Displacement current in parallel plate capacitor

A parallel plate capacitor with circular plates of radius a is charged by a current i. Between the plates there is no conduction current, but the changing electric field gives a displacement current i_d = ε₀ dΦ_E/dt. The total displacement current through the whole plate area equals the charging current i. The displacement current density is uniform:

J_d = i/(πa²)

(a) Value of ∮B·ds at r = a/5

Ampere-Maxwell law: ∮B·ds = μ₀ i_d,enc, where i_d,enc is the displacement current inside the loop of radius r.

i_d,enc = J_d × π r² = i × (r²/a²)
At r = a/5:  i_d,enc = i × (1/25) = i/25

∮B·ds = μ₀ i / 25

Induced magnetic field at r = a/5

B (2π r) = μ₀ i_d,enc
B = μ₀ i_d,enc / (2π r) = μ₀ (i/25) / (2π a/5)
  = μ₀ i / (10 π a)

In general, inside the plates B = μ₀ i r/(2π a²), growing linearly with r, and outside B = μ₀ i/(2π r).

Answer: ∮B·ds = μ₀i/25; B = μ₀ i_d,enc/(2πr) = μ₀i/(10πa), where i_d,enc = i/25.