At an interface without free surface charge, which normal field component is continuous?
A Dn
B En
C Ht
D Et/ε
Answer: A. Dn. Gauss’s electric law requires D2n−D1n=0 when free surface charge is absent.
Across a stationary dielectric boundary, which tangential component is continuous?
A Dt
B Bn/μ
C Et
D Hn/ε
Answer: C. Et. A vanishingly small loop and Faraday’s law give continuous tangential E.
Which normal magnetic component is continuous across any ordinary interface?
A Bn
B Hn always
C En always
D Jt always
Answer: A. Bn. Gauss’s magnetic law gives B2n−B1n=0.
A perfect conductor carries surface current density K. What equals the adjacent tangential H magnitude?
A K/ε
B K
C μK
D Zero always
Answer: B. K. The magnetic boundary condition relates tangential H to surface current when H inside is zero.
With ε1=2ε0 and ε2=4ε0 and no free surface charge, E2n/E1n equals what?
A 1/4
B 2
C 4
D 1/2
Answer: D. 1/2. Continuity of Dn gives ε1E1n=ε2E2n.
On the surface of a stationary perfect conductor, tangential electric field is what?
A Equal to normal E
B Zero
C Infinite
D Equal to surface current
Answer: B. Zero. A nonzero tangential E would drive unbounded current in the perfect conductor.
For charge q above a grounded infinite conducting plane, what image charge reproduces the exterior potential?
A +q at mirror point
B −2q at original point
C −q at mirror point
D No image charge
Answer: C. −q at mirror point. A negative image at the reflected location makes the plane equipotential at zero.
For ideal coaxial conductors of radii a and b, what is capacitance per unit length?
A πεln(b/a)
B εln(b/a)/(2π)
C 2πε/ln(b/a)
D 2πε(b−a)
Answer: C. 2πε/ln(b/a). Gauss’s law and radial voltage integration give C′=2πε/ln(b/a).
Ignoring conductor interior inductance, ideal coaxial-line inductance per unit length is what?
A 2πμ/ln(b/a)
B μ/(2πln(b/a))
C μεln(b/a)
D (μ/2π)ln(b/a)
Answer: D. (μ/2π)ln(b/a). Magnetic flux linkage between a and b gives L′=μln(b/a)/(2π).
For a good conductor, which expression gives skin depth δ?
A √[2/(ωμσ)]
B √[ωμσ/2]
C 2ωμσ
D 1/(ωε)
Answer: A. √[2/(ωμσ)]. A good conductor has δ=√(2/(ωμσ)).
If frequency becomes four times larger with μ and σ fixed, skin depth changes how?
A It doubles
B It quarters
C It stays unchanged
D It halves
Answer: D. It halves. Skin depth is inversely proportional to the square root of frequency.
For a sinusoidal field in a homogeneous conductor, |Jd|/|Jc| equals what?
A ωε/σ
B σ/(ωε)
C ωμ/σ
D εσ/ω
Answer: A. ωε/σ. Displacement current magnitude is ωε|E| and conduction current is σ|E|.
Using conductivity σ and real permittivity ε, the dielectric loss tangent is what?
A ωε/σ
B σ/(ωε)
C σ/(ωμ)
D ε/(ωσ)
Answer: B. σ/(ωε). The ratio of conduction to displacement current magnitudes is σ/(ωε).
In a lossless isotropic medium, electromagnetic wave speed is which expression?
A √(με)
B 1/√(με)
C μ/ε
D 1/(με)
Answer: B. 1/√(με). Maxwell’s equations give v=1/√(με) for a lossless homogeneous medium.
What is the intrinsic impedance of a lossless isotropic medium?
A √(ε/μ)
B √(μ/ε)
C με
D μ+ε
Answer: B. √(μ/ε). For a uniform plane wave, η=|E|/|H|=√(μ/ε).
If μr=1 and εr=4, approximately what is intrinsic impedance?
A 94 Ω
B 377 Ω
C 754 Ω
D 188.5 Ω
Answer: D. 188.5 Ω. η=η0√(μr/εr)=377/2≈188.5 Ω.
A free-space plane wave has electric-field amplitude 120 V/m. Its H amplitude is approximately what?
A 0.032 A/m
B 3.18 A/m
C 45.2 A/m
D 0.318 A/m
Answer: D. 0.318 A/m. H=E/η0≈120/377=0.318 A/m.
For peak E amplitude 100 V/m in free space, what is average power flux approximately?
A 13.3 W/m²
B 6.6 W/m²
C 26.5 W/m²
D 100 W/m²
Answer: A. 13.3 W/m². Sav=E0²/(2η0)=10000/(754)≈13.3 W/m².
At normal incidence from η1=377 Ω to η2=188.5 Ω, what is the E-field reflection coefficient?
A +1/3
B −1/2
C −1/3
D +1/2
Answer: C. −1/3. Γ=(η2−η1)/(η2+η1)=−188.5/565.5=−1/3.
For that same normal-incidence interface with Γ=−1/3, what is transmitted E amplitude relative to incident E?
A 1/3
B 2/3
C 4/3
D 3/2
Answer: B. 2/3. Tangential E is continuous, so transmitted amplitude ratio is 1+Γ=2/3.
A rod of length 0.2 m moves at 4 m/s perpendicular to B=0.5 T. Find motional emf.
A 0.04 V
B 0.8 V
C 4 V
D 0.4 V
Answer: D. 0.4 V. For perpendicular motion and field, emf=BLv=0.5×0.2×4=0.4 V.
A 0.2 m rod rotates about one end at 10 rad/s in a perpendicular 1 T field. Find end-to-end emf.
A 0.2 V
B 0.1 V
C 0.4 V
D 2 V
Answer: A. 0.2 V. Integrating Bωr dr from 0 to L gives emf=½BωL²=0.2 V.
For a long uniform solenoid, which expression approximates inductance?
A μNA/l
B μNl/A
C μN²A/l
D εN²A/l
Answer: C. μN²A/l. Using B≈μNI/l and flux linkage NΦ gives L≈μN²A/l.
With e^(jωt) phasors, Faraday’s differential law becomes which expression?
A ∇×E = +jωB
B ∇·E = −jωB
C ∇×E = −jωB
D ∇×H = −jωB
Answer: C. ∇×E = −jωB. Time differentiation becomes jω, so ∇×E=−∂B/∂t=−jωB.
With e^(jωt) phasors, Ampère–Maxwell law is which expression?
A ∇×H = J+jωD
B ∇×H = J−jωD
C ∇×E = J+jωD
D ∇·B = J+jωD
Answer: A. ∇×H = J+jωD. The total magnetic curl includes conduction J and displacement jωD.