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Radiation pressure of radio wave

Short answer

Radiation pressure on an absorbing surface is S/c = 8.3×10⁻⁸ Pa; on a reflecting surface it is 2S/c = 1.7×10⁻⁷ Pa. Maximum electric field E₀ = √(2S/(ε₀c)) = 137 V/m, and maximum magnetic field B₀ = E₀/c = 4.6×10⁻⁷ T.

A radio wave transmits 25 W/m² of power per unit area. The flat surface area is perpendicular to the direction of propagation of the wave. Calculate the radiation pressure on it and maximum electric and magnetic field associated with the wave.

Given: intensity (average power per unit area) S = 25 W/m², surface perpendicular to the wave. c = 3×10⁸ m/s, ε₀ = 8.85×10⁻¹² F/m, μ₀ = 4π×10⁻⁷ H/m.

Radiation pressure

For a wave completely absorbed by the surface, p = S/c. If it is perfectly reflected, p = 2S/c.

p (absorbed)  = 25 / 3×10⁸ = 8.33×10⁻⁸ Pa
p (reflected) = 2S/c       = 1.67×10⁻⁷ Pa

Maximum electric field

The average intensity is S = E₀²/(2μ₀c) = ½ ε₀ c E₀², so

E₀ = √(2 S / (ε₀ c))
   = √(2 × 25 / (8.85×10⁻¹² × 3×10⁸))
   = √(50 / 2.655×10⁻³)
   = √(1.883×10⁴)
   = 137 V/m

Maximum magnetic field

B₀ = E₀ / c = 137 / 3×10⁸ = 4.58×10⁻⁷ T

Answer: radiation pressure ≈ 8.3×10⁻⁸ N/m² (absorbing surface; 1.7×10⁻⁷ N/m² if reflecting); E₀ ≈ 137 V/m; B₀ ≈ 4.6×10⁻⁷ T

✓ Answer checked by Claude Opus 5.5 (high effort)

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