A rectangular waveguide operates in the dominant TE10 mode with internal dimensions 10 cm × 15 cm. What is the cutoff frequency (Hz) of this dominant mode?

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Solve this multiple-choice question and choose the correct option.

Accepted Answer

Introduction / Context:For rectangular waveguides, the dominant propagating mode is TE10. The cutoff frequency depends only on the larger inside dimension (the broad wall) and the speed of light. This question tests understanding of the TE10 cutoff formula and correct identification of the broad dimension. Given Data / Assumptions: Waveguide inner dimensions: 10 cm × 15 cm. Dominant mode: TE10. Speed of light c ≈ 3 × 10^8 m/s. Broad dimension a = 15 cm = 0.15 m (by convention for TE10, the index 1 is along the broad wall). Concept / Approach:The cutoff frequency for TE10 is fc = c / (2a). Only the broad wall dimension a appears in TE10 cutoff; the narrow wall b does not enter for this mode. Any operating frequency must exceed fc to support propagation in that mode. Step-by-Step Solution: Identify a = 0.15 m (broad wall), b = 0.10 m (narrow wall).Use TE10 cutoff formula: fc = c / (2a).Compute: fc = (3 × 10^8) / (2 × 0.15) = (3 × 10^8) / 0.3 = 1 × 10^9 Hz.Thus, fc = 1 GHz. Verification / Alternative check: Dimensional sanity check: larger a → lower fc; here a = 0.15 m gives fc lower than a 0.10 m broad wall would, consistent with physics. Why Other Options Are Wrong: 10 GHz, 15 GHz, 25 GHz: These would require a to be millimeters, not centimeters.0.5 GHz: Would require a ≈ 0.3 m, larger than given. Common Pitfalls: Mistaking a and b; using the narrow wall dimension in the TE10 formula; forgetting units conversion from centimeters to meters. Final Answer: 1 GHz

A rectangular waveguide operates in the dominant TE10 mode with internal dimensions 10 cm × 15 cm. What is the cutoff frequency (Hz) of this dominant mode?

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