B at centre = μ₀I/(2R). Right-hand thumb rule for direction.

Magnetic Field Due to Current-Carrying Circular Loop

Question

A circular loop of radius $R$ carries a steady current $I$ . Find the magnetic field at the centre of the loop. Also state the direction using the right-hand thumb rule.

Solution — Step by Step

Every small element $d\vec{l}$ of the loop contributes a field $dB$ at the centre. By Biot-Savart:

dB = \frac{\mu_0 I}{4\pi} \cdot \frac{dl \sin\theta}{R^2}

Here, $\theta = 90°$ always — because $d\vec{l}$ is tangential and $\vec{r}$ (from element to centre) is always radial. So $\sin 90° = 1$ throughout.

All $dB$ contributions point in the same direction (into or out of the loop depending on current direction), so they add up directly — no vector cancellation needed.

B = \int dB = \frac{\mu_0 I}{4\pi R^2} \int_0^{2\pi R} dl = \frac{\mu_0 I}{4\pi R^2} \cdot 2\pi R

\boxed{B = \frac{\mu_0 I}{2R}}

This is the formula every JEE Main paper expects you to write directly. Learn it cold.

Curl the fingers of your right hand in the direction of current flow around the loop. Your extended thumb points in the direction of $\vec{B}$ at the centre.

If current flows anticlockwise (viewed from front), $B$ points toward you. Clockwise current → $B$ points away.

Why This Works

The key insight is the geometry: every element of a circular loop is perpendicular to the line joining it to the centre. This means $\sin\theta = 1$ for every single element — no angle correction needed anywhere. That’s what makes circles special compared to, say, a straight wire where $\theta$ varies continuously.

The second reason the integration is clean: all $dB$ contributions are parallel (same direction). For a straight wire, you’d need to project components. Here, symmetry does the heavy lifting for you.

For $N$ turns, the field simply multiplies: $B = \dfrac{\mu_0 N I}{2R}$ . Each turn contributes independently, and all contributions are in the same direction. This appears frequently in JEE Main — don’t forget the $N$ .

Alternative Method — Using Ampere’s Law?

Ampere’s Law cannot be used here directly. That’s worth knowing for MCQs.

Ampere’s Law ( $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$ ) only gives clean answers when you can find an Amperian loop where $B$ is constant and parallel to $d\vec{l}$ everywhere. For a circular loop, no such Amperian path exists at the centre.

Biot-Savart is the only clean route here. Reserve Ampere’s Law for infinite straight wires, solenoids, and toroids — problems with translational or cylindrical symmetry.

Common Mistake

Students often write $B = \dfrac{\mu_0 I}{4\pi R}$ — confusing this with the Biot-Savart formula for a point element. The $4\pi$ cancels during integration: $\dfrac{\mu_0 I}{4\pi R^2} \times 2\pi R = \dfrac{\mu_0 I}{2R}$ . The final formula has a 2 in the denominator, not $4\pi$ . In JEE Main 2023, this exact confusion cost marks in a 1-mark MCQ asking students to identify the correct expression from four options.

Quick Revision

B_{centre} = \frac{\mu_0 I}{2R}

For $N$ turns:

B_{centre} = \frac{\mu_0 N I}{2R}

Direction: Right-hand thumb rule — thumb points along $\vec{B}$ , fingers curl in direction of $I$ .

The ratio to remember: this result is $\pi$ times larger than the field due to a semi-circular loop at its centre ( $B_{semi} = \mu_0 I / 4R$ ). If you’re given a semi-circular loop problem, you’re just integrating half the circumference — the result halves cleanly.