Apply Biot-Savart law to find the magnetic field due to a circular current carrying loop at a point on the axis of the loop.

Consider current carrying loop with radius $\mathrm{R}$ as shown in figure.

Centre point of loop is $\mathrm{O}$ and loop lies on $\mathrm{X}$-axis.

We wish to calculate the magnetic field at the point $P$ on this axis.

Loop is kept perpendicular to the plane of loop.

Let $x$ be the distance of $\mathrm{P}$ from the centre $\mathrm{O}$ of the loop.

The magnitude $d \overrightarrow{\mathrm{B}}$ of the magnetic field due to $d \vec{l}$ is given by the Biot-Savart law, $d \overrightarrow{\mathrm{B}}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I} \vec{d} l \times \vec{r}}{r^{3}}$

Magnitude of this magnetic field is given by,

$|d \overrightarrow{\mathrm{B}}|=\frac{\mu_{0}}{4 \pi} \frac{|\mathrm{I} \vec{d} l \times \vec{r}|}{r^{3}}$

$=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I} d l r \sin \theta^{\prime}}{r^{3}}$

where $\theta^{\prime}=$ angle between $\overrightarrow{d l}$ and $\vec{r}$ but $\overrightarrow{d l} \perp \vec{r}$ so, $\sin \theta^{\prime}=\sin \frac{\pi}{2}=1$

$\therefore|d \overrightarrow{\mathrm{B}}|=\frac{\mu_{0} \mathrm{I}}{4 \pi} \frac{d l}{r^{2}}$

As shown in diagram $r^{2}=x^{2}+\mathrm{R}^{2}$

$|d \overrightarrow{\mathrm{B}}|=\frac{\mu_{0}}{4 \pi} \frac{d l}{\left(x^{2}+\mathrm{R}^{2}\right)}$

Direction of $d \overrightarrow{\mathrm{B}}$ is given by perpendicular to plane of $\overrightarrow{d l}$ and $\vec{r}$. $d \overrightarrow{\mathrm{B}}$ can be divide in two component,

$x$ component $d \mathrm{~B}_{x}=d \mathrm{~B} \cos \theta$

perpendicular component $d \mathrm{~B}_{\perp}=d \mathrm{~B} \sin \theta$