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SAIL MCQs

Practice SAIL-oriented Electronics & Telecommunication Engineering MCQs with answers and explanations for technical exam preparation. Questions are arranged in pages of up to 100.

Question 901

Electromagnetic TheoryMagnetostatics

In a linear isotropic magnetic medium, B and H are related by?

AB = mu H
BB = epsilon H
CB = sigma H
DB = H/mu²

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Correct Answer: A. B = mu H

Explanation: For a linear isotropic medium, magnetic flux density is B=μH.

Reference: Standard electromagnetic theory: magnetostatics

Question 902

Electromagnetic TheoryBiot-Savart Law

The Biot-Savart law is used to calculate magnetic field produced by?

ACurrent elements
BStationary electric charges only
CDielectric polarization only
DScalar potential without current

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Correct Answer: A. Current elements

Explanation: Biot-Savart law gives the magnetic field contribution from electric current elements.

Reference: Standard electromagnetic theory: biot-savart law

Question 903

Electromagnetic TheoryPotential Equations

In a charge-free region with uniform permittivity, the electrostatic potential V satisfies?

ALaplace's equation
BPoisson's equation with nonzero charge density
CWave equation only
DContinuity equation only

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Correct Answer: A. Laplace's equation

Explanation: When volume charge density is zero, Poisson's equation reduces to Laplace's equation, ∇²V=0.

Reference: Standard electromagnetic theory: potential equations

Question 904

Electromagnetic TheoryPotential Equations

Poisson’s equation for electrostatic potential in a homogeneous medium is?

A∇²V = -rho_v/epsilon
B∇²V = rho_v epsilon
C∇×V = -rho_v/epsilon
D∇·V = 0 always

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Correct Answer: A. ∇²V = -rho_v/epsilon

Explanation: Combining Gauss's law with E=-∇V gives ∇²V=-ρv/ε.

Reference: Standard electromagnetic theory: potential equations

Question 905

Electromagnetic TheoryBoundary Conditions

At the boundary between two perfect dielectric media with no free surface charge, which component of electric flux density D is continuous?

ANormal component
BTangential component only
CBoth components must be zero
DNeither component

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Correct Answer: A. Normal component

Explanation: The boundary condition is D1n-D2n=ρs. With no free surface charge, the normal components are equal.

Reference: Standard electromagnetic theory: boundary conditions

Question 906

Electromagnetic TheoryBoundary Conditions

At an interface between two media under electrostatic conditions, the tangential component of electric field E is?

AContinuous
BAlways zero
CAlways discontinuous by epsilon ratio
DEqual to surface charge density

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Correct Answer: A. Continuous

Explanation: Electrostatic boundary conditions require E1t=E2t.

Reference: Standard electromagnetic theory: boundary conditions

Question 907

Electromagnetic TheoryBoundary Conditions

At the surface of a perfect conductor in electrostatic equilibrium, the tangential electric field is?

AZero
BMaximum
CEqual to normal electric field
DEqual to current density

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Correct Answer: A. Zero

Explanation: A nonzero tangential electric field would drive free charges, so electrostatic equilibrium requires Et=0.

Reference: Standard electromagnetic theory: boundary conditions

Question 908

Electromagnetic TheoryCurrent Density

The conduction current density J in an isotropic conductor is related to electric field E by?

AJ = sigma E
BJ = epsilon E
CJ = mu E
DJ = E/sigma²

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Correct Answer: A. J = sigma E

Explanation: Point-form Ohm's law is J=σE, where σ is conductivity.

Reference: Standard electromagnetic theory: current density

Question 909

Electromagnetic TheoryCurrent Density

The SI unit of current density J is?

AA/m²
BA/m
CC/m²
DV/m²

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Correct Answer: A. A/m²

Explanation: Current density is electric current per unit cross-sectional area, measured in amperes per square metre.

Reference: Standard electromagnetic theory: current density

Question 910

Electromagnetic TheoryCurrent Density

A conductor of cross-sectional area 2 mm² carries a uniform current of 4 A. The current density magnitude is?

A2×10^3 A/m²
B2×10^6 A/m²
C8×10^6 A/m²
D0.5×10^-6 A/m²

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Correct Answer: B. 2×10^6 A/m²

Explanation: Area=2×10^-6 m², so J=I/A=4/(2×10^-6)=2×10^6 A/m².

Reference: Standard electromagnetic theory: current density

Question 911

Electromagnetic TheoryContinuity Equation

The general differential form of the charge continuity equation is?

A∇·J = -partial rho_v/partial t
B∇×J = -partial rho_v/partial t
C∇·J = rho_v
D∇×E = J

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Correct Answer: A. ∇·J = -partial rho_v/partial t

Explanation: Charge conservation is expressed locally by ∇·J=-∂ρv/∂t.

Reference: Standard electromagnetic theory: continuity equation

Question 912

Electromagnetic TheoryMagnetic Field

The direction of magnetic field around a long straight current-carrying conductor is determined by the?

ARight-hand rule
BLeft-hand electrostatic rule
CLenz law only
DGauss electric law only

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Correct Answer: A. Right-hand rule

Explanation: Curling the fingers of the right hand around the conductor with the thumb along current gives the magnetic-field direction.

Reference: Standard electromagnetic theory: magnetic field

Question 913

Electromagnetic TheoryAmpere's Law

For an infinitely long straight conductor carrying current I in a medium of permeability mu, the magnetic flux density at radial distance r is?

Amu I/(2 pi r)
Bmu I/(4 pi r²)
CI r/(2 pi mu)
Dmu I r/2

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Correct Answer: A. mu I/(2 pi r)

Explanation: Using Ampere's law around a circular path gives H=I/(2πr), hence B=μI/(2πr).

Reference: Standard electromagnetic theory: ampere's law

Question 914

Electromagnetic TheoryAmpere's Law

A long straight wire carries 10 A. The magnetic field intensity H at a distance 0.1 m from it is approximately?

A1.59 A/m
B15.9 A/m
C50 A/m
D100 A/m

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Correct Answer: B. 15.9 A/m

Explanation: H=I/(2πr)=10/(2π×0.1)≈15.9 A/m.

Reference: Standard electromagnetic theory: ampere's law

Question 915

Electromagnetic TheoryAmpere's Law

Ampere’s circuital law for steady currents states that the closed line integral of H equals?

AEnclosed conduction current
BEnclosed electric charge
CMagnetic flux through the surface
DElectric potential difference

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Correct Answer: A. Enclosed conduction current

Explanation: For magnetostatics, ∮H·dl equals the net current enclosed by the contour.

Reference: Standard electromagnetic theory: ampere's law

Question 916

Electromagnetic TheoryMaxwell Equations Basics

Gauss’s law for magnetism states that the net magnetic flux through any closed surface is?

AZero
BEqual to enclosed current
CEqual to enclosed magnetic charge
DAlways one weber

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Correct Answer: A. Zero

Explanation: Because isolated magnetic monopoles are not part of classical Maxwell theory, ∮B·dS=0.

Reference: Standard electromagnetic theory: maxwell equations basics

Question 917

Electromagnetic TheoryMaxwell Equations Basics

The differential form of Gauss’s law for magnetism is?

A∇·B = 0
B∇×B = 0 always
C∇·B = rho_v
D∇×B = -partial D/partial t

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Correct Answer: A. ∇·B = 0

Explanation: The magnetic flux density is divergence-free: ∇·B=0.

Reference: Standard electromagnetic theory: maxwell equations basics

Question 918

Electromagnetic TheoryLorentz Force

The magnetic force on a charge q moving with velocity v in magnetic flux density B is?

Aq(v × B)
Bq(v · B)
CqB/v
Dqv/B

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Correct Answer: A. q(v × B)

Explanation: The magnetic part of Lorentz force is F=q(v×B).

Reference: Standard electromagnetic theory: lorentz force

Question 919

Electromagnetic TheoryLorentz Force

A charged particle moving exactly parallel to a uniform magnetic field experiences magnetic force of magnitude?

AZero
BqvB
C2qvB
DInfinite

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Correct Answer: A. Zero

Explanation: F=qvB sinθ; for parallel motion θ=0, so the magnetic force is zero.

Reference: Standard electromagnetic theory: lorentz force

Question 920

Electromagnetic TheoryLorentz Force

A positive charge of 2 C moves at 3 m/s perpendicular to a uniform magnetic field of 4 T. The magnetic force magnitude is?

A6 N
B8 N
C12 N
D24 N

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Correct Answer: D. 24 N

Explanation: F=qvB sin90°=2×3×4=24 N.

Reference: Standard electromagnetic theory: lorentz force

Question 921

Electromagnetic TheoryMagnetic Force

The force on a straight conductor of length L carrying current I in a uniform magnetic field B is, in vector form?

AF = I(L × B)
BF = I(L · B)
CF = B/(IL)
DF = IL/B

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Correct Answer: A. F = I(L × B)

Explanation: The magnetic force on a straight current element is F=I(L×B).

Reference: Standard electromagnetic theory: magnetic force

Question 922

Electromagnetic TheoryMagnetic Force

A 0.5 m conductor carries 4 A perpendicular to a uniform 0.2 T magnetic field. The force magnitude is?

A0.04 N
B0.1 N
C0.4 N
D4 N

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Correct Answer: C. 0.4 N

Explanation: F=BIL sin90°=0.2×4×0.5=0.4 N.

Reference: Standard electromagnetic theory: magnetic force

Question 923

Electromagnetic TheoryFaraday's Law

Faraday’s law states that the induced emf around a closed path is equal to?

AThe negative time rate of change of magnetic flux linkage
BThe magnetic flux itself
CThe electric charge enclosed
DThe resistance multiplied by magnetic field

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Correct Answer: A. The negative time rate of change of magnetic flux linkage

Explanation: Faraday's law gives induced emf as the negative time derivative of magnetic flux linkage.

Reference: Standard electromagnetic theory: faraday's law

Question 924

Electromagnetic TheoryFaraday's Law

The negative sign in Faraday’s law is associated with?

ALenz's law
BCoulomb's law
CGauss's electric law
DOhm's law

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Correct Answer: A. Lenz's law

Explanation: The negative sign expresses Lenz's law: the induced effect opposes the change in magnetic flux that produces it.

Reference: Standard electromagnetic theory: faraday's law

Question 925

Electromagnetic TheoryFaraday's Law

A coil of 100 turns experiences a uniform change in flux per turn from 5 mWb to 1 mWb in 0.02 s. The magnitude of average induced emf is?

A5 V
B10 V
C20 V
D40 V

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Correct Answer: C. 20 V

Explanation: |e|=N|ΔPhi|/Δt=100×4×10^-3/0.02=20 V.

Reference: Standard electromagnetic theory: faraday's law

Question 926

Electromagnetic TheoryFaraday's Law

A stationary closed conducting loop is placed in a magnetic field that is constant with time. If the flux through the loop does not change, the transformer-induced emf is?

AZero
BMaximum
CEqual to magnetic flux
DAlways 1 V

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Correct Answer: A. Zero

Explanation: Faraday's law requires a time-varying flux linkage for transformer emf; if flux is constant, induced emf is zero.

Reference: Standard electromagnetic theory: faraday's law

Question 927

Electromagnetic TheoryMotional EMF

Motional emf in a straight conductor of length l moving with speed v perpendicular to a uniform magnetic field B is?

ABlv
BB/(lv)
Clv/B
DB²lv

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Correct Answer: A. Blv

Explanation: For mutually perpendicular B, conductor length and velocity, the motional emf magnitude is e=Blv.

Reference: Standard electromagnetic theory: motional emf

Question 928

Electromagnetic TheoryMotional EMF

A 0.5 m conductor moves at 10 m/s perpendicular to a 0.2 T magnetic field. Its motional emf is?

A0.1 V
B0.5 V
C1 V
D10 V

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Correct Answer: C. 1 V

Explanation: e=Blv=0.2×0.5×10=1 V.

Reference: Standard electromagnetic theory: motional emf

Question 929

Electromagnetic TheoryInductance

Self-inductance of a coil is defined from flux linkage lambda and current I by?

AL = lambda/I for a linear magnetic system
BL = I/lambda
CL = lambda I²
DL = I²/lambda

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Correct Answer: A. L = lambda/I for a linear magnetic system

Explanation: For a linear magnetic system, flux linkage λ=LI, so L=λ/I.

Reference: Standard electromagnetic theory: inductance

Question 930

Electromagnetic TheoryInductance

The SI unit of inductance is?

AHenry
BTesla
CWeber per square metre
DFarad

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Correct Answer: A. Henry

Explanation: Inductance is measured in henries (H).

Reference: Standard electromagnetic theory: inductance

Question 931

Electromagnetic TheoryInductance

The voltage across an ideal inductor of constant inductance L is?

Av = L di/dt
Bv = i/L
Cv = Li
Dv = L/i

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Correct Answer: A. v = L di/dt

Explanation: For a constant linear inductance, the induced terminal voltage is proportional to the rate of change of current: v=L di/dt.

Reference: Standard electromagnetic theory: inductance

Question 932

Electromagnetic TheoryInductance

An inductor of 2 H carries a current increasing at 3 A/s. The magnitude of its induced voltage is?

A1.5 V
B5 V
C6 V
D9 V

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Correct Answer: C. 6 V

Explanation: v=L di/dt=2×3=6 V.

Reference: Standard electromagnetic theory: inductance

Question 933

Electromagnetic TheoryMagnetic Energy

The energy stored in a linear inductor carrying current I is?

A(1/2)LI²
BLI²
C(1/2)L²I
DI²/L

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Correct Answer: A. (1/2)LI²

Explanation: Magnetic energy stored in a linear inductor is W=(1/2)LI².

Reference: Standard electromagnetic theory: magnetic energy

Question 934

Electromagnetic TheoryMagnetic Energy

A 4 H inductor carries a steady current of 2 A. Its stored magnetic energy is?

A2 J
B4 J
C8 J
D16 J

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Correct Answer: C. 8 J

Explanation: W=(1/2)LI²=0.5×4×2²=8 J.

Reference: Standard electromagnetic theory: magnetic energy

Question 935

Electromagnetic TheoryMaxwell Equations

The displacement current density in a linear medium is represented by?

Apartial D/partial t
Bpartial B/partial t
Csigma E only
Drho_v E

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Correct Answer: A. partial D/partial t

Explanation: The displacement current density is Jd=∂D/∂t.

Reference: Standard electromagnetic theory: maxwell equations

Question 936

Electromagnetic TheoryMaxwell Equations

The Maxwell-Ampere equation in differential form is?

Acurl H = J + partial D/partial t
Bcurl E = J + partial D/partial t
Cdiv H = J
Ddiv D = J + partial B/partial t

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Correct Answer: A. curl H = J + partial D/partial t

Explanation: Maxwell-Ampere law is ∇×H=J+∂D/∂t.

Reference: Standard electromagnetic theory: maxwell equations

Question 937

Electromagnetic TheoryMaxwell Equations

Faraday’s law in differential form is?

Acurl E = -partial B/partial t
Bdiv E = -partial B/partial t
Ccurl H = -partial B/partial t
Ddiv B = -partial E/partial t

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Correct Answer: A. curl E = -partial B/partial t

Explanation: The differential Faraday law is ∇×E=-∂B/∂t.

Reference: Standard electromagnetic theory: maxwell equations

Question 938

Electromagnetic TheoryEM Wave Fundamentals

In a source-free lossless homogeneous medium, electromagnetic waves are?

ATransverse, with E and H perpendicular to the direction of propagation
BPurely longitudinal with E parallel to propagation
CPurely electrostatic
DUnable to carry energy

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Correct Answer: A. Transverse, with E and H perpendicular to the direction of propagation

Explanation: Uniform plane electromagnetic waves in a source-free isotropic medium have mutually perpendicular E, H and propagation directions.

Reference: Standard electromagnetic theory: em wave fundamentals

Question 939

Electromagnetic TheoryEM Wave Fundamentals

The speed of an electromagnetic wave in a lossless medium of permeability mu and permittivity epsilon is?

A1/sqrt(mu epsilon)
Bsqrt(mu epsilon)
Cmu/epsilon
Depsilon/mu

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Correct Answer: A. 1/sqrt(mu epsilon)

Explanation: The phase velocity in a lossless homogeneous medium is v=1/sqrt(με).

Reference: Standard electromagnetic theory: em wave fundamentals

Question 940

Electromagnetic TheoryEM Wave Fundamentals

The approximate speed of an electromagnetic wave in free space is?

A3×10^6 m/s
B3×10^8 m/s
C3×10^10 m/s
D9×10^16 m/s

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Correct Answer: B. 3×10^8 m/s

Explanation: Electromagnetic waves propagate in vacuum at approximately 3×10^8 m/s.

Reference: Standard electromagnetic theory: em wave fundamentals

Question 941

Electromagnetic TheoryEM Wave Fundamentals

A free-space electromagnetic wave has frequency 100 MHz. Its wavelength is approximately?

A0.3 m
B3 m
C30 m
D300 m

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Correct Answer: B. 3 m

Explanation: λ=c/f=(3×10^8)/(100×10^6)=3 m.

Reference: Standard electromagnetic theory: em wave fundamentals

Question 942

Electromagnetic TheoryEM Wave Fundamentals

The intrinsic impedance of free space is approximately?

A50 ohm
B75 ohm
C120 pi ohm
D600 ohm

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Correct Answer: C. 120 pi ohm

Explanation: η0=sqrt(mu0/epsilon0)=120π ohm, approximately 377 ohm.

Reference: Standard electromagnetic theory: em wave fundamentals

Question 943

Electromagnetic TheoryEM Wave Fundamentals

For a uniform plane wave propagating in a lossless medium, the magnitude ratio E/H equals the medium’s?

AIntrinsic impedance
BConductivity
CWavelength
DPropagation constant only

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Correct Answer: A. Intrinsic impedance

Explanation: For a uniform plane wave, |E|/|H|=η, the intrinsic impedance of the medium.

Reference: Standard electromagnetic theory: em wave fundamentals

Question 944

Electromagnetic TheoryMutual Inductance

Mutual inductance describes the magnetic coupling between?

ATwo circuits or coils
BOnly two capacitors
CA resistor and a battery only
DTwo isolated static charges

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Correct Answer: A. Two circuits or coils

Explanation: Mutual inductance quantifies how changing current in one circuit induces emf in another magnetically coupled circuit.

Reference: Standard electromagnetic theory: mutual inductance

Question 945

Electromagnetic TheoryMutual Inductance

For two coupled coils with self-inductances L1 and L2 and coupling coefficient k, the mutual inductance magnitude is?

AM = k sqrt(L1 L2)
BM = k(L1+L2)
CM = L1L2/k
DM = k/(L1L2)

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Correct Answer: A. M = k sqrt(L1 L2)

Explanation: The coupling coefficient is defined by k=M/sqrt(L1L2), hence M=k sqrt(L1L2).

Reference: Standard electromagnetic theory: mutual inductance

Question 946

Electromagnetic TheoryMutual Inductance

Two coils have L1=4 H, L2=9 H and coupling coefficient k=0.5. Their mutual inductance is?

A1 H
B2 H
C3 H
D6 H

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Correct Answer: C. 3 H

Explanation: M=k sqrt(L1L2)=0.5×sqrt(36)=3 H.

Reference: Standard electromagnetic theory: mutual inductance

Question 947

Electromagnetic TheoryMaxwell Equations

Maxwell introduced displacement current primarily to extend which law to time-varying fields?

AAmpere's circuital law
BCoulomb's law
CGauss's magnetic law
DBiot-Savart law for static point charges

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Correct Answer: A. Ampere's circuital law

Explanation: The displacement-current term makes Ampere's law consistent with charge conservation in time-varying electromagnetic fields.

Reference: Standard electromagnetic theory: maxwell equations

Question 948

Electromagnetic TheoryPoynting Vector

The Poynting vector represents electromagnetic?

APower-flow density
BCharge density
CMagnetic reluctance
DElectric potential

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Correct Answer: A. Power-flow density

Explanation: The Poynting vector S=E×H gives electromagnetic power flow per unit area.

Reference: Standard electromagnetic theory: poynting vector

Question 949

Electromagnetic TheoryWave Propagation

The phase constant beta and wavelength lambda are related by?

Abeta = 2 pi/lambda
Bbeta = lambda/(2 pi)
Cbeta = 2 pi lambda
Dbeta = 1/lambda²

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Correct Answer: A. beta = 2 pi/lambda

Explanation: A phase change of 2π radians occurs over one wavelength, hence β=2π/λ.

Reference: Standard electromagnetic theory: wave propagation

Question 950

Electromagnetic TheoryWave Propagation

A wave has wavelength 0.5 m. Its phase constant is approximately?

A3.14 rad/m
B6.28 rad/m
C12.57 rad/m
D25.13 rad/m

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Correct Answer: C. 12.57 rad/m

Explanation: β=2π/λ=2π/0.5=4π≈12.57 rad/m.

Reference: Standard electromagnetic theory: wave propagation

Question 951

Electromagnetic TheorySkin Effect

Skin effect causes high-frequency current in a good conductor to concentrate mainly?

ANear the conductor surface
BUniformly throughout the entire cross-section
COnly at the conductor axis
DOutside the conductor with zero internal current

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Correct Answer: A. Near the conductor surface

Explanation: At higher frequency, electromagnetic diffusion causes current density to become concentrated near the surface.

Reference: Standard electromagnetic theory: skin effect

Question 952

Electromagnetic TheorySkin Effect

For a good conductor, skin depth decreases when frequency?

AIncreases
BDecreases
CBecomes zero only
DHas no effect

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Correct Answer: A. Increases

Explanation: For a good conductor, δ≈sqrt(2/(omega mu sigma)), so increasing frequency decreases skin depth.

Reference: Standard electromagnetic theory: skin effect

Question 953

Electromagnetic TheorySkin Effect

For a good conductor, the skin depth is approximately?

Asqrt(2/(omega mu sigma))
Bsqrt(omega mu sigma/2)
Comega mu sigma
D2 omega/(mu sigma)

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Correct Answer: A. sqrt(2/(omega mu sigma))

Explanation: The standard good-conductor approximation is δ=sqrt(2/(ωμσ)).

Reference: Standard electromagnetic theory: skin effect

Question 954

Electromagnetic TheoryReflection & Transmission

The reflection coefficient for electric-field amplitude at normal incidence is defined as?

AReflected electric-field amplitude divided by incident electric-field amplitude
BIncident field divided by reflected field
CTransmitted power divided by incident power only
DReflected magnetic field multiplied by incident field

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Correct Answer: A. Reflected electric-field amplitude divided by incident electric-field amplitude

Explanation: The electric-field reflection coefficient is Γ=Er/Ei at the interface.

Reference: Standard electromagnetic theory: reflection & transmission

Question 955

Electromagnetic TheoryReflection & Transmission

If a wave is incident normally from a medium of intrinsic impedance eta1 onto a medium of impedance eta2, the electric-field reflection coefficient is?

A(eta2-eta1)/(eta2+eta1)
B(eta2+eta1)/(eta2-eta1)
Ceta1/eta2
Deta1+eta2

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Correct Answer: A. (eta2-eta1)/(eta2+eta1)

Explanation: Applying tangential E and H boundary conditions gives Γ=(η2-η1)/(η2+η1).

Reference: Standard electromagnetic theory: reflection & transmission

Question 956

Electromagnetic TheoryReflection & Transmission

When two media have equal intrinsic impedances at normal incidence, the reflection coefficient is?

AZero
BOne
CMinus one
DInfinite

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Correct Answer: A. Zero

Explanation: If η2=η1, Γ=(η2-η1)/(η2+η1)=0, so there is no reflected wave.

Reference: Standard electromagnetic theory: reflection & transmission

Question 957

Electromagnetic TheoryReflection & Transmission

For normal incidence on a perfect electric conductor, the electric-field reflection coefficient is?

A+1
B-1
C0
D+0.5

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Correct Answer: B. -1

Explanation: The tangential total electric field must vanish at a PEC surface, requiring Er=-Ei and Γ=-1.

Reference: Standard electromagnetic theory: reflection & transmission

Question 958

Electromagnetic TheoryStanding Waves

A standing wave is produced by the superposition of?

AIncident and reflected waves traveling in opposite directions
BTwo DC electric fields only
COnly one traveling wave
DTwo unrelated static magnetic fields

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Correct Answer: A. Incident and reflected waves traveling in opposite directions

Explanation: Interference between forward and reflected waves of the same frequency creates a standing-wave pattern.

Reference: Standard electromagnetic theory: standing waves

Question 959

Electromagnetic TheoryPoynting Vector

The SI unit of the Poynting vector is?

AW/m²
BJ/m
CV/m
DA/m²

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Correct Answer: A. W/m²

Explanation: Because the Poynting vector is power per unit area, its SI unit is watt per square metre.

Reference: Standard electromagnetic theory: poynting vector

Question 960

Electromagnetic TheoryStanding Waves

Voltage standing-wave ratio in terms of reflection-coefficient magnitude is?

A(1+|Gamma|)/(1-|Gamma|)
B(1-|Gamma|)/(1+|Gamma|)
C1-|Gamma|²
D|Gamma|/(1+|Gamma|)

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Correct Answer: A. (1+|Gamma|)/(1-|Gamma|)

Explanation: VSWR is the ratio of maximum to minimum field amplitude and equals (1+|Γ|)/(1-|Γ|).

Reference: Standard electromagnetic theory: standing waves

Question 961

Electromagnetic TheoryPoynting Vector

For a uniform plane wave with E=100 V/m and H=0.25 A/m mutually perpendicular and in phase, the instantaneous Poynting-vector magnitude is?

A25 W/m²
B40 W/m²
C100 W/m²
D400 W/m²

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Correct Answer: A. 25 W/m²

Explanation: For perpendicular fields, |S|=|E||H|=100×0.25=25 W/m².

Reference: Standard electromagnetic theory: poynting vector

Question 962

Electromagnetic TheoryStanding Waves

If the magnitude of the reflection coefficient is 0.5, the VSWR is?

A1.5
B2
C3
D4

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Correct Answer: C. 3

Explanation: VSWR=(1+0.5)/(1-0.5)=1.5/0.5=3.

Reference: Standard electromagnetic theory: standing waves

Question 963

Electromagnetic TheoryWave Propagation

The complex propagation constant gamma is conventionally written as?

Aalpha + j beta
Bbeta + j alpha only
Calpha beta
Dj(alpha + beta)

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Correct Answer: A. alpha + j beta

Explanation: The propagation constant is γ=α+jβ, where α is attenuation constant and β is phase constant.

Reference: Standard electromagnetic theory: wave propagation

Question 964

Electromagnetic TheoryStanding Waves

A VSWR of 1 corresponds to?

APerfect matching with no reflection
BTotal reflection
CA reflection coefficient magnitude of 1
DInfinite attenuation

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Correct Answer: A. Perfect matching with no reflection

Explanation: VSWR=1 occurs only when |Γ|=0, corresponding to a matched condition.

Reference: Standard electromagnetic theory: standing waves

Question 965

Electromagnetic TheoryWave Propagation

The attenuation constant alpha is commonly measured in?

ANp/m
Brad/m
CHz/m
DWb/m

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Correct Answer: A. Np/m

Explanation: The attenuation constant describes exponential amplitude decay and is measured in nepers per metre.

Reference: Standard electromagnetic theory: wave propagation

Question 966

Electromagnetic TheoryPolarization

The polarization of an electromagnetic wave is specified by the behavior of its?

AElectric-field vector
BPropagation speed only
CMagnetic permeability only
DFrequency counter reading

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Correct Answer: A. Electric-field vector

Explanation: Wave polarization is conventionally defined by the locus and orientation of the electric-field vector at a fixed point.

Reference: Standard electromagnetic theory: polarization

Question 967

Electromagnetic TheoryWave Propagation

The phase constant beta is commonly measured in?

Arad/m
BNp/m
CS/m
DC/m

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Correct Answer: A. rad/m

Explanation: The phase constant describes phase change per unit distance and is measured in radians per metre.

Reference: Standard electromagnetic theory: wave propagation

Question 968

Electromagnetic TheoryWave Propagation

For a lossless dielectric medium, the attenuation constant alpha is ideally?

AZero
BOne
CEqual to beta
DInfinite

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Correct Answer: A. Zero

Explanation: A lossless medium has no dissipative attenuation, so α=0.

Reference: Standard electromagnetic theory: wave propagation

Question 969

Electromagnetic TheoryPolarization

Two orthogonal field components generally produce elliptical polarization when they have?

ASuitable unequal amplitudes and/or a phase difference other than the special linear/circular cases
BZero frequency only
CNo electric field
DOnly identical phase and identical direction

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Correct Answer: A. Suitable unequal amplitudes and/or a phase difference other than the special linear/circular cases

Explanation: Elliptical polarization is the general case for orthogonal components with relative amplitude and phase that do not reduce to linear or circular polarization.

Reference: Standard electromagnetic theory: polarization

Question 970

Electromagnetic TheoryEM Wave Numericals

A 300 MHz electromagnetic wave propagating in free space has an approximate wavelength of?

A0.1 m
B0.5 m
C1 m
D10 m

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Correct Answer: C. 1 m

Explanation: lambda=c/f=(3×10^8)/(300×10^6)=1 m.

Reference: Standard electromagnetic theory: em wave numericals

Question 971

Electromagnetic TheoryPolarization

For linear polarization, the tip of the electric-field vector at a fixed point traces a?

AStraight line
BCircle
CGeneral ellipse only
DSphere

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Correct Answer: A. Straight line

Explanation: In a linearly polarized wave, the electric field maintains a fixed direction and varies in magnitude, tracing a line.

Reference: Standard electromagnetic theory: polarization

Question 972

Electromagnetic TheoryPolarization

Circular polarization requires two orthogonal electric-field components of equal magnitude with a phase difference of?

A0 degrees
B45 degrees
C90 degrees
D180 degrees

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Correct Answer: C. 90 degrees

Explanation: Equal orthogonal components in phase quadrature produce circular polarization.

Reference: Standard electromagnetic theory: polarization

Question 973

Antenna & Wave PropagationAntenna Fundamentals

The primary function of a transmitting antenna is to convert?

AGuided electromagnetic energy into radiated electromagnetic waves
BDC power directly into mechanical energy
CSound waves into heat only
DDigital data into resistance

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Correct Answer: A. Guided electromagnetic energy into radiated electromagnetic waves

Explanation: A transmitting antenna couples guided RF energy from a transmission system into electromagnetic waves that propagate through space.

Reference: Standard antenna and wave propagation: antenna fundamentals

Question 974

Antenna & Wave PropagationAntenna Fundamentals

A receiving antenna primarily converts incident electromagnetic waves into?

AElectrical signals available at its terminals
BMechanical rotation only
CDC resistance only
DOptical energy only

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Correct Answer: A. Electrical signals available at its terminals

Explanation: A receiving antenna intercepts electromagnetic energy and produces a corresponding electrical signal at its terminals.

Reference: Standard antenna and wave propagation: antenna fundamentals

Question 975

Antenna & Wave PropagationRadiation Pattern

The radiation pattern of an antenna describes the variation of its radiation properties with?

ADirection in space
BSupply voltage only
CConductor resistance only
DTime of day

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Correct Answer: A. Direction in space

Explanation: An antenna radiation pattern shows how radiation intensity, field strength or related quantities vary with direction.

Reference: Standard antenna and wave propagation: radiation pattern

Question 976

Antenna & Wave PropagationRadiation Pattern

The direction in which an antenna radiates maximum power is called the direction of its?

AMain lobe
BNull
CBack lobe only
DMinor axis

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Correct Answer: A. Main lobe

Explanation: The main lobe contains the direction of maximum radiation.

Reference: Standard antenna and wave propagation: radiation pattern

Question 977

Antenna & Wave PropagationRadiation Pattern

A direction in an antenna radiation pattern where radiation is ideally zero is called a?

ANull
BMain beam
CHalf-power point
DFeed point

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Correct Answer: A. Null

Explanation: A null is a direction of zero or very small radiation in the antenna pattern.

Reference: Standard antenna and wave propagation: radiation pattern

Question 978

Antenna & Wave PropagationRadiation Pattern

A lobe other than the major lobe is generally called a?

AMinor lobe
BIsotropic lobe
CFeed lobe
DImpedance lobe

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Correct Answer: A. Minor lobe

Explanation: Side lobes and back lobes are collectively minor lobes because they are not the principal radiation lobe.

Reference: Standard antenna and wave propagation: radiation pattern

Question 979

Antenna & Wave PropagationIsotropic Radiator

An isotropic radiator is a hypothetical antenna that radiates?

AEqually in all directions
BOnly along one axis
COnly in the horizontal plane
DNo power

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Correct Answer: A. Equally in all directions

Explanation: An isotropic radiator is an ideal reference source with uniform radiation intensity in every direction.

Reference: Standard antenna and wave propagation: isotropic radiator

Question 980

Antenna & Wave PropagationDirectivity & Gain

The directivity of an isotropic radiator is?

A1 or 0 dBi
B2 or 3 dBi
C1.64 or 2.15 dBi
DInfinite

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Correct Answer: A. 1 or 0 dBi

Explanation: By definition, an isotropic radiator has directivity 1, corresponding to 0 dBi.

Reference: Standard antenna and wave propagation: directivity & gain

Question 981

Antenna & Wave PropagationDirectivity & Gain

Antenna directivity is the ratio of radiation intensity in a specified direction to?

AAverage radiation intensity over all directions
BInput current only
COhmic resistance
DWavelength squared only

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Correct Answer: A. Average radiation intensity over all directions

Explanation: Directivity compares radiation intensity in a given direction, usually its maximum, with the average radiation intensity.

Reference: Standard antenna and wave propagation: directivity & gain

Question 982

Antenna & Wave PropagationDirectivity & Gain

Antenna gain differs from directivity because gain also accounts for?

AAntenna efficiency
BOnly operating frequency
COnly polarization direction
DSpeed of light

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Correct Answer: A. Antenna efficiency

Explanation: Gain includes the effect of radiation efficiency, whereas directivity is based on the shape of the radiation pattern.

Reference: Standard antenna and wave propagation: directivity & gain

Question 983

Antenna & Wave PropagationDirectivity & Gain

If an antenna has directivity 10 and radiation efficiency 0.8, its power gain is?

A8
B10
C12.5
D80

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Correct Answer: A. 8

Explanation: Gain=efficiency×directivity=0.8×10=8.

Reference: Standard antenna and wave propagation: directivity & gain

Question 984

Antenna & Wave PropagationDirectivity & Gain

A power gain of 10 corresponds to approximately?

A3 dB
B10 dB
C20 dB
D100 dB

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Correct Answer: B. 10 dB

Explanation: Gain in decibels is 10log10(10)=10 dB.

Reference: Standard antenna and wave propagation: directivity & gain

Question 985

Antenna & Wave PropagationBeamwidth

The half-power beamwidth of an antenna is the angular separation between points where the radiated power falls to?

AHalf of its maximum value
BOne-fourth of maximum
CZero
DTwice the maximum

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Correct Answer: A. Half of its maximum value

Explanation: At the half-power points, power is 0.5 of maximum, corresponding to about -3 dB.

Reference: Standard antenna and wave propagation: beamwidth

Question 986

Antenna & Wave PropagationBeamwidth

A narrower main beam generally indicates?

AHigher directivity
BLower directivity in every case
CZero gain
DNo radiation

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Correct Answer: A. Higher directivity

Explanation: Concentrating radiation into a narrower angular region generally increases antenna directivity.

Reference: Standard antenna and wave propagation: beamwidth

Question 987

Antenna & Wave PropagationAntenna Resistance

Radiation resistance is a fictitious resistance that represents?

APower radiated by the antenna
BOnly conductor heating loss
CDielectric breakdown
DFeed-line reactance only

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Correct Answer: A. Power radiated by the antenna

Explanation: Radiation resistance is defined so that the power attributed to it equals the antenna's radiated power for the specified current convention.

Reference: Standard antenna and wave propagation: antenna resistance

Question 988

Antenna & Wave PropagationAntenna Efficiency

Antenna radiation efficiency is commonly defined as radiated power divided by?

AAccepted input power
BIncident power density only
CReactive power only
DReflected power only

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Correct Answer: A. Accepted input power

Explanation: Radiation efficiency is the fraction of power accepted by the antenna that is actually radiated rather than dissipated as loss.

Reference: Standard antenna and wave propagation: antenna efficiency

Question 989

Antenna & Wave PropagationAntenna Efficiency

An antenna accepts 100 W and radiates 80 W. Its radiation efficiency is?

A20%
B50%
C80%
D125%

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Correct Answer: C. 80%

Explanation: Efficiency=Prad/Paccepted=80/100=0.8=80%.

Reference: Standard antenna and wave propagation: antenna efficiency

Question 990

Antenna & Wave PropagationAntenna Impedance

For maximum power transfer between a transmission line and an antenna at a single frequency, the antenna input impedance should ideally be?

AMatched to the line impedance
BZero
CInfinite
DPurely capacitive regardless of the line

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Correct Answer: A. Matched to the line impedance

Explanation: Impedance matching minimizes reflection and permits efficient transfer of power from the line to the antenna.

Reference: Standard antenna and wave propagation: antenna impedance

Question 991

Antenna & Wave PropagationPolarization

The polarization of an antenna is normally defined by the polarization of the wave it?

ARadiates in a specified direction
BAbsorbs as heat only
CReflects from the ground only
DCarries inside a DC cable

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Correct Answer: A. Radiates in a specified direction

Explanation: Antenna polarization is described by the electric-field polarization of its radiated wave in the direction considered.

Reference: Standard antenna and wave propagation: polarization

Question 992

Antenna & Wave PropagationDipole Antennas

A vertically oriented half-wave dipole ideally produces what polarization in its broadside direction?

AVertical linear polarization
BHorizontal linear polarization
CCircular polarization automatically
DUnpolarized radiation

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Correct Answer: A. Vertical linear polarization

Explanation: The electric field of a thin dipole is oriented parallel to the dipole axis, giving vertical linear polarization for a vertical dipole.

Reference: Standard antenna and wave propagation: dipole antennas

Question 993

Antenna & Wave PropagationDipole Antennas

The approximate total physical length of an ideal thin half-wave dipole in free space is?

Alambda/4
Blambda/2
Clambda
D2 lambda

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Correct Answer: B. lambda/2

Explanation: A half-wave dipole is approximately one-half wavelength long; practical resonant lengths are slightly shorter due to end effects.

Reference: Standard antenna and wave propagation: dipole antennas

Question 994

Antenna & Wave PropagationAntenna Numericals

At 300 MHz, the free-space wavelength is approximately?

A0.1 m
B0.5 m
C1 m
D3 m

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Correct Answer: C. 1 m

Explanation: lambda=c/f=3×10^8/(300×10^6)=1 m.

Reference: Standard antenna and wave propagation: antenna numericals

Question 995

Antenna & Wave PropagationDipole Antennas

Ignoring practical shortening effects, the approximate total length of a half-wave dipole for 300 MHz is?

A0.25 m
B0.5 m
C1 m
D1.5 m

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Correct Answer: B. 0.5 m

Explanation: At 300 MHz lambda≈1 m, so a half-wave dipole is approximately lambda/2=0.5 m.

Reference: Standard antenna and wave propagation: dipole antennas

Question 996

Antenna & Wave PropagationMonopole Antennas

A quarter-wave monopole antenna mounted over an ideal conducting ground plane has an approximate physical height of?

Alambda/8
Blambda/4
Clambda/2
Dlambda

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Correct Answer: B. lambda/4

Explanation: A quarter-wave monopole is approximately one-quarter wavelength high and uses the ground plane as its image counterpart.

Reference: Standard antenna and wave propagation: monopole antennas

Question 997

Antenna & Wave PropagationBeamwidth

At a half-power point, the field magnitude relative to its maximum is approximately?

A0.5
B0.707
C1.414
D2

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Correct Answer: B. 0.707

Explanation: Power is proportional to field magnitude squared, so half power corresponds to field magnitude 1/sqrt(2)≈0.707 of maximum.

Reference: Standard antenna and wave propagation: beamwidth

Question 998

Antenna & Wave PropagationDipole Antennas

The radiation pattern of an ideal half-wave dipole has maximum radiation in directions?

AAlong the dipole axis
BPerpendicular to the dipole axis
COnly at 45 degrees to the axis
DEqually in every direction

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Correct Answer: B. Perpendicular to the dipole axis

Explanation: A thin half-wave dipole has maximum radiation broadside to its axis and nulls along the axis.

Reference: Standard antenna and wave propagation: dipole antennas

Question 999

Antenna & Wave PropagationEffective Aperture

An isotropic receiving antenna operating at wavelength 2 m has an effective aperture of approximately?

A0.0796 m²
B0.159 m²
C0.318 m²
D1.273 m²

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Correct Answer: C. 0.318 m²

Explanation: For G=1, Ae=λ²/(4π)=4/(4π)=1/π≈0.318 m².

Reference: Standard antenna and wave propagation: effective aperture

Question 1000

Antenna & Wave PropagationFriis Transmission

The Friis free-space transmission equation is primarily used to estimate?

AReceived power between antennas in free-space far-field conditions
BConductor skin depth
CAntenna mechanical strength
DIonospheric critical frequency only

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Correct Answer: A. Received power between antennas in free-space far-field conditions

Explanation: The Friis equation relates transmitted power, antenna gains, wavelength and separation to received power under ideal free-space far-field conditions.

Reference: Standard antenna and wave propagation: friis transmission