Assertion (A): In a rectangular waveguide operating in TE10 mode, the cutoff frequency is f_c = c / (2a), where a is the broad-wall dimension and c is the speed of light. Reason (R): TE10 is the dominant mode in a rectangular waveguide because it has the lowest cutoff frequency among all allowed modes.

Introduction / Context:A rectangular waveguide supports discrete TE and TM modes. Each mode has a cutoff frequency determined by the waveguide cross-section. Understanding the dominant mode and the exact cutoff formula for TE10 is foundational for single-mode operation, bandwidth planning, and loss estimation.

Given Data / Assumptions:

• Rectangular waveguide with dimensions a (broad wall) and b (narrow wall).
• Electromagnetic wave in TE10 mode.
• Free space inside (for simplicity); speed of light is c.

Concept / Approach:

For a rectangular guide, the cutoff frequency for TEmn is f_c = (c/2) * sqrt((m/a)^2 + (n/b)^2). For TE10 (m = 1, n = 0) this simplifies to f_c = c/(2a). The dominant mode is the one with the lowest cutoff frequency; in standard hollow rectangular guides, that is TE10 because TE00 does not exist and other allowed modes have higher indices.

Step-by-Step Solution:

Verification / Alternative check:

Waveguide handbooks and measurements show TE10 propagates first as frequency increases; the first higher modes (TE20 or TE01) appear only at higher frequencies, confirming the ordering implied by the formula.

Why Other Options Are Wrong:

• Both correct and explanatory: Dominance explains which mode cuts on first, but not the specific equation f_c = c/(2a).
• A correct, R wrong: R is true—TE10 is the dominant mode.
• A wrong, R correct / Both wrong: Conflicts with standard theory.

Common Pitfalls:

Confusing “dominant” with “fundamental TEM” (TEM does not exist in hollow rectangular guides); forgetting that b does not appear for TE10 because n = 0.

Final Answer:

Both A and R are correct but R is not correct explanation of A