Rectangular Waveguide TE10 Mode: Compute phase velocity for an air-filled 8 cm × 10 cm guide at 3 GHz.

Introduction / Context:In microwave engineering, the phase velocity in waveguides exceeds the speed of light in vacuum due to waveguide dispersion. For a rectangular air-filled waveguide, TE10 is the dominant mode, and its cutoff frequency directly influences phase velocity at a given operating frequency.

Given Data / Assumptions:

• Waveguide dimensions: a = 10 cm = 0.10 m, b = 8 cm = 0.08 m (air-filled).
• Mode: TE10 (dominant mode).
• Operating frequency: f = 3 GHz.
• Speed of light: c ≈ 3 × 10^8 m/s.

Concept / Approach:

For a rectangular waveguide, TE10 cutoff frequency is fc = c / (2a). The phase velocity is v_p = c / sqrt(1 − (fc/f)^2) for air (ε_r = 1). When f > fc, the guided wave propagates and v_p ≥ c.

Step-by-Step Solution:

Verification / Alternative check:

The guided wavelength λ_g relates via λ_g = λ / sqrt(1 − (fc/f)^2). Since v_p = f λ_g = (c/f)·f / sqrt(1 − (fc/f)^2) = c / sqrt(1 − (fc/f)^2), consistent with the computed value.

Why Other Options Are Wrong:

• 1.5 × 10^8 m/s and 1.8 × 10^6 m/s: Both are below c and inconsistent with v_p ≥ c in a propagating waveguide mode.
• 5.5 × 10^8 m/s: Too high given the ratio fc/f = 0.5; would require a much smaller denominator.
• 3.00 × 10^8 m/s: Equal to c; occurs only as f ≫ fc, which is not the case at f = 2·fc.

Common Pitfalls:

• Using b (the smaller side) for TE10 cutoff; TE10 depends on a.
• Forgetting that v_p > c in waveguides is permissible because it does not represent signal or energy velocity (group velocity governs energy transport).

Final Answer:

3.46 × 10^8 m/s