---
title: "Poynting numerical, wave in free space"
subject: "Electromagnetism"
chapter: 4
key: "semester_3/electromagnetism:poynting-numerical-em-wave-free-space"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/poynting-numerical-em-wave-free-space
site: BCT Engineering
---

# Poynting numerical, wave in free space

Electromagnetism · Chapter 4: Wave equation and Wave propagation

- Asked in: 2069 Chaitra, 2+2+2 marks
- Video: [Problem 2 | Poynting Vector | Electromagnetic Waves | Field theory | Shiva Panchakshari T G](https://www.youtube.com/watch?v=5DBaBa2p1oQ) — Prof. Shiva panchakshari T G
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/poynting-numerical-em-wave-free-space
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-4

## Question

An EM wave travels in free space with the electric field component E = (10a_y + 5a_z) cos(ωt + 2y − 4z) V/m. Find (a) ω and λ (b) the magnetic field component (c) the time average power in the wave.

## Short answer

Angular frequency ω = β c = 1.34×10⁹ rad/s, wavelength λ = 2π/β = 1.405 m. Magnetic field H = (1/η₀) a_k × E = −0.0297 cos(ωt + 2y − 4z) a_x A/m. Time-average power density is 0.166 W/m² directed along a_k.

## Answer

Given: free space, E = (10 a_y + 5 a_z) cos(ωt + 2y − 4z) V/m.

The phase is ωt + k·r with k = 2 a_y − 4 a_z, so the wave travels along −k, i.e. along a_k = (−2 a_y + 4 a_z)/√20. Check that E ⟂ k: (10)(2) + (5)(−4) = 0 ✓.

**(a) ω and λ**

```text
β = |k| = √(2² + 4²) = √20 = 4.472 rad/m
λ = 2π/β = 6.2832/4.472 = 1.405 m
ω = β c = 4.472 × 3×10⁸ = 1.342×10⁹ rad/s     (f = 213.5 MHz)
```

**(b) Magnetic field.** H = (1/η₀) a_k × E, with η₀ = 377 Ω and a_k = −0.4472 a_y + 0.8944 a_z.

```text
a_k × (10 a_y + 5 a_z)
 a_x: (−0.4472)(5) − (0.8944)(10) = −2.236 − 8.944 = −11.18
 a_y: 0,  a_z: 0
⇒ a_k × E = −11.18 a_x

H = (−11.18/377) cos(ωt + 2y − 4z) a_x
  = −29.66 cos(1.342×10⁹ t + 2y − 4z) a_x   mA/m
```

**(c) Time-average power.** The magnitude |E| = √(10² + 5²) = √125 = 11.18 V/m.

```text
P_avg = |E|²/(2η₀) = 125 / (2 × 377) = 0.1658 W/m²
```

Direction: along a_k (the direction of propagation), i.e. P_avg = 0.166 (−0.4472 a_y + 0.8944 a_z) = −0.0741 a_y + 0.1483 a_z W/m².

**Answer:** ω = 1.34×10⁹ rad/s, λ = 1.405 m; H = −29.7 cos(ωt + 2y − 4z) a_x mA/m; P_avg ≈ 0.166 W/m² (166 mW/m²).

## Check yourself

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

1. E = (10a_y + 5a_z) cos(wt + 2y - 4z). How is w found in free space?
   A. w = (2 + 4)*c = 1.8x10^9 rad/s
   B. w = 2*c = 6x10^8 rad/s
   C. beta = |k| = sqrt(2^2 + 4^2) = 4.472 rad/m, so w = beta*c = 1.34x10^9 rad/s
   D. w = 4.472 rad/s
   Answer: C. The phase constant is the magnitude of the vector k = 2a_y - 4a_z, and w = beta*c in free space. (Key point: Angular frequency)

2. Phase is (wt + 2y - 4z). Along which direction does the wave travel?
   A. (2a_y - 4a_z)/sqrt(20), along k
   B. a_k = (-2a_y + 4a_z)/sqrt(20), opposite to k
   C. a_x
   D. (10a_y + 5a_z)/sqrt(125), along E
   Answer: B. With wt + k.r the wave moves along -k. The check E.k = 20 - 20 = 0 confirms E is perpendicular to it. (Key point: Direction of travel)

3. What is the time-average power density?
   A. |E|^2/eta0 = 0.33 W/m^2
   B. |E|/(2 eta0) = 0.0148 W/m^2
   C. 10^2/(2 eta0) = 0.133 W/m^2 (y-component only)
   D. |E|^2/(2 eta0) = 125/754 = 0.166 W/m^2
   Answer: D. Use the full magnitude |E|^2 = 10^2 + 5^2 = 125 and the factor 1/2 for the time average. (Key point: Time-average power)

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