Poynting numerical, wave in free space

Given: free space, E = (10 a_y + 5 a_z) cos(ωt + 2y − 4z) V/m.

The phase is ωt + k·r with k = 2 a_y − 4 a_z, so the wave travels along −k, i.e. along a_k = (−2 a_y + 4 a_z)/√20. Check that E ⟂ k: (10)(2) + (5)(−4) = 0 ✓.

(a) ω and λ

β = |k| = √(2² + 4²) = √20 = 4.472 rad/m
λ = 2π/β = 6.2832/4.472 = 1.405 m
ω = β c = 4.472 × 3×10⁸ = 1.342×10⁹ rad/s     (f = 213.5 MHz)

(b) Magnetic field. H = (1/η₀) a_k × E, with η₀ = 377 Ω and a_k = −0.4472 a_y + 0.8944 a_z.

a_k × (10 a_y + 5 a_z)
 a_x: (−0.4472)(5) − (0.8944)(10) = −2.236 − 8.944 = −11.18
 a_y: 0,  a_z: 0
⇒ a_k × E = −11.18 a_x

H = (−11.18/377) cos(ωt + 2y − 4z) a_x
  = −29.66 cos(1.342×10⁹ t + 2y − 4z) a_x   mA/m

(c) Time-average power. The magnitude |E| = √(10² + 5²) = √125 = 11.18 V/m.

P_avg = |E|²/(2η₀) = 125 / (2 × 377) = 0.1658 W/m²

Direction: along a_k (the direction of propagation), i.e. P_avg = 0.166 (−0.4472 a_y + 0.8944 a_z) = −0.0741 a_y + 0.1483 a_z W/m².

Answer: ω = 1.34×10⁹ rad/s, λ = 1.405 m; H = −29.7 cos(ωt + 2y − 4z) a_x mA/m; P_avg ≈ 0.166 W/m² (166 mW/m²).