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
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Aa + b
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Bπ(a + b)
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Cπa
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Dπb
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E2π(b − a)
Answer
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)