The magnetic field due to a current carrying square loop of side a at a point located symmetrically at a distance of a/2 from its centre (as shown is)

Let ${\text{B}}_{\text{0}}$ be the field at P due to each side

${\text{B}}_{\text{net}}=4\text{B}\sin 45°$

$=2\sqrt{2}{\text{B}}_{\text{0}}$

${\text{B}}_{\text{net}}=2\sqrt{2}\frac{{\mu }_{0}i}{4\pi \text{d}}\left(\sin {\theta }_1+\sin {\theta }_2\right)$

$x=\sqrt{\frac{{\text{a}}^{\text{2}}}{4}+\frac{{\text{a}}^{\text{2}}}{\text{2}}}=\sqrt{3}\frac{\text{a}}{\text{2}}$

$\sin {\theta }_1+\sin {\theta }_2=\frac{\text{a/2}}{\sqrt{3}\text{a/2}}=\frac{1}{\sqrt{3}}$

${\text{B}}_{\text{net}}=\frac{2{\mu }_{0}i}{\sqrt{3}\pi \text{a}}$