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Question

A capacitor in a circuit has a rate of change of electric flux of \( 4 \times 10^{11} \, \text{Vm/s}
\). What is the displacement current? (Given \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \))

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Explanation

**Ampere-Maxwell law** ∮ B·dl = μ₀(I_c + ε₀ dΦ_E/dt) generalizes Ampere's law, displacement current arises from time-varying electric field, source of magnetic field like conduction current. For rate of change of flux 2×10¹¹ V·m/s, I_d = ε₀×2×10¹¹ =8.85×10⁻¹²×2×10¹¹=1.77 A. Displacement current i_d = ε₀ (d Φ_E/dt) . Substituting the values, i_d = (8.85 × 10⁻¹²) × (4 × 10¹¹) = 3.54 A . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 3.54 A, illustrating EM wave transverse nature and Maxwell's displacement current concept.

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