Induced magnetic field, displacement current

Given: dE/dt = 1.5×10¹² V/(m·s), plate diameter 110 mm, so R = 55 mm = 0.055 m. ε₀ = 8.85×10⁻¹² F/m, μ₀ = 4π×10⁻⁷ H/m.

The changing field E between the plates gives a displacement current I_d = ε₀ dΦ_E/dt = ε₀ A dE/dt, which induces a magnetic field.

Displacement current

A   = π R² = π (0.055)² = 9.503×10⁻³ m²
I_d = ε₀ A dE/dt
    = 8.85×10⁻¹² × 9.503×10⁻³ × 1.5×10¹²
    = 0.126 A

Induced magnetic field

Ampere-Maxwell law on a circle of radius r ≤ R: B(2πr) = μ₀ε₀ (πr²) dE/dt, so B = μ₀ε₀ r/2 · dE/dt. At the edge of the plates (r = R, the maximum):

B = μ₀ ε₀ R (dE/dt) / 2
  = (4π×10⁻⁷)(8.85×10⁻¹²)(0.055)(1.5×10¹²)/2
  = 4.59×10⁻⁷ T
(check: μ₀ I_d /(2πR) = 4π×10⁻⁷ × 0.126 /(2π × 0.055) = 4.59×10⁻⁷ T)

Answer: I_d ≈ 0.126 A; B ≈ 4.6×10⁻⁷ T at the plate edge (r = 55 mm). Inside, B grows linearly with r.