Derive E and H of plane wave

A uniform plane wave is one whose E and H are constant over every plane perpendicular to the direction of travel (the fields have no variation in the transverse plane), and the fields are perpendicular to each other and to the direction of propagation (transverse electromagnetic, TEM).

Wave equation for a perfect dielectric (σ = 0, ρv = 0, J = 0)

Maxwell's equations:

∇×E = −μ ∂H/∂t
∇×H =  ε ∂E/∂t
∇·E = 0,   ∇·H = 0

Take the curl of the first and use the second:

∇×(∇×E) = −μ ∂(∇×H)/∂t = −με ∂²E/∂t²
∇×(∇×E) = ∇(∇·E) − ∇²E = −∇²E
⇒ ∇²E = με ∂²E/∂t²

and likewise ∇²H = με ∂²H/∂t². For time-harmonic fields e^{jωt}: ∇²E_s = −ω²με E_s = −β² E_s, with β = ω√(με).

Solution for a wave travelling in +z

Let E = E_x(z, t) a_x only (E_x does not vary with x or y). The equation reduces to

∂²E_x/∂z² = με ∂²E_x/∂t²

Its solution is a forward wave plus a backward wave. For the forward wave:

E = E₀ cos(ωt − βz) a_x V/m

with

β = ω √(με)  (rad/m),   v = ω/β = 1/√(με),   λ = 2π/β

In free space v = 1/√(μ₀ε₀) = c = 3×10⁸ m/s and β = ω/c.

Magnetic field

Use ∇×E = −μ ∂H/∂t. Only E_x(z, t) exists, so ∇×E = (∂E_x/∂z) a_y, with ∂E_x/∂z = E₀ β sin(ωt − βz). Hence H has only a y-component:

−μ ∂H_y/∂t = E₀ β sin(ωt − βz)
H_y = (E₀ β / μω) cos(ωt − βz)

Since β/(ωμ) = √(με)/μ = √(ε/μ) = 1/η:

H = (E₀/η) cos(ωt − βz) a_y A/m

where η = √(μ/ε) Ω is the intrinsic impedance of the medium. For free space η₀ = √(μ₀/ε₀) = 120π ≈ 377 Ω.

Properties

• E and H are in time phase, mutually perpendicular, and both perpendicular to the z-direction (a_x × a_y = a_z, the direction of E × H).
• Ratio E/H = η.
• Phase velocity v = 1/√(με); the wave moves without attenuation in a lossless dielectric.
• In phasor form: E_s = E₀ e^{−jβz} a_x, H_s = (E₀/η) e^{−jβz} a_y.