Eight-cavity magnetron – Total phase shift around the ring in π-mode For a magnetron with 8 identical cavity resonators, what is the total phase shift around the periphery in the π-mode (adjacent cavities 180° out of phase)?
-
A± 8π radians around the periphery
-
B± 4π radians around the periphery
-
C± 2π radians around the periphery
-
D± 16π radians around the periphery
-
EZero because the fields are all in phase
Answer
Correct Answer: ± 8π radians around the periphery
Explanation
Introduction:Multi-cavity magnetrons support discrete phase modes around their circular cavity arrays. The π-mode is preferred for stable, efficient operation and places adjacent cavities 180° out of phase.
Given Data / Assumptions:
- Eight cavities equally spaced.
- π-mode: phase step of π between neighbors.
- Steady periodic boundary condition around the ring.
Concept / Approach:
With N cavities and a phase advance Δφ between adjacent cavities, the total phase shift around the periphery is N * Δφ. In π-mode, Δφ = π.
Step-by-Step Solution:
1) N = 8 cavities.2) Δφ = π radians (adjacent cavities out of phase).3) Total phase shift = N * Δφ = 8 * π = 8π radians (± indicating direction of traversal).Verification / Alternative check:
Mode charts for magnetrons label the π-mode with alternating polarity across adjacent cavities and total 8π phase around an 8-cavity ring.
Why Other Options Are Wrong:
4π or 2π correspond to other fractional modes; 16π doubles the correct value; zero would mean all in phase (0-mode), not π-mode.
Common Pitfalls:
Confusing total phase around the ring with the phase relative to a fixed cavity; forgetting to multiply by the number of cavities.
Final Answer:
± 8π radians around the periphery.