Rectangular waveguide cutoff condition — TE20 mode Let a be the broad-wall width and b be the narrow-wall height of a rectangular waveguide. What is the cutoff wavelength λ_c for the TE20 mode?

Introduction / Context:
Each TEmn mode in a rectangular waveguide has a specific cutoff wavelength and frequency determined by the cross-sectional dimensions a (broad wall) and b (narrow wall). Recognizing these standard results is essential for band selection and single-mode operation.

Given Data / Assumptions:

• Metal rectangular waveguide with dimensions a and b.
• Mode: TE20 (m = 2, n = 0).
• Lossless ideal walls for cutoff calculation.

Concept / Approach:
The general formula is λ_c = 2 / sqrt((m/a)^2 + (n/b)^2). For TE m0, this simplifies to λ_c = 2a/m. Substituting m = 2 gives λ_c = a.

Step-by-Step Solution:

Verification / Alternative check:
Special cases: λ_c(TE10) = 2a and λ_c(TE01) = 2b; by the same pattern, TE20 → a. Handbooks and mode charts confirm this relationship.

Why Other Options Are Wrong:

• a + b or a − b: no linear combination arises from the correct formula.
• a2 + b2: does not match the inverse-root dependence on dimensions.

Common Pitfalls:
Confusing TE10 dominance with TE20 cutoff; forgetting dependence on the m index across the broad wall.

Final Answer:
a