Applications of Biot-Savart's Law > Magnetic Field at the Axis of a Circular Current-carrying Loop

You already know that a current-carrying straight wire produces magnetic field lines as concentric circles around it, following the right-hand thumb rule. Now imagine bending that straight wire into a loop. Something interesting happens: as you move from the outer part of the wire toward the centre of the loop, those "concentric circles" of magnetic field become so large that, near the centre, they look almost like straight lines.

Every tiny segment of the loop contributes its own small magnetic field at the centre, and — because of the loop's symmetry — all these contributions point in the same direction. This is why a full loop produces a much stronger, more concentrated field at its centre than a straight wire does at an equivalent distance. If you stack multiple loops together (a coil of N turns), the field simply multiplies: N loops produce N times the field of one loop.

Picture a circular loop of radius R, lying flat in the y-z plane, centred at origin O, carrying a steady current I. We want to find the magnetic field at a point P sitting on the axis of the loop (the x-axis), at a distance x from the centre.

Step 1: By the Biot–Savart law, each tiny wire segment dl contributes a small field:

where r² = x² + R².

Step 2: This dB has two parts — one along the axis and one sideways (perpendicular to the axis). Now consider the element diametrically opposite on the loop. Its sideways component points in exactly the opposite direction, so the two cancel out. This happens for every pair of opposite elements around the loop, so all perpendicular components cancel completely, leaving only the axial components to add up.

Step 3: Using the geometry, \[\cos\theta = \dfrac{R}{\sqrt{x^2+R^2}}\], the axial component of each element's field is:

Step 4: Adding up (integrating) dB_x over the entire loop just means adding up all the dl segments, which together equal the loop's circumference, 2πR.

Step 5: Substituting this gives the total magnetic field at point P along the axis:

This is the master formula for the field anywhere on the axis of a current loop.

Set x = 0 in Equation A (point P is now exactly at the centre of the loop):

For a coil of N tightly wound turns, multiply by N:

This is the strongest point along the axis — the field weakens as you move away from the centre in either direction.

Curl the fingers of your right hand in the direction the current flows around the loop; your thumb then points in the direction of the magnetic field along the axis. One face of the loop behaves like a magnetic north pole, and the opposite face behaves like a south pole.

A wire carrying a 12 A current is bent into a semicircular arc of radius 2.0 cm. Find the field at the centre.

• The two straight segments contribute nothing, because for a straight segment, the direction from each point to the centre is along the wire itself, making dl × r = 0.
• The curved (semicircular) part behaves like "half" a full loop, so its field is exactly half of what a complete circular loop of the same radius and current would produce: B = 1.9 × 10⁻⁴ T, directed into the plane of the paper.
• If the arc is bent the opposite way, the magnitude stays the same, but the direction flips.

A tightly wound coil has 100 turns, radius 10 cm, carrying 1 A. Find the field at the centre.

Using Equation B with N = 100:

Common mistake to avoid: Students often forget to multiply by N when the coil has multiple turns, or mix up R (loop radius) with x (axial distance) — always double-check which one the question is asking about.

This is exactly the principle behind an electric doorbell or a simple electromagnet — winding wire into a coil dramatically boosts magnetic strength compared to a straight wire, which is why coils (not straight wires) are used in speakers, transformers, and MRI machines.

• A circular current loop produces a magnetic field whose axial value is B = \[\frac{\mu_0IR^2}{2(x^2+R^2)^{3/2}}\].
• At the centre of the loop (x = 0), this simplifies to B₀ = \[\frac {μ_0I}{2R}\]​, and for N turns, B₀ = \[\frac {μ_0NI}{2R}\].
• Perpendicular field components from opposite points on the loop cancel; only axial components add up.
• Direction follows the right-hand thumb rule; one face of the loop acts as a north pole, the other as a south pole.
• Straight wire segments (as in a semicircular arc problem) contribute zero field at a point lying on the line of the wire itself.