Field distribution in TE10 mode — location of maximum E-field For the dominant TE10 mode in a rectangular waveguide, where does the electric field have its maximum magnitude?

Difficulty: Easy

Correct Answer: along centre line of broad wall (i.e. x = a/2)

Explanation:


Introduction / Context:
Knowing where fields peak inside a waveguide is critical for breakdown assessment, probe placement, and coupling design. In TE10, the dominant transverse electric field varies sinusoidally across the broad wall dimension.


Given Data / Assumptions:

  • Rectangular waveguide with width a (broad wall) and height b (narrow wall).
  • Mode: TE10 (m = 1, n = 0).


Concept / Approach:
The principal field component varies as a cosine function across the a-dimension and is uniform across b. Boundary conditions enforce zero tangential electric field at the conducting walls (x = 0 and x = a), with a maximum at the midpoint x = a/2. Thus, the electric field maximum lies along the center line of the broad wall.


Step-by-Step Solution:

Model transverse field: E(x) ∝ cos(π x / a).At x = 0 and x = a → conductor boundary condition forces field components to zero.Maximum occurs at x = a/2 → center line across the broad wall.


Verification / Alternative check:
Standard field plots and coupling-probe placements show strongest coupling at x = a/2 for TE10, confirming the location of the E-field maximum.


Why Other Options Are Wrong:

  • B: The field is zero at the side walls, not maximum.
  • C/D: Contradict the sinusoidal distribution and boundary conditions.


Common Pitfalls:
Confusing TE10 with higher-order modes; misapplying boundary conditions at conductor walls.


Final Answer:
along centre line of broad wall (i.e. x = a/2)

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