---
title: "Circular waveguide TM field strength and dominant mode"
subject: "RF and Microwave Engineering"
chapter: 4
key: "semester_7/rf_and_microwave_engineering:circular-waveguide-tm-field"
source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/circular-waveguide-tm-field
site: BCT Engineering
---

# Circular waveguide TM field strength and dominant mode

RF and Microwave Engineering · Chapter 4: RF and Microwave Components and Devices

- Asked in: 2075 Bhadra, 8+2 marks
- Video: [Circular waveguide TRANSVERSE MAGENTIC Mode](https://www.youtube.com/watch?v=CWjQWZbK8PM) — Dream Life
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/circular-waveguide-tm-field
- Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-4

## Question

Derive the expression for the field strength for TM waves for an air-filled circular waveguide. Check the dominant mode in TE and TM modes.

## Short answer

TM field components are proportional to Jn(kcρ) with kc = pnm/a from Jn(kc a)=0; transverse fields follow from derivatives of Ez. The overall dominant mode is TE11 (cutoff from Jn' root 1.841, λc=3.41a); the lowest TM mode is TM01 (p01=2.405, λc=2.61a).

## Answer

#### TM field expressions for an air-filled circular waveguide
For a circular guide of radius *a* (coordinates ρ, φ, z), a **TM mode** has Hz = 0 and Ez ≠ 0, with fields varying as e^(j(ωt − βz)).

```text
        .-------.
      /     ρ     \
     |   o----->   |   radius a, air (μ0, ε0)
      \           /    propagation along z
        '-------'
```

**Step 1 – wave equation for Ez:**
```text
 ∂²Ez/∂ρ² + (1/ρ)∂Ez/∂ρ + (1/ρ²)∂²Ez/∂φ² + kc² Ez = 0
 kc² = k² - β²,  k = ω√(μ0ε0)
```
**Step 2 – separation of variables:** Ez = R(ρ)·Φ(φ)·e^(−jβz) gives
```text
 Φ'' + n²Φ = 0          ->  Φ = A sin nφ + B cos nφ
 ρ²R'' + ρR' + (kc²ρ² - n²)R = 0   (Bessel's equation)
 R = C·Jn(kcρ) + D·Yn(kcρ)
```
Yn is infinite at ρ = 0, so D = 0:
```text
 Ez = (A sin nφ + B cos nφ) Jn(kcρ) e^(-jβz)
```
**Step 3 – boundary condition:** tangential Ez = 0 at the wall ρ = a:
```text
 Jn(kc a) = 0   ->   kc = pnm / a
```
where pnm is the m-th zero of Jn(x).

**Step 4 – transverse fields** from Maxwell's equations (with Hz = 0):
```text
 Eρ = -(jβ/kc²) ∂Ez/∂ρ        Hρ =  (jωε/(kc²ρ)) ∂Ez/∂φ
 Eφ = -(jβ/(kc²ρ)) ∂Ez/∂φ     Hφ = -(jωε/kc²) ∂Ez/∂ρ
```
giving the **TMnm field strengths**:
```text
 Ez = (A sin nφ + B cos nφ) Jn(kcρ) e^(-jβz)
 Eρ = -(jβ/kc) (A sin nφ + B cos nφ) J'n(kcρ) e^(-jβz)
 Eφ = -(jβn/(kc²ρ)) (A cos nφ - B sin nφ) Jn(kcρ) e^(-jβz)
 Hρ =  (jωε0 n/(kc²ρ)) (A cos nφ - B sin nφ) Jn(kcρ) e^(-jβz)
 Hφ = -(jωε0/kc) (A sin nφ + B cos nφ) J'n(kcρ) e^(-jβz)
 Hz = 0
```
**Characteristic relations (air):**
```text
 fc = pnm·c/(2πa),   λc = 2πa/pnm
 β = √(k² - (pnm/a)²)
 Z_TM = Eρ/Hφ = β/(ωε0) = η0 √(1 - (fc/f)²)
```

#### Dominant modes in TE and TM
- **TM:** the smallest zero of Jn is p01 = 2.405, so **TM01** is the lowest TM mode (fc = 2.405c/(2πa), λc = 2.61a).
- **TE:** the smallest zero of J'n is p'11 = 1.841, so **TE11** is the lowest TE mode (λc = 3.41a).
- Since 1.841 < 2.405, **TE11 is the dominant mode of the circular waveguide**; TM01 is the second mode. Example: a = 1 cm gives fc(TE11) = 8.79 GHz and fc(TM01) = 11.48 GHz.

## Check yourself

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

1. In solving for Ez in a circular guide, why is the Yn(kcρ) term dropped (D = 0)?
   A. Yn is zero at ρ = a
   B. Yn is infinite at ρ = 0, but the field must be finite on the axis
   C. Yn does not satisfy Bessel's equation
   D. Yn only applies to TE modes
   Answer: B. Yn diverges at ρ = 0, and the field at the centre must be finite, so only Jn remains. (Key point: Bessel solution choice)

2. What boundary condition gives the cut-off wave number of TM modes in a circular guide?
   A. J'n(kc a) = 0
   B. Hz = 0 at the axis
   C. Jn(kc a) = 0, since tangential Ez = 0 at ρ = a
   D. Eρ = 0 at the axis
   Answer: C. Ez is tangential to the wall and must vanish there, so Jn(kc a) = 0 and kc = pnm/a. (Key point: TM boundary condition)

3. For an air-filled circular guide of radius a, the cut-off frequency of TMnm is:
   A. fc = pnm c/(2πa), pnm being the m-th zero of Jn
   B. fc = p'nm c/(2πa), p'nm being the m-th zero of J'n
   C. fc = c/(2a)
   D. fc = pnm c/a
   Answer: A. kc = pnm/a and fc = kc c/(2π) = pnm c/(2πa). (Key point: Cut-off frequency of TM modes)

4. Why is TE11 the dominant mode of the circular guide and not TM01?
   A. p'11 = 1.841 is smaller than p01 = 2.405
   B. TM modes have Hz = 0
   C. TM01 has no Ez component
   D. TE11 has a larger radius
   Answer: A. The lowest TE root 1.841 is below the lowest TM root 2.405, so TE11 has the lowest cut-off; TM01 is the second mode. (Key point: Dominant mode)

## Related in RF and Microwave Engineering

- [TE vs TM mode in circular waveguide](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Acircular-waveguide-te-vs-tm) — RF and Microwave Engineering, chapter 4
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