---
title: "Helmholtz's equation for EM wave"
subject: "Electromagnetism"
chapter: 4
key: "semester_3/electromagnetism:helmholtz-equation-establish"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/helmholtz-equation-establish
site: BCT Engineering
---

# Helmholtz's equation for EM wave

Electromagnetism · Chapter 4: Wave equation and Wave propagation

- Asked in: 2076 Asoj, 5 marks
- Video: [Wave Equation in Electromagnetic Waves Explained | EM Waves | Electromagnetics Theory](https://www.youtube.com/watch?v=RuQuZNk_0dA) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/helmholtz-equation-establish
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-4

## Question

Establish the relation for Helmholtz's equation for electromagnetic wave propagation.

## 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).

## Answer

**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

```text
∇×E = −μ ∂H/∂t        ∇×H = σE + ε ∂E/∂t
∇·E = 0               ∇·H = 0
```

Take the curl of the first equation and substitute the second:

```text
∇×(∇×E) = −μ ∂(∇×H)/∂t = −μσ ∂E/∂t − με ∂²E/∂t²
```

Use the identity ∇×(∇×E) = ∇(∇·E) − ∇²E = −∇²E (because ∇·E = 0):

```text
∇²E = μσ ∂E/∂t + με ∂²E/∂t²      (general wave equation)
```

For fields varying as e^{jωt}, ∂/∂t → jω and ∂²/∂t² → −ω²:

```text
∇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
- **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.
- **Free space:** k = ω/c = ω√(μ₀ε₀).
- **Good conductor:** γ ≈ (1 + j)√(πfμσ).

The solutions are travelling waves, e.g. E_s = E₀ e^{−γz} for the +z direction.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Helmholtz's equation is:
   A. The static Laplace equation
   B. The wave equation for time-harmonic fields in phasor form
   C. Gauss's law for B
   D. The continuity equation
   Answer: B. It is the wave equation written for sinusoidal fields with angular frequency w, using phasors. (Key point: What Helmholtz's equation is)

2. The general wave equation in a conducting medium with div E = 0 is:
   A. Laplacian E = mu epsilon d^2E/dt^2 only
   B. Laplacian E = mu sigma dE/dt + mu epsilon d^2E/dt^2
   C. Laplacian E = -mu epsilon d^2E/dt^2
   D. Laplacian E = sigma E
   Answer: B. Curl of curl E = -mu sigma dE/dt - mu epsilon d^2E/dt^2, and curl curl E = -Laplacian E when div E = 0. (Key point: General wave equation)

3. With e^(jwt) dependence, the equation becomes:
   A. Laplacian E_s - gamma^2 E_s = 0, gamma^2 = jw mu (sigma + jw epsilon)
   B. Laplacian E_s + gamma E_s = 0
   C. Laplacian E_s = w E_s
   D. Laplacian E_s = sigma epsilon E_s
   Answer: A. Replacing d/dt by jw and d^2/dt^2 by -w^2 gives gamma^2 = jw mu (sigma + jw epsilon). (Key point: Phasor form)

4. In a lossless dielectric (sigma = 0), the equation reduces to:
   A. Laplacian E_s - k^2 E_s = 0
   B. Laplacian E_s = 0
   C. Laplacian E_s + k^2 E_s = 0, k = w sqrt(mu epsilon)
   D. Laplacian E_s = jw E_s
   Answer: C. gamma = j beta, so gamma^2 = -w^2 mu epsilon and the equation becomes Laplacian E_s + k^2 E_s = 0. (Key point: Lossless special case)

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