---
title: "Radiation pressure of radio wave"
subject: "Engineering Physics"
chapter: 9
key: "semester_1/engineering_physics:radiation-pressure-radio-wave"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/radiation-pressure-radio-wave
site: BCT Engineering
---

# Radiation pressure of radio wave

Engineering Physics · Chapter 9: Electromagnetic Waves

- Asked in: 2074 Chaitra, 5 marks
- Video: [Radiation Pressure | Physics with Professor Matt Anderson | M25-17](https://www.youtube.com/watch?v=T7aLj3hxVxs) — Physics with Professor Matt Anderson
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/radiation-pressure-radio-wave
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-9

## Question

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.

## 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.

## Answer

**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.

```text
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

```text
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
```text
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**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For a wave of intensity S completely absorbed by a perpendicular surface, the radiation pressure is:
   A. p = S/c
   B. p = 2S/c
   C. p = Sc
   D. p = S/c²
   Answer: A. Absorbed radiation transfers momentum S/c per unit time per area. (Key point: Pressure for absorbing surface)

2. Compared with an absorbing surface, a perfectly reflecting surface experiences radiation pressure that is:
   A. Half as large
   B. Twice as large, 2S/c
   C. The same
   D. Zero
   Answer: B. Reflection reverses the momentum, giving a change of twice the momentum. (Key point: Reflecting surface)

3. Which relation gives E₀ from the average intensity S?
   A. S = ε₀ c E₀, so E₀ = S/(ε₀c)
   B. S = E₀²/c
   C. S = ½ ε₀ c E₀², so E₀ = √(2S/(ε₀c))
   D. S = ½ ε₀ E₀², so E₀ = √(2S/ε₀)
   Answer: C. The time-averaged Poynting vector is ½ ε₀ c E₀². (Key point: Intensity-field relation)

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