Derive the expression for magnetic field on the axis of a circular current loop

Question

Derive the expression for the magnetic field at a point on the axis of a circular loop of radius $R$ carrying current $I$ , at a distance $x$ from the centre.

(NCERT Class 12, Chapter 4)

Solution — Step by Step

Consider a small current element $d\vec{l}$ on the loop. The point $P$ is on the axis at distance $x$ from the centre.

The distance from $d\vec{l}$ to $P$ is $r = \sqrt{R^2 + x^2}$ .

By the Biot-Savart law: $d\vec{B} = \frac{\mu_0}{4\pi} \frac{I\,d\vec{l} \times \hat{r}}{r^2}$

Since $d\vec{l} \perp \hat{r}$ (the element is on the loop, the line to $P$ is along the slant), $|d\vec{l} \times \hat{r}| = dl$ .

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

Each $d\vec{B}$ has two components: one along the axis ( $dB_x$ ) and one perpendicular to the axis ( $dB_\perp$ ).

By symmetry, the perpendicular components from diametrically opposite elements cancel. Only the axial components survive.

dB_x = dB \cos\alpha = dB \cdot \frac{R}{\sqrt{R^2 + x^2}}

B = \int dB_x = \frac{\mu_0 I R}{4\pi(R^2 + x^2)^{3/2}} \int dl = \frac{\mu_0 I R}{4\pi(R^2 + x^2)^{3/2}} \cdot 2\pi R

\boxed{B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}}

directed along the axis (use the right-hand rule for direction).

At the centre ( $x = 0$ ): $B = \frac{\mu_0 I}{2R}$

Far from the loop ( $x \gg R$ ): $B \approx \frac{\mu_0 I R^2}{2x^3} = \frac{\mu_0}{4\pi} \cdot \frac{2m}{x^3}$

where $m = I\pi R^2$ is the magnetic dipole moment. This is the dipole field behaviour.

Why This Works

The Biot-Savart law gives the field from each tiny current element. The clever part is recognising that the perpendicular components cancel by symmetry — for every element on one side of the loop, there’s a matching element on the opposite side whose perpendicular field points the other way.

Only the axial component, which involves the factor $\cos\alpha = R/\sqrt{R^2 + x^2}$ , adds up. The integration around the full loop just gives $2\pi R$ (the circumference), since every element contributes the same axial component.

Alternative Method — Using the magnetic dipole formula directly

For points far from the loop ( $x \gg R$ ), treat the loop as a magnetic dipole with moment $\vec{m} = I \cdot \pi R^2 \hat{n}$ :

B_{axial} = \frac{\mu_0}{4\pi} \cdot \frac{2m}{x^3}

This is faster for “far-field” problems but doesn’t work near the loop.

This derivation is a 5-mark CBSE favourite. The diagram showing the current element, the point $P$ on the axis, the angle $\alpha$ , and the component resolution is critical. Without a clear labelled diagram, you’ll lose 1-2 marks even if the math is correct.

Common Mistake

Students often forget to resolve $dB$ into components and directly integrate $dB$ as if it were along the axis. The field $d\vec{B}$ from each element is NOT along the axis — it points at an angle. You must multiply by $\cos\alpha = R/\sqrt{R^2 + x^2}$ to get the axial component. Skipping this step gives $B = \mu_0 I R / [2(R^2 + x^2)]$ — missing the $3/2$ exponent in the denominator.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the magnetic dipole formula directly

Common Mistake

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