---
title: "Cut-off of TM11 and dominant mode"
subject: "Electromagnetism"
chapter: 6
key: "semester_3/electromagnetism:waveguide-tm11-cutoff-dielectric-1-07"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/waveguide-tm11-cutoff-dielectric-1-07
site: BCT Engineering
---

# Cut-off of TM11 and dominant mode

Electromagnetism · Chapter 6: Wave guides

- Asked in: 2068 Chaitra, 2+2 marks
- Video: [Rectangular Waveguide Problem 2 - Guided Waves - Electromagnetic Field and Wave Theory](https://www.youtube.com/watch?v=mH4Ne_iAdC4) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/waveguide-tm11-cutoff-dielectric-1-07
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-6

## Question

Consider a rectangular waveguide with ε_r = 2, μ = μ0 with dimensions a = 1.07 cm, b = 0.43 cm. Find the cut-off frequency for TM11 mode and the dominant mode.

## Short answer

With ε_r=2, speed u = c/√2. Dominant TE10 cutoff f_c = u/(2a) = 9.91 GHz. TM11 cutoff using (u/2)√[(1/a)²+(1/b)²] yields 26.58 GHz.

## Answer

Given: ε_r = 2, μ = μ₀, a = 1.07 cm, b = 0.43 cm.

Speed in the filling: u = c/√ε_r = 3×10⁸/√2 = 2.121×10⁸ m/s.

f_c = (u/2) √[(m/a)² + (n/b)²], with u/2 = 1.0607×10⁸.

#### TM₁₁ mode

```text
(1/a)² = (1/0.0107)² = 8734.4
(1/b)² = (1/0.0043)² = 54083.8
sum    = 62818.2,  √ = 250.64 m⁻¹

f_c(TM₁₁) = 1.0607×10⁸ × 250.64 = 26.58×10⁹ Hz
```

#### Dominant mode (TE₁₀)

```text
f_c(TE₁₀) = u/(2a) = 2.121×10⁸ / (2 × 0.0107) = 9.91×10⁹ Hz
```

**Answer: f_c(TM₁₁) = 26.58 GHz; dominant mode TE₁₀ with f_c = 9.91 GHz.**

## Check yourself

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

1. What speed is used in the cut-off formula for ε_r = 2, μ = μ0?
   A. 3×10^8/2 = 1.5×10^8 m/s
   B. 3×10^8×√2 = 4.24×10^8 m/s
   C. 3×10^8/√2 = 2.121×10^8 m/s
   D. 3×10^8 m/s
   Answer: C. u = c/√ε_r = 2.121×10^8 m/s. (Key point: Speed in dielectric)

2. What is the dominant-mode cut-off?
   A. 26.58 GHz
   B. 9.91 GHz
   C. 19.8 GHz
   D. 14.0 GHz
   Answer: B. f_c = u/(2a) = 2.121×10^8/(2×0.0107) = 9.91 GHz. (Key point: Dominant mode cut-off)

3. What is f_c of TM11?
   A. 9.91 GHz
   B. 13.3 GHz
   C. 53.2 GHz
   D. 26.58 GHz
   Answer: D. (u/2)√[(1/a)^2 + (1/b)^2] = 1.0607×10^8 × 250.64 = 26.58 GHz. (Key point: TM11 cut-off)

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