Skip to main content

Displacement current density at 60 Hz

Short answer

Displacement current density J_d = (ε/σ) dJ_c/dt. For J_c = sin(377t - 117.1z) MA/m^2, dJ_c/dt = 377 cos(...) MA/m^2. Amplitude = (ε0 ω/σ) * 1e6 = (8.854e-12 * 377 / 5.8e7) * 1e6 = 5.76e-11 A/m^2.

Find the amplitude of the displacement current density in a metallic conductor at 60 Hz, if ε = ε0, μ = μ0, σ = 5.8×10⁷ S/m and J = sin(377t − 117.1z) a_x MA/m².

The conduction current density is J_c = σE and the displacement current density is J_d = ∂D/∂t = ε ∂E/∂t. Given J = J_c = sin(377t − 117.1z) a_x MA/m², so

E = J_c / σ
J_d = ε ∂E/∂t = (ε/σ) ∂J_c/∂t

Step 1: differentiate. ω = 377 rad/s (60 Hz).

∂J_c/∂t = 377 cos(377t − 117.1z) a_x   MA/m²

Step 2: substitute ε = ε₀ = 8.854×10⁻¹² F/m, σ = 5.8×10⁷ S/m.

ε₀ ω / σ = (8.854×10⁻¹² × 377) / (5.8×10⁷)
         = 3.338×10⁻⁹ / 5.8×10⁷
         = 5.755×10⁻¹⁷

J_d = 5.755×10⁻¹⁷ cos(377t − 117.1z) a_x  MA/m²
    = 5.755×10⁻¹¹ cos(377t − 117.1z) a_x  A/m²

Step 3: ratio check. J_d/J_c = ωε/σ = 5.755×10⁻¹⁷ ≈ 10⁻¹⁷, so displacement current is completely negligible in a metal at power frequency.

Answer: amplitude of J_d = 5.755×10⁻¹⁷ MA/m² = 5.76×10⁻¹¹ A/m² (leading J_c by 90°).

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Electromagnetism chapter 3 questions →