Let us consider a circular loop of radius a with centre $C$ . Let the plane of the coil be perpendicular to the plane of the paper and current $I$ be flowing in the direction shown. Suppose $P$ is any point on the axis at direction $r$ from thecentre. Let us consider a current element $d l$ on top (L) where, current comes out of paper normally whereas at bottom $(M)$ enters into the plane paper normally. $LP \perp d l$Also MP $\perp d l$ $LP = MP = \sqrt{r^2+a^2}$Now, magnetic field at $P$ due to current element at L according to Biot-Savart Law, $dB = \; \frac{\mu_0}{4\pi} \cdot \; \frac{I d l \sin 90^{\circ}}{(r^2+a^2)}$Where, $a =$ radius of circular loop. $r = \text{distance of point} \; P \; \text{from centre}$along the axis. $d B \cos \phi$ components balance each other and net magnetic field is given by $B = \oint d B \sin \phi$ $\; = \oint \; \frac{\mu_o}{4\pi} \left[ \; \frac{I d l}{r^2+a^2} \right] \cdot \; \frac{a}{\sqrt{r^2+a^2}}$ $\; \left[ \therefore \sin \phi = \frac{a}{\sqrt{r^2+a^2}} \; \text{In} \; \triangle PCM \right]$ $= \; \frac{\mu_o}{4\pi} \; \frac{I a}{(r^2+a^2)^{\frac{3}{2}}} \oint d l$ $B = \; \frac{\mu_o}{4\pi} \; \frac{I a}{(r^2+a^2)^{\frac{3}{2}}} \oint d l$ $B = \; \frac{\mu_o I^2 a^2}{2 (r^2+a^2)^{\frac{3}{2}}} (2\pi a)$ For $n$ turns, $B = \; \frac{\mu_o n I^2 a^2}{2 (r^2+a^2)^{\frac{3}{2}}}$