---
title: "Plane wave in free space, find β and H"
subject: "Electromagnetism"
chapter: 4
key: "semester_3/electromagnetism:z-polarized-wave-find-beta-h"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/z-polarized-wave-find-beta-h
site: BCT Engineering
---

# Plane wave in free space, find β and H

Electromagnetism · Chapter 4: Wave equation and Wave propagation

- Asked in: 2067 Mangsir, 6 marks
- Video: [Numerical Problems on Electromagnetic Wave Propagation- Part 1](https://www.youtube.com/watch?v=ml__SVxWKMs) — M Classes
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/z-polarized-wave-find-beta-h
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-4

## Question

A z-polarized uniform plane wave propagating in free space has E = 2 cos(10⁷πt − βz) a_x, determine β and H.

## Short answer

β = ω/c = 10⁷π/(3×10⁸) ≈ 0.105 rad/m. H = (E₀/η₀) cos(ωt−βz) a_y = (2/377) cos(10⁷πt−0.105z) a_y ≈ 5.31 cos(10⁷πt−0.105z) a_y mA/m.

## Answer

Given: E = 2 cos(10⁷πt − βz) a_x V/m in free space. The wave moves in +z with E along a_x (the question's "z-polarized" refers to travel along z).

#### Phase constant

```text
ω = 10⁷π rad/s
β = ω/c = 10⁷π / (3×10⁸) = π/30
β ≈ 0.105 rad/m
```

(Frequency f = ω/2π = 5 MHz, λ = 2π/β = 60 m.)

#### Magnetic field

For a uniform plane wave, H = (1/η) a_k × E, and in free space η₀ = 120π ≈ 377 Ω. The direction of travel is a_z:

```text
a_z × a_x = a_y
H = (E₀/η₀) cos(ωt − βz) a_y
  = (2/377) cos(10⁷πt − 0.105z) a_y
  = 5.31 cos(10⁷πt − 0.105z) a_y   mA/m
```

Check: E × H = a_x × a_y = a_z, so power flows along +z, as it should.

**Answer: β = 0.105 rad/m; H = 5.31 cos(10⁷πt − 0.105z) a_y mA/m.**

## Check yourself

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

1. E = 2 cos(10^7 pi t - beta z) a_x in free space. What is beta?
   A. w/c = 10^7 pi/(3x10^8) = pi/30 = 0.105 rad/m
   B. 10^7 pi rad/m
   C. w*c = 9.4x10^15 rad/m
   D. 2 pi f = 3.14x10^7 rad/m
   Answer: A. In free space beta = w/c, giving 0.105 rad/m (lambda = 60 m). (Key point: Phase constant)

2. What is the amplitude of H?
   A. E0*eta0 = 754 A/m
   B. E0/eta0 = 2/120 = 16.7 mA/m
   C. E0 = 2 A/m
   D. E0/eta0 = 2/377 = 5.31 mA/m
   Answer: D. H amplitude is E0 divided by eta0 = 377 ohm. (Key point: Magnetic field amplitude)

3. Why is H along a_y?
   A. Because H is always parallel to E
   B. Because a_x x a_z = a_y
   C. Because a_z x a_x = a_y, from H = (1/eta) a_k x E for travel along +z
   D. Because H must point along the direction of travel
   Answer: C. H = (1/eta) a_k x E with a_k = a_z gives a_z x a_x = a_y; then E x H = a_z, so power flows along +z. (Key point: Direction of H)

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