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Electromagnetic Theory Practice Hub

Practice electromagnetic theory with question sets on Maxwell equations, wave propagation, transmission lines, and GATE-level problems for ECE students.

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BeginnerGauss law formIntermediateTransmission line Z0AdvancedGATE SWR problem

Hub intro

Electromagnetic theory is mathematically demanding and conceptually deep. This hub covers electrostatics, magnetostatics, Maxwell's equations, plane wave propagation, and transmission line theory. Beginner questions address Coulomb's law and Gauss's law. Intermediate questions require applying boundary conditions and calculating wave parameters. Advanced questions follow GATE patterns on standing wave ratio, skin depth, and waveguide cutoff.

Difficulty levels

BeginnerIntermediateAdvanced

Practice sets

Beginner: Electrostatics and Fundamental Laws
Beginner
State Gauss's law in integral form and identify each quantity in the expression.
Answer: ∮ D · dS = Q_enclosed, where D is the electric flux density (C/m²), dS is the outward normal surface area element, and Q_enclosed is the total free charge enclosed by the Gaussian surface.
Gauss's law relates the total electric flux through a closed surface to the enclosed free charge. It follows from Coulomb's law for point charges and the superposition principle. It is most useful for computing E (or D) when the charge distribution has spherical, cylindrical, or planar symmetry. In differential form: ∇·D = ρv, where ρv is the volume charge density.
Write all four of Maxwell's equations in differential form.
Answer: ∇×E = -∂B/∂t (Faraday). ∇×H = J + ∂D/∂t (Ampere-Maxwell). ∇·D = ρv (Gauss electric). ∇·B = 0 (Gauss magnetic, no magnetic monopoles).
Faraday's law says a time-varying magnetic field induces a circulating electric field. Ampere's law (with Maxwell's displacement current ∂D/∂t) says currents and time-varying electric fields produce circulating magnetic fields. The two Gauss laws describe divergence: electric field lines originate at charges, magnetic field lines have no sources or sinks. Together, these four equations completely describe classical electromagnetics.
What is the skin depth of a good conductor and what does it represent physically?
Answer: Skin depth δ = √(2/(ωµσ)), where ω is the angular frequency, µ is the permeability, and σ is the conductivity. It is the depth at which the field amplitude in the conductor falls to 1/e (≈ 37%) of its surface value.
At high frequencies, currents in a conductor concentrate near the surface due to the opposing fields induced by Faraday's law. The current density decreases exponentially with depth as J = J0 × e^(-z/δ). At depth δ, the amplitude is 1/e of the surface value. Skin depth decreases as frequency or conductivity increases. For copper at 1 MHz, δ ≈ 66 µm; at 1 GHz, δ ≈ 2.1 µm.
State Poynting's theorem and identify the Poynting vector S.
Answer: Poynting's theorem: the rate of decrease of electromagnetic energy in a volume equals the power dissipated as heat plus the power flowing out through the surface. S = E × H (W/m²) is the power flow density (energy flux).
The Poynting vector S = E × H represents the instantaneous power per unit area flowing in the direction of S. For a plane wave in free space, S points in the direction of propagation. For time-averaged power, use the phasor form: S_avg = (1/2) Re(E × H*). Poynting's theorem is derived from Maxwell's equations and represents energy conservation in electromagnetic fields.
Define the intrinsic impedance of a medium and state its value for free space.
Answer: Intrinsic impedance η = √(µ/ε), where µ is the permeability and ε is the permittivity of the medium. For free space: η0 = √(µ0/ε0) ≈ 377 Ω.
The intrinsic impedance relates the electric and magnetic field amplitudes of a plane wave: η = |E|/|H|. For a lossless medium, η is real; for a lossy medium, η is complex. µ0 = 4π×10^-7 H/m, ε0 = 8.854×10^-12 F/m. η0 = √(4π×10^-7 / 8.854×10^-12) = √(1.131×10^5) ≈ 377 Ω. This value is important in antenna feed matching and transmission line design.
Intermediate: Wave Propagation and Transmission Lines
Intermediate
A plane wave with E = 100 cos(10^9 t - βz) V/m travels in a lossless medium with εr = 4 and µr = 1. Calculate the phase velocity, wavelength, and intrinsic impedance.
Answer: v = c/√(εr·µr) = 3×10^8/2 = 1.5×10^8 m/s. β = ω/v = 10^9/1.5×10^8 ≈ 6.67 rad/m. λ = 2π/β ≈ 0.942 m. η = η0/√εr = 377/2 ≈ 188.5 Ω.
In a lossless non-magnetic medium (µr=1): v = c/√εr = 3×10^8/√4 = 1.5×10^8 m/s. β = ω/v = 10^9/1.5×10^8 = 6.667 rad/m. λ = 2π/β = 2π/6.667 ≈ 0.9425 m. η = η0/√εr = 377/2 = 188.5 Ω. The magnetic field amplitude: H = E/η = 100/188.5 ≈ 0.531 A/m. The factor √(εr·µr) reduces both the phase velocity and the intrinsic impedance relative to free space.
A lossless transmission line has characteristic impedance Z0 = 50 Ω and is terminated in a load ZL = 100 + j50 Ω. Calculate the reflection coefficient Γ at the load.
Answer: Γ = (ZL - Z0)/(ZL + Z0) = (100+j50-50)/(100+j50+50) = (50+j50)/(150+j50).
Numerator: 50+j50 = 50√2 ∠45°. Denominator: 150+j50 = √(150²+50²) ∠arctan(50/150) = √25000 ∠18.43° = 158.11∠18.43°. |Γ| = (50√2)/158.11 = 70.71/158.11 ≈ 0.447. angle(Γ) = 45° - 18.43° = 26.57°. Γ ≈ 0.447∠26.57°. The standing wave ratio SWR = (1+|Γ|)/(1-|Γ|) = (1+0.447)/(1-0.447) = 1.447/0.553 ≈ 2.62.
Calculate the cutoff frequency of the TE10 mode in a rectangular waveguide with inner dimensions a = 4 cm and b = 2 cm. (c = 3×10^8 m/s)
Answer: fc(TE10) = c/(2a) = 3×10^8/(2×0.04) = 3.75 GHz.
For the TE_mn mode in a rectangular waveguide: fc = (c/2)×√((m/a)²+(n/b)²). For TE10 (m=1, n=0): fc = (c/2)×(1/a) = c/(2a) = 3×10^8/(0.08) = 3.75×10^9 Hz = 3.75 GHz. The TE10 mode is the dominant mode (lowest cutoff frequency) for a > b. The guide must be operated above 3.75 GHz for TE10 propagation. At 3.75 GHz, the wavelength in free space is λ = c/fc = 0.08 m = 8 cm = 2a.
A 50 Ω transmission line is terminated in a short circuit. Using the standing wave pattern, find the impedance at a distance λ/4 from the short.
Answer: Zin = jZ0 tan(βl) = j50 tan(β×λ/4) = j50 tan(π/2) → infinity. The input impedance is an open circuit at λ/4 from a short.
For a lossless transmission line with load ZL: Zin = Z0(ZL + jZ0 tan βl)/(Z0 + jZL tan βl). For ZL=0 (short circuit): Zin = jZ0 tan βl. At l=λ/4: βl = (2π/λ)×(λ/4) = π/2. tan(π/2) → ∞, so Zin → ∞ (open circuit). A quarter-wave section transforms a short to an open and vice versa. This is the principle of a quarter-wave transformer, used for impedance matching.
State the boundary conditions for the tangential components of E and H at the interface between two lossless media.
Answer: The tangential component of E is continuous across the boundary: Et1 = Et2. The tangential component of H is continuous (for a surface without free surface current): Ht1 = Ht2.
These boundary conditions follow from Faraday's law (Et continuous) and Ampere's law (Ht continuous for no surface current). The normal boundary conditions (from Gauss's laws) are: Dn1 - Dn2 = ρs (surface charge density) and Bn1 = Bn2. For a perfect conductor, Et = 0 and Dn = ρs at the surface, and all fields inside are zero. These conditions govern the reflection and transmission of waves at interfaces.
Advanced: GATE-Style Problems
Advanced
A plane wave in free space is incident normally on a dielectric with εr = 9, µr = 1. Calculate the transmission coefficient τ and verify that the power reflection coefficient |Γ|² and power transmission coefficient |τ|²(η1/η2) add to 1.
Answer: η1 = 377 Ω, η2 = 377/3 = 125.7 Ω. Γ = (η2-η1)/(η2+η1) = (125.7-377)/(125.7+377) = -251.3/502.7 = -0.5. τ = 2η2/(η2+η1) = 251.4/502.7 = 0.5. Power check: |Γ|² = 0.25. (η1/η2)|τ|² = 3×0.25 = 0.75. 0.25+0.75=1.
Reflection coefficient: Γ = (η2-η1)/(η2+η1). Transmission coefficient: τ = 1+Γ = 2η2/(η2+η1). Note τ is for the E field, not power. Power reflection = |Γ|² = 0.25 (25% reflected). Power transmission = 1 - |Γ|² = 0.75 (75% transmitted). The factor η1/η2 = η1/η2 accounts for the different wave impedances: transmitted power = (1/2)|E_t|²/η2 × A, while incident power = (1/2)|E_i|²/η1 × A. So power transmission = (η1/η2)|τ|² = (377/125.7)×0.25 = 3×0.25 = 0.75. Consistent.
In a coaxial transmission line, the inner conductor radius is a = 2 mm and the outer conductor inner radius is b = 8 mm. The dielectric has εr = 2.25. Calculate the characteristic impedance Z0.
Answer: Z0 = (60/√εr) × ln(b/a) = (60/1.5) × ln(4) = 40 × 1.386 = 55.45 Ω.
For a coaxial line: Z0 = (1/(2π)) × √(µ/ε) × ln(b/a) = (η/(2π)) × ln(b/a). With µ=µ0, ε=εr·ε0: η = 377/√εr = 377/1.5 = 251.3 Ω. Z0 = (251.3/2π) × ln(8/2) = (251.3/6.2832) × ln(4) = 40.0 × 1.3863 = 55.45 Ω. A common approximation is Z0 = (60/√εr) × ln(b/a), confirming the result: 60/1.5 × 1.386 = 40 × 1.386 ≈ 55.4 Ω.
The standing wave ratio (SWR) on a 50 Ω transmission line is measured as 3. Find the magnitude of the reflection coefficient and the maximum and minimum impedances on the line.
Answer: |Γ| = (SWR-1)/(SWR+1) = 2/4 = 0.5. Zmax = Z0 × SWR = 150 Ω. Zmin = Z0/SWR = 16.67 Ω.
SWR = (1+|Γ|)/(1-|Γ|). Solving: |Γ| = (SWR-1)/(SWR+1) = (3-1)/(3+1) = 2/4 = 0.5. At voltage maximum positions, Zin = Z0(1+|Γ|)/(1-|Γ|) = Z0 × SWR = 50 × 3 = 150 Ω (real, resistive). At voltage minimum positions: Zin = Z0(1-|Γ|)/(1+|Γ|) = Z0/SWR = 50/3 ≈ 16.67 Ω (also real). GATE trap: confuse the positions of Zmax and Zmin with voltage maximum and minimum; they coincide with voltage maxima and minima respectively.
An electromagnetic wave travels in a medium with µr = 1, σ/ωε = 1 (loss tangent = 1). Is this medium a good conductor, good dielectric, or neither? Calculate the attenuation constant α.
Answer: Loss tangent = 1 means neither a good conductor nor a good dielectric (boundary case). α = ω√(µε/2) × √(√(1+(σ/ωε)²) - 1) = ω√(µε/2) × √(√2 - 1) ≈ ω√(µε/2) × 0.6436.
A good dielectric has σ/ωε << 1; a good conductor has σ/ωε >> 1. At σ/ωε = 1, both conduction and displacement currents are equal; neither approximation applies. The exact expression for α is α = ω√(µε/2)×√(√(1+(σ/ωε)²)-1). With σ/ωε=1: α = ω√(µε/2)×√(√2-1) = ω√(µε/2)×√(0.4142) = ω√(µε/2)×0.6436. The GATE trap is applying the good-conductor formula α≈√(ωµσ/2) for loss tangent=1, which overestimates α; the exact formula must be used.
A magnetic field H = H0·cos(ωt - βz)·x̂ exists inside a conductor. Using Faraday's law, derive the direction of the induced electric field.
Answer: E is in the ŷ direction. By Faraday's law, ∇×E = -∂B/∂t = -µ∂H/∂t = µωH0·sin(ωt-βz)·x̂. For E = E_y·ŷ: ∇×E = ∂E_y/∂z·(-x̂) + ... = -∂E_y/∂z·x̂. So -∂E_y/∂z = µωH0·sin(ωt-βz), giving E_y = (µω/β)H0·cos(ωt-βz)·ŷ = η·H0·cos(ωt-βz)·ŷ.
For a plane wave propagating in the z-direction with H in the x-direction, the electric field must be in the y-direction (E, H, and propagation direction form a right-handed orthogonal set: ŷ × x̂ = -ẑ, so the wave propagates in the +z direction for E=Ey·ŷ and H=Hx·x̂ since ŷ × (-x̂) = ẑ... check: E×H = ŷ × x̂ = -ẑ. Not +ẑ. So E should be in the -ŷ direction for propagation in +z: (-ŷ)×x̂ = ẑ. Yes, E = -Ey·ŷ and H = Hx·x̂ gives propagation in +ẑ. The Poynting vector S = E×H = -Ey·Hx (ŷ×x̂) = -Ey·Hx×(-ẑ) = Ey·Hx·ẑ > 0 for Ey < 0 and Hx > 0 (or appropriate sign). Always verify the direction using the right-hand rule.

Lab exercises

  • Measure the standing wave ratio on a coaxial transmission line using a slotted line and verify the reflection coefficient: /labs/slotted-line-swr-measurement-lab
  • Calculate and verify the cutoff frequency of a rectangular waveguide by measuring the S-parameters of a WR-90 waveguide section: /labs/waveguide-cutoff-frequency-lab
  • Simulate a plane wave incident on a dielectric interface in MATLAB and plot the transmitted and reflected field amplitudes versus angle of incidence: /labs/plane-wave-dielectric-interface-lab

Revision checklist

  • Can you write all four Maxwell equations in both differential and integral forms without looking them up?
  • Can you calculate skin depth for copper at a given frequency and explain what it means physically?
  • Do you know the formula for characteristic impedance of a coaxial line and a parallel-plate transmission line?
  • Can you derive the relationship SWR = (1+|Γ|)/(1-|Γ|) from the definition of the standing wave pattern?
  • Can you apply the boundary conditions for E and H at a conductor surface to determine the surface current density?