Magnetic field on axis of circular coil

Biot–Savart law. The magnetic field dB at a point due to a small current element I dl is directly proportional to I, dl and sinθ (θ = angle between dl and the line joining the element to the point), and inversely proportional to the square of the distance r:

dB = (μ0/4π) · I dl sinθ / r²

In vector form, dB = (μ0/4π) I (dl × r̂)/r².

Field on the axis of a circular coil

Coil of radius a, current I; point P on the axis at distance x from the centre O.

          dl
           o
          /|\
      r  / |a \  dB
        /  |    \
   ----o---O-----o----
           |
           |  x
           P

1. Take an element dl on the ring. It is perpendicular to the line r joining it to P, with r = √(a² + x²). So θ = 90° and

dB = (μ0/4π) I dl / (a² + x²)

2. dB is perpendicular to r, making angle α with the axis, where sinα = a/r. Resolve it:
   • dB sinα along the axis
   • dB cosα perpendicular to the axis
3. By symmetry, the perpendicular components of diametrically opposite elements cancel. Only the axial components add:

B = ∮ dB sinα = (μ0 I/4π)(1/(a² + x²))(a/√(a² + x²)) ∮dl

4. ∮dl = 2πa, so

B = μ0 I a² / [ 2 (a² + x²)^(3/2) ]

For N turns multiply by N. At the centre (x = 0): B = μ0 I/(2a). For x ≫ a: B ≈ μ0 I a²/(2x³).