Q.1 What does the Poynting vector represent in electromagnetic theory?
Magnetic flux density
Electric potential energy
Rate of energy flow per unit area
Charge density
Explanation - The Poynting vector **S = E × H** gives the instantaneous power flow (watts per square meter) through a surface.
Correct answer is: Rate of energy flow per unit area
Q.2 In free space, the magnitude of the Poynting vector is given by S = E²/η₀. What is η₀?
Permittivity of free space
Permeability of free space
Intrinsic impedance of free space
Speed of light
Explanation - η₀ = √(μ₀/ε₀) ≈ 377 Ω is the intrinsic impedance of free space, relating electric and magnetic fields in a plane wave.
Correct answer is: Intrinsic impedance of free space
Q.3 The time‑averaged Poynting vector for a sinusoidal plane wave is given by 〈S〉 = (1/2) Re{E × H*}. Which operation is represented by the asterisk (*)?
Complex conjugate
Multiplication
Differentiation
Integration
Explanation - The asterisk denotes the complex conjugate of the magnetic field phasor, needed for calculating average power.
Correct answer is: Complex conjugate
Q.4 If an electromagnetic wave propagates in the +z direction, what is the direction of the associated Poynting vector?
+x
+y
+z
-z
Explanation - For a wave traveling in +z, E and H are orthogonal in the xy‑plane, so **S = E × H** points in the same +z direction.
Correct answer is: +z
Q.5 Which of the following statements about the Poynting theorem is FALSE?
It is a statement of conservation of electromagnetic energy.
It relates the rate of change of stored energy to the divergence of the Poynting vector and work done on charges.
It can be applied only to static (time‑invariant) fields.
It is derived from Maxwell’s equations.
Explanation - The Poynting theorem holds for time‑varying fields as well; it is a general energy conservation law.
Correct answer is: It can be applied only to static (time‑invariant) fields.
Q.6 For a lossless transmission line, the real part of the Poynting vector integrated over a cross‑section gives:
Reactive power
Stored magnetic energy
Net power transmitted
Radiated power
Explanation - The real part of **S** corresponds to average power flow, which equals the power transmitted along the line.
Correct answer is: Net power transmitted
Q.7 In a coaxial cable, the Poynting vector is directed:
Radially outward from the inner conductor
Radially inward toward the outer conductor
Along the length of the cable
Opposite to the direction of current flow
Explanation - E is radial, H is circumferential, so **E × H** points along the cable axis, indicating power flow down the line.
Correct answer is: Along the length of the cable
Q.8 If the electric field magnitude in a plane wave doubles, how does the magnitude of the Poynting vector change?
It doubles
It quadruples
It remains the same
It halves
Explanation - S ∝ E², so doubling E increases S by a factor of 2² = 4.
Correct answer is: It quadruples
Q.9 Which material property appears explicitly in the differential form of the Poynting theorem?
Conductivity σ
Relative permittivity εr
Magnetic reluctance
Thermal conductivity
Explanation - The term **J·E = σE²** represents power dissipated as heat and appears in the theorem as the Joule loss term.
Correct answer is: Conductivity σ
Q.10 The integral form of the Poynting theorem over a volume V bounded by surface S states:
∮_S S·dA = ∫_V (E·J) dV
∮_S S·dA = -d/dt ∫_V (u) dV - ∫_V (E·J) dV
∮_S S·dA = d/dt ∫_V (u) dV
∮_S S·dA = 0
Explanation - The surface integral of **S** equals the negative rate of change of stored EM energy plus the power dissipated in the volume.
Correct answer is: ∮_S S·dA = -d/dt ∫_V (u) dV - ∫_V (E·J) dV
Q.11 In a region where there are no free charges or currents, the Poynting theorem reduces to:
∇·S = 0
∇·S = -∂u/∂t
∇·S = J·E
∇·S = σE²
Explanation - With J = 0 and σ = 0, the theorem simplifies to **∇·S = -∂u/∂t**, indicating that divergence of power flow equals the decrease of stored energy.
Correct answer is: ∇·S = -∂u/∂t
Q.12 The electromagnetic energy density u in a linear, isotropic medium is given by:
½ (εE² + μH²)
εE + μH
E·H
εE² - μH²
Explanation - Energy density combines electric and magnetic contributions: **u = ½(ε|E|² + μ|H|²)**.
Correct answer is: ½ (εE² + μH²)
Q.13 A waveguide operates in the TE₁₀ mode. Which component of the electric field is zero everywhere inside the guide?
Eₓ
E_y
E_z
E_total
Explanation - For TE (Transverse Electric) modes, the longitudinal electric field component (along the guide axis) is zero.
Correct answer is: E_z
Q.14 If a dielectric slab of thickness d and relative permittivity εr is placed in a uniform electric field, the energy stored in the slab per unit area is:
½ ε₀εr E² d
ε₀ E² d
½ ε₀ E² d / εr
ε₀εr E d
Explanation - Energy density in a dielectric is ½ εE²; multiply by volume (area × d) gives the stored energy per unit area.
Correct answer is: ½ ε₀εr E² d
Q.15 Which of the following best describes the physical meaning of the term **∂u/∂t** in the Poynting theorem?
Power dissipated as heat
Rate of change of electromagnetic energy density
Magnetic flux density
Electric field intensity
Explanation - **∂u/∂t** quantifies how fast the stored EM energy density changes with time at a point.
Correct answer is: Rate of change of electromagnetic energy density
Q.16 For a perfectly conducting surface, the tangential component of the electric field is:
Maximum
Zero
Equal to the magnetic field
Equal to the surface charge density
Explanation - Boundary condition on a perfect conductor forces **E_tangential = 0**, which also makes the Poynting vector parallel to the surface.
Correct answer is: Zero
Q.17 Consider a rectangular waveguide with width a and height b. Which dimension determines the cutoff frequency of the TE₁₀ mode?
a
b
√(a² + b²)
a + b
Explanation - For TE₁₀, the cutoff wavenumber is **π/a**, so the wider dimension a sets the cutoff frequency.
Correct answer is: a
Q.18 In a resonant cavity, the total time‑averaged stored energy is split equally between electric and magnetic fields. This is a consequence of:
Ohm’s law
Kirchhoff’s voltage law
Energy equipartition theorem
Maxwell’s displacement current
Explanation - At resonance, the average electric and magnetic energies are equal, reflecting equipartition of energy.
Correct answer is: Energy equipartition theorem
Q.19 The net electromagnetic power flowing out of a closed surface is zero if:
The fields are static
The surface encloses no sources and the fields are lossless
The material inside is conductive
The magnetic field is uniform
Explanation - With no sources and no losses, the divergence of **S** is zero, yielding zero net power flow through the surface.
Correct answer is: The surface encloses no sources and the fields are lossless
Q.20 A plane wave in a non‑magnetic dielectric (μ = μ₀) has an electric field amplitude of 200 V/m. What is the magnitude of its magnetic field amplitude?
0.53 A/m
1.68 A/m
75.5 A/m
377 A/m
Explanation - In free space, **H = E/η₀**. So H = 200 V/m ÷ 377 Ω ≈ 0.53 A/m.
Correct answer is: 0.53 A/m
Q.21 Which term in the Poynting theorem accounts for energy conversion to heat in a conducting medium?
∂u/∂t
∇·S
J·E
ε∂E/∂t·E
Explanation - **J·E** (current density dot electric field) represents the power per unit volume dissipated as Joule heating.
Correct answer is: J·E
Q.22 In a waveguide, the group velocity is always less than the speed of light in vacuum because:
Energy is stored in the fields
The wave is attenuated
The waveguide walls are conductive
Magnetic permeability is high
Explanation - Part of the energy oscillates between electric and magnetic fields, reducing the net propagation speed (group velocity) below **c**.
Correct answer is: Energy is stored in the fields
Q.23 For a lossless dielectric, the time‑averaged Poynting vector is parallel to the direction of:
Electric field
Magnetic field
Wave propagation
Surface normal
Explanation - In a lossless medium, **S** points along the direction of energy transport, which coincides with the wave’s propagation direction.
Correct answer is: Wave propagation
Q.24 If a uniform plane wave encounters a perfectly absorbing (matched) load, what happens to the reflected Poynting vector?
It doubles
It becomes zero
It reverses direction
It remains unchanged
Explanation - A matched load eliminates reflections, so only the transmitted (absorbed) Poynting vector exists; reflected component is zero.
Correct answer is: It becomes zero
Q.25 The surface integral of the Poynting vector over a closed surface equals:
Total charge inside the surface
Total electromagnetic energy inside the surface
Net electromagnetic power flowing out of the surface
Magnetic flux through the surface
Explanation - ∮_S **S·dA** gives the total power crossing the surface, positive outward.
Correct answer is: Net electromagnetic power flowing out of the surface
Q.26 Which of the following statements is true for the time‑average Poynting vector of a standing wave?
It is zero everywhere
It points from node to antinode
It is non‑zero only at antinodes
It has the same magnitude as a traveling wave
Explanation - In an ideal standing wave, the instantaneous **S** oscillates back and forth, giving zero average net power flow.
Correct answer is: It is zero everywhere
Q.27 A rectangular loop of wire with area A is placed in a time‑varying magnetic field B(t) = B₀ cos(ωt) × µ̂. The induced emf around the loop is:
-A ω B₀ sin(ωt)
A ω B₀ sin(ωt)
-A B₀ cos(ωt)
0
Explanation - Faraday’s law: **ε = -dΦ/dt = -A dB/dt = -A (-ω B₀ sin ωt) = -A ω B₀ sin ωt**.
Correct answer is: -A ω B₀ sin(ωt)
Q.28 In a waveguide, the power transmitted is given by P = (1/2) Re{∫_S (E × H*)·µ̂ dS}. What does the vector µ̂ represent?
Unit vector normal to the waveguide walls
Unit vector along the direction of propagation
Unit vector of the electric field
Unit vector of the magnetic field
Explanation - The dot product with **µ̂** extracts the component of **S** along the propagation direction, giving total transmitted power.
Correct answer is: Unit vector along the direction of propagation
Q.29 Which physical quantity is conserved according to the integral form of the Poynting theorem?
Total charge
Electromagnetic momentum
Electromagnetic energy
Magnetic flux
Explanation - The theorem expresses the conservation of electromagnetic energy, balancing stored energy, power flow, and dissipated work.
Correct answer is: Electromagnetic energy
Q.30 A wave traveling in a lossy dielectric has complex propagation constant γ = α + jβ. The attenuation constant α primarily affects:
Phase velocity
Wavelength
Amplitude decay
Impedance matching
Explanation - α (neper/m) determines exponential attenuation of the wave amplitude with distance.
Correct answer is: Amplitude decay
Q.31 If the time‑average Poynting vector over a surface is 5 W/m² and the surface area is 0.2 m², what is the total average power crossing the surface?
1 W
0.25 W
10 W
0.5 W
Explanation - P = ⟨S⟩·A = 5 W/m² × 0.2 m² = 1 W.
Correct answer is: 1 W
Q.32 For a uniform plane wave in free space, the ratio of electric field magnitude to magnetic field magnitude is:
μ₀
ε₀
η₀ (≈377 Ω)
c (≈3×10⁸ m/s)
Explanation - In free space, **E/H = η₀**, the intrinsic impedance of free space (~377 Ω).
Correct answer is: η₀ (≈377 Ω)
Q.33 When a wave impinges on a dielectric interface at normal incidence, the reflected Poynting vector magnitude depends on:
Difference of permittivities only
Sum of permeabilities only
Reflection coefficient magnitude squared
Angle of incidence
Explanation - Reflected power = |Γ|² times incident power, where Γ is the Fresnel reflection coefficient.
Correct answer is: Reflection coefficient magnitude squared
Q.34 In a coaxial cable, the characteristic impedance Z₀ is given by:
√(L/C)
L/C
1/(√(LC))
√(μ/ε) * ln(b/a)
Explanation - For a coaxial line, **Z₀ = (1/2π) √(μ/ε) ln(b/a)**, where a and b are inner and outer radii.
Correct answer is: √(μ/ε) * ln(b/a)
Q.35 A perfect magnetic conductor (PMC) enforces which boundary condition on the magnetic field?
Normal component of H = 0
Tangential component of H = 0
Normal component of B = 0
Tangential component of E = 0
Explanation - PMC surfaces force **H_tangential = 0**, the dual of a perfect electric conductor (PEC) which forces **E_tangential = 0**.
Correct answer is: Tangential component of H = 0
Q.36 Which of the following expressions correctly gives the average power delivered to a resistor R by a sinusoidal voltage source V(t)=V₀cos(ωt)?
V₀² / (2R)
V₀² / R
V₀² / (4R)
V₀ / (2R)
Explanation - Average power for sinusoidal voltage is **P_avg = (V_rms)² / R = (V₀/√2)² / R = V₀²/(2R)**.
Correct answer is: V₀² / (2R)
Q.37 In an optical fiber, the core has a higher refractive index than the cladding. This condition ensures:
Total internal reflection and guided modes
Zero attenuation
Higher group velocity than in free space
Uniform Poynting vector across the cross‑section
Explanation - A higher core index causes total internal reflection at the core‑cladding interface, confining light to the core.
Correct answer is: Total internal reflection and guided modes
Q.38 For a TEM mode in a transmission line, the relationship between electric and magnetic fields is:
E ⟂ H and both are transverse to the direction of propagation
E // H
E has a longitudinal component
H has a longitudinal component
Explanation - TEM (Transverse Electromagnetic) mode means **E**, **H**, and propagation direction are mutually orthogonal, with no longitudinal components.
Correct answer is: E ⟂ H and both are transverse to the direction of propagation
Q.39 When computing the total radiated power of a small dipole antenna, which integral is used?
∫_V J·E dV
∮_S S·dA over a sphere
∫_L I dl
∮_C H·dl
Explanation - Radiated power equals the surface integral of the Poynting vector over a closed surface (often a sphere) surrounding the antenna.
Correct answer is: ∮_S S·dA over a sphere
Q.40 The stored electric energy in a parallel‑plate capacitor (area A, separation d) is:
½ ε₀ A V² / d
½ ε₀ A d V²
ε₀ A V / d
½ ε₀ A d / V²
Explanation - Energy U = ½ C V², and C = ε₀ A / d, giving U = ½ ε₀ A V² / d.
Correct answer is: ½ ε₀ A V² / d
Q.41 If the magnetic field in a region doubles while the electric field stays the same, the magnitude of the Poynting vector:
Doubles
Quadruples
Halves
Remains unchanged
Explanation - S = E × H; doubling H doubles the magnitude of **S**, assuming E is unchanged and orthogonal.
Correct answer is: Doubles
Q.42 In a lossy dielectric, the complex permittivity is expressed as ε_c = ε' - jε''. Which term ε'' is related to:
Stored electric energy
Magnetic loss
Dielectric loss (heat)
Conductivity
Explanation - The imaginary part **ε''** quantifies energy dissipation as heat (dielectric loss) in the material.
Correct answer is: Dielectric loss (heat)
Q.43 For a wave incident on a perfectly conducting plane, the reflected wave has a Poynting vector that:
Points in the same direction as the incident wave
Is zero everywhere
Points opposite to the incident wave
Has twice the magnitude of the incident Poynting vector
Explanation - A perfect conductor reflects the wave with a phase reversal, causing the reflected Poynting vector to be opposite the incident direction.
Correct answer is: Points opposite to the incident wave
Q.44 Which of the following best describes the physical significance of the term **∇·S** in the differential form of the Poynting theorem?
Rate of change of magnetic flux
Net power flowing out of an infinitesimal volume
Stored electric energy density
Current density divergence
Explanation - **∇·S** represents the net outward power per unit volume (the divergence of the power flow density).
Correct answer is: Net power flowing out of an infinitesimal volume
Q.45 A rectangular aperture of area 0.01 m² is illuminated by a uniform plane wave with an average Poynting vector of 10 W/m². Assuming all power passes through, what is the power transmitted through the aperture?
0.1 W
1 W
10 W
100 W
Explanation - P = ⟨S⟩·A = 10 W/m² × 0.01 m² = 0.1 W.
Correct answer is: 0.1 W
Q.46 In the context of antenna theory, the radiation resistance is defined using:
Stored magnetic energy
Poynting vector flux through a far‑field sphere
Current distribution on the antenna
Capacitance of the antenna
Explanation - Radiation resistance **R_r = 2P_rad / I_rms²**, where **P_rad** is obtained by integrating the far‑field Poynting vector over a sphere.
Correct answer is: Poynting vector flux through a far‑field sphere
Q.47 The time‑averaged power density in a lossy medium can be expressed as ½ σ|E|². This term represents:
Stored magnetic energy
Power radiated
Joule heating loss
Displacement current power
Explanation - In a conducting medium, **P_loss = ½ σ|E|²**, the average power dissipated as heat per unit volume.
Correct answer is: Joule heating loss
Q.48 Which of the following boundary conditions is true at the interface between two dielectrics for the normal component of the electric displacement field D?
D₁⊥ = D₂⊥
D₁⊥ = ε₁E₁⊥
D₁⊥ - D₂⊥ = σ_f (free surface charge)
D₁⊥ + D₂⊥ = 0
Explanation - The discontinuity in **D⊥** equals the free surface charge density at the interface.
Correct answer is: D₁⊥ - D₂⊥ = σ_f (free surface charge)
Q.49 If a coaxial line is lossless, the total electromagnetic power transmitted is equal to:
The sum of electric and magnetic stored energies
The product of voltage and current at any point
Zero, because there is no loss
The integral of σE² over the cross‑section
Explanation - In a lossless line, **P = V·I** (real power) is constant along the line and equals the integral of the Poynting vector across the cross‑section.
Correct answer is: The product of voltage and current at any point
Q.50 For a sinusoidal steady‑state field, the complex Poynting vector is defined as **S_c = ½ E × H***. What does the real part of **S_c** represent?
Instantaneous power density
Average power density
Reactive power density
Stored energy density
Explanation - The real part of the complex Poynting vector gives the time‑averaged (real) power flow; the imaginary part corresponds to reactive power.
Correct answer is: Average power density
Q.51 The speed of energy propagation in a waveguide (group velocity) can be expressed as:
v_g = c
v_g = c √(1 - (f_c/f)²)
v_g = c / √(1 - (f_c/f)²)
v_g = f λ
Explanation - Group velocity in a waveguide is **v_g = c √(1 - (f_c/f)²)**, where **f_c** is the cutoff frequency.
Correct answer is: v_g = c √(1 - (f_c/f)²)
Q.52 A uniform plane wave in a non‑magnetic dielectric with relative permittivity ε_r has an intrinsic impedance:
η₀ / √ε_r
η₀ √ε_r
η₀ ε_r
η₀ / ε_r
Explanation - Intrinsic impedance in a dielectric: **η = η₀ / √ε_r** (since μ = μ₀).
Correct answer is: η₀ / √ε_r
Q.53 In a resonant LC circuit, the time‑averaged stored electric and magnetic energies are:
Equal at all times
Equal over a full cycle
Never equal
Proportional to the resistance
Explanation - At resonance, energy oscillates between electric (capacitor) and magnetic (inductor) forms, being equal on average over a cycle.
Correct answer is: Equal over a full cycle
Q.54 When a plane wave propagates in a waveguide below cutoff, the fields decay exponentially. The corresponding Poynting vector:
Is purely real
Has zero average power flow
Is directed opposite to propagation
Is infinite
Explanation - Below cutoff, the fields are evanescent; the time‑averaged **S** is zero because there is no net power transport.
Correct answer is: Has zero average power flow
Q.55 Which of the following is a direct consequence of the conservation of electromagnetic momentum?
Poynting theorem
Faraday’s law
Lorentz force law
Snell’s law
Explanation - The Poynting theorem can be derived from the momentum conservation equation combined with Maxwell’s equations.
Correct answer is: Poynting theorem
Q.56 A plane wave with electric field **E = E₀ cos(kz - ωt) × µ̂_x** propagates in free space. What is the direction of the associated magnetic field **H**?
+y
-y
+z
-z
Explanation - Using **H = (1/η₀) µ̂_k × µ̂_E**, with **k** along +z and **E** along +x, **H** points along +y.
Correct answer is: +y
Q.57 If the complex permittivity of a material is ε_c = ε' (1 - j tanδ), the loss tangent tanδ is defined as:
ε''/ε'
ε'/ε''
σ/(ωε')
ωε'/σ
Explanation - Loss tangent tanδ = ε''/ε' = σ/(ωε'), quantifying dielectric losses.
Correct answer is: ε''/ε'
Q.58 Which of the following integrals yields the total electromagnetic energy stored in a volume V?
∫_V (E·D + B·H) dV
∮_S S·dA
∫_V (J·E) dV
∮_C E·dl
Explanation - Stored EM energy density is **u = ½ (E·D + B·H)**; integrating over V gives total energy.
Correct answer is: ∫_V (E·D + B·H) dV
Q.59 In an optical waveguide, the power carried by a mode is proportional to:
The square of the electric field amplitude
The square of the magnetic field amplitude
The product of electric and magnetic field amplitudes
The integral of the Poynting vector over the mode’s cross‑section
Explanation - Mode power equals **∫_A S·µ̂ dA**, integrating the longitudinal component of the Poynting vector across the waveguide cross‑section.
Correct answer is: The integral of the Poynting vector over the mode’s cross‑section
Q.60 Which of the following statements about the Poynting vector in a lossy dielectric is correct?
Its direction is always parallel to the electric field
Its magnitude is independent of conductivity
It includes both real (power) and imaginary (reactive) components
It is zero inside the material
Explanation - In lossy media, the complex Poynting vector has a real part (average power flow) and an imaginary part (reactive power).
Correct answer is: It includes both real (power) and imaginary (reactive) components
Q.61 For a cylindrical waveguide operating in the TE₁₁ mode, the dominant field component is:
E_z
H_z
E_φ
H_φ
Explanation - In TE modes, the longitudinal electric field is zero; the dominant longitudinal component is **H_z**.
Correct answer is: H_z
Q.62 The radiated power of a small (Hertzian) dipole antenna of length l << λ, carrying current I₀, is proportional to:
I₀² l² / λ²
I₀² l⁴ / λ²
I₀ l / λ
I₀² λ / l
Explanation - Radiated power **P_rad ∝ (I₀ l)² (k⁴) = I₀² l⁴ / λ⁴**, but after constants, the dominant dependence is **I₀² l⁴ / λ²** (since k = 2π/λ).
Correct answer is: I₀² l⁴ / λ²
Q.63 A waveguide is terminated with its characteristic impedance. What happens to the reflected Poynting vector?
It becomes equal in magnitude to the incident one
It vanishes
It reverses direction
It doubles in magnitude
Explanation - Impedance matching eliminates reflections, so the reflected power (and thus reflected Poynting vector) is zero.
Correct answer is: It vanishes
Q.64 If an electromagnetic wave propagates in a medium with conductivity σ, the attenuation constant α (in nepers per meter) is approximately:
σ / (2√(με))
√(π f μ σ)
ω√(με)
σ / (√(με))
Explanation - For a good conductor (σ ≫ ωε), **α ≈ √(π f μ σ) ≈ σ / (2√(με))**; the simpler low‑frequency approximation is σ/(2√(με)).
Correct answer is: σ / (2√(με))
Q.65 The quantity **S·n̂ dA**, where **n̂** is the outward normal to a surface element, represents:
Electric flux
Magnetic flux
Differential power crossing the surface
Charge enclosed
Explanation - **S·n̂ dA** is the infinitesimal power flowing through the differential area **dA**.
Correct answer is: Differential power crossing the surface
Q.66 Which of the following best describes why the Poynting vector can be used to calculate the radiation pressure exerted by light on a surface?
Because it measures electric field intensity
Because its magnitude equals the momentum flux density
Because it is proportional to magnetic permeability
Because it represents charge density
Explanation - Radiation pressure = power flux / c = **S/c**, linking the Poynting vector to momentum transfer.
Correct answer is: Because its magnitude equals the momentum flux density
Q.67 In a coaxial line, if the inner conductor radius is a and the outer conductor radius is b, the capacitance per unit length is:
2π ε / ln(b/a)
π ε / (b - a)
ε ln(b/a) / (2π)
ε / (2π ln(b/a))
Explanation - Capacitance per unit length **C' = 2π ε / ln(b/a)** for a coaxial geometry.
Correct answer is: 2π ε / ln(b/a)
Q.68 If the phase velocity v_p of a waveguide mode exceeds the speed of light, does this violate relativity?
Yes, because information travels faster than light
No, because phase velocity does not convey information
Yes, because energy flows at v_p
No, because v_p is always less than c
Explanation - Phase velocity can exceed **c** without violating causality; only group velocity (information speed) is limited to **c**.
Correct answer is: No, because phase velocity does not convey information
Q.69 The reactive power associated with the imaginary part of the complex Poynting vector is measured in:
Watts
Volt‑amps reactive (VAR)
Amperes
Henries
Explanation - Imaginary part of complex power corresponds to reactive power, expressed in VAR.
Correct answer is: Volt‑amps reactive (VAR)
Q.70 A waveguide is filled with a material of relative permeability μ_r = 4 and relative permittivity ε_r = 1. The wave impedance inside the guide is:
η₀ √(μ_r / ε_r)
η₀ √(ε_r / μ_r)
η₀ / √(μ_r ε_r)
η₀ √(μ_r ε_r)
Explanation - Intrinsic impedance **η = η₀ √(μ_r / ε_r)**; substituting the values yields **η = η₀ √4 = 2η₀**.
Correct answer is: η₀ √(μ_r / ε_r)
Q.71 Which term in the Poynting theorem represents the rate at which electromagnetic energy is converted to mechanical work on charges?
∇·S
∂u/∂t
J·E
σE²
Explanation - **J·E** quantifies the power per unit volume delivered to charges, i.e., mechanical work (or heating).
Correct answer is: J·E
Q.72 For a TM mode in a rectangular waveguide, which field component is zero?
E_z
H_z
E_x
H_y
Explanation - TM (Transverse Magnetic) modes have zero longitudinal magnetic field component (**H_z = 0**).
Correct answer is: H_z
Q.73 If the electric field of a plane wave is polarized at 45° to the x‑axis in the xy‑plane, the Poynting vector will be:
Along the x‑axis
Along the y‑axis
Along the propagation direction, independent of polarization
At 45° to the propagation direction
Explanation - Regardless of polarization, **S = E × H** points along the wave’s propagation direction.
Correct answer is: Along the propagation direction, independent of polarization
Q.74 A wave propagating in a waveguide has a phase constant β = 2π/λ_g, where λ_g is the guided wavelength. The group velocity is given by:
v_g = ω / β
v_g = dω/dβ
v_g = λ_g f
v_g = c² / v_p
Explanation - Group velocity is the derivative of angular frequency with respect to phase constant: **v_g = dω/dβ**.
Correct answer is: v_g = dω/dβ
Q.75 The time‑averaged power transmitted by a uniform plane wave through a circular aperture of radius r is proportional to:
r²
r⁴
r
1/r
Explanation - Aperture area A = π r², so transmitted power = ⟨S⟩·A ∝ r².
Correct answer is: r²
Q.76 In a lossy transmission line, the power attenuation per unit length α (in nepers/m) is related to the line resistance per unit length R and characteristic impedance Z₀ by:
α = R / (2 Z₀)
α = 2 R Z₀
α = R Z₀²
α = √(R / Z₀)
Explanation - For a low‑loss line, **α ≈ R / (2 Z₀)** gives the attenuation constant.
Correct answer is: α = R / (2 Z₀)
Q.77 Which of the following statements about the Poynting vector in a static (electrostatic) field is correct?
It is zero everywhere
It points from high to low potential
It equals ε₀ E × B
It represents magnetic flux
Explanation - In static fields, there is no time‑varying magnetic field, so **H = 0** and the Poynting vector **S = E × H = 0**.
Correct answer is: It is zero everywhere
Q.78 A waveguide is filled with a dielectric of ε_r = 4. Its cutoff frequency for the TE₁₀ mode (width a) becomes:
f_c = c / (2a)
f_c = c / (2a√ε_r)
f_c = c √ε_r / (2a)
f_c = c / (a√ε_r)
Explanation - Cutoff frequency **f_c = (1/2a)·(c/√ε_r)** for TE₁₀ in a dielectric‑filled guide.
Correct answer is: f_c = c / (2a√ε_r)
Q.79 For a sinusoidal electromagnetic wave, the instantaneous Poynting vector can be expressed as:
S(t) = E(t)·H(t)
S(t) = E(t) × H(t)
S(t) = ½ Re{E × H*}
S(t) = ε₀ E(t)²
Explanation - Instantaneous power density is given by the vector cross product **S = E × H**.
Correct answer is: S(t) = E(t) × H(t)
Q.80 The magnitude of the time‑averaged Poynting vector for a plane wave in free space is equal to:
E₀² / η₀
E₀² η₀
½ E₀² / η₀
½ E₀² η₀
Explanation - Average power density **⟨S⟩ = (1/2) (E₀² / η₀)** for a sinusoidal plane wave.
Correct answer is: ½ E₀² / η₀
Q.81 When a wave propagates from medium 1 (impedance η₁) to medium 2 (impedance η₂), the reflected power coefficient (|Γ|²) is given by:
(η₂ - η₁)² / (η₂ + η₁)²
(η₁ - η₂)² / (η₁ + η₂)²
(η₁ η₂)²
η₁ / η₂
Explanation - Reflection coefficient **Γ = (η₂ - η₁)/(η₂ + η₁)**; reflected power = |Γ|².
Correct answer is: (η₂ - η₁)² / (η₂ + η₁)²
Q.82 The electromagnetic momentum density **g** in a lossless medium is related to the Poynting vector by:
g = S / c²
g = S / c
g = S × c
g = c S
Explanation - Momentum density **g = S / c²**, reflecting the relationship between energy flux and momentum in electromagnetic fields.
Correct answer is: g = S / c²
Q.83 A uniform plane wave in a medium with conductivity σ experiences power loss per unit volume of:
σ E²
½ σ |E|²
σ H²
½ σ |H|²
Explanation - Average Joule loss in a sinusoidal field is **P_loss = ½ σ |E|²**.
Correct answer is: ½ σ |E|²
Q.84 Which of the following is true about the energy flow in a standing wave formed by two equal‑amplitude traveling waves?
Net average energy flow is zero
Energy flows only towards the nodes
Energy flows only towards the antinodes
Average energy flow equals twice that of a single traveling wave
Explanation - In a standing wave the instantaneous power oscillates back and forth, resulting in zero net average energy transport.
Correct answer is: Net average energy flow is zero
Q.85 The power transmitted by a lossless TEM transmission line is given by:
P = V² / Z₀
P = I² Z₀
P = (V I)/2
Both A and B
Explanation - For a TEM line, **P = V I = V² / Z₀ = I² Z₀**; all three expressions are equivalent.
Correct answer is: Both A and B
Q.86 A waveguide operates at a frequency well above its cutoff. Which of the following statements about its phase velocity v_p is correct?
v_p < c
v_p = c
v_p > c
v_p = 0
Explanation - Above cutoff, the phase velocity in a waveguide exceeds the speed of light, while the group velocity remains below c.
Correct answer is: v_p > c
Q.87 The average energy density stored in the magnetic field of a wave is:
½ μ |H|²
μ |H|²
½ ε |E|²
ε |E|²
Explanation - Magnetic energy density is **u_m = ½ μ |H|²**.
Correct answer is: ½ μ |H|²
Q.88 In a rectangular waveguide, the dominant TE₁₀ mode has an electric field that is primarily oriented:
Along the direction of propagation
Across the wider dimension (width a)
Across the narrower dimension (height b)
Radially outward
Explanation - For TE₁₀, **E_y** (across the height b) is the dominant component; the field varies sinusoidally across the width a.
Correct answer is: Across the narrower dimension (height b)
Q.89 The Poynting vector in a lossy dielectric can be expressed as **S = S_real + j S_imag**. The imaginary part **S_imag** is associated with:
Radiated power
Stored (reactive) power
Resistive loss
Magnetic flux
Explanation - The imaginary component of complex power density represents reactive power, which oscillates between fields without net transfer.
Correct answer is: Stored (reactive) power
Q.90 A plane wave in a medium with relative permittivity ε_r and relative permeability μ_r has a wavelength λ = λ₀ / √(ε_r μ_r). If ε_r = 4 and μ_r = 1, the wavelength is:
λ₀ / 2
λ₀ / 4
2 λ₀
4 λ₀
Explanation - λ = λ₀ / √(4·1) = λ₀ / 2.
Correct answer is: λ₀ / 2
Q.91 Which of the following integrals yields the total electromagnetic force exerted on a surface by an incident wave?
∮_S (S·n̂) dA
∮_S (T·n̂) dA where T is the Maxwell stress tensor
∫_V J·E dV
∮_C E·dl
Explanation - The force on a surface is obtained by integrating the Maxwell stress tensor over that surface.
Correct answer is: ∮_S (T·n̂) dA where T is the Maxwell stress tensor
Q.92 In a waveguide filled with a material having loss tangent tanδ = 0.01, the power lost per unit length is primarily due to:
Conduction loss
Radiation loss
Dielectric loss
Magnetic loss
Explanation - A small loss tangent indicates dielectric loss dominates; conduction and magnetic losses are negligible.
Correct answer is: Dielectric loss
Q.93 The time‑averaged Poynting vector over a perfectly conducting surface is:
Zero
Maximum
Equal to the incident wave’s magnitude
Perpendicular to the surface
Explanation - On a perfect conductor, the tangential electric field is zero, making **S** parallel to the surface and its normal component (power crossing) zero.
Correct answer is: Zero
Q.94 A waveguide is excited in the TE₁₁ mode. The cutoff frequency f_c is proportional to:
1 / a
1 / b
√(1/a² + 1/b²)
a + b
Explanation - For TE_mn, **f_c = (c/2) √[(m/a)² + (n/b)²]**; for TE₁₁, m=n=1.
Correct answer is: √(1/a² + 1/b²)
Q.95 When an EM wave is incident on a dielectric slab at normal incidence, the transmitted power is maximized when the slab thickness is:
Zero
An integer multiple of half wavelengths inside the material
An odd multiple of quarter wavelengths inside the material
Equal to the wavelength in free space
Explanation - Constructive interference (no reflection) occurs when the slab thickness equals **n λ/2** inside the material.
Correct answer is: An integer multiple of half wavelengths inside the material
Q.96 The quantity **∮_S (E × H)·dA** over a closed surface surrounding a radiating antenna is equal to:
Total radiated power
Stored energy in the antenna
Charge on the antenna
Magnetic flux through the surface
Explanation - Surface integral of the Poynting vector over a sphere surrounding the antenna gives the total radiated power.
Correct answer is: Total radiated power
Q.97 In a TEM transmission line, the characteristic impedance Z₀ is independent of:
Frequency
Geometry
Material permittivity
All of the above
Explanation - For lossless TEM lines, **Z₀ = √(L/C)** depends only on geometry and material constants, not on frequency.
Correct answer is: Frequency
Q.98 The power flow through a surface element dA is given by **dP = S·n̂ dA**. If S = 3 W/m² and n̂ is opposite to S, the power crossing dA is:
3 W
-3 W
0 W
Depends on dA size
Explanation - Negative sign indicates power flowing opposite to the outward normal (into the volume).
Correct answer is: -3 W
Q.99 A wave traveling in a waveguide has a phase velocity v_p = 2c. Its group velocity v_g must be:
c/2
2c
c
√3 c
Explanation - For waveguide modes, **v_p·v_g = c²**; thus **v_g = c² / v_p = c² / (2c) = c/2**.
Correct answer is: c/2
Q.100 The complex power **P = ½ ∫_S (E × H*)·dS** can be written as **P = P_real + j P_imag**. What does **P_imag** represent?
Average transmitted power
Reactive power
Resistive loss
Stored kinetic energy
Explanation - Imaginary part of complex power corresponds to reactive (non‑real) power, measured in VAR.
Correct answer is: Reactive power
Q.101 In the Poynting theorem, the term **-∇·S** is interpreted as:
Power generated inside the volume
Power entering the volume
Power leaving the volume
Power stored in the volume
Explanation - Negative divergence indicates net outward flow of power from the volume.
Correct answer is: Power leaving the volume
Q.102 If an EM wave propagates in a medium with permittivity ε and permeability μ, the speed of propagation is:
c
1/√(ε μ)
√(ε/μ)
√(μ/ε)
Explanation - Wave speed **v = 1/√(ε μ)**, reducing to **c** in free space.
Correct answer is: 1/√(ε μ)
Q.103 A rectangular aperture of dimensions a × b is illuminated by a uniform plane wave. The total power transmitted is proportional to:
a + b
ab
a² + b²
√(a² + b²)
Explanation - Aperture area = a·b; transmitted power = ⟨S⟩·area ∝ ab.
Correct answer is: ab
Q.104 The Poynting vector inside a perfect electric conductor (PEC) is:
Zero
Maximum
Parallel to the surface
Perpendicular to the surface
Explanation - Inside a PEC, both **E** and **H** are zero, giving **S = 0**.
Correct answer is: Zero
Q.105 For a uniform plane wave, the ratio of the time‑averaged magnetic power density to the electric power density is:
η₀
1/η₀
η₀²
1
Explanation - Average magnetic power density **⟨S_H⟩ = ⟨H⟩·⟨E⟩ = ⟨E⟩² / η₀**, so the ratio **⟨S_H⟩ / ⟨S_E⟩ = 1/η₀**.
Correct answer is: 1/η₀
Q.106 A waveguide is filled with a magnetic material (μ_r > 1) but ε_r = 1. Compared with an empty waveguide, the cutoff frequency for the TE₁₀ mode:
Increases
Decreases
Remains unchanged
Becomes zero
Explanation - Cutoff frequency **f_c = (c/2a)·1/√(ε_r μ_r)**; increasing μ_r lowers **f_c**.
Correct answer is: Decreases
Q.107 Which quantity has the same units as the Poynting vector?
Energy density (J/m³)
Power (W)
Power per unit area (W/m²)
Electric field (V/m)
Explanation - The Poynting vector **S** has units of watts per square meter (W/m²).
Correct answer is: Power per unit area (W/m²)
Q.108 The stored electromagnetic energy in a lossless resonant cavity is given by:
U = ½ ∫_V (ε|E|² + μ|H|²) dV
U = ∫_V (J·E) dV
U = ∮_S S·dA
U = ½ ∫_V (σ|E|²) dV
Explanation - Energy stored in electric and magnetic fields adds up to **U = ½∫ (ε|E|² + μ|H|²) dV**.
Correct answer is: U = ½ ∫_V (ε|E|² + μ|H|²) dV
Q.109 If the incident power on a surface is 10 W and the reflected power is 2 W, the net power absorbed by the surface is:
12 W
8 W
2 W
10 W
Explanation - Net absorbed power = incident - reflected = 10 W - 2 W = 8 W.
Correct answer is: 8 W
Q.110 In a waveguide, the transverse electric field component for the TE₁₀ mode varies sinusoidally across the width a as:
sin(π x / a)
cos(π x / a)
sin(2π x / a)
cos(2π x / a)
Explanation - TE₁₀ has **E_y ∝ sin(π x / a)**, with a half‑wave variation across the width.
Correct answer is: sin(π x / a)
Q.111 The instantaneous Poynting vector **S(t) = E(t) × H(t)** has units of:
V·A
W/m²
J/m³
V/m
Explanation - Cross product of **E (V/m)** and **H (A/m)** yields **W/m²**, the unit of power flux density.
Correct answer is: W/m²
Q.112 In a medium with loss tangent tanδ << 1, the power lost per cycle is:
Proportional to tanδ
Independent of tanδ
Inversely proportional to tanδ
Zero
Explanation - For small loss tangent, loss per cycle ≈ tanδ times the stored energy, so proportional to tanδ.
Correct answer is: Proportional to tanδ
Q.113 The Poynting vector for a circularly polarized wave has:
Rotating magnitude
Constant magnitude and rotating direction
Zero magnitude
Constant magnitude and fixed direction
Explanation - Circular polarization yields a constant |S|, but the instantaneous direction of **E** and **H** rotates, causing **S** to rotate as well while maintaining constant magnitude.
Correct answer is: Constant magnitude and rotating direction
Q.114 For an electromagnetic wave in a dispersive medium, the group velocity is defined as:
v_g = dβ/dω
v_g = ω/β
v_g = dω/dβ
v_g = β/ω
Explanation - Group velocity is the derivative of angular frequency with respect to phase constant: **v_g = dω/dβ**.
Correct answer is: v_g = dω/dβ
Q.115 If a waveguide is terminated with a short circuit, the reflected wave will have a phase shift of:
0°
90°
180°
270°
Explanation - A short reflects the wave with a 180° phase reversal (Γ = -1).
Correct answer is: 180°
Q.116 The net electromagnetic force on a current‑carrying loop in a magnetic field can be derived from which theorem?
Poynting theorem
Gauss's law
Stokes' theorem
Lorentz force law
Explanation - Applying the Poynting theorem to the volume enclosing the loop and using the Maxwell stress tensor yields the net force.
Correct answer is: Poynting theorem