The cutoff frequency (in GHz) for the dominant T E₁₀ mode of an air-filled rectangular waveguide of inner dimension 0.28 inch × 0.14 inch is ______. (rounded off to two decimal places).

The correct answer is 21.09.

Step-by-step Explanation:

For an air-filled rectangular waveguide with inner dimensions $a$ (width) and $b$ (height), the dominant mode is the TE₁₀ mode (since $a\; >\; b$).

1. Identify the given parameters:
Width of the waveguide, $a\; =\; 0.28$ inches
Height of the waveguide, $b\; =\; 0.14$ inches
Velocity of light in free space (air), $c\; =\; 3\; \backslash times\; 10^8$ m/s

2. Convert the dimension $a$ from inches to meters:
Since 1 inch is equal to $0.0254$ meters, we have:
$$a=0.28\times 0.0254\text{m}=0.007112\text{m}$$

3. Use the formula for the cutoff frequency of the dominant TE₁₀ mode:
The cutoff frequency $f\_c$ for the TE₁₀ mode is given by:
$$f_c=\frac{c}{2a}$$

4. Calculate the cutoff frequency:
Substitute the values into the formula:
$$f_c=\frac{3\times {10}^8}{2\times 0.007112}$$

$$f_c=\frac{3\times {10}^8}{0.014224}$$

$$f_c\approx 2.10911\times {10}^{10}\text{Hz}$$
Converting to gigahertz (GHz), where $1\backslash text\{\; GHz\}\; =\; 10^9\backslash text\{\; Hz\}$:
$$f_c\approx 21.0911\text{GHz}$$
Rounding off to two decimal places yields the cutoff frequency as 21.09 GHz.