Magnetic Field on the Axis of a Circular Current-Carrying Loop

When a current flows through a circular conductor, it generates a magnetic field around it. The magnetic field lines form closed loops around the wire and are densest at the centre of the circular loop. On the axis, contributions from all current elements add up only along the axial direction — the transverse components cancel pairwise by symmetry.

Analogy: Think of the current loop like a bar magnet. The face from which field lines emerge is the North pole; the face into which they enter is the South pole. At large distances, the axial field of a current loop and a bar magnet follows the same formula.

\[\vec{B}=\frac{\mu_0IR^2}{2(x^2+R^2)^{3/2}}\hat{i}\]

Where:

• I = current
• R = radius of loop
• x = distance from centre along axis
• μ0 = permeability of free space

F = IL × B

Step 1: Apply Biot–Savart Law to one element

The magnetic field due to a small current element Idl at point P:

(Since dl ⊥ r always, |dl × r̂| = dl, so sinθ = 1)

Step 2: Resolve dB into components

Each dB has two components:

• dBₓ → along the x-axis (axial)
• dB⊥ → perpendicular to x-axis

For diametrically opposite elements, dB⊥ components are equal and opposite → net perpendicular field = 0.

Step 3: Find dBₓ

Step 4: Integrate over the full loop

Summing dl over the entire circumference: ∮dl = 2πR

Final Result

Rule: Curl the fingers of your right hand in the direction of current flow around the loop. Your thumb points in the direction of the magnetic field inside and along the axis of the loop.

• Face from which B emerges → acts as the North pole of the equivalent magnet
• Face into which B enters → acts as the South pole
• Reversing the current direction reverses the field direction

Setup: A straight wire carrying I = 12 A is bent so that the middle portion forms a semicircular arc of radius R = 2.0 cm. The straight ends extend outward from both sides of the arc. Find the magnetic field B at the centre of the arc.

Part (a): Contribution from the straight segments

For any element dl on a straight segment, dl is directed radially — i.e., parallel to r (the displacement vector from dl to the centre).

By Biot–Savart Law, |dB| ∝ |\[d\vec l\] × \[\hat r\]| = 0

Straight segments contribute zero magnetic field at the centre.

Part (b): Contribution from the semicircular arc

For every element of the arc, dl ⊥ r always (just like a full circle), so |\[d\vec l\] × \[\hat r\]| = dl. All contributions point in the same direction (into the plane of the paper by the right-hand rule), so they all add up.

A full circular loop gives: Bcircle = \[\frac{\mu_0I}{2R}\]

A semicircle is half of a full loop, so:

Part (c): Arc bent in the opposite direction

The magnitude of B remains the same (1.9 × 10⁻⁴ T), but the direction reverses — now out of the plane of the paper.

Setup: A tightly wound coil has N = 100 turns, radius R = 10 cm = 0.1 m, carrying current I = 1 A. Find B at the centre.

Key reasoning: Since the coil is tightly wound, all turns have effectively the same radius R. Each turn independently contributes \[\frac{\mu_0I}{2R}\]​ at the centre, and all contributions are in the same direction, so they simply add up:

Substituting values: