---
title: "Energy transfer rate of em wave"
subject: "Engineering Physics"
chapter: 9
key: "semester_1/engineering_physics:em-wave-energy-transfer-intensity"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/em-wave-energy-transfer-intensity
site: BCT Engineering
---

# Energy transfer rate of em wave

Engineering Physics · Chapter 9: Electromagnetic Waves

- Asked in: 2068 Chaitra, 5 marks
- Video: [Electromagnetic Energy | Physics with Professor Matt Anderson | M25-06](https://www.youtube.com/watch?v=OKRCoKEZUvY) — 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/em-wave-energy-transfer-intensity
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-9

## Question

Obtain an expression for energy transfer rate by electromagnetic wave. From your result show that I ∝ E²rms, where I is the intensity of the em wave and Erms is the root mean square value of electric field.

## Short answer

The Poynting vector S = (E×B)/μ₀ gives instantaneous energy flux; for a plane wave S = E²/(μ₀c). Time-averaging yields intensity I = E₀²/(2μ₀c) = E_rms²/(μ₀c), so I ∝ E_rms².

## Answer

#### Energy transfer rate (Poynting vector)
Energy flows with the wave. The rate of energy transfer per unit area is the Poynting vector

S = (E × B)/μ₀

For a plane wave E and B are perpendicular, so the magnitude is S = EB/μ₀.

1. In an EM wave E/B = c, so B = E/c.
2. Substitute: **S = E²/(μ₀ c)** (instantaneous energy flow per unit area, W/m²).
3. For E = E₀ sin(kx − ωt), the instantaneous S = (E₀²/μ₀c) sin²(kx − ωt).
4. The time average of sin² over a cycle is ½. So the **intensity** (average S) is

   I = S_avg = E₀²/(2μ₀c)
5. The rms value of E is E_rms = E₀/√2, so E₀² = 2E_rms². Therefore

**I = E_rms² / (μ₀ c)**

Since μ₀ and c are constants, **I ∝ E_rms²**.

Equivalent forms: I = ½ε₀cE₀² = ε₀ c E_rms², and I = c × (average energy density), since u = ε₀E² for the wave with u_E = u_B.

Example: for I = 1 kW/m², E_rms = √(μ₀cI) = √(4π×10⁻⁷ × 3×10⁸ × 1000) ≈ 614 V/m.

## Check yourself

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

1. The rate of energy flow per unit area of an EM wave is given by:
   A. S = μ₀ E × B
   B. S = (E × B)/μ₀
   C. S = E·B
   D. S = (E × B)ε₀
   Answer: B. The Poynting vector is E × B divided by μ₀. (Key point: Poynting vector)

2. Why does the intensity have a factor 1/2 relative to E₀²/(μ₀c)?
   A. Half the energy is reflected
   B. B is half of E
   C. The time average of sin² over a cycle is 1/2
   D. The wave travels at c/2
   Answer: C. S ∝ sin²(kx − ωt), whose average is 1/2. (Key point: Averaging sin²)

3. Using E_rms = E₀/√2, the intensity becomes:
   A. I = E_rms/(μ₀c), so I ∝ E_rms
   B. I = 2E_rms/(μ₀c)
   C. I = E_rms²μ₀c
   D. I = E_rms²/(μ₀c), so I ∝ E_rms²
   Answer: D. E₀² = 2E_rms² turns E₀²/(2μ₀c) into E_rms²/(μ₀c). (Key point: I in terms of Erms)

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