---
title: "Cut-off of dominant and TE11, 2.5×1.2 cm"
subject: "Electromagnetism"
chapter: 6
key: "semester_3/electromagnetism:waveguide-2-5x1-2-cutoff-te11"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/waveguide-2-5x1-2-cutoff-te11
site: BCT Engineering
---

# Cut-off of dominant and TE11, 2.5×1.2 cm

Electromagnetism · Chapter 6: Wave guides

- Asked in: 2075 Asoj, 4 marks
- Video: [Important Problem solved in Rectangular Waveguide](https://www.youtube.com/watch?v=kVYDgThPYdM) — Padmasri Naban
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/waveguide-2-5x1-2-cutoff-te11
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-6

## Question

A rectangular waveguide has a cross-section of 2.5 cm × 1.2 cm. Find the cut-off frequencies at dominant mode and TE(1,1).

## Short answer

Using f_c = (c/2)√[(m/a)²+(n/b)²] with a=2.5 cm, b=1.2 cm: dominant TE10 has f_c = c/(2a) = 6 GHz; TE11 has f_c = 13.87 GHz.

## Answer

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

f_c = (c/2) √[(m/a)² + (n/b)²]

#### Dominant mode (TE₁₀)

```text
f_c(10) = c/(2a) = 3×10⁸ / (2 × 0.025) = 6×10⁹ Hz = 6 GHz
```

#### TE₁₁ mode

```text
(1/a)² = (1/0.025)² = 1600
(1/b)² = (1/0.012)² = 6944.4
sum    = 8544.4,   √sum = 92.44 m⁻¹

f_c(11) = 1.5×10⁸ × 92.44 = 13.87×10⁹ Hz
```

**Answer: f_c(TE₁₀) = 6 GHz; f_c(TE₁₁) = 13.87 GHz.**

## Check yourself

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

1. Which mode is dominant for a = 2.5 cm, b = 1.2 cm, and what is its cut-off?
   A. TE01, 12.5 GHz
   B. TE10, 6 GHz
   C. TE10, 12 GHz
   D. TE11, 13.87 GHz
   Answer: B. Since a > b, TE10 is dominant with f_c = c/(2a) = 3×10^8/0.05 = 6 GHz. (Key point: Dominant mode)

2. Which setup correctly gives f_c of TE11?
   A. (c/2)√[(1/a)^2 + (1/b)^2]
   B. (c/2)(1/a + 1/b)
   C. c/(2a) × c/(2b)
   D. (c/2)√[(1/a)^2 − (1/b)^2]
   Answer: A. Cut-off is (c/2) times the root of the sum of squares of m/a and n/b, with m = n = 1. (Key point: TE11 formula)

3. What is f_c(TE11)?
   A. 6.94 GHz
   B. 8.54 GHz
   C. 13.87 GHz
   D. 92.44 GHz
   Answer: C. √(1600 + 6944.4) = 92.44 m^-1; times 1.5×10^8 gives 13.87 GHz. (Key point: TE11 value)

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