GATE Electrical Engineering (EE) 2020 Solved Paper

Given data,$\mathit{M}=\frac{\mathit{\phi }}{{\mathit{i}}_1}$
where, $\mathit{\varphi }$ is the flux linked with square loop due to current in the wire.
Let, magnetic field intensity due to long wire be,
$\overrightarrow{\mathit{H}}=\frac{\mathit{I}}{2\mathit{\pi }\mathit{\rho }}{\stackrel{\wedge }{\mathit{a}}}_{\mathit{\phi }}$
$\overrightarrow{\mathit{B}}=\frac{\mathit{\mu }\mathit{I}}{2\mathit{\pi }\mathit{\rho }}{\stackrel{\wedge }{\mathit{a}}}_{\mathit{\phi }}$
Now flux linked with the elemental strip,
$\mathit{d}\mathit{\varphi }=\overrightarrow{\mathit{B}}.\mathit{d}\mathit{s}$
$=\frac{\mathit{\mu }\mathit{I}}{2\mathit{\pi }\mathit{\rho }}\left(1.\mathit{d}\mathit{\rho }\right)$
$\mathit{\varphi }=\frac{\mathit{\mu }\mathit{I}}{2\mathit{\pi }}\underset{1}{\overset{2}{\int }}\frac{\mathit{d}\mathit{\rho }}{\mathit{\rho }}$
$=\frac{\mathit{\mu }\mathit{I}}{2\mathit{\pi }}\ln 2$
Now,
$\mathit{M}=\frac{\mathit{\varphi }}{{\mathit{i}}_1}=\left(\frac{\mathit{\mu }\mathit{I}}{2\mathit{\pi }}\ln 2\right)\times \frac{1}{\mathit{I}}$
$=\frac{4\mathit{\pi }\times {10}^{-7}}{2\mathit{\pi }}\times 0.693$
$=1.386\times {10}^{-7}\mathit{H}$
$=138.6\times 10-9\mathit{H}=138.6\mathit{n}\mathit{H}$