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

2024
Views:

70

Pages:

7
Author:

Jabin F.

# EM wave propagation through vacuum: Permeability: μ = μ₀ Permittivity: ε = ε₀ Electrical Conductivity: σ = 0 # Most generalized form of Maxwell's equations: ∇⋅D = ρ (i) ∇⋅B = 0 (ii) ∇×E = -∂B/∂t (iii) ∇×B = μ₀J + μ₀ε₀∂E/∂t (iv) In free space: μ = μ₀, ε = ε₀, σ = 0 ρ = 0, Jf = 0, B = μ₀H, M = 0 Derivations: ∇⋅B = 0 → ∇⋅(μ₀H) = 0 → ∇⋅H = 0 (vii) ∇×E = -∂B/∂t → ∇×E = -(∂/∂t)(μ₀H) = -μ₀(∂H/∂t) (vii) ∇×H = Jf + ∂D/∂t → ∇×H = ε₀(∂E/∂t) (vii) We need to convert these equations into two second-order differential equations. # Take curl of equation (iii): ∇×(∇×E) = -∂/∂t(∇×B) ∇(∇⋅E) - ∇²E = -μ₀ε₀(∂²/∂t²)(∇×E) ∇²E - μ₀ε₀(∂²E/∂t²) = 0 Similarly, from equation (iv): ∇×(∇×B) = μ₀ε₀(∂/∂t)(∇×E) ∇(∇⋅B) - ∇²B = μ₀ε₀(∂²/∂t²)(∇×E) ∇²B - μ₀ε₀(∂²B/∂t²) = 0 From equation (4): ∇²E - μ₀ε₀(∂²E/∂t²) = 0 ∇²E = (iR)(iR)E = -R²E ∇²E = -k²E From equation (1): ∇²E = (-iω)(-iω)E = -ω²E Putting in equation (1): -k²E + ω²μ₀ε₀E = 0 k² = ω²μ₀ε₀ k = ±√(μ₀ε₀)ω Wave velocity (v): v = ω/k = 1/√(μ₀ε₀) v = c (speed of light in vacuum) From equation (4) and (5): Same solution will be obtained for B. Coupled oscillations of E and B are traveling with the speed of light in vacuum # Ē = Ē₀e^(i(kz - ωt)) B = Ḃ₀e^(i(kz - ωt)) These solutions should also satisfy the Maxwell equations. ∇⋅Ē = 0, ∇⋅Ḃ = 0 ⇒ (iR)⋅Ē₀ = 0, (iR)⋅Ḃ₀ = 0 ⇒ Ē⋅R = 0, Ḃ⋅R = 0 (14) R = wave vector = Direction of wave propagation. R₁ = 2π/λ, λ = wavelength of wave ∇⋅Ē = 0 → ∇⋅(Ē₀e^(i(kz - ωt))) = 0 → (iR)⋅Ē₀ = 0 ∇⋅Ḃ = 0 → ∇⋅(Ḃ₀e^(i(kz - ωt))) = 0 → (iR)⋅Ḃ₀ = 0 From (14), these show that Ē and Ḃ are perpendicular to the wave vector (R). ∇×Ē = -∂Ḃ/∂t → ∇×(Ē₀e^(i(kz - ωt))) = -iω(Ḃ₀e^(i(kz - ωt))) → (iR)×Ē₀ = -iωḂ₀ → Ē₀ = (ω/R) × Ḃ₀ (16) |Ē₀| = (ω/R) |Ḃ₀| |Ē₀| = c