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Circular coil field and magnetic dipole

Short answer

B = μ0 I R^2 / [2(R^2 + z^2)^(3/2)] along axis. For z >> R, B ≈ (μ0/4π) * 2m/z^3 with m = IπR^2, matching the dipole field form, so the coil acts as a magnetic dipole.

Use the Biot-Savart law to find the magnetic field on the axis of a current carrying circular coil (radius R, at distance z from the center). Show that the circular current carrying coil behaves as a magnetic dipole for a large distance.

Biot–Savart law: the field due to a current element I dl at displacement r from it is dB = (μ0/4π) I (dl × r̂)/r².

Field on the axis

Coil of radius R, current I; point P on the axis at distance z from the centre.

            dl (element)
              o
             /|\
        s  /  | \   dB
         /    |R  \
   -----o-----O-----o-----
              |
              |  z
              P
  1. Each element dl is perpendicular to s, where s = √(R² + z²). So dB = (μ0 I dl)/(4π s²).
  2. dB is perpendicular to both dl and s. Components perpendicular to the axis cancel over the full ring. The axial component is dB_z = dB sinα, with sinα = R/s.
  3. B = ∫ dB_z = (μ0 I/4π s²)(R/s) ∮dl = (μ0 I/4π)(R/s³)(2πR)

B = μ0 I R² / [ 2 (R² + z²)^(3/2) ]

Behaviour as a magnetic dipole

  1. For z ≫ R: (R² + z²)^(3/2) ≈ z³, so

B ≈ μ0 I R² / (2 z³) = (μ0/4π) · 2 (I π R²)/z³

  1. The magnetic dipole moment of the coil is m = I A = I πR². So

B = (μ0/4π) · 2m / z³

  1. This has the same form as the axial field of an electric dipole, E = (1/4πε0)·2p/z³, and of a bar magnet. The field falls as 1/z³ and is along the axis, in the direction of m.

Hence a current-carrying loop acts as a magnetic dipole at large distance, with north pole on the side from which the current looks anticlockwise.

✓ Answer checked by Claude Opus 5.5 (high effort)

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