---
title: "Why no TEM mode; first four modes"
subject: "Electromagnetism"
chapter: 6
key: "semester_3/electromagnetism:waveguide-no-tem-four-modes"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/waveguide-no-tem-four-modes
site: BCT Engineering
---

# Why no TEM mode; first four modes

Electromagnetism · Chapter 6: Wave guides

- Asked in: 2072 Chaitra, 2+4 marks
- Video: [Why TEM Mode does not exist in Rectangular Waveguide? | E and H Field equations of Waveguide](https://www.youtube.com/watch?v=uDGKJ19gxBM) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/waveguide-no-tem-four-modes
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-6

## Question

Why does a hollow rectangular waveguide not support TEM mode? A rectangular air-filled waveguide has a cross-section of 45×90 mm. Find the cut-off frequencies of the first four propagating modes.

## Short answer

A hollow waveguide lacks a second conductor to carry return current for transverse fields, so TEM cannot exist. For 45×90 mm, f_c values: TE10 = 1.667 GHz, TE20 and TE01 both 3.333 GHz, TE11/TM11 = 3.727 GHz.

## Answer

#### Why a hollow waveguide cannot support TEM

A TEM wave has E_z = H_z = 0, so both E and H lie in the transverse plane and the field is like a plane wave. Inside a hollow, single-conductor guide:

- The transverse H lines must form closed loops in a transverse plane (∇·B = 0). Maxwell's equations then require a longitudinal conduction or displacement current inside the guide for them to circulate around.
- There is no centre conductor to carry axial conduction current, and a TEM wave would have no E_z for displacement current either.
- Equivalently, a TEM field would need a transverse electrostatic potential between two conductors, but a single closed conductor is at one potential, so the transverse E would be zero.

So only TE or TM modes exist (H_z or E_z ≠ 0), and each has a cut-off frequency.

#### Cut-off frequencies (45 × 90 mm)

Take a = 90 mm = 0.09 m (the larger side), b = 45 mm = 0.045 m.

```text
f_c = 1.5×10⁸ √[(m/0.09)² + (n/0.045)²]

TE₁₀:  1.5×10⁸ × 11.11 = 1.667 GHz
TE₂₀:  1.5×10⁸ × 22.22 = 3.333 GHz
TE₀₁:  1.5×10⁸ × 22.22 = 3.333 GHz
TE₁₁/TM₁₁: 1.5×10⁸ × √(123.5 + 493.8) = 1.5×10⁸ × 24.85 = 3.727 GHz
```

| Mode | f_c (GHz) |
|---|---|
| TE₁₀ | 1.667 |
| TE₂₀ | 3.333 |
| TE₀₁ | 3.333 (equal to TE₂₀ as a = 2b) |
| TE₁₁ and TM₁₁ | 3.727 |

**Answer: 1.667 GHz, 3.333 GHz (TE₂₀ and TE₀₁), and 3.727 GHz (TE₁₁/TM₁₁).**

## Check yourself

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

1. Why can a hollow single-conductor waveguide not support a TEM mode?
   A. Because the walls are lossy
   B. Because the guide is too small
   C. Because TEM needs a frequency below cut-off
   D. Because H loops would need an axial current, and a single conductor has no potential difference to give transverse E
   Answer: D. Transverse H must circulate around a longitudinal current, which is absent, and a single conductor at one potential gives no transverse E. (Key point: No TEM in hollow guide)

2. For the 45 × 90 mm guide, which is taken as a, and what is the TE10 cut-off?
   A. a = 90 mm, 1.667 GHz
   B. a = 45 mm, 3.333 GHz
   C. a = 90 mm, 3.333 GHz
   D. a = 45 mm, 1.667 GHz
   Answer: A. a is the larger side, so f_c = c/(2×0.09) = 1.667 GHz. (Key point: Choosing a and b)

3. Why are TE20 and TE01 both 3.333 GHz here?
   A. Both are TM modes
   B. a = 2b
   C. a = b
   D. Both have n = 0
   Answer: B. c/a and c/(2b) are equal when a = 2b. (Key point: Equal cut-offs)

4. Which mode(s) give the fourth cut-off and its value?
   A. TE02 at 6.67 GHz
   B. TE21 at 4.71 GHz
   C. TE11/TM11 at 3.727 GHz
   D. TM11 only at 1.667 GHz
   Answer: C. 1.5×10^8 × √(123.5 + 493.8) = 3.727 GHz for TE11 and TM11. (Key point: Fourth mode)

## Related in Electromagnetism

- [First four modes, 2.5×1.2 cm guide](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Afirst-four-modes-air-copper-waveguide) — Electromagnetism, chapter 6
- [Dominant cut-off, dielectric-filled guide](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Awaveguide-dielectric-filled-dominant-cutoff) — Electromagnetism, chapter 6
- [Single-mode range, air-filled 2×1 cm](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Awaveguide-2x1cm-single-mode-range) — Electromagnetism, chapter 6
- [Cut-off and single-mode range, 4×2 cm](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Awaveguide-4x2cm-cutoff-single-mode-range) — Electromagnetism, chapter 6
- [Cut-off of TM11 and dominant mode](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Awaveguide-tm11-cutoff-dielectric-1-07) — Electromagnetism, chapter 6
- [Cut-off of dominant and TE11, 2.5×1.2 cm](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Awaveguide-2-5x1-2-cutoff-te11) — Electromagnetism, chapter 6

## Related in other subjects

- [Air-filled 3 cm waveguide at 6 GHz; coupler parameters](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Awaveguide-3cm-6ghz-directional-coupler) — RF and Microwave Engineering, chapter 4
- [Rectangular waveguide cut-off and phase velocity](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Arectangular-waveguide-assumed-dimensions) — RF and Microwave Engineering, chapter 4
- [Rectangular waveguide modes and critical parameters](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Arectangular-waveguide-modes-parameters) — RF and Microwave Engineering, chapter 4
- [Rectangular waveguide TM mode field equations](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Arectangular-waveguide-tm-equations) — RF and Microwave Engineering, chapter 4, asked 11×