Q.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 centre along its axis.

Concept: Biot-Savart Law + Symmetry

Consider a circular coil of $N$ turns, radius $r$, carrying current $I$. We find the field at point P on the axis, distance $x$ from the center.

Take a small element $d\vec{l}$ on the loop. By Biot-Savart, it produces $d\vec{B} = \frac{\mu_0 I}{4\pi} \frac{d\vec{l} \times \vec{r'}}{r'^3}$ where $\vec{r'}$ points from the element to P. The distance is $r' = \sqrt{r^2 + x^2}$.

Each $d\vec{B}$ has components perpendicular and parallel to the axis. By symmetry, perpendicular components from diametrically opposite elements cancel. Only the axial component survives: $dB_{\text{axial}} = dB \cos\theta = dB \cdot \frac{r}{\sqrt{r^2+x^2}}$. …

The magnetic field produced by a current-carrying conductor is governed by the Biot-Savart law. For a circular coil, the geometry is particularly elegant: every current element sits at the same distance from any point on the axis, and symmetry eliminates all but the axial component of the field. This makes the circular coil one of the few configurations where we can derive an exact, closed-form expression for $B$.

The strategy is to consider a tiny element of the coil, find its contribution using Biot-Savart, recognize that perpendicular components from diametrically opposite elements cancel, and integrate around the loop.

Derivation

1. Set up the geometry. Place the circular coil of radius $r$ in the $yz$-plane with its centre at the origin, and take the axis along $x$. We want the field at a point $P$ on the axis at distance $x$ from the centre. Consider a small current element $I\,d\vec{\ell}$ at some position on the loop; the distance from this element to point $P$ is

This distance is the same for every element on the loop — a key simplification.

2. Apply the Biot-Savart law. The field contribution from the element is

where $\hat{R}$ is the unit vector from the element to $P$. Now note the geometry: $d\vec{\ell}$ is tangent to the loop, while $\vec{R}$ lies entirely in the plane containing the loop's radius at that point and the axis — both the radial direction and the axial direction are perpendicular to the tangent. So $d\vec{\ell} \perp \vec{R}$, the angle between them is $90^\circ$, and $|d\vec{\ell} \times \hat{R}| = d\ell$. Hence

3. Resolve into components. Each $d\vec{B}$ is perpendicular to $\vec{R}$ and has two components: one along the axis and one perpendicular to it. The perpendicular components from diametrically opposite elements cancel by symmetry; only the axial component survives. From the right triangle with sides $r$, $x$ and hypotenuse $R$, the axial component of $d\vec{B}$ carries the factor

where $\alpha$ is the angle between $d\vec{B}$ and the axis. Therefore

4. Integrate around the loop. …