Normal incidence on perfect conductor

Given: f = 100 MHz in air, wave travels in +x, E along a_z, amplitude E₀ = 3 mV/m, perfect conductor at x = 0.

ω = 2π×10⁸ rad/s
β = ω/c = 2π×10⁸ / 3×10⁸ = 2.094 rad/m  (2π/3)
η₀ = 120π = 376.7 Ω
E₀/η₀ = 3×10⁻³/376.7 = 7.96 μA/m

For a perfect conductor η₂ = 0, so Γ = (η₂ − η₁)/(η₂ + η₁) = −1: all power is reflected and the reflected E is the negative of the incident E at x = 0 (tangential E must vanish).

The direction of H follows from H = (1/η) a_k × E.

(a) Incident wave (travelling in +x)

E_i (phasor)  = 3×10⁻³ e^(−j2.094x) a_z   V/m
E_i (instant) = 3×10⁻³ cos(2π×10⁸t − 2.094x) a_z   V/m

a_x × a_z = −a_y
H_i (phasor)  = −7.96×10⁻⁶ e^(−j2.094x) a_y   A/m
H_i (instant) = −7.96×10⁻⁶ cos(2π×10⁸t − 2.094x) a_y   A/m

(b) Reflected wave (travelling in −x), Γ = −1

E_r (phasor)  = −3×10⁻³ e^(+j2.094x) a_z   V/m
E_r (instant) = −3×10⁻³ cos(2π×10⁸t + 2.094x) a_z   V/m

(−a_x) × (−a_z) = a_x × a_z = −a_y
H_r (phasor)  = −7.96×10⁻⁶ e^(+j2.094x) a_y   A/m
H_r (instant) = −7.96×10⁻⁶ cos(2π×10⁸t + 2.094x) a_y   A/m

Check at x = 0: E_i + E_r = 0 (boundary condition satisfied), and H total = −15.9 μA/m a_y, twice the incident H, which equals the surface current density J_s = n × H.

The total field in x < 0 is a standing wave: E = −j6×10⁻³ sin(βx) a_z, H = −15.9×10⁻⁶ cos(βx) a_y (phasors).