Understanding TE10 in a rectangular waveguide Which statement best describes the physical nature of TE10 propagation in a hollow rectangular waveguide?

Introduction / Context:
Waveguides do not support TEM modes unless multiple conductors are present. Rectangular waveguides are single-conductor enclosures; their propagating solutions are TE or TM, each with cutoff. The TE10 mode is the dominant mode in a standard rectangular waveguide and is a genuine TE solution to Maxwell’s equations with boundary conditions, not a superposition of TEM modes.

Given Data / Assumptions:

• Single hollow PEC rectangular guide, air-filled.
• TE10 is dominant with cutoff f_c = c/(2a).
• No center conductor → no true TEM mode exists.

Concept / Approach:

TEM waves require at least two conductors to define a transverse E field with zero longitudinal components everywhere. In a single-conductor hollow guide, boundary conditions enforce solutions with nonzero longitudinal components (either Ez or Hz). TE10 has Hz ≠ 0 with Ez = 0, exhibiting a standing variation across the broad wall and propagating along z with phase constant β above cutoff.

Step-by-Step Solution:

Verification / Alternative check:

Mode charts and derivations show Ez = 0 for TE modes, Ez ≠ 0 for TM modes, and neither equals the TEM condition (Ez = Hz = 0).

Why Other Options Are Wrong:

Options A–C attempt to reinterpret TE10 as combinations of TEM, which is physically incorrect in a single-conductor guide.

Common Pitfalls:

Assuming “all waves are TEM” or importing transmission-line intuition to waveguides without considering boundary conditions and cutoff behavior.

Final Answer:

None of the above descriptions are correct because TE10 is a true TE mode, not a combination of TEM waves.