H in coaxial cable regions

Setup. Let the inner conductor have radius a, carrying a uniformly distributed current I along +z. The outer conductor has inner radius b and outer radius c, and carries −I (uniformly distributed). By symmetry H has only an a_φ component that depends on ρ. Apply Ampere's circuital law ∮H·dl = I_enc on a circle of radius ρ, where ∮H·dl = H_φ(2πρ).

 cross-section
        ___________
      /  .-------.  \      c = outer radius of outer conductor
     |  /  .---.  \  |     b = inner radius of outer conductor
     | |  ( I ⊙ )  | |     a = radius of inner conductor
     |  \  '---'  /  |     inner: current I out of page
      \  '-------'  /      outer: current −I (into page)
        '-----------'

Region 1: 0 ≤ ρ < a (inside the inner conductor)

The current enclosed is the fraction of the uniform current within radius ρ: I_enc = I (πρ²)/(πa²) = Iρ²/a².

H_φ (2πρ) = Iρ²/a²
H = Iρ/(2πa²)  a_φ

Region 2: a < ρ < b (the space between conductors)

I_enc = I.

H = I/(2πρ)  a_φ

Region 3: b < ρ < c (inside the outer conductor)

The enclosed current is I plus the portion of −I within radius ρ:

I_enc = I − I (ρ² − b²)/(c² − b²) = I (c² − ρ²)/(c² − b²)

H = I/(2πρ) · (c² − ρ²)/(c² − b²)  a_φ

Region 4: ρ > c (outside the cable)

I_enc = I − I = 0, so

H = 0

Summary

H rises linearly from zero at the axis to I/(2πa) at ρ = a, falls as 1/ρ in the gap, decreases to zero across the outer conductor, and is zero outside, so the coaxial cable has no external magnetic field (shielding).