---
title: "Dominant cut-off, dielectric-filled guide"
subject: "Electromagnetism"
chapter: 6
key: "semester_3/electromagnetism:waveguide-dielectric-filled-dominant-cutoff"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/waveguide-dielectric-filled-dominant-cutoff
site: BCT Engineering
---

# Dominant cut-off, dielectric-filled guide

Electromagnetism · Chapter 6: Wave guides

- Asked in: 2074 Asoj, 2 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/waveguide-dielectric-filled-dominant-cutoff
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-6

## Question

A rectangular waveguide with cross-section 2 cm × 1 cm is filled with a lossless medium characterized by ε = 4ε0 and μ_r = 1. Calculate the cut-off frequency of the dominant mode.

## Short answer

With ε_r=4, wave speed u = c/√ε_r = 1.5×10⁸ m/s. Dominant TE10 cutoff f_c = u/(2a) = 1.5×10⁸/(2×0.02) = 3.75 GHz.

## Answer

Given: a = 2 cm, b = 1 cm, filled with ε = 4ε₀, μ_r = 1.

The wave speed in the filling is

```text
u = 1/√(με) = c/√ε_r = 3×10⁸/√4 = 1.5×10⁸ m/s
```

The dominant mode is TE₁₀ (a > b):

```text
f_c = u/(2a) = 1.5×10⁸ / (2 × 0.02) = 3.75×10⁹ Hz
```

**Answer: f_c = 3.75 GHz** (half the 7.5 GHz of the air-filled guide, since √ε_r = 2).

## Check yourself

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

1. For ε = 4ε0, μ_r = 1, what replaces c in the cut-off formula?
   A. u = 3×10^8 × 2 = 6×10^8 m/s
   B. u = c/ε_r = 0.75×10^8 m/s
   C. u = c/√ε_r = 1.5×10^8 m/s
   D. u = c
   Answer: C. Speed in the medium is 1/√(με) = c/√ε_r = 1.5×10^8 m/s. (Key point: Wave speed in filling)

2. Which mode is dominant for a = 2 cm, b = 1 cm?
   A. TE01
   B. TM11
   C. TE20
   D. TE10
   Answer: D. For a > b the lowest cut-off belongs to TE10. (Key point: Dominant mode choice)

3. What is the dominant-mode cut-off?
   A. 3.75 GHz
   B. 7.5 GHz
   C. 1.875 GHz
   D. 15 GHz
   Answer: A. f_c = u/(2a) = 1.5×10^8/0.04 = 3.75 GHz, half the air value because √ε_r = 2. (Key point: Cut-off result)

## Related in Electromagnetism

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