Understanding TE01 vs. TE10 in a Rectangular Waveguide Which statement best describes the TE01 mode in a rectangular waveguide relative to the dominant TE10 mode?

Introduction / Context:
Rectangular waveguides support TE and TM modes indexed by two integers corresponding to standing-wave variations along the broad (a) and narrow (b) walls. Understanding how TE01 relates to TE10 helps with cutoff and field-pattern intuition.

Given Data / Assumptions:

• Uniform rectangular waveguide with dimensions a (broad) and b (narrow).
• Standard boundary conditions for perfect electric conductors.
• Air-filled guide for simplicity.

Concept / Approach:

TE10 has one half-sine variation across the a-dimension and none across b. TE01 has the complementary pattern: one half-sine across the b-dimension and none across a. If one physically rotates the guide by 90°, the axes swap and the field pattern of TE01 corresponds to TE10 in the rotated frame. Thus, TE01 is effectively TE10 rotated by 90°, although cutoffs differ if a ≠ b.

Step-by-Step Solution:

Verification / Alternative check:

Cutoff frequencies fc(mn) = (c/2) * sqrt((m/a)^2 + (n/b)^2). Swapping a and b interchanges (m, n) = (1, 0) and (0, 1), consistent with a 90° rotation.

Why Other Options Are Wrong:

180° rotation (B) leaves the same mode unchanged; (C) “the same” ignores the axis swap; (D) cannot be right since (B) is wrong; (E) confuses rectangular-guide TE01 with circular-guide TE11.

Common Pitfalls:

Assuming TE01 is another higher-order version of TE10 without considering geometry; forgetting that if a ≠ b, the cutoff frequencies differ even though patterns are rotationally related.

Final Answer:

is merely the TE10 mode rotated 90°