Electromagnetic Theory Questions and Answers - Spherical Coordinate System
Electromagnetic Theory Questions and Answers - Spherical Coordinate System
Electromagnetic Theory Questions and Answers - Spherical Coordinate System
Answer: a
Explanation: r = √(x2+y2+z2) = √50 = 7.07
Θ = cos-1(z/r) = cos-1(5/5√2) = 45⁰
Φ = tan-1(y/x) = tan-1(4/3) = 53⁰.
Answer: a
Explanation: From a point charge +Q, the electric field spreads in all 360 degrees. The
calculation of electric field in this case will be spherical system. Thus it is charge in the space.
Answer: b
Explanation: There is no waveguide designed spherically to avoid absorption, rather than
propagation.
Answer: b
Explanation: r = √(x2+y2+z2) = √14 = 3.74
Θ = cos-1(z/r) = cos-1(-1/3.74) = 105.5⁰
Φ = tan-1(y/x) = tan-1(3/2) = 56.31⁰.
Answer: b
Explanation: x = r sin θ cos φ = 4 sin25⁰ cos 120⁰ = -0.845
y = r sin θ sin φ = 4 sin 25⁰ sin 120⁰ = 1.462
z = r cos θ = 4 cos 25⁰ = 3.625.
7. The area of sphere can be computed from the sphere volume. State True/False.
a) True
b) False
View Answer
Answer: a
Explanation: On double integrating the differential volume, the area can be computed for a
sphere.
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8. Given B= (10/r)i+( rcos θ) j+k in spherical coordinates. Find Cartesian points at (-3,4,0)
a) -2i + j
b) 2i + k
c) i + 2j
d) –i – 2k
View Answer
Answer: a
Explanation: r = √(x2+y2+z2) = √25 = 5
Θ = cos-1(z/r) = 1
Φ = tan-1(y/x) = tan-1(-4/3)
Thus, B = -2i + j.
Answer: a
Explanation: The radius varies from unity to infinity, the plane angle from zero to 360 ⁰ and the z
plane from (-∞, ∞) .
Answer: b
Explanation: r = √(x2+y2+z2) = 3.74
Θ = cos-1(z/r) = cos-1(3/3.74) = 36.7⁰
Φ = tan-1(y/x) = tan-1(2/1) = 63.4⁰
A = (4 sin θ cos φ – 2 sin θ sin φ – 4cos θ)i + (4 cos θ cos φ – 2 cos θ sin φ + 4 sin θ)j + (-4 sin φ
– 2 cos φ)k
On substituting r, θ, φ, A = -3.197i + 2.393j – 4.472k.
1. When a wave is incident from the more dense into a less dense medium at an angle equal to or
exceeding the critical angle, the wave suffers total internal __________ .
A. reflection
B. refraction
C. transmission
D. none of the above
Answer: B
Answer: C
A. microwave frequency
B. medium frequency
C. low frequency
D. high frequency
Answer: A
Answer: A
5. The potential that appears at a point in space due to the current which caused it is called
potential
A. accelerating
B. retardation
C. oscillating
D. lagging the current
Answer: B
A. unity
B. constant
C. both (a) and (b)
D. none of the above
Answer: D
7. It is best to use for coupling a coaxial line to a parallel wire transmission line.
A. directional Coupler
B. balun
C. slotted line
D. tee
Answer: B
A. totally retarded
B. retarded
C. accelerated
D. neither accelerated nor retarded
Answer: B
Answer: C
10. In case of antenna the ratio of the power radiated in the desired direction to the power
radiated in the opposite direction is known as
A. transmission efficiency
B. front to back ratio
C. loss coefficient
D. none of the above
Answer: B
11. The radiation pattern of Hertzian dipole in the plane perpendicular to dipole is a
A. null
B. circle
C. figure of eight
D. none of the above
Answer: C
12. The transmitting antennas for lower frequencies (below 500 kHz) are generally
Answer: A
13. A satellite that simply reflect back the signals from one region of the earth to the other region
is known as
A. orbiting satellite
B. geostationary satellite
C. active satellite
D. passive satellite
Answer: D
A. True
B. False
Answer: B
15. Which of the following will increase the antenna radiation efficiency?
Answer: C
A. phase shifter
B. an attenuator
C. an isolator
D. none
Answer: B
A. Short-circuited stub
B. A quarter wave line
C. Balun transformer
D. A half wave line
Answer: B
18. In case of surface waves the field strength at a point is directly proportional to
A. antenna height
B. wave frequency
C. current of the antenna
D. distance of the point from the antenna
Answer: D
Answer: C
A. HF
B. VHF
C. UHF
D. VLF
Answer: C
A. only
B. with a reflector
C. with one or more directors
D. with a reflector and one or more directors
Answer: D
22. The minimum height of outer atmosphere is
A. 100 km
B. 150 km
C. 200 km
D. 400 km
Answer: D
A. radio astronomy
B. tv broadcast
C. point to point communication
D. all of the above
Answer: D
Answer: D
25. The power gain of a half-wave dipole with respect to an isotropic radiator is
A. 2.15 db
B. 3 db
C. 4.15 db
D. 6 db
Answer: A
26. In an impedance smith chart, a clockwise movement along a constant resistance circle gives
rise to
27. When the phase velocity of an EM wave depends on frequency in any medium the
phenomenon is called
A. scattering
B. polarisation
C. absorption
D. dispersion
Answer: B
A. poor directivity
B. narrow bandwidth
C. high standing wave ratio
D. low directional coupling
Answer: B
A. omnidirectional
B. a figure of directional
C. highly directional
D. none of the above
Answer: C
30. In microwave communication links, when fading due to rain attenuation occurs then
technique adopted for solving the problem would include
Answer: D
31. Due to which of the following reasons a lightning conductor on top of a building is made into
a pointed spike?
Answer: B
A. equal to
B. less than
C. higher than
D. no relation
Answer: C
33. The divergence of a vector is a scalar, while the curl of a vector is another
A. scalar
B. vector
C. unit vector
D. none of the above
Answer: B
34. An open wire transmission line having a characteristic impedance of 600 ohms is terminated
by a resistive load of 900 O. The standing wave ratio will be
A. 1.5
B. 2.5
C. 3.5
D. 4.5
Answer: A
35. The skip distance is the maximum distance upto which ionospheric reflection is possible.
A. True
B. False
Answer: B
A. Hertz antenna
B. Marconi antenna
C. Folded dipole
D. None of the above
Answer: C
A. circle
B. semicircle
C. elliptical
D. cardoid or Limacon
Answer: D
A. True
B. False
Answer: B
A. very high
B. very low
C. zero
D. infinite
Answer: B
40. When a current carrying conductor is brought into magnetic field, the force that moves the
conductor depends upon
Answer: A
A. TV broadcast
B. Point-to-point communication
C. Radio astronomy
D. All of the above
Answer: D
42. According to maximum power transfer theorem, the maximum power is absorbed by one
network from another network when
Answer: B
A. zero
B. unity
C. infinity
D. none of the above
Answer: A
44. In a perfect conductor the incident and reflected wave combine to produce
Answer: C
A. True
B. False
Answer: A
A. Square loop
B. Log-periodic
C. Conical horn
D. Folded dipole
Answer: B
Answer: C
A. a = 3b
B. a = 2b
C. a = b
D. a = b/4
Answer: B
A. 3 to 30 kHz
B. 300 to 3000 kHz
C. 3000 to 30000 MHz
D. 30000 to 300000 MHz
Answer: C
A. radar
B. direction finding
C. satellite communication
D. all of the above
Answer: B
1. ZL = 200 Ω and it is desired that Zi = 50 Ω The quarter wave transformer should have a
characteristic impedance of
A
100 Ω
.
B.40 Ω
C.10000 Ω
D
4Ω
.
Answer: Option A
Explanation:
Z0 = Zi . ZL
2. A broadside array consisting of 200 cm wavelength with 10 half-wave dipole spacing 10 cm.
And if each array element feeding with 1 amp. current and operating at same frequency then
find the half power beamwidth
A
4°
.
B.2°
C.10°
D
15°
.
Answer: Option B
Explanation:
3. Refractive index of glass is 1.5. Find the wavelength of a beam of light with a frequency of
1014 Hz in glass. Assume velocity of light is 3 x 108 m/sec in vacuum.
A
4 μm
.
B.3 μm
C.2 μm
D
1 μm
.
Answer: Option C
Explanation:
Explanation:
f(x - vot) represents in +ve direction while f(x + vot) represents a progressive wave in -ve
direction.
5. A broadside array operating at 100 cm wavelength consist of 4 half-wave dipoles spaced 50
cm apart. Each element carries radio frequency current in the same phase and of magnitude
0.5 A. The radiated power will be
A
196 W
.
B.73 W
C.36.5 W
D
18.25 W
.
Answer: Option A
Explanation:
n = 4.
This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on
“Magnetostatic Properties”.
1. The magnetostatics highly relies on which property?
a) Resistance
b) Capacitance
c) Inductance
d) Moment
View Answer
Answer: c
Explanation: The magnetostatics highly relies on the inductance of the magnetic materials, which
decides its behavior in the influence of magnetic field.
Answer: b
Explanation: The inductance is a property of an electric conductor/coil which measures the
amount of emf generated by passing current through the coil.
3. Find the total flux in a coil of magnetic flux density 12 units and area 7 units.
a) 0.84
b) 0.96
c) 8.4
d) 9.6
View Answer
Answer: a
Explanation: The total flux in a coil is defined by φ = BA, where B = 12 and A = 0.07. On
substituting these values, we get φ = 12 x 0.07 = 0.84 units.
4. Find the energy of a coil of inductance 18mH and current passing through it 1.25A.(in 10-3
order)
a) 14.06
b) 61
c) 46.1
d) 28.12
View Answer
Answer: a
Explanation: The magnetic energy possessed by a coil is given by E = 0.5 x LI2. Put L = 18 x 10-3
and I = 1.25, thus we get E = 0.5 x 18 x 10-3 x 1.252 = 14.06 x 10-3 units.
5. Using Maxwell equation which of the following cannot be calculated directly?
a) B
b) D
c) A
d) H
View Answer
Answer: c
Explanation: The Maxwell equations can be used to compute E,H,D,B and J directly. It is not
possible to find the magnetic vector potential A directly.
Answer: d
Explanation: The magnetic flux density is the product the permeability and the magnetic field
intensity. This statement is always true for any material (permeability).
7. The permeability and permittivity of air or free space is unity. State true/false.
a) True
b) False
View Answer
Answer: b
Explanation: The permeability and permittivity of free space or air is always unity. This implies
that the air is always ready to store electric or magnetic charges subjected to it.
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Answer: c
Explanation: For any magnetic field, the magnetic field intensity will be the negative gradient of
the potential of the field. This is given by H = -Grad(V).
9. Find the magnetic field when the magnetic vector potential is a unit vector.
a) 1
b) -1
c) 0
d) 2
View Answer
Answer: c
Explanation: We know that H = -Grad(V), where is a unit vector. The gradient of a constant/unit
vector will be zero. Thus the magnetic field intensity will be zero.
Answer: d
Explanation: The electromagnetic wave experiences Lorentz force which is the combination of
the electrostatic force and magneto static force. It is given by F = QE + Q(V X B).
Answer: c
Explanation: The electric field intensity is the force per unit charge on a test charge, i.e, q1 = 1C.
E = F/Q = Q/(4∏εr2).
Answer: c
Explanation: Force is the product of charge and electric field.
F = q X E = 2 X 1 = 2 N.
3. Find the electric field intensity of two charges 2C and -1C separated by a distance 1m in air.
a) 18 X 109
b) 9 X 109
c) 36 X 109
d) -18 X 109
View Answer
Answer: b
Explanation: F = q1q2/(4∏εor2) = -2 X 9/(10-9 X 12) = -18 X 109
E = F/q = 18 X 109/2 = 9 X 109.
4. What is the electric field intensity at a distance of 20cm from a charge 2 X 10-6 C in vacuum?
a) 250,000
b) 350,000
c) 450,000
d) 550,000
View Answer
Answer: c
Explanation: E = Q/ (4∏εor2)
= (2 X 10-6)/(4∏ X εo X 0.22) = 450,000 V/m.
5. Determine the charge that produces an electric field strength of 40 V/cm at a distance of 30cm
in vacuum(in 10-8C)
a) 4
b) 2
c) 8
d) 6
View Answer
Answer: a
Explanation: E = Q/ (4∏εor2)
Q = (4000 X 0.32)/ (9 X 109) = 4 X 10-8 C.
6. The field intensity of a charge defines the impact of the charge on a test charge placed at a
distance. It is maximum at d = 0cm and minimises as d increases. State True/False
a) True
b) False
View Answer
Answer: a
Explanation: If a test charge +q is situated at a distance r from Q, the test charge will experience
a repulsive force directed radially outward from Q. Since electric field is inversely proportional
to distance, thus the statement is true.
Answer: a
Explanation: The electric field intensity of an infinitely long conductor is given by, E = λ/(4πεh).
(sin α2 – sin α1)i + (cos α2 + cos α1)j
For an infinitely long conductor, α = 0. E = λ/(4πεh).(cos 0 + cos 0) = λ/(2πεh).aN.
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Answer: d
Explanation: E = σ/2ε.(1- cos α), where α = h/(√(h2+a2))
Here, h is the distance of the sheet from point P and a is the radius of the sheet. For infinite sheet,
α = 90. Thus E = σ/2ε.
Answer: c
Explanation: E = Q/ (4∏εor2)
When distance d is infinity, the electric field will be zero, E= 0.
10. In electromagnetic waves, the electric field will be perpendicular to which of the following?
a) Magnetic field intensity
b) Wave propagation
c) Both H and wave direction
d) It propagates independently
View Answer
Answer: c
Explanation: In an electromagnetic wave, the electric field and magnetic field will be
perpendicular to each other. Both of these fields will be perpendicular to the wave propagation.
This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on
“Electric Field Density”.
Answer: b
Explanation: Electric flux density is given by the ratio between number of flux lines crossing a
surface normal to the lines and the surface area. The direction of D at a point is the direction of
the flux lines at that point.
Answer: a
Explanation: D= εE, where ε=εoεr is the permittivity of electric field and E is the electric field
intensity. Thus electric flux density is the product of permittivity and electric field intensity.
Answer: d
Explanation: The electric flux passing through any closed surface is equal to the total charge
enclosed by that surface. In other words, electric flux per unit volume leaving a point (vanishing
small volume), is equal to the volume charge density.
Answer: b
Explanation: Electric field intensity of infinite sheet of charge E = σ/2ε.
Thus D = εE = σ/2 = 25/2 = 12.5.
7. A uniform surface charge of σ = 2 μC/m2, is situated at z = 2 plane. What is the value of flux
density at P(1,1,1)m?
a) 10-6
b) -10-6
c) 106
d) -106
View Answer
Answer: b
Explanation: The flux density of any field is independent of the position (point). D = σ/2 = 2 X
10-6(-az)/2 = -10-6.
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8. Find the flux density of line charge of radius (cylinder is the Gaussian surface) 2m and charge
density is 3.14 units?
a) 1
b) 0.75
c) 0.5
d) 0.25
View Answer
Answer: d
Explanation: The electric field of a line charge is given by, E = λ/(2περ), where ρ is the radius of
cylinder, which is the Gaussian surface and λ is the charge density. The density D = εE = λ/(2πρ)
= 3.14/(2π X 2) = 1/4 = 0.25.
9. If the radius of a sphere is 1/(4π)m and the electric flux density is 16π units, the total flux is
given by,
a) 2
b) 3
c) 4
d) 5
View Answer
Answer: c
Explanation: Total flux leaving the entire surface is, ψ = 4πr2D from Gauss law. Ψ = 4π(1/16π2)
X 16π = 4.
10. Find the electric field intensity of transformer oil (εr = 2 approx) with density 1/4π (in 109
units)
a) 2.5
b) 3.5
c) 4.5
d) 5.5
View Answer
Answer: c
Explanation: D = εE. E = (1/4π)/(2Xεo) = 4.5 X 109 units.
Q. Which form of Maxwell's equation specifies the fundamental relationship between the
electric and magnetic fields in time varying field?
- Published on 05 Oct 15
a. Point form
b. Integral form
c. Exponential form
d. None of the above
a. Inductive reactance
b. Capacitive susceptance
c. Shunt conductance
d. Series admittance
Answer: d
Explanation: The divergence of the magnetic flux density is always zero. This is because of the
non existence of magnetic monopoles in a magnetic field.
2. Find the charge density when the electric flux density is given by 2x i + 3y j + 4z k.
a) 10
b) 9
c) 24
d) 0
View Answer
Answer: b
Explanation: The charge density is the divergence of the electric flux density by Maxwell’s
equation. Thus ρ = Div (D) and Div (D) = 2 + 3 + 4 = 9. We get ρ = 9 units.
Answer: c
Explanation: From the Faraday’s law and Lenz law, using Stoke’s theorem, we get Curl(E) =
-dB/dt. This is the Maxwell’s first law of electromagnetics.
Answer: c
Explanation: From the current density definition and Ohm’s law, the Ampere circuital law
Curl(H) = J can be derived. This is Maxwell’s second law of electromagnetics.
Answer: a
Explanation: The stationary loop in a varying magnetic field results in an induced emf due to the
change in the flux linkage of the loop. This emf is called as induced or transformer EMF.
Answer: a
Explanation: Maxwell equations can be represented in differential/point form and integral form
alternatively. Sometimes, it can be represented by time varying fields called harmonic form.
Answer: b
Explanation: The charge in the capacitor is due to displacement current. It is the current in the
presence of the dielectric placed between two parallel metal plates.
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9. Find the flux enclosed by a material of flux density 12 units in an area of 80cm.
a) 9.6
b) 12/80
c) 80/12
d) 12/0.8
View Answer
Answer: a
Explanation: The total flux in a material is the product of the flux density and the area. It is given
by flux = 12 x 0.8= 9.6 units.
10. Find the electric flux density of a material with charge density 16 units in unit volume.
a) 1/16
b) 16t
c) 16
d) 162
View Answer
Answer: c
Explanation: The electric flux density from Maxwell’s equation is given by D = ∫ ρ dv. On
substituting ρ = 16 and ∫dv = 1, we get D = 16 units
Answer: a
Explanation: The second Maxwell equation is based on Ampere law. It states that the field
intensity of a system is same as the current enclosed by it, i.e, Curl(H) = J.
2. The Maxwell second equation that is valid in any conductor is
a) Curl(H) = Jc
b) Curl(E) = Jc
c) Curl(E) = Jd
d) Curl(H) = Jd
View Answer
Answer: a
Explanation: For conductors, the conductivity parameter σ is significant and only the conduction
current density exists. Thus the component J = Jc and Curl(H) = Jc.
Answer: a
Explanation: In dielectric medium conductivity σ will be zero. So the current density has only
the displacement current density. Thus the Maxwell equation will be Curl(H) = Jd.
4. Find the displacement current density of a material with flux density of 5sin t
a) 2.5cos t
b) 2.5sin t
c) 5cos t
d) 5sin t
View Answer
Answer: c
Explanation: The displacement current density is the derivative of the flux density. Thus Jd =
dD/dt. Put D = 5sin t in the equation, we get Jd = 5cos t units.
5. Find the conduction current density of a material with conductivity 200units and electric field
1.5 units.
a) 150
b) 30
c) 400/3
d) 300
View Answer
Answer: d
Explanation: The conduction current density is given by Jc = σE, where σ = 200 and E = 1.5.
Thus we get, Jc = 200 x 1.5 = 300 units.
6. Calculate the conduction density of a material with resistivity of 0.02 units and electric
intensity of 12 units.
a) 300
b) 600
c) 50
d) 120
View Answer
Answer: b
Explanation: The conduction density is given by Jc = σE, where σ is the inverse of resistivity and
it is 1/0.02 = 50. Thus we get, Jc = 50 x 12 = 600 units.
7. In the conversion of line integral of H into surface integral, which theorem is used?
a) Green theorem
b) Gauss theorem
c) Stokes theorem
d) It cannot be converted
View Answer
Answer: c
Explanation: To convert line integral to surface integral, i.e, in this case from line integral of H to
surface integral of J, we use the Stokes theorem. Thus the Maxwell second equation can be
written as ∫H.dl = ∫∫J.ds.
Answer: a
Explanation: The continuity equation indicates the current density in conductors. This is the
product of the conductivity of the conductor and the electric field subjected to it. Thus J = σE is
the implication of the continuity equation for conductors.
Answer: a
Explanation: The displacement current density is Jd = dD/dt. Since D = εE and in frequency
domain d/dt = jw, thus we get Jd = jwεE.
10. The total current density is given as 0.5i + j – 1.5k units. Find the curl of the magnetic field
intensity.
a) 0.5i – 0.5j + 0.5k
b) 0.5i + j -1.5k
c) i – j + k
d) i + j – k
View Answer
Answer: b
Explanation: By Maxwell second equation, the curl of H is same as the sum of conduction
current density and displacement current density. Thus Curl(H) = J = 0.5i + j – 1.5k units.
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Answer: a
Explanation: The DC field refers to zero frequency. The conduction current is independent of the
frequency, whereas the displacement current density is dependent on the frequency, i.e, Jd =
jwεE. Thus at DC field, the displacement current density will be zero.
12. Both the conduction and displacement current densities coexist in which medium?
a) Only conductors in air
b) Only dielectrics in air
c) Conductors placed in any dielectric medium
d) Both the densities can never coexist
View Answer
Answer: c
Explanation: Conduction density exists only for good conductors and displacement density is for
dielectrics in any medium at high frequency. Thus both coexist when a conductor is placed in a
dielectric medium.