---
title: "First four modes, 2.5×1.2 cm guide"
subject: "Electromagnetism"
chapter: 6
key: "semester_3/electromagnetism:first-four-modes-air-copper-waveguide"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/first-four-modes-air-copper-waveguide
site: BCT Engineering
---

# First four modes, 2.5×1.2 cm guide

Electromagnetism · Chapter 6: Wave guides

- Asked in: 2073 Shrawan, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/first-four-modes-air-copper-waveguide
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-6

## Question

Calculate the cut-off frequencies of the first four propagating modes for an air-filled copper waveguide with dimension a = 2.5 cm, b = 1.2 cm.

## Short answer

Apply f_c = (c/2)√[(m/a)²+(n/b)²] with a=2.5 cm, b=1.2 cm. The first four cutoff frequencies are TE10: 6.00 GHz, TE20: 12.00 GHz, TE01: 12.50 GHz, and TE11/TM11: 13.87 GHz.

## Answer

Given: a = 2.5 cm, b = 1.2 cm, air-filled, c = 3×10⁸ m/s.

f_c(mn) = (c/2) √[(m/a)² + (n/b)²] = 1.5×10⁸ √[(m/0.025)² + (n/0.012)²]

```text
TE₁₀:  1.5×10⁸ × 40            = 6.00 GHz
TE₂₀:  1.5×10⁸ × 80            = 12.00 GHz
TE₀₁:  1.5×10⁸ × 83.33         = 12.50 GHz
TE₁₁ / TM₁₁:  1.5×10⁸ × 92.44  = 13.87 GHz
```

(Next: TE₂₁/TM₂₁ = 17.33 GHz.)

| Order | Mode | f_c (GHz) |
|---|---|---|
| 1 | TE₁₀ (dominant) | 6.00 |
| 2 | TE₂₀ | 12.00 |
| 3 | TE₀₁ | 12.50 |
| 4 | TE₁₁ and TM₁₁ (degenerate) | 13.87 |

**Answer: 6 GHz, 12 GHz, 12.5 GHz and 13.87 GHz.** Because the walls are copper (a good conductor), they can be treated as perfect conductors when finding the cut-off frequencies.

## Check yourself

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

1. For a = 2.5 cm, b = 1.2 cm, what is the order of the lowest cut-offs?
   A. TE10, TE01, TE20, TE11/TM11
   B. TE10, TE20, TE11/TM11, TE01
   C. TE10, TE20, TE01, TE11/TM11
   D. TE10, TE11, TE20, TE01
   Answer: C. Values are 6, 12, 12.5 and 13.87 GHz respectively. (Key point: Ordering of modes)

2. What is f_c of TE20?
   A. 12 GHz
   B. 6 GHz
   C. 24 GHz
   D. 13.87 GHz
   Answer: A. f_c = c/a = 3×10^8/0.025 = 12 GHz. (Key point: TE20 cut-off)

3. What is f_c of TE01?
   A. 6 GHz
   B. 25 GHz
   C. 13.87 GHz
   D. 12.5 GHz
   Answer: D. f_c = c/(2b) = 3×10^8/0.024 = 12.5 GHz. (Key point: TE01 cut-off)

4. What is meant by TE11 and TM11 being degenerate?
   A. They have the same cut-off frequency (13.87 GHz)
   B. They have the same field pattern
   C. Neither can propagate
   D. They both have E_z = 0
   Answer: A. TE_mn and TM_mn with the same m, n share the cut-off formula, so they start at the same frequency. (Key point: Degenerate modes)

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