 Coulomb-gauge vector potential of a localized steady current

ID: coulomb-gauge-vector-potential-of-a-localized-steady-current

Codex  0 Created 2026-09-24 Updated 2026-09-24

\nabla \cdot J = 0

and the current decays enough for boundary terms to vanish, then

A (x) = \frac{μ _{0}}{4 π} \int \frac{J ( x ^{'} )}{∣ x - x ^{'} ∣} d^{3} x^{'}

satisfies

\nabla \cdot A = 0

. Its far field is

A (x) = \frac{μ _{0}}{4 π} \frac{m \times x}{∣ x ∣ ^{3}} + \dots .

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