Final answer:
The dominant mode with the lowest cutoff frequency in a WR-90 rectangular waveguide is the TE10 mode. The cutoff frequency for this mode can be calculated using the given dimensions of the waveguide and is approximately 8.288 GHz.
Step-by-step explanation:
A WR-90 rectangular waveguide has cross-sectional dimensions of 22.86 mm x 10.16 mm. The lowest cutoff frequency (dominant) mode occurs when the TE10 mode is excited, which has the lowest cutoff frequency among all possible modes in this waveguide.
In a rectangular waveguide, the cutoff frequency is inversely proportional to the dimensions of the waveguide. The cutoff frequency for the TE10 mode can be calculated using the formula:
Cutoff frequency = (c / 2π) * sqrt((m/a)^2 + (n/b)^2)
where c is the speed of light, a and b are the width and height of the waveguide, and m and n are the mode numbers. For the TE10 mode, m = 1 and n = 0.
Using the given dimensions of the waveguide, the cutoff frequency for the TE10 mode can be calculated as:
Cutoff frequency = (3e8 / 2π) * sqrt((1/0.02286)^2 + (0/0.01016)^2)
Calculating the above expression gives the cutoff frequency as approximately 8.288 GHz.