H of infinite filament via vector potential

The vector magnetic potential A is defined by B = ∇×A, with the integral form for a line current:

A = ∫ μ₀ I dl / (4π R)

Step 1: A of a filament of length 2L (then let L → ∞)

Place the filament along the z-axis from z = −L to z = L, carrying I in the +a_z direction. The field point is P(ρ, φ, 0), and R = √(ρ² + z²). The current element is dl = dz a_z, so A has only a z-component:

A_z = (μ₀ I / 4π) ∫₋L^L dz / √(ρ² + z²)
    = (μ₀ I / 4π) [ln(z + √(ρ² + z²))]₋L^L
    = (μ₀ I / 4π) ln[ (√(ρ²+L²) + L) / (√(ρ²+L²) − L) ]

For L ≫ ρ, √(ρ²+L²) ≈ L, so the argument is ≈ 2L·2L/ρ² = 4L²/ρ²:

A_z ≈ (μ₀ I / 4π) ln(4L²/ρ²) = (μ₀ I / 2π) ln(2L/ρ)

As L → ∞ this grows without limit, but only the ρ-dependence matters for B. Writing A_z = −(μ₀ I / 2π) ln ρ + C, where C is a constant (reference level):

A = −(μ₀ I / 2π) ln ρ a_z + C a_z

Step 2: find B and H

A depends only on ρ, so in cylindrical coordinates:

B = ∇×A = −(∂A_z/∂ρ) a_φ
  = (μ₀ I / 2πρ) a_φ

Step 3: H

H = B/μ₀ = (I / 2πρ) a_φ   A/m

This is the same result as from the Biot-Savart law and Ampere's circuital law. The constant C disappears on differentiation, confirming that A is defined only up to a constant.