Question Analysis

We are given the cutoff frequency of a dielectric-filled circular waveguide for the dominant TE11 mode. We are also given the dielectric constant of the material. We need to find the radius of the waveguide.

Given Data
- Cutoff frequency for TE11 mode = 6 GHz
- Dielectric constant = 2.56

Solution

Step 1: Calculate the wavelength in the waveguide
The cutoff frequency is the frequency below which the guide will not propagate any energy for the given mode. The cutoff frequency is related to the wavelength as follows:

c = λf

Where:
c = speed of light in free space
λ = wavelength
f = frequency

Rearranging the equation, we get:

λ = c/f

Substituting the values, we have:

λ = (3 x 10^8 m/s) / (6 x 10^9 Hz)
= 0.05 m

Step 2: Calculate the wave number in the waveguide
The wave number is defined as the number of wavelengths in one unit length. It is given by:

k = (2π) / λ

Substituting the value of λ, we get:

k = (2π) / 0.05
= 40π

Step 3: Calculate the cutoff wave number in the waveguide
The cutoff wave number is the wave number at the cutoff frequency. It is given by:

kc = k0 √(εr - 1)

Where:
k0 = free space wave number = 2π/λ
εr = relative permittivity of the dielectric

Substituting the values, we have:

kc = (2π/0.05) √(2.56 - 1)
= 40π √1.56
≈ 98.37

Step 4: Calculate the radius of the waveguide
The relationship between the cutoff wave number and the radius of the waveguide for the dominant TE11 mode is given by:

kc = (π/a) √(εr - 1)

Where:
a = radius of the waveguide

Substituting the values, we have:

98.37 = (π/a) √(2.56 - 1)

Simplifying the equation, we get:

a = π / (98.37 / √(2.56 - 1))
≈ π / (98.37 / √1.56)
≈ π / (98.37 / 1.25)
≈ π / 78.70
≈ 0.0398 m
≈ 3.98 cm

Therefore, the radius of the waveguide is approximately 3.98 cm.

Answer
The radius of the waveguide in cm is approximately 3.98 cm.