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Proof that TE10 is the dominant mode

Short answer

For a>b, cutoff is fc,mn=(c/2)sqrt((m/a)^2+(n/b)^2). TE10 gives c/2a, TE01 gives c/2b, and since a>b, TE10 has the lowest cutoff; higher modes and TM11 have larger fc. Hence TE10 is dominant, with λc=2a.

Show that the dominant mode in rectangular waveguides operating in TE mode is TE10, when b>a.

The dominant mode is the mode with the lowest cut-off frequency. The condition "b > a" in the question is taken as a misprint for the usual convention a > b, where a is the wider (broad) wall and the index m counts half-wave variations along it.

For an air-filled rectangular guide, the cut-off frequency of the TEmn mode is

 fc,mn = (c/2) √((m/a)² + (n/b)²)

For TE modes m and n can be 0, 1, 2 … but not both zero (TE00 has no field).

The candidates for the lowest cut-off are:

 TE10: fc = c/(2a)
 TE01: fc = c/(2b)
 TE11: fc = (c/2)√(1/a² + 1/b²)

Since a > b, 1/a < 1/b, so c/(2a) < c/(2b) < (c/2)√(1/a² + 1/b²). Any higher m or n only increases fc, and TM modes start from TM11, which also has a higher fc. Hence TE10 has the lowest cut-off frequency and is the dominant mode, with λc = 2a.

(If the indices were assigned the other way, i.e. the wider wall were along y, the same argument would make TE01 the dominant mode; physically it is always the mode with one half-wave across the wider wall.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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