Derive an expression for the magnetic field B, due to a circular coil of N turns, each of radius r carrying current I, at a distance ' x ' from the…

The magnitude of the magnetic field $\displaystyle d\mathrm{B}$ due to the current element $\displaystyle I d l$ is $\displaystyle d \mathrm{B}=\dfrac{\mu_{0}}{4 \pi} \dfrac{I|d \vec{l} \times \hat{a}|}{a^{3}}$

$\displaystyle d \mathrm{B}=\dfrac{\mu_{0}}{4 \pi} \dfrac{I d l}{\left(\mathrm{r}^{2}+\mathrm{x}^{2}\right)}$

The components of the magnetic field perpendicular to the x-axis $\displaystyle \left(d \mathrm{B}_{\perp}\right)$ are summed over and get cancelled.

The components of the magnetic field along the x-axis contribute towards the net magnetic field: $\displaystyle \mathrm{B}_{\mathrm{x}}=\int d \mathrm{B} \cos \theta=\int \dfrac{\mu_{0}}{4 \pi} \dfrac{I d l}{\left(\mathrm{x}^{2}+\mathrm{r}^{2}\right)} \cdot \dfrac{\mathrm{r}}{\sqrt{\mathrm{x}^{2}+\mathrm{r}^{2}}}$

$\displaystyle \overrightarrow{\mathrm{B}}=\mathrm{B}_{\mathrm{x}} \hat{\imath}=\dfrac{\mu_{0} I \mathrm{r}^{2}}{2\left(\mathrm{x}^{2}+\mathrm{r}^{2}\right)^{3 / 2}}(\hat{\imath})$

For $\displaystyle N$ number of turns, net magnetic field $\displaystyle \overrightarrow{\mathrm{B}}_{\text{net}}=\dfrac{\mu_{0} \mathrm{NIr}^{2}}{2\left(\mathrm{x}^{2}+\mathrm{r}^{2}\right)^{3 / 2}}(\hat{\mathrm{i}})$