To solve this problem, we'll break it down into two parts: first, calculating the cut-off frequency, and then finding the wavelength in the guide for the given frequency of 3 GHz.

Part 1: Cut-off Frequency

The cut-off frequency $f_c$ for a TE$_{mn}$ mode in a circular waveguide is given by the formula:

$f_c = \frac{X_{mn}}{2 \pi a \sqrt{\mu \varepsilon}}$

Where:

• $X_{mn}$ is the m-th root of the Bessel function (given as 1.842 for TE$_{11}$),
• $a$ is the radius of the waveguide,
• $\mu$ is the permeability of the medium (for air, $\mu \approx \mu_0 = 4 \pi \times 10^{-7} \, \text{H/m}$),
• $\varepsilon$ is the permittivity of the medium (for air, $\varepsilon \approx \varepsilon_0 = 8.854 \times 10^{-12} \, \text{F/m}$).

First, let's calculate the radius of the waveguide:

• The diameter is given as 10 cm, so the radius is $a = 5 \, \text{cm} = 0.05 \, \text{m}$.

Now, we can substitute the values into the formula for $f_c$:

$f_c = \frac{1.842}{2 \pi (0.05) \sqrt{(4 \pi \times 10^{-7})(8.854 \times 10^{-12})}}$

Simplifying:

$f_c = \frac{1.842}{2 \pi (0.05) \times 3 \times 10^8} = \frac{1.842 \times 3 \times 10^8}{2 \pi \times 0.05}$

$f_c \approx \frac{1.842 \times 3 \times 10^8}{0.314} \approx 1.76 \, \text{GHz}$

Thus, the cut-off frequency $f_c \approx 1.76 \, \text{GHz}$.

Part 2: Wavelength in the Guide at 3 GHz

The wavelength in the waveguide ($\lambda_g$) is related to the frequency and the cut-off frequency by the following formula:

$\lambda_g = \frac{\lambda_0}{\sqrt{1 - \left(\frac{f_c}{f}\right)^2}}$

Where:

• $\lambda_0 = \frac{c}{f}$ is the free-space wavelength,
• $c = 3 \times 10^8 \, \text{m/s}$ is the speed of light,
• $f$ is the operating frequency (given as 3 GHz).

First, calculate the free-space wavelength $\lambda_0$:

$\lambda_0 = \frac{3 \times 10^8}{3 \times 10^9} = 0.1 \, \text{m}$

Now, substitute the values into the formula for $\lambda_g$:

$\lambda_g = \frac{0.1}{\sqrt{1 - \left(\frac{1.76}{3}\right)^2}} = \frac{0.1}{\sqrt{1 - 0.3456}} = \frac{0.1}{\sqrt{0.6544}}$

$\lambda_g \approx \frac{0.1}{0.8089} \approx 0.1236 \, \text{m}$

Thus, the wavelength in the guide $\lambda_g \approx 0.1236 \, \text{m} = 12.36 \, \text{cm}$.

Final Answers:

• The cut-off frequency is approximately 1.76 GHz.
• The wavelength in the guide at 3 GHz is approximately 12.36 cm.

Would you like further details on any step or clarification?

Here are 5 related questions:

1. What happens if the operating frequency is below the cut-off frequency?
2. How does the mode (TE$_{mn}$) affect the cut-off frequency in different waveguides?
3. What is the significance of the Bessel function root $X_{mn}$ in waveguide analysis?
4. How would the cut-off frequency change if the waveguide were filled with a dielectric material?
5. What is the group velocity of the wave in the waveguide at 3 GHz?

Tip: Always check whether the operating frequency is above the cut-off frequency to ensure wave propagation in a waveguide.