---
title: "EM wave equations in vacuum"
subject: "Engineering Physics"
chapter: 9
key: "semester_1/engineering_physics:maxwell-differential-free-space-wave-equations"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/maxwell-differential-free-space-wave-equations
site: BCT Engineering
---

# EM wave equations in vacuum

Engineering Physics · Chapter 9: Electromagnetic Waves

- Asked in: 2073 Chaitra, 5 marks; 2068 Shrawan, 5 marks (asked 2 times)
- Video: [L-3b || EM || Maxwell’s electromagnetic wave equations in vacuum](https://www.youtube.com/watch?v=3TjlE804NBQ) — Dr. Randhir Singh
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/maxwell-differential-free-space-wave-equations
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-9

## Question

Write Maxwell equations in differential form in free space. Derive electromagnetic wave equations in vacuum. Find their plane wave solutions.

## Short answer

Free-space wave equations: ∇²E = μ₀ε₀ ∂²E/∂t², ∇²B = μ₀ε₀ ∂²B/∂t², with speed c = 1/√(μ₀ε₀). Plane solution: E = E₀ sin(kx−ωt) ŷ, B = B₀ sin(kx−ωt) ẑ, E₀/B₀ = c, transverse.

## Answer

#### Maxwell's equations in free space (ρ = 0, J = 0)
1. ∇·E = 0
2. ∇·B = 0
3. ∇×E = −∂B/∂t
4. ∇×B = μ₀ε₀ ∂E/∂t

#### Wave equations
1. Take the curl of (3): ∇×(∇×E) = −∂(∇×B)/∂t.
2. Left side: ∇×(∇×E) = ∇(∇·E) − ∇²E = −∇²E (using ∇·E = 0).
3. Right side: use (4): −∂/∂t (μ₀ε₀ ∂E/∂t) = −μ₀ε₀ ∂²E/∂t².
4. So:

   **∇²E = μ₀ε₀ ∂²E/∂t²**

Taking the curl of (4) and using (3) gives in the same way:

   **∇²B = μ₀ε₀ ∂²B/∂t²**

These have the form ∇²ψ = (1/c²)∂²ψ/∂t², so the speed is **c = 1/√(μ₀ε₀) = 3×10⁸ m/s**.

#### Plane wave solutions
Take a wave travelling along +x. The fields depend on x and t only.

```text
∂²E/∂x² = (1/c²) ∂²E/∂t²
E = E₀ sin(kx − ωt)      (along y)
B = B₀ sin(kx − ωt)      (along z)
k = 2π/λ,  ω = 2πf,  ω/k = c
```

From ∇·E = 0 and ∇·B = 0, E_x = B_x = 0, so the wave is transverse. From Faraday's law, ∂E_y/∂x = −∂B_z/∂t, which gives kE₀ = ωB₀, so **E₀/B₀ = c**. E, B and the direction of propagation are mutually perpendicular, with E and B in phase.

## Check yourself

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

1. Which statement is correct for Maxwell's equations in free space?
   A. ∇·E = 0 and ∇×B = μ₀ε₀ ∂E/∂t
   B. ∇·E = ρ/ε₀ and ∇×B = μ₀J
   C. ∇·B = μ₀J and ∇×E = 0
   D. ∇×B = 0 and ∇·E = μ₀ε₀
   Answer: A. Free space has ρ = 0 and J = 0, so only the displacement term sources the curl of B. (Key point: Free-space equations)

2. Which steps give ∇²E = μ₀ε₀ ∂²E/∂t²?
   A. Take divergence of Faraday's law and set ∇×E = 0
   B. Take curl of Faraday's law, use the curl-curl identity with ∇·E = 0, and substitute ∇×B from Ampere-Maxwell
   C. Add Gauss's two laws
   D. Integrate Ampere's law over time only
   Answer: B. Curl of ∇×E gives −∇²E, and ∂(∇×B)/∂t supplies μ₀ε₀ ∂²E/∂t². (Key point: Obtaining the wave equation)

3. The wave equations have the form ∇²ψ = (1/c²)∂²ψ/∂t². What is c?
   A. √(μ₀ε₀)
   B. 1/(μ₀ε₀)
   C. 1/√(μ₀ε₀) = 3×10⁸ m/s
   D. μ₀/ε₀
   Answer: C. Matching coefficients gives 1/c² = μ₀ε₀. (Key point: Speed of wave in vacuum)

4. From ∇·E = 0, ∇·B = 0 and Faraday's law, what do we conclude about the plane wave?
   A. It is longitudinal with E along x
   B. E and B are 90° out of phase with E₀/B₀ = 1
   C. E₀/B₀ = 1/c and E is parallel to B
   D. It is transverse, E and B are perpendicular and in phase, and E₀/B₀ = c
   Answer: D. The divergence conditions remove x components, and kE₀ = ωB₀ gives E₀/B₀ = ω/k = c. (Key point: Nature of plane wave)

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