- 2076 Asoj · 5 marks
Helmholtz's equation for EM wave
Short answer
Taking curl of Maxwell's equations and using ∇·E=0 gives ∇²E = μσ ∂E/∂t + με ∂²E/∂t². For time-harmonic fields (e^{jωt}) this becomes the Helmholtz equation ∇²E_s − γ²E_s = 0 with γ² = jωμ(σ+jωε). In a lossless medium (σ=0), γ = jβ, giving ∇²E_s + β²E_s = 0, β = ω√(με).
Establish the relation for Helmholtz's equation for electromagnetic wave propagation.
Helmholtz's equation is the wave equation for time-harmonic fields, written in phasor form. It describes how E and H vary in space for a sinusoidal wave of angular frequency ω.
Derivation
Consider a homogeneous, linear, isotropic medium with ε, μ, σ and no free charge (ρv = 0). Maxwell's equations are
∇×E = −μ ∂H/∂t ∇×H = σE + ε ∂E/∂t
∇·E = 0 ∇·H = 0
Take the curl of the first equation and substitute the second:
∇×(∇×E) = −μ ∂(∇×H)/∂t = −μσ ∂E/∂t − με ∂²E/∂t²
Use the identity ∇×(∇×E) = ∇(∇·E) − ∇²E = −∇²E (because ∇·E = 0):
∇²E = μσ ∂E/∂t + με ∂²E/∂t² (general wave equation)
For fields varying as e^{jωt}, ∂/∂t → jω and ∂²/∂t² → −ω²:
∇²E_s = jωμ(σ + jωε) E_s = γ² E_s
So
∇²E_s − γ²E_s = 0, with γ² = jωμ(σ + jωε)
and in the same way ∇²H_s − γ²H_s = 0. Here γ = α + jβ is the propagation constant.
Special cases
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Lossless dielectric (σ = 0): γ = jβ, γ² = −ω²με, so
∇²E_s + k²E_s = 0, k = β = ω√(με)
This is the usual form of Helmholtz's equation, with k the wave number.
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Free space: k = ω/c = ω√(μ₀ε₀).
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Good conductor: γ ≈ (1 + j)√(πfμσ).
The solutions are travelling waves, e.g. E_s = E₀ e^{−γz} for the +z direction.
✓ Answer checked by Claude Opus 5.5 (high effort)
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