---
title: "Phasor E and H of wave in −y direction"
subject: "Electromagnetism"
chapter: 4
key: "semester_3/electromagnetism:plane-wave-minus-y-phasor-expressions"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/plane-wave-minus-y-phasor-expressions
site: BCT Engineering
---

# Phasor E and H of wave in −y direction

Electromagnetism · Chapter 4: Wave equation and Wave propagation

- Asked in: 2075 Chaitra, 3+5 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/plane-wave-minus-y-phasor-expressions
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-4

## Question

A uniform plane wave in free space is propagating in the −a_y direction at a frequency of 5 MHz. If E = 200 cos(ωt + βy) a_z V/m, write the expressions for electric and magnetic fields, i.e., E_s(x, y, z) and H_s(x, y, z) respectively in phasor forms.

## Short answer

In free space, ω=10⁷π rad/s, β=ω/c=0.1047 rad/m. Phasor electric field: E_s = 200 e^{j0.1047y} a_z V/m. Propagation direction −a_y gives magnetic field: H_s = (1/η₀)(−a_y)×E_s = −0.5305 e^{j0.1047y} a_x A/m.

## Answer

Given: free space, wave travels in −a_y, f = 5 MHz, E = 200 cos(ωt + βy) a_z V/m.

#### Step 1: constants
```text
ω = 2πf = 2π × 5×10⁶ = 3.1416×10⁷ rad/s
β = ω/c = 3.1416×10⁷ / 3×10⁸ = 0.1047 rad/m   (= π/30)
η₀ = 120π = 377 Ω
```

#### Step 2: electric field phasor (3 marks)
cos(ωt + βy) = Re{e^{jβy} e^{jωt}}, so

**E_s = 200 e^{j0.1047 y} a_z  V/m**

(it is independent of x and z, so E_s(x, y, z) has only a y-dependence).

#### Step 3: magnetic field phasor (5 marks)
For a plane wave, H = (1/η) a_k × E, where a_k is the direction of propagation. Here a_k = −a_y:

```text
a_k × E = (−a_y) × a_z = −a_x
H_s = (200/377) e^{jβy} (−a_x)
```

Check: E × H = a_z × (−a_x) = −a_y, the direction of travel. ✓

```text
H_s = −0.5305 e^{j0.1047 y} a_x   A/m
```

In real time: H = −0.5305 cos(3.1416×10⁷ t + 0.1047 y) a_x A/m.

**Answer:** E_s = 200 e^{j0.1047y} a_z V/m; H_s = −0.5305 e^{j0.1047y} a_x A/m.

## Check yourself

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

1. For f = 5 MHz in free space, beta is:
   A. 0.1047 rad/m
   B. 1.047 rad/m
   C. 0.0105 rad/m
   D. 31.4 rad/m
   Answer: A. beta = w/c = 3.1416 x 10^7/3 x 10^8 = 0.1047 rad/m. (Key point: Phase constant)

2. For E = 200 cos(wt + beta y) a_z, the phasor E_s is:
   A. 200 e^(-j0.1047 y) a_z
   B. 200 e^(+j0.1047 y) a_z
   C. 200 e^(j0.1047 z) a_y
   D. 200 a_z
   Answer: B. cos(wt + beta y) = Re{e^(j beta y) e^(jwt)}, so E_s = 200 e^(j beta y) a_z. (Key point: E phasor)

3. For a wave travelling in -a_y with E along a_z, H is along:
   A. +a_x
   B. -a_z
   C. +a_y
   D. -a_x
   Answer: D. H = (1/eta) a_k x E = (1/eta)(-a_y) x a_z = -a_x/eta, and E x H = -a_y agrees with the travel direction. (Key point: Direction of H)

4. The amplitude of H_s is:
   A. 200 x 377 A/m
   B. 0.53 A/m (200/377)
   C. 1.89 A/m
   D. 0.1047 A/m
   Answer: B. |H| = E/eta0 = 200/377 = 0.5305 A/m, so H_s = -0.5305 e^(j0.1047 y) a_x. (Key point: H magnitude)

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