Nepal Engineering Council · Electronics & Communication Engineering · Chapter 6
Electromagnetic Waves and Propagation
Tap an option to check it. Wrong picks show the right answer and the hint.
183 questions in 6 syllabus topics.
6.1 Electric field
31 questions · AExE0601
1. Physically, the divergence of a vector field at a point represents:
Show hintHide hint
Think of a source or sink of field lines.
Show answerHide answer
Answer: D. The net outward flux per unit volume as the volume shrinks to that point
Divergence is defined as the limit of the net outward flux through a closed surface divided by the enclosed volume; it measures the source strength at the point.
2. The divergence theorem relates:
Show hintHide hint
Surface on one side, volume on the other.
Show answerHide answer
Answer: B. The closed-surface integral of a vector to the volume integral of its divergence
Divergence theorem: ∮ A·dS = ∫ (∇·A) dv over the volume enclosed by the surface. The closed-line/curl relation is Stokes' theorem.
3. Gauss's law in point (differential) form is:
Show hintHide hint
Flux density diverges from free charge.
Show answerHide answer
Answer: C. ∇·D = ρv
Applying the divergence theorem to ∮D·dS = Qenc gives ∇·D = ρv. In terms of E it would be ∇·E = ρv/ε, not ρv.
4. A point charge of 2 nC is in free space. The electric field intensity at a distance of 3 m from it is approximately:
Show hintHide hint
Field falls as the square of distance.
Show answerHide answer
Answer: C. 2 V/m
E = Q/(4πε0 r²) = 9×10⁹ × 2×10⁻⁹ / 3² = 18/9 = 2 V/m.
5. In free space the electric field intensity at a point is 100 V/m. The electric flux density there is about:
Show hintHide hint
D = ε0E in free space.
Show answerHide answer
Answer: B. 0.885 nC/m²
D = ε0E = 8.854×10⁻¹² × 100 = 8.85×10⁻¹⁰ C/m² ≈ 0.885 nC/m².
6. A closed surface encloses charges of +5 nC and −2 nC, while a charge of +4 nC lies outside it. The total electric flux (Ψ = ∮D·dS) leaving the surface is:
Show hintHide hint
Charges outside the surface do not count.
Show answerHide answer
Answer: A. 3 nC
By Gauss's law, only enclosed charge counts: Ψ = 5 − 2 = 3 nC. The outside charge contributes zero net flux.
7. Given D = x²y ax + yz ay C/m², the volume charge density at the point (1, 2, 3) m is:
Show hintHide hint
Take ∂Dx/∂x + ∂Dy/∂y.
Show answerHide answer
Answer: A. 7 C/m³
ρv = ∇·D = ∂(x²y)/∂x + ∂(yz)/∂y = 2xy + z = 2(1)(2) + 3 = 7 C/m³.
8. The potential in a region is V = x²y volts (x, y in metres). The electric field intensity at (1, 2) m is:
Show hintHide hint
E is the negative gradient of V.
Show answerHide answer
Answer: D. −4 ax − ay V/m
E = −∇V = −(2xy ax + x² ay). At (1, 2): −(4 ax + 1 ay) = −4 ax − ay V/m.
9. The electric field intensity at a point is directed:
Show hintHide hint
Note the minus sign in E = −∇V.
Show answerHide answer
Answer: C. Along the direction of maximum decrease of potential
Since E = −∇V and the gradient points toward maximum increase, E points toward maximum decrease of V and is normal to equipotential surfaces.
10. For a static electric field, ∮E·dl = 0 around any closed path. This means:
Show hintHide hint
Work done moving a charge round a loop.
Show answerHide answer
Answer: A. The electrostatic field is conservative, i.e. ∇×E = 0
A zero closed-path line integral means the work done is path-independent, so E is conservative (curl-free) and can be written as −∇V.
11. The energy density stored in an electrostatic field in a linear dielectric of permittivity ε is:
Show hintHide hint
Compare with ½CV² for a capacitor.
Show answerHide answer
Answer: C. ½ εE² J/m³
wE = ½ D·E = ½ εE² joules per cubic metre.
12. The electric field in air is uniform at 1 MV/m. The energy stored per unit volume is about:
Show hintHide hint
Square the field before multiplying.
Show answerHide answer
Answer: A. 4.43 J/m³
w = ½ ε0E² = 0.5 × 8.854×10⁻¹² × (10⁶)² = 4.43 J/m³.
13. The work required to move a 3 µC charge through a potential difference of 200 V is:
Show hintHide hint
Work = charge × potential difference.
Show answerHide answer
Answer: B. 0.6 mJ
W = QΔV = 3×10⁻⁶ × 200 = 6×10⁻⁴ J = 0.6 mJ.
14. In a polarized dielectric with polarization P, the bound volume charge density is:
Show hintHide hint
Mind the sign; P·an is the surface version.
Show answerHide answer
Answer: A. ρb = −∇·P
Bound volume charge density is ρb = −∇·P; the bound surface charge density is ρsb = P·an.
15. A dielectric with εr = 5 is in a uniform electric field of 10 kV/m. The polarization P is about:
Show hintHide hint
Use the susceptibility χe = εr − 1.
Show answerHide answer
Answer: B. 0.354 µC/m²
P = ε0(εr − 1)E = 8.854×10⁻¹² × 4 × 10⁴ = 3.54×10⁻⁷ C/m² ≈ 0.354 µC/m².
16. The relative permittivity εr of a linear dielectric is related to its electric susceptibility χe by:
Show hintHide hint
Free space has χe = 0 and εr = 1.
Show answerHide answer
Answer: B. εr = 1 + χe
D = ε0E + P = ε0(1 + χe)E, so ε = ε0(1 + χe) and εr = 1 + χe.
17. For charges +Q and −Q separated by a distance d, the electric dipole moment is:
Show hintHide hint
Convention: the vector points toward the positive charge.
Show answerHide answer
Answer: B. Qd, directed from −Q to +Q
The dipole moment p = Qd, where d is the vector from the negative charge to the positive charge.
18. At a large distance r from an electric dipole, the potential and the field intensity vary respectively as:
Show hintHide hint
One power of r faster than a point charge.
Show answerHide answer
Answer: A. 1/r² and 1/r³
Dipole potential V = p cosθ/(4πε0 r²); taking the gradient adds one more power of r, so E ∝ 1/r³.
19. An electric dipole of moment 1 nC·m is in free space. The potential at a point 1 m away, at 60° from the dipole axis, is about:
Show hintHide hint
Use cosθ, measured from the dipole axis.
Show answerHide answer
Answer: C. 4.5 V
V = p cosθ/(4πε0 r²) = 9×10⁹ × 10⁻⁹ × cos60° / 1² = 9 × 0.5 = 4.5 V.
20. Under electrostatic conditions, at the surface of a perfect conductor:
Show hintHide hint
Field lines leave a conductor perpendicularly.
Show answerHide answer
Answer: C. Tangential E is zero and normal D equals the surface charge density
Inside a conductor E = 0; at its surface Et = 0 and Dn = ρs, so the field leaves the surface normally.
21. Two dielectrics with εr1 = 2 and εr2 = 6 meet at a plane boundary with no free surface charge. If the normal component of E in medium 1 is 30 V/m, the normal component of E in medium 2 is:
Show hintHide hint
Which component is continuous: normal D or normal E?
Show answerHide answer
Answer: B. 10 V/m
Normal D is continuous: εr1E1n = εr2E2n, so E2n = 2 × 30 / 6 = 10 V/m.
22. At the interface of two perfect dielectrics carrying no free surface charge, which quantity is continuous?
Show hintHide hint
Comes from ∮E·dl = 0 around a thin loop.
Show answerHide answer
Answer: B. Tangential component of E
Boundary conditions: E1t = E2t and D1n = D2n (with no free surface charge). Tangential D and normal E change by the permittivity ratio.
23. A uniform current density of 10 A/m² flows normally through a cross-section of area 2 cm². The current is:
Show hintHide hint
Convert cm² to m² first.
Show answerHide answer
Answer: D. 2 mA
I = J × A = 10 × 2×10⁻⁴ = 2×10⁻³ A = 2 mA.
24. The equation of continuity (conservation of charge) in point form is:
Show hintHide hint
Outflow of current means charge inside decreases.
Show answerHide answer
Answer: C. ∇·J = −∂ρv/∂t
Current flowing out of a volume equals the rate of decrease of the charge inside, giving ∇·J = −∂ρv/∂t.
25. In a region the current density is J = 5x ax A/m² (x in metres). The time rate of change of the volume charge density is:
Show hintHide hint
Apply the continuity equation.
Show answerHide answer
Answer: D. −5 C/m³ per second
∂ρv/∂t = −∇·J = −∂(5x)/∂x = −5 C/m³·s⁻¹.
26. The relaxation time of a material is:
Show hintHide hint
Check the units of ε/σ.
Show answerHide answer
Answer: D. ε/σ, the time for charge placed inside it to decay to 1/e of its initial value
Combining the continuity equation, Ohm's law and Gauss's law gives ρv = ρ0 e^(−t/Tr) with Tr = ε/σ, the 1/e (≈36.8%) decay time.
27. A material has εr = 4 and σ = 10⁻⁴ S/m. Its relaxation time is about:
Show hintHide hint
Tr = ε0εr/σ.
Show answerHide answer
Answer: C. 0.354 µs
Tr = ε/σ = 4 × 8.854×10⁻¹² / 10⁻⁴ = 3.54×10⁻⁷ s ≈ 0.354 µs.
28. Poisson's equation for the electrostatic potential in a linear, homogeneous medium is:
Show hintHide hint
Combine Gauss's law with E = −∇V.
Show answerHide answer
Answer: A. ∇²V = −ρv/ε
From ∇·D = ρv and E = −∇V: ∇·(−ε∇V) = ρv, so ∇²V = −ρv/ε. Laplace's equation ∇²V = 0 is the charge-free special case.
29. Two large parallel plates at x = 0 and x = d = 5 mm are held at 0 V and 100 V, with no charge between them. Solving Laplace's equation, the potential at x = d/4 is:
Show hintHide hint
Laplace's equation in one dimension gives a straight line.
Show answerHide answer
Answer: A. 25 V
In 1-D, d²V/dx² = 0 gives a linear potential V = 100·x/d, so at x = d/4, V = 25 V (field 20 kV/m).
30. The uniqueness theorem in electrostatics states that:
Show hintHide hint
It justifies the method of images.
Show answerHide answer
Answer: D. A solution of Laplace's or Poisson's equation that satisfies the given boundary conditions is the only solution
If any method (guessing, images, separation of variables) yields a potential that satisfies the equation and all boundary conditions, it is the unique solution.
31. Free charges differ from bound charges in that free charges:
Show hintHide hint
Think conductor versus dielectric.
Show answerHide answer
Answer: D. Can move over macroscopic distances, like conduction electrons in a metal
Bound charges are displaced only slightly within atoms or molecules (polarization); free charges such as conduction electrons can move through the material.
6.2 Magnetic field
30 questions · AExE0602
32. According to the Biot-Savart law, the magnetic field intensity due to a current element I dl at distance R is:
Show hintHide hint
H is perpendicular to both dl and R.
Show answerHide answer
Answer: B. dH = I dl × aR / (4πR²)
The Biot-Savart law gives dH = I dl × aR/(4πR²): a cross product (H is perpendicular to both dl and R) with inverse-square dependence.
33. The SI unit of magnetic flux density B is:
Show hintHide hint
Flux per unit area.
Show answerHide answer
Answer: B. Tesla (Wb/m²)
B is flux per unit area: 1 T = 1 Wb/m². A/m is the unit of H, Wb is the unit of flux and H/m is the unit of permeability.
34. A long straight wire in air carries 10 A. The magnetic flux density at 5 cm from it is:
Show hintHide hint
Use B = µ0I/(2πr).
Show answerHide answer
Answer: A. 40 µT
B = µ0I/(2πr) = 4π×10⁻⁷ × 10 / (2π × 0.05) = 4×10⁻⁵ T = 40 µT.
35. A circular loop of radius 10 cm carries a current of 5 A. The magnetic field intensity H at its centre is:
Show hintHide hint
The loop-centre formula differs from the long-wire formula.
Show answerHide answer
Answer: C. 25 A/m
At the centre of a circular loop H = I/(2a) = 5/(2 × 0.1) = 25 A/m.
36. A long air-cored solenoid has 1000 turns per metre and carries 2 A. The flux density inside it is about:
Show hintHide hint
Inside a long solenoid H = nI.
Show answerHide answer
Answer: B. 2.51 mT
B = µ0nI = 4π×10⁻⁷ × 1000 × 2 = 2.51×10⁻³ T.
37. A toroid with 500 turns carries 1 A. The magnetic field intensity at a mean radius of 10 cm inside the core is about:
Show hintHide hint
The Amperian path is a circle of radius ρ.
Show answerHide answer
Answer: C. 796 A/m
By Ampere's law, H × 2πρ = NI, so H = 500 × 1/(2π × 0.1) = 796 A/m.
38. A long solid conductor of radius 2 mm carries 10 A distributed uniformly. The magnetic field intensity at 1 mm from its axis is about:
Show hintHide hint
Only part of the current is enclosed.
Show answerHide answer
Answer: C. 398 A/m
Inside the wire the enclosed current is I(ρ/a)², so H = Iρ/(2πa²) = 10 × 10⁻³/(2π × 4×10⁻⁶) = 398 A/m.
39. Ampere's circuital law in integral form states that:
Show hintHide hint
Watch whether it is H or B.
Show answerHide answer
Answer: A. ∮H·dl equals the net current enclosed by the path
Ampere's circuital law: the line integral of H around any closed path equals the current enclosed, ∮H·dl = Ienc (for B it would be µ0Ienc).
40. Ampere's circuital law is most convenient for finding H when:
Show hintHide hint
When can H be pulled out of the integral?
Show answerHide answer
Answer: D. The current distribution is highly symmetric, as in an infinite wire, solenoid or toroid
Ampere's law is always true, but it gives H easily only when symmetry lets H be taken out of the integral along a suitable path; otherwise Biot-Savart is used.
41. The point (differential) form of Ampere's law for steady currents is:
Show hintHide hint
Curl relates to circulation.
Show answerHide answer
Answer: A. ∇×H = J
Applying Stokes' theorem to ∮H·dl = ∫J·dS gives ∇×H = J (equivalently ∇×B = µ0J).
42. The physical significance of the curl of a vector field at a point is:
Show hintHide hint
Imagine a small paddle wheel placed in the field.
Show answerHide answer
Answer: D. The maximum circulation of the field per unit area at that point, indicating rotation
Curl is the limiting circulation per unit area, oriented for maximum value; it measures how much the field swirls about the point.
43. Stokes' theorem states that:
Show hintHide hint
Line integral on one side, open surface on the other.
Show answerHide answer
Answer: D. The circulation of a vector around a closed path equals the surface integral of its curl over any surface bounded by that path
Stokes' theorem: ∮A·dl = ∫(∇×A)·dS, with the surface bounded by the closed path.
44. In a region H = y ax − x ay A/m. The current density there is:
Show hintHide hint
Only the z-component of the curl survives here.
Show answerHide answer
Answer: B. −2 az A/m²
J = ∇×H; the z-component is ∂Hy/∂x − ∂Hx/∂y = −1 − 1 = −2, so J = −2 az A/m².
45. The Maxwell equation ∇·B = 0 implies that:
Show hintHide hint
Gauss's law for magnetism.
Show answerHide answer
Answer: B. Magnetic flux lines always close on themselves; isolated magnetic poles do not exist
Zero divergence of B means no sources or sinks of magnetic flux, i.e. no magnetic monopoles; the net flux through any closed surface is zero.
46. An electron (charge 1.6×10⁻¹⁹ C) moves at 10⁶ m/s perpendicular to a uniform magnetic field of 0.1 T. The magnitude of the magnetic force on it is:
Show hintHide hint
F = qvB when v is perpendicular to B.
Show answerHide answer
Answer: D. 1.6×10⁻¹⁴ N
F = qvB sin90° = 1.6×10⁻¹⁹ × 10⁶ × 0.1 = 1.6×10⁻¹⁴ N.
47. A static magnetic field acting on a moving charged particle:
Show hintHide hint
Is the force ever along the velocity?
Show answerHide answer
Answer: A. Changes the direction of its velocity but not its kinetic energy
F = q(v × B) is always perpendicular to v, so it does no work: it only bends the path and the speed stays constant.
48. A straight conductor 0.2 m long carrying 4 A is placed at 30° to a uniform field of 0.5 T. The force on it is:
Show hintHide hint
Use the sine of the angle between the conductor and B.
Show answerHide answer
Answer: B. 0.2 N
F = BIL sinθ = 0.5 × 4 × 0.2 × sin30° = 0.4 × 0.5 = 0.2 N.
49. Two long parallel wires carry currents in the same direction. They:
Show hintHide hint
Like currents behave opposite to like charges.
Show answerHide answer
Answer: A. Attract each other
Each wire lies in the field of the other; F = I L × B gives a force toward the other wire when the currents are in the same direction.
50. Two long parallel wires 10 cm apart in air carry 10 A and 20 A. The force per metre length between them is:
Show hintHide hint
F/L = µ0I1I2/(2πd).
Show answerHide answer
Answer: C. 0.4 mN/m
F/L = µ0I1I2/(2πd) = 4π×10⁻⁷ × 10 × 20 / (2π × 0.1) = 4×10⁻⁴ N/m.
51. A single square loop of side 10 cm carries a current of 2 A. Its magnetic dipole moment is:
Show hintHide hint
Moment = current × area.
Show answerHide answer
Answer: D. 0.02 A·m²
m = IA = 2 × (0.1)² = 0.02 A·m², directed normal to the loop by the right-hand rule.
52. A 50-turn coil of area 0.01 m² carries 0.5 A in a uniform field of 0.2 T. The plane of the coil is parallel to B. The torque on the coil is:
Show hintHide hint
Find the angle between the coil's normal and B.
Show answerHide answer
Answer: A. 0.05 N·m
With the plane parallel to B, the moment (normal to the plane) is perpendicular to B, so T = NIAB = 50 × 0.5 × 0.01 × 0.2 = 0.05 N·m (maximum).
53. The torque on a magnetic dipole of moment m in a uniform field B is:
Show hintHide hint
Torque is a vector that tends to align m with B.
Show answerHide answer
Answer: A. T = m × B
Torque is the cross product T = m × B; −m·B is the dipole's potential energy, not the torque.
54. At distances much larger than its size, a small current loop behaves as:
Show hintHide hint
Compare it with a bar magnet.
Show answerHide answer
Answer: A. A magnetic dipole whose field falls off as 1/r³
A small loop is the magnetic dipole; its far field has the same form as an electric dipole's, varying as 1/r³.
55. Magnetization M of a material is defined as:
Show hintHide hint
It is the magnetic analogue of polarization P.
Show answerHide answer
Answer: C. Magnetic dipole moment per unit volume
M is the net magnetic dipole moment per unit volume (A/m), and B = µ0(H + M).
56. A linear magnetic material with µr = 101 has H = 50 A/m inside it. The magnetization M is:
Show hintHide hint
χm = µr − 1.
Show answerHide answer
Answer: C. 5000 A/m
M = χmH = (µr − 1)H = 100 × 50 = 5000 A/m.
57. A diamagnetic material is characterized by:
Show hintHide hint
Diamagnetic materials are weakly repelled by a magnet.
Show answerHide answer
Answer: D. A small negative susceptibility, so µr is slightly less than 1
Diamagnetic materials (e.g. copper, bismuth) have χm ≈ −10⁻⁵, so µr is slightly below 1; paramagnetic ones are slightly positive and ferromagnetic ones very large.
58. The bound (magnetization) volume current density in a magnetized material is:
Show hintHide hint
M × an is the surface version.
Show answerHide answer
Answer: B. Jb = ∇×M
The bound volume current density is Jb = ∇×M; the bound surface current density is Kb = M × an.
59. At the boundary between two magnetic media, which quantity is always continuous?
Show hintHide hint
It follows from ∇·B = 0.
Show answerHide answer
Answer: C. The normal component of B
From ∮B·dS = 0, B1n = B2n always. Tangential H is continuous only when there is no surface current (H1t − H2t = K).
60. Region 1 (µr1 = 4) and region 2 (µr2 = 1) share a boundary carrying no surface current. In region 1, H1 = 5 at + 3 an A/m (at tangential, an normal). In region 2, H2 is:
Show hintHide hint
Apply Ht continuity and Bn continuity separately.
Show answerHide answer
Answer: D. 5 at + 12 an A/m
Tangential H is continuous (5). Normal B is continuous: µr1H1n = µr2H2n, so H2n = 4 × 3/1 = 12. Hence H2 = 5 at + 12 an.
61. An infinite plane sheet carries a uniform surface current density K (A/m). The magnetic field intensity it produces is:
Show hintHide hint
Use a rectangular Amperian loop crossing the sheet.
Show answerHide answer
Answer: B. K/2 in magnitude, independent of distance from the sheet
Applying Ampere's law to a rectangular path straddling the sheet gives 2Hl = Kl, so H = K/2 on either side, independent of distance.
6.3 Wave equation and wave propagation
31 questions · AExE0603
62. Maxwell added the displacement current density ∂D/∂t to Ampere's law mainly to:
Show hintHide hint
Take the divergence of both sides of ∇×H = J.
Show answerHide answer
Answer: B. Make it consistent with the continuity equation for time-varying fields
Taking the divergence of ∇×H = J gives ∇·J = 0, which contradicts ∇·J = −∂ρv/∂t for time-varying charge; adding ∂D/∂t removes the contradiction.
63. A capacitor is connected to an AC source. The current through the dielectric between its plates is:
Show hintHide hint
Current continuity must hold around the circuit.
Show answerHide answer
Answer: C. Displacement current, equal in magnitude to the conduction current in the leads
No charge crosses the gap; the changing D between the plates constitutes a displacement current ε dE/dt × A, equal to the lead current, which keeps Ampere's law consistent.
64. In free space E = 10 sin(10⁹ t) ax V/m. The amplitude of the displacement current density is about:
Show hintHide hint
Differentiate E with respect to time and multiply by ε0.
Show answerHide answer
Answer: A. 0.0885 A/m²
Jd = ε0 ∂E/∂t = ε0 × 10 × 10⁹ cos(10⁹t), so amplitude = 8.854×10⁻¹² × 10¹⁰ = 0.0885 A/m².
65. Sea water has σ = 4 S/m and εr = 81. At 1 MHz the ratio of conduction current to displacement current density (σ/ωε) is about:
Show hintHide hint
This ratio is the loss tangent.
Show answerHide answer
Answer: A. 888
σ/(ωε) = 4/(2π × 10⁶ × 81 × 8.854×10⁻¹²) ≈ 888, so sea water behaves as a good conductor at 1 MHz.
66. Which Maxwell equation in point form expresses Faraday's law of induction?
Show hintHide hint
Look for the curl of E.
Show answerHide answer
Answer: D. ∇×E = −∂B/∂t
Faraday's law states that a time-varying magnetic field produces a circulating electric field: ∇×E = −∂B/∂t.
67. The integral form of the Maxwell equation ∇·B = 0 is:
Show hintHide hint
Divergence goes with a closed surface.
Show answerHide answer
Answer: C. ∮B·dS = 0
By the divergence theorem, ∫(∇·B)dv = ∮B·dS = 0: the net magnetic flux leaving any closed surface is zero.
68. In a source-free, lossless, homogeneous medium, the electric field satisfies the wave equation:
Show hintHide hint
It should contain a second time derivative.
Show answerHide answer
Answer: B. ∇²E = µε ∂²E/∂t²
Taking the curl of Faraday's law and using Ampere-Maxwell with J = 0 gives ∇²E = µε ∂²E/∂t², with wave speed 1/√(µε).
69. A uniform plane wave is a transverse electromagnetic (TEM) wave, which means:
Show hintHide hint
Both fields are transverse.
Show answerHide answer
Answer: A. E and H are perpendicular to each other and to the direction of propagation
In a uniform plane wave neither E nor H has a component along the direction of propagation; E, H and the direction of propagation are mutually perpendicular.
70. A plane wave in free space has a magnetic field amplitude of 1 A/m. The electric field amplitude is about:
Show hintHide hint
Use the intrinsic impedance of free space.
Show answerHide answer
Answer: C. 377 V/m
E = η0H with η0 = 120π ≈ 377 Ω, so E ≈ 377 V/m.
71. The intrinsic impedance of a lossless non-magnetic dielectric with εr = 4 is about:
Show hintHide hint
η scales as 1/√εr.
Show answerHide answer
Answer: C. 188 Ω
η = η0√(µr/εr) = 377/√4 ≈ 188.5 Ω.
72. The phase velocity of a plane wave in a lossless non-magnetic dielectric with εr = 9 is:
Show hintHide hint
Divide c by the refractive index.
Show answerHide answer
Answer: B. 1×10⁸ m/s
vp = c/√(µrεr) = 3×10⁸/3 = 1×10⁸ m/s.
73. A 300 MHz plane wave travels in a lossless non-magnetic medium with εr = 4. Its wavelength in the medium is:
Show hintHide hint
Find λ0 first, then divide by √εr.
Show answerHide answer
Answer: C. 0.5 m
Free-space λ0 = c/f = 1 m; in the medium λ = λ0/√εr = 1/2 = 0.5 m.
74. In free space E = 50 cos(10⁸ t − βz) ax V/m. The phase constant β is about:
Show hintHide hint
β = ω√(µ0ε0).
Show answerHide answer
Answer: C. 0.333 rad/m
β = ω/c = 10⁸/(3×10⁸) = 0.333 rad/m.
75. For a uniform plane wave in a lossless dielectric:
Show hintHide hint
With σ = 0 the intrinsic impedance is real.
Show answerHide answer
Answer: A. There is no attenuation and E and H are in time phase
With σ = 0, α = 0 and η is purely real, so E and H are in phase and the amplitude stays constant.
76. For a plane wave in a lossy dielectric (σ ≠ 0):
Show hintHide hint
Compute η with σ included.
Show answerHide answer
Answer: D. The intrinsic impedance is complex, so E and H are out of time phase
η = √(jωµ/(σ + jωε)) becomes complex, so H lags E by the angle of η, and α > 0 causes attenuation.
77. A medium behaves as a good conductor at a given frequency when:
Show hintHide hint
Compare conduction and displacement currents.
Show answerHide answer
Answer: D. σ ≫ ωε
When the loss tangent σ/(ωε) ≫ 1, conduction current dominates displacement current and the medium is a good conductor.
78. For a plane wave in a good conductor, the magnetic field lags the electric field by about:
Show hintHide hint
Look at the angle of √j.
Show answerHide answer
Answer: B. 45°
In a good conductor η ≈ √(ωµ/σ) ∠45°, so H lags E by 45°.
79. The skin depth of copper (σ = 5.8×10⁷ S/m, µr = 1) at 1 MHz is about:
Show hintHide hint
δ = 1/√(πfµσ).
Show answerHide answer
Answer: D. 66 µm
δ = 1/√(πfµσ) = 1/√(π × 10⁶ × 4π×10⁻⁷ × 5.8×10⁷) ≈ 6.6×10⁻⁵ m = 66 µm.
80. If the frequency of a wave incident on a good conductor is increased four times, the skin depth:
Show hintHide hint
Skin depth varies inversely with the square root of frequency.
Show answerHide answer
Answer: A. Becomes half
δ ∝ 1/√f, so a 4× increase in f reduces δ by √4 = 2.
81. After a plane wave travels a distance of one skin depth into a good conductor, its field amplitude falls to about:
Show hintHide hint
e⁻¹.
Show answerHide answer
Answer: D. 36.8% of its surface value
Amplitude varies as e^(−z/δ); at z = δ it is e⁻¹ ≈ 0.368 of the surface value (power falls to e⁻² ≈ 13.5%).
82. A plane wave in free space has an electric field amplitude of 10 V/m. Its time-average power density is about:
Show hintHide hint
Don't forget the factor ½ for peak amplitudes.
Show answerHide answer
Answer: A. 0.133 W/m²
Pavg = E0²/(2η0) = 100/(2 × 377) ≈ 0.133 W/m².
83. The Poynting vector P = E × H represents:
Show hintHide hint
Check the units of E × H.
Show answerHide answer
Answer: A. The instantaneous power flow per unit area, in W/m²
E × H has units (V/m)(A/m) = W/m² and gives the direction and density of electromagnetic power flow.
84. A plane wave in air is normally incident on a lossless non-magnetic dielectric with εr = 4. The reflection coefficient for E is:
Show hintHide hint
Γ = (η2 − η1)/(η2 + η1).
Show answerHide answer
Answer: B. −1/3
η2 = 377/2; Γ = (η2 − η1)/(η2 + η1) = (0.5 − 1)/(0.5 + 1) = −1/3. (The transmission coefficient is 1 + Γ = 2/3.)
85. A plane wave in air is normally incident on a lossless non-magnetic dielectric with εr = 9. The fraction of incident power reflected is:
Show hintHide hint
Power fraction is |Γ|².
Show answerHide answer
Answer: A. 25%
η2 = η0/3, Γ = (1/3 − 1)/(1/3 + 1) = −0.5, so reflected power fraction = |Γ|² = 0.25.
86. When a plane wave is normally incident on a perfect conductor:
Show hintHide hint
Tangential E must vanish at a perfect conductor.
Show answerHide answer
Answer: D. Γ = −1, the wave is totally reflected and a standing wave with an E-field null at the surface forms
For a perfect conductor η2 = 0, so Γ = −1: total reflection, tangential E = 0 at the surface, and a pure standing wave in front of it.
87. A plane wave in air is obliquely incident on a non-magnetic dielectric with εr = 3. The Brewster angle is:
Show hintHide hint
tan θB = √(εr2/εr1).
Show answerHide answer
Answer: C. 60°
tan θB = √(ε2/ε1) = √3, so θB = 60°.
88. For non-magnetic dielectrics, a Brewster angle (zero reflection) exists for:
Show hintHide hint
Polarizing sunglasses exploit this.
Show answerHide answer
Answer: D. Parallel (p) polarization only
For µ1 = µ2 the reflection coefficient vanishes at θB only when E lies in the plane of incidence (parallel polarization); this is why reflected light is partly polarized.
89. A wave travels from a non-magnetic dielectric with εr = 4 toward air. The critical angle for total internal reflection is:
Show hintHide hint
sin θc = n2/n1.
Show answerHide answer
Answer: C. 30°
sin θc = √(ε2/ε1) = √(1/4) = 0.5, so θc = 30°.
90. A wave E = E0(ax cos ωt + ay sin ωt), with equal x and y amplitudes and 90° phase difference, is:
Show hintHide hint
Trace the tip of E over one period.
Show answerHide answer
Answer: B. Circularly polarized
Two orthogonal components with equal amplitude in phase quadrature make the E vector rotate with constant magnitude: circular polarization.
91. A plane wave meets a boundary where the magnitude of the reflection coefficient is 0.5. The standing wave ratio (SWR) is:
Show hintHide hint
SWR = (1 + |Γ|)/(1 − |Γ|).
Show answerHide answer
Answer: B. 3
SWR = (1 + |Γ|)/(1 − |Γ|) = 1.5/0.5 = 3.
92. The propagation constant γ = α + jβ of a plane wave in a general (lossy) medium is:
Show hintHide hint
Do not confuse it with the intrinsic impedance.
Show answerHide answer
Answer: B. √(jωµ(σ + jωε))
γ = √(jωµ(σ + jωε)). The ratio form √(jωµ/(σ + jωε)) is the intrinsic impedance; ω√(µε) is β for the lossless case.
6.4 Wave-guides and antenna
30 questions · AExE0604
93. The dominant mode of a rectangular waveguide with broad dimension a greater than narrow dimension b is:
Show hintHide hint
The mode with the lowest cutoff frequency.
Show answerHide answer
Answer: C. TE10
TE10 has the lowest cutoff frequency, fc = c/(2a), so it is the dominant mode when a > b.
94. The lowest-order TM mode that can exist in a rectangular waveguide is:
Show hintHide hint
What happens to Ez if m or n is zero?
Show answerHide answer
Answer: B. TM11
For TM modes Ez ∝ sin(mπx/a) sin(nπy/b); if m or n is zero all fields vanish, so the lowest TM mode is TM11.
95. A TEM wave cannot propagate inside a hollow rectangular waveguide because:
Show hintHide hint
Compare with a coaxial cable.
Show answerHide answer
Answer: D. A hollow guide has only one conductor, and TEM waves need at least two conductors
A TEM mode needs a static-like transverse potential difference between two separate conductors, which a single hollow conductor cannot support.
96. In a transverse electric (TE) mode of a waveguide:
Show hintHide hint
'Transverse electric' says which field has no axial component.
Show answerHide answer
Answer: A. Ez = 0 and Hz ≠ 0
TE means the electric field is entirely transverse (no component along the guide axis), while the magnetic field has a longitudinal component.
97. When a waveguide is operated below the cutoff frequency of a mode, that mode:
Show hintHide hint
A waveguide behaves like a high-pass filter.
Show answerHide answer
Answer: D. Is evanescent: its fields decay exponentially along the guide and carry no real power
Below cutoff the propagation constant is purely real (attenuation), so the waveguide behaves as a high-pass filter for that mode.
98. An air-filled rectangular waveguide has a = 2.286 cm and b = 1.016 cm. The cutoff frequency of the TE10 mode is about:
Show hintHide hint
TE10 cutoff depends only on the broad dimension.
Show answerHide answer
Answer: C. 6.56 GHz
fc(TE10) = c/(2a) = 3×10⁸/(2 × 0.02286) ≈ 6.56 GHz.
99. The cutoff wavelength of the TE10 mode in a rectangular waveguide of broad dimension a is:
Show hintHide hint
Half a wavelength must fit across the broad wall.
Show answerHide answer
Answer: C. 2a
λc = 2/√((m/a)² + (n/b)²); for m = 1, n = 0 this gives λc = 2a.
100. An air-filled rectangular waveguide has a = 5 cm and b = 2.5 cm. The cutoff frequency of the TM11 mode is about:
Show hintHide hint
Use both m = 1 and n = 1 in the cutoff formula.
Show answerHide answer
Answer: D. 6.71 GHz
fc = (c/2)√((1/a)² + (1/b)²) = 1.5×10⁸ × √(400 + 1600) ≈ 6.71 GHz.
101. An air-filled waveguide with a = 5 cm (TE10 cutoff 3 GHz) is operated at 5 GHz. The guide wavelength is:
Show hintHide hint
Guide wavelength is longer than the free-space wavelength.
Show answerHide answer
Answer: A. 7.5 cm
λ0 = 6 cm; λg = λ0/√(1 − (fc/f)²) = 6/√(1 − 0.36) = 6/0.8 = 7.5 cm.
102. In the same air-filled guide (TE10 cutoff 3 GHz) at 5 GHz, the group velocity is:
Show hintHide hint
Group velocity is always less than c in an air-filled guide.
Show answerHide answer
Answer: B. 2.4×10⁸ m/s
vg = c√(1 − (fc/f)²) = 3×10⁸ × 0.8 = 2.4×10⁸ m/s (and vp·vg = c²).
103. In the same air-filled guide (TE10 cutoff 3 GHz) at 5 GHz, the TE10 wave impedance is about:
Show hintHide hint
TE wave impedance exceeds η0.
Show answerHide answer
Answer: A. 471 Ω
ZTE = η0/√(1 − (fc/f)²) = 377/0.8 ≈ 471 Ω.
104. For a propagating mode in an air-filled waveguide:
Show hintHide hint
Energy cannot travel faster than light.
Show answerHide answer
Answer: C. Phase velocity exceeds c while group velocity is less than c, with vp·vg = c²
vp = c/√(1 − (fc/f)²) > c and vg = c√(1 − (fc/f)²) < c; energy travels at vg, so no signal exceeds c.
105. An air-filled waveguide has a = 5 cm and b = 2.5 cm. At 7 GHz, how many TE/TM modes can propagate?
Show hintHide hint
Remember TE11 and TM11 share the same cutoff.
Show answerHide answer
Answer: A. 5
Cutoffs: TE10 3 GHz; TE20 and TE01 6 GHz; TE11 and TM11 6.71 GHz; next (TE30, TE21, TM21) are higher than 7 GHz. So 5 modes propagate.
106. Electromagnetic radiation from an antenna is produced by:
Show hintHide hint
Static sources produce static fields.
Show answerHide answer
Answer: B. Time-varying (accelerating or decelerating) currents or charges
Static charges produce only electrostatic fields and steady currents only magnetostatic fields; radiation requires time-varying current, i.e. accelerated charge.
107. The reciprocity theorem applied to antennas implies that:
Show hintHide hint
Can the same dish be used to transmit and receive?
Show answerHide answer
Answer: B. An antenna's radiation pattern and impedance are the same whether it transmits or receives
Because Maxwell's equations are reciprocal in linear, isotropic media, an antenna's transmit and receive properties (pattern, gain, impedance) are identical.
108. A 1 m diameter dish antenna operates at 3 GHz. The minimum distance for far-field (Fraunhofer) measurements is about:
Show hintHide hint
R = 2D²/λ.
Show answerHide answer
Answer: B. 20 m
λ = 0.1 m; far-field distance R = 2D²/λ = 2 × 1/0.1 = 20 m.
109. In the reactive near-field region of an antenna:
Show hintHide hint
Think of the terms that fall faster than 1/r.
Show answerHide answer
Answer: C. Energy is mainly stored and exchanged with the antenna rather than radiated
Very close to the antenna the reactive (stored) field terms (∝ 1/r², 1/r³) dominate; the radiating far-field terms (∝ 1/r) dominate only far away.
110. The directivity of an antenna is the ratio of:
Show hintHide hint
Compare with an isotropic radiator of equal total power.
Show answerHide answer
Answer: A. Its maximum radiation intensity to the average radiation intensity over all directions
D = Umax/Uavg = 4πUmax/Prad, i.e. compared with an isotropic source radiating the same total power. Losses do not enter directivity.
111. An antenna has directivity 10 and radiation efficiency 80%. Its gain is about:
Show hintHide hint
Multiply first, then convert to dB.
Show answerHide answer
Answer: C. 9.0 dBi
G = ηD = 0.8 × 10 = 8, and 10 log10(8) ≈ 9.03 dBi.
112. A Hertzian (short) dipole has length λ/20. Its radiation resistance is about:
Show hintHide hint
Rr = 80π²(dl/λ)².
Show answerHide answer
Answer: A. 1.97 Ω
Rr = 80π²(dl/λ)² = 80 × 9.87 × (1/20)² ≈ 1.97 Ω.
113. An antenna has radiation resistance 73 Ω and loss resistance 2 Ω. Its radiation efficiency is about:
Show hintHide hint
Efficiency compares radiated power with total input power.
Show answerHide answer
Answer: A. 97.3%
η = Rr/(Rr + RL) = 73/75 ≈ 0.973.
114. The directivity of a thin half-wave dipole is about:
Show hintHide hint
Slightly more directive than a short dipole.
Show answerHide answer
Answer: D. 1.64 (2.15 dBi)
A half-wave dipole has D ≈ 1.64 (2.15 dBi) and Rr ≈ 73 Ω; 1.5 is the Hertzian dipole's directivity.
115. A Hertzian dipole (D = 1.5) operates at a wavelength of 1 m. Its maximum effective aperture is about:
Show hintHide hint
Ae = Gλ²/(4π).
Show answerHide answer
Answer: C. 0.119 m²
Ae = Dλ²/(4π) = 1.5 × 1/(4π) ≈ 0.119 m².
116. The half-power beamwidth (HPBW) of an antenna is the angular width of the main lobe between the points where:
Show hintHide hint
Half power, not half field.
Show answerHide answer
Answer: D. The radiated power density falls to half (−3 dB) of its maximum
HPBW is measured between the −3 dB (half-power, 0.707 field) points; the angle between the first nulls is FNBW.
117. An antenna has half-power beamwidths of 20° in both principal planes. Using D ≈ 41 253/(θE·θH), its directivity is about:
Show hintHide hint
Beamwidths are in degrees in this formula.
Show answerHide answer
Answer: B. 103 (about 20 dBi)
D ≈ 41 253/(20 × 20) ≈ 103, i.e. about 20.1 dBi.
118. A transmitter delivers 10 W to an antenna with 20 dBi gain (lossless feed). The EIRP is:
Show hintHide hint
Convert dBi to a ratio before multiplying.
Show answerHide answer
Answer: B. 1000 W (30 dBW)
20 dBi = 100; EIRP = PtGt = 10 × 100 = 1000 W = 30 dBW.
119. The SI unit of radiation intensity of an antenna is:
Show hintHide hint
Intensity is per unit solid angle.
Show answerHide answer
Answer: A. Watt per steradian
Radiation intensity U is the power radiated per unit solid angle (W/sr); power density is W/m², and U = r² × power density in the far field.
120. The polarization of an antenna is defined by:
Show hintHide hint
Wave polarization always refers to one field.
Show answerHide answer
Answer: D. The orientation of the radiated electric field in the direction of maximum radiation
Antenna polarization is the polarization of the wave it radiates, described by the path traced by the E-field vector, normally in the direction of maximum gain.
121. The front-to-back ratio of a directional antenna is:
Show hintHide hint
Front versus back direction.
Show answerHide answer
Answer: B. The ratio of power radiated in the main-beam direction to that radiated in the opposite direction
Front-to-back ratio, usually in dB, compares radiation in the forward (maximum) direction with radiation 180° away.
122. The radiation resistance of an antenna is:
Show hintHide hint
It represents useful radiated power.
Show answerHide answer
Answer: D. A fictitious resistance that would dissipate the same power the antenna radiates, for the same input current
Rr = 2Prad/|I0|²; it accounts for power leaving as radiation, as opposed to loss resistance, which accounts for heat.
6.5 Antenna's classification
30 questions · AExE0605
123. An isotropic antenna is:
Show hintHide hint
It is the reference for dBi.
Show answerHide answer
Answer: A. A hypothetical lossless point source that radiates equally in all directions
The isotropic radiator cannot be built; its spherical pattern serves as the 0 dBi reference for gain and directivity.
124. An omnidirectional antenna, such as a vertical dipole, has a radiation pattern that is:
Show hintHide hint
Think of the doughnut shape.
Show answerHide answer
Answer: D. Uniform in one plane (e.g. azimuth) but directional in the perpendicular plane
An omnidirectional antenna radiates equally in all directions of one plane (a circle in the H-plane) while its E-plane pattern has nulls, giving a doughnut-shaped 3-D pattern.
125. The radiation pattern of a thin vertical half-wave dipole has:
Show hintHide hint
A dipole does not radiate off its ends.
Show answerHide answer
Answer: D. Maximum radiation broadside (perpendicular) to the wire and nulls along its axis
The dipole pattern varies roughly as sin θ (more precisely cos(π/2 cosθ)/sinθ), maximum at θ = 90° and zero along the axis.
126. The theoretical (free-space) length of a half-wave dipole for 100 MHz is:
Show hintHide hint
Find λ first.
Show answerHide answer
Answer: B. 1.5 m
λ = c/f = 3 m, so λ/2 = 1.5 m (practical dipoles are made about 5% shorter because of end effect).
127. The input impedance of a two-wire folded half-wave dipole is about:
Show hintHide hint
Impedance step-up equals the square of the number of conductors.
Show answerHide answer
Answer: B. 292 Ω
With two equal conductors the current divides equally, so the impedance is stepped up by 2² = 4: about 4 × 73 ≈ 292 Ω (commonly quoted as 300 Ω).
128. A quarter-wave vertical monopole for 150 MHz is about how long?
Show hintHide hint
Monopole = λ/4.
Show answerHide answer
Answer: C. 0.5 m
λ = 3×10⁸/150×10⁶ = 2 m; λ/4 = 0.5 m.
129. The radiation resistance of a quarter-wave monopole over a perfectly conducting ground plane is about:
Show hintHide hint
Use image theory.
Show answerHide answer
Answer: D. 36.5 Ω
By image theory the monopole with its image forms a half-wave dipole, but it radiates into only a half space, so Rr is half of 73 Ω ≈ 36.5 Ω.
130. A travelling-wave (non-resonant) antenna such as a long wire terminated in a matched load:
Show hintHide hint
What does the matched termination remove?
Show answerHide answer
Answer: A. Has no standing wave on it, a unidirectional pattern and a wide bandwidth
The matched termination absorbs the forward wave, so there is no reflected wave; the current is a travelling wave giving a unidirectional pattern toward the load and broad bandwidth.
131. A rhombic antenna is mainly used for:
Show hintHide hint
It needs a large area and wire lengths of several wavelengths.
Show answerHide answer
Answer: B. Long-distance HF point-to-point sky-wave links
The terminated rhombic is a large, broadband, directive travelling-wave antenna made of long wires, suited to HF point-to-point communication via the ionosphere.
132. A V antenna consists of:
Show hintHide hint
The name describes its shape.
Show answerHide answer
Answer: A. Two long wires arranged in a V and fed at the apex
Two long-wire radiators at a suitable apex angle are fed in antiphase at the apex; their major lobes add along the bisector of the V.
133. A large flat conducting sheet is placed λ/4 behind a half-wave dipole that is parallel to the sheet. The effect is that:
Show hintHide hint
Use the image of the dipole in the sheet.
Show answerHide answer
Answer: C. Forward radiation is reinforced and backward radiation is largely eliminated
The image current is reversed (180°); the extra λ/2 round trip adds another 180°, so the reflected and direct waves add in phase in the forward direction while the sheet blocks the back.
134. In a 90° corner reflector antenna, the number of images of the driven element formed by the two plane sheets is:
Show hintHide hint
Think of two perpendicular plane mirrors.
Show answerHide answer
Answer: B. 3
For a corner angle of 180°/n the number of images is 2n − 1; for 90° (n = 2) there are 3 images, which with the driven element form a four-element array.
135. The key geometric property of a paraboloidal reflector is that:
Show hintHide hint
This is why dish feeds sit at the focus.
Show answerHide answer
Answer: A. Rays from a source at the focus are reflected parallel to the axis, with equal path lengths to the aperture plane
All paths from the focus to the aperture plane via the parabola are equal, so a spherical wave from the focal feed becomes a plane wave, producing a narrow pencil beam.
136. A parabolic dish 3 m in diameter operates at 10 GHz with aperture efficiency 0.55. Its gain is about:
Show hintHide hint
G = η(πD/λ)².
Show answerHide answer
Answer: B. 47.3 dBi
λ = 0.03 m; G = η(πD/λ)² = 0.55 × (π × 100)² ≈ 5.43×10⁴, i.e. 47.3 dBi.
137. Using HPBW ≈ 70λ/D degrees, a 3 m dish at 10 GHz has a half-power beamwidth of about:
Show hintHide hint
Find λ in metres first.
Show answerHide answer
Answer: A. 0.7°
λ = 0.03 m; HPBW ≈ 70 × 0.03/3 = 0.7°.
138. A parabolic dish has an aperture diameter of 2 m and a depth of 0.25 m. Its focal length is:
Show hintHide hint
f = D²/(16 × depth).
Show answerHide answer
Answer: A. 1.0 m
For a parabola f = D²/(16d) = 4/(16 × 0.25) = 1 m.
139. A Cassegrain reflector antenna uses:
Show hintHide hint
Gregorian uses the other conic section.
Show answerHide answer
Answer: B. A hyperboloidal subreflector with the feed near the vertex of the main paraboloid
In the Cassegrain system a convex hyperbolic subreflector shares a focus with the paraboloid and redirects energy from a feed near the vertex; a Gregorian system uses a concave ellipsoidal subreflector instead.
140. The main drawback of a spherical (circular) reflector compared with a paraboloid is:
Show hintHide hint
A sphere has no true single focal point.
Show answerHide answer
Answer: C. Spherical aberration: rays from a point feed are not all reflected parallel
A sphere has no single focus, so a point source produces phase errors (spherical aberration); its advantage is that the beam can be scanned by moving the feed.
141. A parabolic cylinder (cylindrical parabolic) reflector is normally fed by:
Show hintHide hint
Its focus is a line, not a point.
Show answerHide answer
Answer: A. A line source, such as a linear array of dipoles placed along its focal line
A parabolic cylinder focuses in only one plane, so its feed is a line source along the focal line, producing a fan beam.
142. A horn antenna can be regarded as:
Show hintHide hint
What happens if a waveguide is simply left open?
Show answerHide answer
Answer: D. A flared waveguide that matches the guide impedance gradually to free space
A horn is an aperture antenna: flaring the end of a waveguide provides a smooth transition to free-space impedance and a larger aperture, giving higher gain and low VSWR.
143. In a Yagi-Uda antenna, compared with the driven element:
Show hintHide hint
The beam points toward the directors.
Show answerHide answer
Answer: D. The reflector is slightly longer and the directors are slightly shorter
A longer (inductive) reflector and shorter (capacitive) directors set the currents' phases so the radiation is concentrated toward the directors.
144. The main advantage of a log-periodic dipole array is that it:
Show hintHide hint
Look at how its element lengths and spacings scale.
Show answerHide answer
Answer: A. Has nearly constant impedance and pattern over a very wide frequency range
Its element lengths and spacings scale by a constant factor τ, so its properties repeat periodically with log f, making it frequency-independent over a wide band.
145. For an end-fire uniform linear array with element spacing λ/4, the magnitude of the progressive phase shift between adjacent elements is:
Show hintHide hint
Phase shift equals kd for end-fire.
Show answerHide answer
Answer: B. 90°
End-fire needs β = −kd = −(2π/λ)(λ/4) = −π/2, i.e. 90° in magnitude.
146. A broadside uniform linear array has:
Show hintHide hint
Feed all elements in phase.
Show answerHide answer
Answer: C. Zero progressive phase shift, with maximum radiation perpendicular to the array axis
With all elements fed in phase, their fields add in phase in directions perpendicular to the line of the array.
147. A 4-element broadside uniform array has spacing λ/2. Its first null occurs at what angle from the array axis?
Show hintHide hint
Set Nψ/2 = π with ψ = kd cosθ.
Show answerHide answer
Answer: D. 60°
Nulls occur at Nψ/2 = π, so ψ = kd cosθ = π cosθ = π/2, giving cosθ = 0.5 and θ = 60° from the axis (30° from broadside).
148. A small loop antenna (circumference ≪ λ) has:
Show hintHide hint
It acts as a magnetic dipole.
Show answerHide answer
Answer: C. Maximum radiation in the plane of the loop and a null along its axis, making it useful for direction finding
A small loop is a magnetic dipole: its pattern is like a short dipole's with the axis perpendicular to the loop, and its sharp null is used for direction finding.
149. For axial-mode operation, the circumference of a helical antenna should be about one wavelength. At 1.5 GHz the helix diameter is about:
Show hintHide hint
Diameter = circumference/π.
Show answerHide answer
Answer: B. 6.4 cm
λ = 0.2 m; C ≈ λ, so D = λ/π = 0.2/π ≈ 0.064 m.
150. A helical antenna operating in the axial mode radiates:
Show hintHide hint
Used for satellite and space links.
Show answerHide answer
Answer: D. A circularly polarized beam along the axis of the helix
In axial (end-fire) mode with C ≈ λ, the helix produces a directive, circularly polarized beam along its axis, which is why it is used in satellite links.
151. Ignoring fringing, the resonant length of a rectangular microstrip patch at 2.4 GHz on a substrate with εr = 4.4 is about:
Show hintHide hint
A patch is about half a wavelength in the dielectric.
Show answerHide answer
Answer: C. 3.0 cm
λ0 = 12.5 cm; L ≈ λ0/(2√εr) = 12.5/(2 × 2.10) ≈ 2.98 cm.
152. Compared with most other antennas, a microstrip patch antenna characteristically has:
Show hintHide hint
Think of the antenna inside a mobile phone.
Show answerHide answer
Answer: C. Low profile and easy fabrication but narrow bandwidth
Patches are thin, cheap and conformal (printed-circuit technology), but their resonant nature gives narrow bandwidth, low power handling and modest efficiency.
6.6 Propagation and radio frequency spectrum
31 questions · AExE0606
153. Ground (surface) wave propagation is mainly used in which frequency range?
Show hintHide hint
Attenuation of the surface wave grows with frequency.
Show answerHide answer
Answer: D. LF and MF, e.g. medium-wave AM broadcasting
Ground-wave attenuation rises rapidly with frequency, so it is useful mainly below about 2–3 MHz (VLF, LF, MF).
154. Ground waves travel farthest over:
Show hintHide hint
Which surface has the highest conductivity?
Show answerHide answer
Answer: C. Sea water
Ground-wave attenuation decreases with increasing ground conductivity; sea water (σ ≈ 4 S/m) is the best conductor among these.
155. Ground-wave signals are vertically polarized because:
Show hintHide hint
Recall the boundary condition at a conductor.
Show answerHide answer
Answer: A. Horizontally polarized E fields are largely short-circuited by the conducting earth
A horizontal E field parallel to a conducting earth is cancelled by the induced ground currents, so surface waves need vertical polarization.
156. Space-wave propagation, used at VHF and above, consists of:
Show hintHide hint
It travels in the lower atmosphere above the ground.
Show answerHide answer
Answer: B. A direct wave plus a wave reflected from the ground, limited roughly to line of sight
The space wave travels through the troposphere as the direct ray and the ground-reflected ray; its range is set by the radio horizon.
157. Using d ≈ 4.12(√ht + √hr) km (heights in metres, standard atmosphere), the radio-horizon range between antennas 100 m and 16 m high is about:
Show hintHide hint
Add the horizon distances of both antennas.
Show answerHide answer
Answer: D. 57.7 km
d = 4.12(√100 + √16) = 4.12 × 14 ≈ 57.7 km.
158. With the standard k-factor of 4/3, the effective earth radius used for radio propagation is about:
Show hintHide hint
Multiply the true earth radius by k.
Show answerHide answer
Answer: D. 8 495 km
Effective radius = k × actual radius = (4/3) × 6371 ≈ 8495 km; this accounts for the downward bending of rays in a standard atmosphere.
159. In a standard troposphere, radio waves bend slightly toward the earth because:
Show hintHide hint
Rays bend toward the region of higher refractive index.
Show answerHide answer
Answer: B. The refractive index of air decreases with height
Air refractivity falls with altitude, so the upper part of a wavefront travels faster and the ray curves downward, extending the radio horizon beyond the optical horizon.
160. Duct propagation (super-refraction) occurs when:
Show hintHide hint
Associated with temperature inversion in the troposphere.
Show answerHide answer
Answer: C. A temperature inversion makes the refractive index fall steeply with height, trapping VHF/UHF/microwave signals
In a duct the ray curvature exceeds the earth's curvature, so signals are guided between a layer and the ground (or between layers) over unusually long distances.
161. Tropospheric scatter propagation is used for:
Show hintHide hint
It needs high power because only a tiny fraction of energy is scattered.
Show answerHide answer
Answer: D. Beyond-the-horizon UHF/microwave links using scattering from irregularities in the troposphere
Small fluctuations of refractive index in the troposphere scatter a little energy forward, allowing links of a few hundred kilometres beyond the horizon with high power and large antennas.
162. Sky-wave propagation is mainly used for long-distance communication in the:
Show hintHide hint
Short-wave radio.
Show answerHide answer
Answer: B. HF band (3–30 MHz)
HF waves are refracted back to earth by the ionosphere, allowing ranges of thousands of kilometres through single or multiple hops.
163. Which ionospheric layer mainly absorbs MF and lower-HF signals in the daytime and almost disappears at night?
Show hintHide hint
Absorption is strongest where the air is densest.
Show answerHide answer
Answer: A. D layer
The D layer (about 50–90 km) exists mainly in daylight; it absorbs rather than reflects, which is why distant MF stations are heard mostly at night.
164. The ionospheric layer most important for long-distance HF communication, being the highest and most ionized, is the:
Show hintHide hint
Highest layer gives the longest hop.
Show answerHide answer
Answer: B. F2 layer
The F2 layer (about 250–400 km) has the highest electron density and persists at night, so it supports the longest single-hop sky-wave paths.
165. The critical frequency of an ionospheric layer is:
Show hintHide hint
Vertical incidence.
Show answerHide answer
Answer: A. The highest frequency reflected back to earth when the wave is sent vertically upward
fc is the maximum frequency returned at vertical incidence; for oblique paths the maximum usable frequency is higher.
166. An ionospheric layer has a maximum electron density of 10¹² electrons/m³. Using fc = 9√Nmax Hz, its critical frequency is:
Show hintHide hint
√(10¹²) = 10⁶.
Show answerHide answer
Answer: B. 9 MHz
fc = 9√(10¹²) = 9 × 10⁶ Hz = 9 MHz.
167. A layer has a critical frequency of 8 MHz. For a wave incident on it at 60° from the vertical (flat-earth secant law), the maximum usable frequency is:
Show hintHide hint
MUF = fc/cosθ.
Show answerHide answer
Answer: A. 16 MHz
MUF = fc secθi = 8/cos60° = 16 MHz.
168. The maximum usable frequency (MUF) is:
Show hintHide hint
It depends on the distance between stations.
Show answerHide answer
Answer: C. The highest frequency returned to earth by the ionosphere for a given angle of incidence (path length)
MUF depends on the path: MUF = fc secθi. The optimum working frequency is usually taken as about 85% of the MUF.
169. For a flat-earth layer at virtual height 300 km, if MUF/fc = 2, the skip distance is about:
Show hintHide hint
D = 2h√((f/fc)² − 1).
Show answerHide answer
Answer: C. 1039 km
D = 2h√((MUF/fc)² − 1) = 2 × 300 × √3 ≈ 1039 km.
170. The skip distance in sky-wave propagation is:
Show hintHide hint
Inside it you may hear nothing from the sky wave.
Show answerHide answer
Answer: C. The shortest distance from the transmitter at which a sky wave of given frequency returns to earth
Waves sent at steeper angles penetrate the layer, so a given frequency first returns at the skip distance; between the ground-wave range and the skip distance lies the silent (skip) zone.
171. The ionosphere has N = 10¹² electrons/m³. Using n = √(1 − 81N/f²), its refractive index for an 18 MHz wave is about:
Show hintHide hint
Compute 81N/f² first.
Show answerHide answer
Answer: C. 0.866
81N/f² = 81×10¹²/(3.24×10¹⁴) = 0.25; n = √0.75 ≈ 0.866.
172. A vertically transmitted ionosonde pulse returns after 2 ms. The virtual height of the reflecting layer is:
Show hintHide hint
The pulse travels up and back.
Show answerHide answer
Answer: A. 300 km
h' = ct/2 = 3×10⁸ × 2×10⁻³/2 = 3×10⁵ m = 300 km.
173. The virtual height of an ionospheric layer is:
Show hintHide hint
Virtual height assumes the wave travels at c all the way.
Show answerHide answer
Answer: B. Greater than the actual height of reflection, because the wave slows down inside the layer
Virtual height assumes travel at c along straight lines to a mirror; the actual wave is gradually bent at lower group velocity, so the actual reflection height is below the virtual height.
174. Using FSPL(dB) = 32.44 + 20 log10(d km) + 20 log10(f MHz), the free-space path loss over 10 km at 1 GHz is about:
Show hintHide hint
Use f = 1000 MHz.
Show answerHide answer
Answer: D. 112.4 dB
FSPL = 32.44 + 20 + 60 = 112.44 dB.
175. A link has Pt = 30 dBm, Gt = 20 dBi, Gr = 10 dBi and free-space path loss 112.4 dB. The received power is about:
Show hintHide hint
Add the gains, subtract the loss.
Show answerHide answer
Answer: A. −52.4 dBm
Friis in dB: Pr = 30 + 20 + 10 − 112.4 = −52.4 dBm.
176. In free-space propagation, if the distance between transmitter and receiver is doubled, the received power:
Show hintHide hint
Inverse-square law.
Show answerHide answer
Answer: A. Falls by 6 dB
Received power varies as 1/d², so doubling d reduces it to 1/4, i.e. −6 dB.
177. In the plane-earth (two-ray) model at large distances, the received power varies with distance d as:
Show hintHide hint
Faster than free space.
Show answerHide answer
Answer: C. 1/d⁴
Pr ≈ PtGtGr(hthr)²/d⁴: the direct and ground-reflected rays nearly cancel, giving a 40 dB/decade fall-off.
178. In the plane-earth model, doubling the transmitting antenna height (all else fixed) changes the received power by about:
Show hintHide hint
Pr depends on (ht hr)².
Show answerHide answer
Answer: B. +6 dB
Pr ∝ ht², so doubling ht multiplies Pr by 4, i.e. +6 dB.
179. An isotropic antenna radiates 100 W. In free space the power density at 1 km is about:
Show hintHide hint
Spread the power over a sphere.
Show answerHide answer
Answer: D. 7.96 µW/m²
S = Pt/(4πr²) = 100/(4π × 10⁶) ≈ 7.96×10⁻⁶ W/m².
180. An isotropic antenna radiates 100 W in free space. The rms electric field at 1 km (E = √(30Pt)/r) is about:
Show hintHide hint
Take the square root of 30Pt.
Show answerHide answer
Answer: B. 54.8 mV/m
E = √(30 × 100)/1000 = 54.77/1000 ≈ 54.8 mV/m.
181. Fading in sky-wave reception is mainly caused by:
Show hintHide hint
Multipath.
Show answerHide answer
Answer: D. Interference between waves arriving over different paths whose lengths keep changing
Variations in the ionosphere change the relative phases of multipath (and different-mode) signals, producing random rises and falls in signal strength.
182. A 15 MHz signal has a free-space wavelength of:
Show hintHide hint
λ = c/f, and recall the HF limits.
Show answerHide answer
Answer: C. 20 m, in the HF band
λ = c/f = 3×10⁸/15×10⁶ = 20 m; 3–30 MHz is HF (decametric waves).
183. Satellite links use frequencies well above about 30 MHz mainly because such frequencies:
Show hintHide hint
Compare with the MUF of sky-wave links.
Show answerHide answer
Answer: A. Pass through the ionosphere instead of being reflected back to earth
Frequencies above the maximum frequency the ionosphere can return penetrate it, so VHF, UHF and microwaves are used to reach satellites.