Rectangular Waveguide at 9 GHz with a = 2.0 cm: Compute the group velocity for the dominant TE10 mode (air-filled guide).

Introduction / Context:In rectangular waveguides, the dominant mode is TE10 with cutoff frequency fc = c/(2a). Group velocity in a dispersive guide depends on the operating frequency relative to cutoff. This exercise checks your understanding of TE modes and waveguide dispersion relations.

Given Data / Assumptions:

• Air-filled rectangular waveguide.
• Broad wall dimension a = 2 cm = 0.02 m (dominant TE10).
• Frequency f = 9 GHz.
• Speed of light c ≈ 3 × 10^8 m/s.

Concept / Approach:

For TE10 in air: fc = c/(2a). Group velocity: v_g = c * sqrt(1 − (fc/f)^2). v_g is always less than c and increases as f moves far above fc.

Step-by-Step Solution:

Verification / Alternative check:

Since f > fc but not by a large margin, v_g should be well below c; 1.8×10^8 m/s is reasonable and closest to the calculated value ~1.66×10^8 m/s.

Why Other Options Are Wrong:

• 3×10^8 m/s and 5×10^8 m/s: exceed or equal c, impossible for group velocity in a guide near cutoff.
• 1.5×10^8 m/s and 9×10^7 m/s: lower than the computed value; less accurate than 1.8×10^8 m/s.

Common Pitfalls:

• Using phase velocity formula instead of group velocity.
• Taking a as 1 cm instead of 2 cm, which changes fc.

Final Answer:

1.8 × 10^8 m/s