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Displacement current: B at 50 mm

Short answer

At r = 50 mm, B = μ₀ J_d r/2 = 6.3×10⁻⁷ T. The rate of change of electric field is dE/dt = J_d/ε₀ = 2.3×10¹² V m⁻¹ s⁻¹.

As a parallel plate capacitor with circular plates 20cm in diameter is being charged, the current density of the displacement current in the region between the plates is uniform and has a magnitude of 20A/m². Calculate the magnitude of magnetic field (B) at a distance r = 50mm from the axis of symmetry of this region. Also calculate dE/dt in this region.

Given: plate diameter 20 cm, so R = 0.10 m; displacement current density J_d = 20 A/m² (uniform); r = 50 mm = 0.05 m (inside the plates since r < R).

Magnetic field at r = 50 mm

Ampere-Maxwell law (no conduction current between the plates): ∮B·dl = μ₀ I_d,enc.

B (2π r) = μ₀ J_d (π r²)
B = μ₀ J_d r / 2
  = (4π×10⁻⁷ × 20 × 0.05) / 2
  = 6.28×10⁻⁷ T

dE/dt

J_d = ε₀ dE/dt, so

dE/dt = J_d / ε₀ = 20 / 8.85×10⁻¹²
      = 2.26×10¹² V/(m·s)

Answer: B ≈ 6.3×10⁻⁷ T; dE/dt ≈ 2.3×10¹² V m⁻¹s⁻¹

✓ Answer checked by Claude Opus 5.5 (high effort)

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