Assertion–Reason (circular waveguide TE01): Assertion (A): In a circular waveguide, the attenuation of the TE01 mode decreases as the operating frequency increases (within its passband). Reason (R): In TE01, the wall currents flow predominantly circumferentially, minimizing axial current and reducing ohmic loss.

Difficulty: Medium

Correct Answer: Both A and R are correct and R is correct explanation of A

Explanation:


Introduction / Context:
The TE01 mode in circular waveguide is popular for low-loss, high-power transmission. Its unique wall-current distribution leads to lower attenuation than many other modes, particularly at higher microwave frequencies within the passband.



Given Data / Assumptions:

  • Circular hollow waveguide operated in TE01 mode above cutoff.
  • Conductor (ohmic) loss dominates.
  • Surface resistance increases with √f, but field distribution also changes with frequency.


Concept / Approach:

Attenuation depends on electric and magnetic fields at the walls and their associated surface currents. In TE01, currents are primarily circumferential, yielding minimal longitudinal electric field at the wall. As frequency increases, the power flow grows faster than the incremental wall loss for this mode, so attenuation per unit length decreases within the useful band.



Step-by-Step Solution:

Identify current pattern: mainly circumferential, little axial current.Relate wall fields to conductor loss via surface resistance.Note TE01’s favorable scaling of wall fields with frequency.Conclude that both A and R are correct and R explains A.


Verification / Alternative check:

Attenuation charts for circular guides show TE01 with decreasing attenuation versus frequency over a practical range, matching theoretical conductor-loss expressions for this mode.



Why Other Options Are Wrong:

  • Claiming R unrelated ignores the mode’s current distribution as the cause of the trend.
  • Denying A or R contradicts established waveguide-loss analyses.


Common Pitfalls:

Mixing TE01 with TM01 or assuming all modes behave the same at higher frequencies.



Final Answer:

Both A and R are correct and R is correct explanation of A

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