\mu}_{0}I}{\piL}$

Magnetic flux linked with the small square loop,

$ϕ = BA = B{l}^{2} = \dfrac{2\sqrt{2}{\mu}_{0}I{l}^{2}}{\piL}$

Mutual inductance of the coil,

$M = \dfrac{ϕ}{I} = \dfrac{2\sqrt{2}{\mu}_{0}{l}^{2}}{\piL}$.