In an air-filled guide with broad side 3 cm, TE₁₀ cutoff is?
A 5 GHz
B 2.5 GHz
C 10 GHz
D 15 GHz
Answer: A. 5 GHz. For TE₁₀, fc = c/(2a); with a=0.03 m this is 5 GHz.
At 6 GHz in that guide, guide wavelength is about?
A 3.02 cm
B 18.1 cm
C 9.05 cm
D 5.00 cm
Answer: C. 9.05 cm. λg = λ₀/√(1−(fc/f)²), giving about 0.0905 m.
Why can a hollow single-conductor guide not support TEM propagation?
A No boundary walls
B No two-conductor path
C No magnetic field
D No electric charge
Answer: B. No two-conductor path. A TEM mode requires two conductors to sustain a transverse voltage.
Below waveguide cutoff, field magnitude along the guide shows what?
A Linear growth
B Constant amplitude
C Periodic increase
D Exponential decay
Answer: D. Exponential decay. Below cutoff, the longitudinal propagation constant is imaginary and fields decay.
Which indices must be nonzero for rectangular-waveguide TM modes?
A Only n
B Neither index
C Both m and n
D Only m
Answer: C. Both m and n. TM modes require variation in both transverse dimensions.
Average power density for a 10 V/m peak wave in air is?
A 0.133 W/m²
B 1.33 W/m²
C 0.265 W/m²
D 0.0265 W/m²
Answer: A. 0.133 W/m². For peak fields, S_avg = E₀²/(2η₀), about 0.133 W/m².
At 1 MHz, copper skin depth is closest to what value?
A 6.6 μm
B 0.66 mm
C 6.6 mm
D 66 μm
Answer: D. 66 μm. Using δ = √(2/(ωμσ)) for copper gives approximately 66 micrometres.
Given σ=0.01 S/m, εr=4, f=1 GHz, tanδ is about?
A 0.45
B 0.045
C 4.5
D 0.0045
Answer: B. 0.045. tanδ=σ/(ωε), which is approximately 0.045 for these values.
Coaxial radii 1 mm and 4 mm in εr=2.25 give capacitance per unit length near?
A 90 pF/m
B 900 pF/m
C 9 pF/m
D 180 pF/m
Answer: A. 90 pF/m. C′=2πε/ln(b/a), which evaluates to about 90 pF/m.
If an 80-turn coil sees flux changing at 0.2 Wb/s, |emf| is?
A 4 V
B 80 V
C 16 V
D 160 V
Answer: C. 16 V. Faraday’s law gives |emf|=N|dΦ/dt|=80×0.2=16 V.
Equal orthogonal E components with 90° phase difference produce?
A Unpolarized light
B No polarization
C Linear polarization
D Circular polarization
Answer: D. Circular polarization. Equal magnitudes in quadrature make the electric-field tip trace a circle.
When quadrature E components have unequal amplitudes, polarization is?
A Unpolarized
B Elliptical
C Circular
D Linear
Answer: B. Elliptical. Unequal orthogonal components in quadrature trace an ellipse.
Two orthogonal E components in phase produce what polarization?
A Circular
B Elliptical
C Linear
D Unpolarized
Answer: C. Linear. In-phase orthogonal components combine along a fixed line.
With surface current present, which field component has a boundary jump?
A Tangential H
B Tangential D
C Normal B
D Normal E
Answer: A. Tangential H. Surface current density determines the discontinuity in tangential H.
Given D=2x ax+3y ay C/m², free charge density is?
A 6 C/m³
B 1 C/m³
C 3 C/m³
D 5 C/m³
Answer: D. 5 C/m³. ρfree=∇·D=2+3=5 C/m³.
At x=2 m, V=5x²+2y² gives what Ex?
A +20 V/m
B −20 V/m
C −10 V/m
D +10 V/m
Answer: B. −20 V/m. Ex=−∂V/∂x=−10x, so Ex=−20 V/m.
Inside a uniformly charged solid sphere, E at radius r is?
A ρr/ε₀
B ρR²/(3ε₀r)
C ρr/(3ε₀)
D ρR/(3ε₀)
Answer: C. ρr/(3ε₀). Gauss’s law gives enclosed charge proportional to r³ and E=ρr/(3ε₀).
Line charge λ=2 nC/m in air gives E at r=0.1 m about?
A 360 V/m
B 36 kV/m
C 36 V/m
D 3.6 kV/m
Answer: A. 360 V/m. E=λ/(2πε₀r), which is approximately 360 V/m.
Just outside a charged perfect conductor in vacuum, normal E equals?
A ε₀σₛ
B μ₀σₛ
C σₛ/μ₀
D σₛ/ε₀
Answer: D. σₛ/ε₀. The normal-E boundary condition gives Eₙ=σₛ/ε₀ in vacuum.
Peak phasors E₀=20 V/m and H₀=0.05 A/m are in phase; S_avg is?
A 0.05 W/m²
B 0.5 W/m²
C 2 W/m²
D 1 W/m²
Answer: B. 0.5 W/m². Average power density is one half E₀H₀ for in-phase peak fields.
In a lossless waveguide, the product vp·vg equals?
A c²
B Zero
C c
D 2c
Answer: A. c². For a lossless guide, phase and group velocities satisfy vpvg=c².
TE₂₀ cutoff in an air guide with broad side 4 cm is?
A 3.75 GHz
B 5 GHz
C 7.5 GHz
D 15 GHz
Answer: C. 7.5 GHz. For TE₂₀, fc=c/a; with a=0.04 m this is 7.5 GHz.
When |Γ|=0.5, what fraction of incident power is reflected?
A 0.75
B 0.25
C 1
D 0.5
Answer: B. 0.25. Power reflectance is |Γ|², so the fraction is 0.25.
If field amplitude falls by e⁻¹ over 2 cm, attenuation constant is?
A 200 Np/m
B 0.5 Np/m
C 2 Np/m
D 50 Np/m
Answer: D. 50 Np/m. Since e^(−αz)=e⁻¹ at z=0.02 m, α=1/0.02=50 Np/m.
With H=3x ay A/m and zero conduction current, displacement current density is?
A −3 az A/m²
B 3 ax A/m²
C 3 az A/m²
D Zero
Answer: C. 3 az A/m². Ampère–Maxwell law gives ∂D/∂t=∇×H=3 az A/m².