Q.Obtain a relation for the magnetic field at a point along the axis of a circular coil carrying current.

Magnetic Field on the Axis of a Current Loop

Imagine a circular wire carrying a steady current. You want to know the magnetic field not at the centre, but at some point along the line that passes through the centre and is perpendicular to the plane of the loop — that's the axis.

Why would the field be along the axis at all? Because of symmetry. For every tiny segment of the loop, there is an opposite segment on the other side. Their perpendicular components of the magnetic field cancel out, leaving only the component along the axis. So the net field points straight along the axis, either towards or away from the loop depending on the current direction.

The Intuition

At the centre of the loop ($x = 0$), every segment is at the same distance $R$ from the centre, and the field is strongest. As you move away along the axis, two things happen: the distance from each current element to your observation point increases, and the angle at which the field points along the axis becomes less favourable. So the field drops off.

Far away from the loop, the loop looks like a tiny magnetic dipole — a small bar magnet. The field falls off as $1/x^3$, exactly like a dipole field.

The Precise Statement

For a circular loop of radius $R$, carrying a steady current $I$, the magnitude of the magnetic field at a point on the axis at a distance $x$ from the centre is:

where $\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}$ is the permeability of free space.

The direction of $\mathbf{B}$ is along the axis, given by the right-hand rule: curl the fingers of your right hand in the direction of the current, and your thumb points in the direction of the magnetic field on the axis.

Special Cases

At the centre ($x = 0$):

Far away ($x \gg R$):

The denominator $(R^2 + x^2)^{3/2} \approx x^3$, so

This is exactly the field of a magnetic dipole of moment $m = I \cdot (\pi R^2) = I A$, where $A$ is the area of the loop. So a current loop behaves like a magnetic dipole at large distances.

Why the $3/2$ Power?

The $3/2$ comes from the vector sum of contributions from all current elements. Each element contributes a field that falls as $1/r^2$ (where $r = \sqrt{R^2 + x^2}$), but only the axial component survives, which introduces an extra factor of $\sin\theta$ or $\cos\theta$ that brings in another $R/r$ factor. The product gives $1/r^3$, and since $r = (R^2 + x^2)^{1/2}$, you get $(R^2 + x^2)^{-3/2}$.

Final answer: $B = \frac{\mu_0 I R^2}{2\,(R^2 + x^2)^{3/2}}$ along the axis, falling as $1/x^3$ far away, behaving like a magnetic dipole of moment $I \pi R^2$.

The axial magnetic field of a current loop is a formula-heavy NCERT Class 12 Physics topic, frequently searched as magnetic field on the axis of a circular loop formula or magnetic dipole moment of a current loop class 12. Its far-field dipole limit connects directly to magnetic-moment concepts tested in both CBSE boards and JEE Main physics.