A square loop of wire, edge length $\alpha$, carries a current $\mathcal{i}$. Compute the magnitude of the magnetic field produced at a point on the axis of the loop at a distance $\mathcal{x}$ from the centre.

A point on the axis of the loop is on perpendicular bisector of each of the loop sides. The figure shows field of a single wire.

$B=\frac{{\mu }_0\mathcal{i}\alpha }{\pi \sqrt{4{\mathcal{x}}^2+{\alpha }^2\sqrt{4{\mathcal{x}}^2+2{\alpha }^2}}}$

When the fields of all the four sides are considered, the horizontal components add to zero, so the total field is given by

$B_R=4B\cos \left(\theta \right)=\frac{4B\alpha }{2r}=\frac{4B\alpha }{\sqrt{4{\mathcal{x}}^2+{\alpha }^2}}$

      $=\frac{4{\mu }_0\mathcal{i}{\alpha }^2}{\pi (4{\mathcal{x}}^2+{\alpha }^2)\sqrt{4{\mathcal{x}}^2+2{\alpha }^2}}$

For $\mathcal{x}=0$ the expression reduces to

$B_R=\frac{4{\mu }_0\mathcal{i}{\alpha }^2}{\pi {\alpha }^2\sqrt{2\alpha }}=\frac{2\sqrt{2}{\mu }_0\mathcal{i}}{\pi \alpha }$