∇×H in coaxial cable regions

By Maxwell's (Ampere's) law for steady fields, ∇×H = J, so we find H from Ampere's circuital law and then take the curl (or directly the current density).

Set-up: the inner conductor (radius a) carries +I in a_z, uniformly distributed. The outer conductor (inner radius b, outer radius c) carries −I (return current). By symmetry H = H_φ(r) a_φ, and Ampere's law gives H_φ · 2πr = I_enc.

Region 0 ≤ r ≤ a

J = I/(πa²) a_z, so I_enc = I r²/a².

H_φ = I r / (2π a²)
∇×H = (1/r) d(r H_φ)/dr a_z = (1/r) d/dr [ I r² / (2π a²) ] a_z
    = I / (π a²) a_z

∇×H = I/(πa²) a_z, which equals J, as it must.

Region a ≤ r ≤ b

I_enc = I (the whole inner conductor current).

H_φ = I / (2π r)
r H_φ = I / 2π = constant
∇×H = (1/r) d(I/2π)/dr a_z = 0

∇×H = 0 (no current in the dielectric).

Region b ≤ r ≤ c

The outer conductor carries −I spread over the area π(c² − b²). The part of it enclosed within radius r is −I (r² − b²)/(c² − b²).

I_enc = I − I (r² − b²)/(c² − b²) = I (c² − r²)/(c² − b²)
H_φ   = I (c² − r²) / [2π r (c² − b²)]
r H_φ = I (c² − r²) / [2π (c² − b²)]
∇×H   = (1/r) d/dr [ I (c² − r²) / (2π (c² − b²)) ] a_z
      = (1/r) × [ −2 r I / (2π (c² − b²)) ] a_z
      = −I / [π (c² − b²)] a_z

∇×H = −I/[π(c² − b²)] a_z, which equals the current density J of the outer conductor (negative because the current returns in −a_z).

Summary

For r > c, I_enc = 0, so H = 0 and ∇×H = 0 (the cable has no external field).