Explanation

(d) From Biot-Savart’s law, magnetic field due to current carrying conductor is
$\ \ \ \ B = \ \frac{\mu_{0}}{4\pi}\frac{i}{r}\ \left( \sin\phi_{1} + \sin\phi_{2} \right)$
Where, $\phi_{1} = 45{^\circ},\phi_{2} = 45{^\circ}$
$\ \ \ \ B = \frac{\mu_{0}}{4\pi}\ .\frac{i}{r}\left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) = \frac{\mu_{0}i\sqrt{2}}{4\pi r}$
Since, square consists of four sides, total magnetic field is
$\ \ \ B^{'} = 4B = \frac{4}{4\pi r}\ .\ \mu_{0}i\sqrt{2} = \frac{\mu_{0}i\sqrt{2}}{\pi r}$
From figure, $r = \sqrt{\left( \frac{L\sqrt{2}}{2} \right) - \left( \frac{L}{2} \right)^{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \sqrt{\frac{2L^{2}}{4} - \frac{L^{2}}{4}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \sqrt{\frac{L^{2}}{4}} = \frac{L}{2}$
$\therefore B = \frac{\mu_{0}I\sqrt{2}}{\pi\frac{L}{2}} = \frac{\mu_{0}I.2\sqrt{2}}{\pi L}$