University:
Santa Fe College
Course:
ETS 2527C | Electromechanical Components and Mechanisms
Academic year:

2024
Views:

237

Pages:

6
Author:

Jabin F.

# Displacement Current and modified Ampere's Law Before Maxwell, there were four equations defined in E.D 1. ∇.E = ρ/ε₀ [Gauss Law of Electrostatics] 2. ∇.B = 0 3. ∂E/∂t = -∇xB [Faraday's Law] 4. ∇xB = μ₀J [Ampere's Law] To check Consistency of these equations with Electrodynamics from eqn(3) ∇xE = -∂B/∂t => ∇.(∇xE) = -∂/∂t(∇.B) => ∇.(∇xE) = 0 from eqn(4) ∇xB = μ₀J => ∇.(∇xB) = μ₀(∇.J) => 0 = μ₀(∇.J) => ∇.J = 0 (magnetostatic) ∇.J ≠ 0 (electrostatic)] ∇xB = μ₀J [Ampere's law]x It is valid in magneto static but not valid in electrodynamics. ∴ ∇.J = -∂ρ/∂t => Continuity eqn (5) ∂ρ/∂t = 0 (magnetostatic) ∂ρ/∂t ≠ 0 [Electrodynamics] ∴ ∇.J ≠ 0 but ∇.J + ∂ρ/∂t = 0 from eqn (D) (Gauss law) => ∇.J + ε₀(∂E/∂t) = 0 (continuity eqn with Gauss law) J should be replaced by [J + ε₀(∂E/∂t)] So now from Ampere's law modified to- ∇xB = μ₀[J + ε₀(∂E/∂t)] => ∇xB = μ₀J + μ₀ε₀(∂E/∂t) This is modified Ampere's Law ε₀(∂E/∂t) = ∂D/∂t = Jd (Displacement current density) D = Displacement vector ε₀(∂E/∂t) = Jd => ∇xB = μ₀Jc + μ₀Jd Jc = Conduction Current (Because of flow of charge in wire) Jd = Displacement Current exp: I = Conduction current if I = steady current (Time independent) then Jd = 0 -> ∇.E = ρ/ε₀ - (i) ∇.B = 0 - (ii) ∇xE = -∂B/∂t - (iii) ∇xB = μ₀J + μ₀ε₀(∂E/∂t) - (iv) These are known as Maxwell's eqn Differential form Integral form 1. ∇.E = ρ/ε₀ 1. ∮E.ds = q/ε₀ 2. ∇.B = 0 2. ∮B.ds = 0 3. ∇xE = -∂B/∂t 3. ∮E.dl = -∫(∂B/∂t).ds 4. ∇xB = μ₀J + μ₀ε₀(∂E/∂t) 4. ∮B.dl = μ₀I + μ₀ε₀∫(∂E/∂t).ds Conduction displacement Current Current