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An air filled rectangular waveguide of dimension 7 × 3.5 cm2 operates in the dominant TE10 mode. The value of phase velocity of the wave in the guide at a frequency of 3.5 GHz is given by:
1. 3.78 × 108 m/s
2. 6.78 × 105 m/s
3. 5.78 × 108 m/s 
4. 4.78 × 108 m/s

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Correct Answer - Option 1 : 3.78 × 108 m/s

Concept:

Phase velocity is given by:

\({V_P} = \frac{c}{{\cos \theta }}\)

Where:

\(\cos \theta = \sqrt {1 - {{\left( {\frac{{{f_c}}}{f}} \right)}^2}} \)

fc = cut-off frequency

f = frequency of operation

c = speed of light

Calculation:

For the given TE10 mode, the cut-off frequency will be:

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

With a = 7 cm (Given)

\({f_c} = \frac{{3 \times {{10}^{10}}}}{{2 \times 7}} = 2.142\;GHz\)

f = 3.5 GHz

\(\cos \theta = \sqrt {1 - {{\left( {\frac{{2.142}}{{3.5}}} \right)}^2}} = 0.79\)

\({V_p} = \frac{{3 \times {{10}^8}}}{{0.79}}\)

\(= 3.78 \times {10^8}\frac{m}{s}\)

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