Difficulty: Medium
Correct Answer: P_max = (a * b / 4) * (E_d^2 / Z_TE)
Explanation:
Introduction / Context:
Waveguide power handling is limited by the maximum electric field before breakdown. For a rectangular waveguide operating in the dominant TE10 mode, there is a known relationship between the peak electric field amplitude and the time-average power flow. This item asks you to identify the correct formula using Z_TE and breakdown strength E_d.
Given Data / Assumptions:
Concept / Approach:
For TE10, the transverse electric field varies as cos(pi x/a) across the broad wall, with a maximum field magnitude E_max. The average power flow P in a lossless guide can be written in terms of E_max as P = (a * b / 4) * (E_max^2 / Z_TE). Setting E_max = E_d at the threshold gives the maximum permissible power P_max. The 1/4 factor arises from spatial averaging of the field distribution across the guide cross-section.
Step-by-Step Solution:
Verification / Alternative check:
Dimensional analysis: E^2 / Z_TE has dimensions of power density; multiplying by area ab yields power. The quarter factor is consistent with integrating cos^2(pi x/a) over the width.
Why Other Options Are Wrong:
Common Pitfalls:
Forgetting the spatial averaging factor; mixing up Z_TE in numerator vs denominator; ignoring mode dependence of field distribution.
Final Answer:
P_max = (a * b / 4) * (E_d^2 / Z_TE)
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