Magnetic Field on the Axis of a Circular Current Loop

The setup: loop, axis, and the point P

Consider a circular loop of radius $R$ carrying a steady current $I$, placed in free space with its centre at the origin $O$ and its plane perpendicular to the $x$-axis. The $x$-axis is the axis of the loop. We wish to find the magnetic field at a point $P$ lying on this axis at a distance $x$ from the centre. The strategy, exactly as in NCERT Section 4.5, is to sum the effect of every infinitesimal current element $I\,d\mathbf{l}$ around the loop using the Biot-Savart law.

Two geometric facts make the integral tractable. First, every element of the loop is perpendicular to the line joining it to $P$, so the cross product simplifies. Second, the distance from any element to $P$ is the same, $r = \sqrt{x^2 + R^2}$, because all elements lie on a circle of radius $R$ at the same axial distance $x$. Symmetry then forces the perpendicular contributions to cancel, leaving only an axial field.

Deriving the axial field from Biot-Savart

The magnitude of the field due to a single element $d\mathbf{l}$, by the Biot-Savart law, is

$$ dB = \frac{\mu_0}{4\pi}\,\frac{I\,|d\mathbf{l} \times \mathbf{r}|}{r^3}. $$

Now $r^2 = x^2 + R^2$. Since each element of the loop is perpendicular to the displacement vector from the element to the axial point, $|d\mathbf{l} \times \mathbf{r}| = r\,dl$. Hence

$$ dB = \frac{\mu_0}{4\pi}\,\frac{I\,dl}{(x^2 + R^2)}. $$

The direction of $d\mathbf{B}$ is perpendicular to the plane formed by $d\mathbf{l}$ and $\mathbf{r}$. It has an $x$-component $dB_x$ and a component perpendicular to the axis, $dB_\perp$. When the perpendicular components are summed over the loop, they cancel out and we obtain a null result: the $dB_\perp$ contribution due to $d\mathbf{l}$ is cancelled by the contribution from the diametrically opposite element. Thus only the $x$-component survives, obtained by integrating $dB_x = dB\cos\theta$ over the loop, where

$$ \cos\theta = \frac{R}{(x^2 + R^2)^{1/2}}. $$

Combining the two relations,

$$ dB_x = \frac{\mu_0 I}{4\pi}\,\frac{R\,dl}{(x^2 + R^2)^{3/2}}. $$

The summation of elements $dl$ over the loop yields $2\pi R$, the circumference. Therefore the magnetic field at $P$ due to the entire circular loop is

$$ \mathbf{B} = \frac{\mu_0 I R^2}{2(x^2 + R^2)^{3/2}}\,\hat{\mathbf{i}}. $$

For a tightly wound coil of $N$ turns, each turn contributes equally, so the result is multiplied by $N$: $B = \mu_0 N I R^2 / 2(x^2 + R^2)^{3/2}$. The field points along the axis, in the direction set by the right-hand rule described below.

Field at the centre of the loop

The centre of the loop is the special case $x = 0$. Setting $x = 0$ in the axial formula collapses $(x^2 + R^2)^{3/2}$ to $R^3$, giving

$$ \mathbf{B} = \frac{\mu_0 I}{2R}\,\hat{\mathbf{i}}. $$

This is the maximum field on the axis — the loop concentrates its field most strongly at its own centre. NIOS Section 18.3.1 reaches the same result by a direct centre-only calculation: each element subtends $90^\circ$ to $r = R$, so $dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl}{R^2}$, and summing $\sum dl = 2\pi R$ again returns $B = \mu_0 I/2R$ (with $\mu_0 N I/2R$ for $N$ turns). Two independent routes, one number.

How the field varies along the axis

As the observation point moves out along the axis, the denominator $(x^2 + R^2)^{3/2}$ grows, so the field falls monotonically from its peak at the centre. The curve is symmetric about $x = 0$ (the field at $-x$ equals that at $+x$ in magnitude and direction) and has a smooth, rounded maximum rather than a sharp spike — a hallmark shape NEET sometimes asks students to recognise.

Three regions are worth distinguishing. Near $x = 0$, the field is essentially flat and equal to its peak. For $x$ comparable to $R$, both terms matter and the full $(x^2 + R^2)^{3/2}$ must be kept. For $x \gg R$, the loop radius becomes negligible and a clean power law emerges, treated next.

The far-field limit: a magnetic dipole

When $x \gg R$, the term $R^2$ is negligible compared with $x^2$ in the denominator, so $(x^2 + R^2)^{3/2} \to x^3$ and

$$ B \approx \frac{\mu_0 I R^2}{2x^3}. $$

Introducing the magnetic moment of the loop, $m = I A = I\pi R^2$, the area $A = \pi R^2$ lets us write $I R^2 = m/\pi$, giving the dipole form

$$ B \approx \frac{\mu_0\,(2m)}{4\pi x^3}, \qquad m = IA. $$

This is structurally identical to the axial field of an electric dipole, $E = \dfrac{2p}{4\pi\varepsilon_0 x^3}$, with $m$ replacing $p$ and $\mu_0$ replacing $1/\varepsilon_0$. Far away, a current loop is indistinguishable from a magnetic dipole of moment $m = IA$ — the same moment that governs the torque on a current loop in an external field. The axial field falling as $1/x^3$ (not $1/x^2$ as for a wire, nor $1/x$) is the dipole signature.

Direction and field lines of a loop

The magnetic field lines due to a circular wire form closed loops. The direction of the field is given by a right-hand thumb rule: curl the palm of your right hand around the circular wire with the fingers pointing in the direction of the current; the right-hand thumb then gives the direction of the magnetic field along the axis. The face of the loop from which field lines emerge behaves as the north pole and the opposite face as the south pole — so a current loop is the elementary magnet of the chapter.

Three regimes at a glance

The single formula $B = \mu_0 I R^2 / 2(x^2 + R^2)^{3/2}$ contains all the limiting cases. Reading them off saves time in the exam.