Microwave coaxial line (inner radius = a, inner radius of outer conductor = b): the approximate cutoff wavelength for the first TE mode (TE11) is closest to

Difficulty: Medium

Correct Answer: π(a + b)

Explanation:


Introduction:
Although coax supports a TEM mode without cutoff, higher-order modes (TE/TM) have finite cutoff frequencies. Knowing the TE11 cutoff helps set the upper frequency limit for single-mode coaxial operation.


Given Data / Assumptions:

  • Coax geometry: inner conductor radius a, inner radius of outer conductor b.
  • We consider the first higher-order TE mode (TE11).
  • Air dielectric; standard boundary conditions.


Concept / Approach:

Analytical solutions give an approximate TE11 cutoff wavelength λc ≈ π(a + b), which is a widely used design rule. Operation should remain well below this cutoff frequency (i.e., at wavelengths ≪ λc) to avoid multimoding.


Step-by-Step Solution:

1) Recall approximate TE11 cutoff relation: fc ≈ c / (π(a + b)).2) Therefore λc = c / fc ≈ π(a + b).3) Compare with choices: π(a + b) matches directly.


Verification / Alternative check:

More exact expressions involve solving characteristic equations with Bessel functions; the π(a + b) rule closely tracks the true cutoff and is sufficient for design checks.


Why Other Options Are Wrong:

  • a + b, πa, πb: ignore the dependence on both radii with the correct coefficient.
  • 2π(b − a): relates to circumference gap, not TE11 cutoff.


Common Pitfalls:

Confusing TEM (no cutoff) with TE11 (finite cutoff); the upper frequency limit for single-mode coax is set by TE11.


Final Answer:

π(a + b)

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