Skip to main content

Plane wave in non-magnetic medium

Short answer

Propagation direction is −a_z. Phase constant β = 2 rad/m, wavelength λ = 3.14 m, velocity v_p = 5×10⁷ m/s, relative permittivity ε_r = 36, intrinsic impedance η = 62.8 Ω. Magnetic field H = 0.796 cos(10⁸t + 2z) a_x A/m.

A uniform plane wave in a non-magnetic medium has E = 50 cos(10⁸t + 2z) a_y V/m. Find: (i) the direction of propagation; (ii) phase constant β, wavelength λ, velocity v_p, relative permittivity ε_r, intrinsic impedance η; (iii) H.

Given: E = 50 cos(10⁸t + 2z) a_y V/m in a non-magnetic medium (μ_r = 1). So ω = 10⁸ rad/s, E₀ = 50 V/m.

(i) Direction of propagation. The phase is (ωt + 2z), with a plus sign between t and z, so the wave travels in the −z direction (−a_z).

(ii) Parameters.

β = 2 rad/m                         (coefficient of z)
λ = 2π/β = 2π/2 = π = 3.142 m
v_p = ω/β = 10⁸/2 = 5×10⁷ m/s

Non-magnetic: v_p = c/√ε_r, so

ε_r = (c/v_p)² = (3×10⁸ / 5×10⁷)² = 6² = 36
η = η₀/√ε_r = 377/6 = 62.83 Ω

(iii) H. For a wave travelling along a_k = −a_z, H = (1/η) a_k × E:

a_k × E = (−a_z) × a_y = a_x          (since a_z × a_y = −a_x)
H = (50/62.83) cos(10⁸t + 2z) a_x
  = 0.796 cos(10⁸t + 2z) a_x   A/m

Check: E × H = a_y × a_x = −a_z ✓ (travelling in −z).

Answer: (i) −a_z; (ii) β = 2 rad/m, λ = 3.14 m, v_p = 5×10⁷ m/s, ε_r = 36, η = 62.8 Ω; (iii) H = 0.796 cos(10⁸t + 2z) a_x A/m.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Electromagnetism chapter 4 questions →