Let us consider a circular loop of radius a with centre $\mathrm{C}$ . Let the plane of the coil be perpendicular to the plane of the paper and current $\mathit{I}$ be flowing in the direction shown. Suppose $\mathrm{P}$ is any point on the axis at direction $\mathit{r}$ from thecentre.
Let us consider a current element $\mathit{d}\mathit{l}$ on top (L) where, current comes out of paper normally whereas at bottom $\left(\mathrm{M}\right)$ enters into the plane paper normally.
$\mathrm{LP}\perp \mathit{d}\mathit{l}$
Also MP $\perp \mathit{d}\mathit{l}$
$\mathrm{LP}=\mathrm{MP}=\sqrt{{\mathit{r}}^2+{\mathit{a}}^2}$
Now, magnetic field at $\mathrm{P}$ due to current element at L according to Biot-Savart Law,
$\mathit{d}\mathit{B}=\frac{{\mathit{\mu }}_0}{4\mathit{\pi }}\cdot \frac{\mathit{I}\mathit{d}\mathit{l}\mathit{s}\mathit{i}\mathit{n}{90}^{\circ }}{\left({\mathit{r}}^2+{\mathit{a}}^2\right)}$
Where, $\mathit{a}=$ radius of circular loop.

along the axis.
$\mathit{d}\mathit{B}\cos \mathit{\varphi }$ components balance each other and net magnetic field is given by
$\mathit{B}=\oint \mathit{d}\mathit{B}\mathit{s}\mathit{i}\mathit{n}\mathit{\varphi }$
$=\oint \frac{{\mathit{\mu }}_{\mathit{o}}}{4\mathit{\pi }}\left[\frac{\mathit{I}\mathit{d}\mathit{l}}{{\mathit{r}}^2+{\mathit{a}}^2}\right]\cdot \frac{\mathit{a}}{\sqrt{{\mathit{r}}^2+{\mathit{a}}^2}}$

$=\frac{{\mathit{\mu }}_{\mathit{o}}}{4\mathit{\pi }}\frac{\mathit{I}\mathit{a}}{{\left({\mathit{r}}^2+{\mathit{a}}^2\right)}^{\frac{3}{2}}}\oint \mathit{d}\mathit{l}$
$\mathit{B}=\frac{{\mathit{\mu }}_{\mathit{o}}}{4\mathit{\pi }}\frac{\mathit{I}\mathit{a}}{{\left({\mathit{r}}^2+{\mathit{a}}^2\right)}^{\frac{3}{2}}}\oint \mathit{d}\mathit{l}$
$\mathit{B}=\frac{{\mathit{\mu }}_{\mathit{o}}{\mathit{I}}^2{\mathit{a}}^2}{2{\left({\mathit{r}}^2+{\mathit{a}}^2\right)}^{\frac{3}{2}}}\left(2\mathit{\pi }\mathit{a}\right)$
For $\mathit{n}$ turns, $\mathit{B}=\frac{{\mathit{\mu }}_{\mathit{o}}\mathit{n}{\mathit{I}}^2{\mathit{a}}^2}{2{\left({\mathit{r}}^2+{\mathit{a}}^2\right)}^{\frac{3}{2}}}$