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Phasor E and H of wave in −y direction

Short answer

In free space, ω=10⁷π rad/s, β=ω/c=0.1047 rad/m. Phasor electric field: E_s = 200 e^{j0.1047y} a_z V/m. Propagation direction −a_y gives magnetic field: H_s = (1/η₀)(−a_y)×E_s = −0.5305 e^{j0.1047y} a_x A/m.

A uniform plane wave in free space is propagating in the −a_y direction at a frequency of 5 MHz. If E = 200 cos(ωt + βy) a_z V/m, write the expressions for electric and magnetic fields, i.e., E_s(x, y, z) and H_s(x, y, z) respectively in phasor forms.

Given: free space, wave travels in −a_y, f = 5 MHz, E = 200 cos(ωt + βy) a_z V/m.

Step 1: constants

ω = 2πf = 2π × 5×10⁶ = 3.1416×10⁷ rad/s
β = ω/c = 3.1416×10⁷ / 3×10⁸ = 0.1047 rad/m   (= π/30)
η₀ = 120π = 377 Ω

Step 2: electric field phasor (3 marks)

cos(ωt + βy) = Re{e^{jβy} e^{jωt}}, so

E_s = 200 e^{j0.1047 y} a_z V/m

(it is independent of x and z, so E_s(x, y, z) has only a y-dependence).

Step 3: magnetic field phasor (5 marks)

For a plane wave, H = (1/η) a_k × E, where a_k is the direction of propagation. Here a_k = −a_y:

a_k × E = (−a_y) × a_z = −a_x
H_s = (200/377) e^{jβy} (−a_x)

Check: E × H = a_z × (−a_x) = −a_y, the direction of travel. ✓

H_s = −0.5305 e^{j0.1047 y} a_x   A/m

In real time: H = −0.5305 cos(3.1416×10⁷ t + 0.1047 y) a_x A/m.

Answer: E_s = 200 e^{j0.1047y} a_z V/m; H_s = −0.5305 e^{j0.1047y} a_x A/m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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