Question

The dominant mode in a rectangular wave guide is TE₁₀, because this mode has
1. No attenuation
2. No cut off
3. No magnetic field component
4. The highest cut-off wavelength

Answer 1

Correct Answer - Option 4 : The highest cut-off wavelength

Concept:

The dominant mode in a particular waveguide is the mode having the lowest cut-off frequency.

So the dominant mode in a particular waveguide is the mode having the highest cut-off wavelength.

The cut-off frequency for a rectangular waveguide with dimension ‘a (length)’ and ‘b (width)’ is given as:

\({{f}_{c\left( mn \right)}}=\frac{c}{2}\sqrt{{{\left( \frac{m}{a} \right)}^{2}}+{{\left( \frac{n}{b} \right)}^{2}}}\)

'm' and 'n' represents the possible modes.

c = speed of light = 3 × 1010 cm/s

Calculation:

TE10 mode means m = 1, n = 0

The cut - off frequency of the dominant mode TE10 of the rectangular waveguide is:

\({f_c} = \frac{c}{{2a}}\)

λ_c = 2a 

Where a is the dimension of the inner broad wall.