Questions: Electromagnetic Field Solutions in Cavities

Question 1

A rectangular cavity has dimensions a × b × d. An engineer doubles the length d while keeping a and b fixed. What happens to the resonant frequencies?

• A. All resonant frequencies are halved, since the cavity is twice as large
• B. Resonant frequencies of modes with p ≠ 0 shift to lower values; modes with p = 0 are unaffected
• C. All resonant frequencies are unchanged, since only a and b determine resonance
• D. All resonant frequencies double, since the cavity's resonant condition scales with volume

Answer: (B)

The resonant frequency formula ωₙₘₚ = cπ√[(n/a)² + (m/b)² + (p/d)²] shows that doubling d replaces (p/d)² with (p/2d)², reducing the contribution of that term by a factor of 4. Modes with p = 0 have no dependence on d and are unaffected; modes with p ≠ 0 shift lower.

Question 2

Why are resonant frequencies in a cavity discrete rather than forming a continuum?

• A. Because TE and TM modes cannot coexist, limiting available frequencies
• B. Because the boundary condition E_tan = 0 on all conducting walls requires integer numbers of half-wavelengths to fit along each dimension simultaneously
• C. Because electromagnetic energy dissipates at non-resonant frequencies before establishing a standing wave
• D. Because the cavity material has a discrete permittivity spectrum

Answer: (B)

Every conducting wall requires the tangential electric field to vanish. This forces an integer number of half-wavelengths to fit along each direction, just like a string fixed at both ends. Only specific frequency triplets (n, m, p) satisfy all six boundary conditions simultaneously — making the allowed frequencies a discrete set.

Question 3

The dominant mode (lowest resonant frequency) of a rectangular cavity resonator is determined primarily by the cavity's largest dimension.

• T. True
• F. False

Answer: True

ωₙₘₚ is minimized when the term under the square root is smallest. The largest dimension contributes the smallest (p/L)² term, so the mode that places one half-wavelength along the longest axis is the lowest-frequency resonant mode.

Question 4

A TE mode in a cavity resonator has no magnetic field component along the propagation axis.

• T. True
• F. False

Answer: False

It is the ELECTRIC field that has no component along the propagation axis in a TE (transverse electric) mode — Hz ≠ 0 but Ez = 0. TM (transverse magnetic) modes have no magnetic field component along the propagation axis. Confusing the two is a very common error.

Question 5

Explain why sealing both ends of a waveguide to form a cavity produces discrete resonant frequencies, using an analogy with a vibrating string.

Model answer: A string fixed at both ends can only sustain standing waves where an integer number of half-wavelengths fit between the endpoints; other frequencies produce destructive interference and die out. A closed cavity is the 3D electromagnetic analog: sealing the ends forces the field to reflect and interfere. Constructive standing waves exist only when integer half-wavelengths fit along all three dimensions simultaneously, selecting a discrete set of resonant frequencies.

The key insight is that boundary conditions at the ends impose a quantization condition. In both the string and the cavity, the discreteness arises from the requirement that the wave 'fit' in the confined space with nodes at every boundary — a count of half-wavelengths.