Derivation of Magnetic Field on the Axis of a Circular Current Loop

Step 1 (Formula / Governing Law & Given Data): [1 Mark]

According to the Biot-Savart Law, the magnetic field $d\vec{B}$ due to a small current element $Id\vec{l}$ at a point $P$ is given by:

$$d\vec{B} = \frac{\mu_0}{4\pi} \frac{I(d\vec{l} \times \vec{r})}{r^3}$$

For a circular coil of radius $R$ (using $R$ for radius to avoid confusion with distance $r$) with $N$ turns, the distance from the element to the point $P$ on the axis at distance $x$ is $s = \sqrt{R^2 + x^2}$.

Step 2 (Step-by-Step Mathematical Working): [1.5 Marks]

• The magnitude of the magnetic field due to a small element $dl$ is: $$dB = \frac{\mu_0}{4\pi} \frac{I dl}{R^2 + x^2}$$
• By symmetry, the components of $dB$ perpendicular to the axis ($dB \sin\theta$) cancel out for every diametrically opposite element. Only the axial components ($dB \cos\theta$) add up.
• The total magnetic field $B$ is the integral of the axial components for $N$ turns: $$B = \int dB \cos\theta = \int \left( \frac{\mu_0}{4\pi} \frac{I dl}{R^2 + x^2} \right) \cos\theta$$
• From the geometry of the triangle, $\cos\theta = \frac{R}{\sqrt{R^2 + x^2}}$. Substituting this: $$B = \frac{\mu_0 I}{4\pi} \frac{R}{(R^2 + x^2)^{3/2}} \int_0^{2\pi R} N dl$$
• Integrating $dl$ over the circumference $2\pi R$: $$B = \frac{\mu_0 I N R}{4\pi (R^2 + x^2)^{3/2}} \cdot (2\pi R)$$

Step 3 (Final Concluded Answer): [0.5 Mark]

Simplifying the expression, we get the final formula for the magnetic field along the axis:

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

The direction of $\vec{B}$ is along the axis, determined by the Right-Hand Thumb Rule.

💡 EL10 Concept Insight: Imagine the loop as a ring of tiny magnets; because the ring is circular, all the "sideways" magnetic forces cancel each other out, leaving only the forces pointing straight along the center line to add up and create a strong field.