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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

51.

The expression for refractive index is given by(a) N = v/c(b) N = c/v(c) N = cv(d) N = 1/cvI got this question in examination.This interesting question is from Refractive Index and Numerical Aperture topic in portion EM Wave Propagation of Electromagnetic Theory

Answer»

The correct option is (b) N = c/v

For explanation: The refractive index is DEFINED as the RATIO of the VELOCITY of light in a vacuum to its velocity in a specified medium. It is GIVEN by n = c/v. It is constant for a particular MATERIAL.

52.

Numerical aperture is expressed as the(a) NA = sin θa(b) NA = cos θa(c) NA = tan θa(d) NA = sec θaThe question was posed to me in an online interview.I would like to ask this question from Refractive Index and Numerical Aperture topic in division EM Wave Propagation of Electromagnetic Theory

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Right ANSWER is (a) NA = sin θa

Easy explanation: The numerical aperture is the measure of how MUCH LIGHT the fiber can collect. It is the sine of the acceptance angle, the angle at which the light must be transmitted in order to get MAXIMUM reflection. THUS it is given by NA = sin θa.

53.

The transmission coefficient is given by 0.65. Find the return loss of the wave.(a) 9.11(b) 1.99(c) 1.19(d) 9.91I had been asked this question in examination.The origin of the question is Power, Power Loss and Return Loss in portion EM Wave Propagation of Electromagnetic Theory

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The correct ANSWER is (a) 9.11

To explain I would say: The transmission coefficient is the reverse of the reflection coefficient, i.e, T + R = 1. When T = 0.65, we get R = 0.35. THUS the RETURN loss RL = -20log R = -20log 0.35 = 9.11 DECIBEL.

54.

The return loss is given as 12 decibel. Calculate the reflection coefficient.(a) 0.35(b) 0.55(c) 0.25(d) 0.75The question was asked in an interview for internship.My enquiry is from Power, Power Loss and Return Loss in division EM Wave Propagation of Electromagnetic Theory

Answer»

Right choice is (C) 0.25

For explanation I would say: The return loss is given by RL = -20log R. The REFLECTION COEFFICIENT can be CALCULATED as R = 10^(-RL/20), by anti LOGARITHM property. For the given return loss RL = 12, we get R = 10^(-12/20) = 0.25.

55.

The radiation resistance of an antenna having a power of 120 units and antenna current of 5A is(a) 4.8(b) 9.6(c) 3.6(d) 1.8This question was addressed to me during an interview.My doubt is from Power, Power Loss and Return Loss topic in section EM Wave Propagation of Electromagnetic Theory

Answer»

The correct choice is (a) 4.8

Easy explanation: The POWER of an ANTENNA is GIVEN by Prad = Ia^2 Rrad, where Ia is the antenna current and Rrad is the radiation resistance. On SUBSTITUTING the given data, we get Rrad = Prad/Ia^2 = 120/5^2 = 4.8 ohm.

56.

The current reflection coefficient is given by -0.75. Find the voltage reflection coefficient.(a) -0.75(b) 0.25(c) -0.25(d) 0.75This question was addressed to me in class test.My query is from Power, Power Loss and Return Loss in portion EM Wave Propagation of Electromagnetic Theory

Answer»

The CORRECT choice is (d) 0.75

The best I can explain: The VOLTAGE REFLECTION coefficient is the NEGATIVE of the CURRENT reflection coefficient. For a current reflection coefficient of -0.75, the voltage reflection coefficient will be 0.75.

57.

The reflection coefficient is 0.5. Find the return loss.(a) 12.12(b) -12.12(c) 6.02(d) -6.02The question was posed to me in semester exam.The above asked question is from Power, Power Loss and Return Loss in section EM Wave Propagation of Electromagnetic Theory

Answer» CORRECT option is (c) 6.02

Explanation: The return LOSS is given by RL = -20log R, where is the reflection COEFFICIENT. It is given as 0.5. Thus the return loss will be RL = -20 log 0.5 = 6.02 decibel.
58.

The attenuation is given by 20 units. Find the power loss in decibels.(a) 13.01(b) 26.02(c) 52.04(d) 104.08The question was asked in examination.I would like to ask this question from Power, Power Loss and Return Loss topic in chapter EM Wave Propagation of Electromagnetic Theory

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Correct OPTION is (a) 13.01

The best I can explain: The ATTENUATION refers to the power loss. THUS the power loss is given by 20 units. The power loss in dB will be 10 LOG 20 = 13.01 decibel.

59.

The incident and the reflected voltage are given by 15 and 5 respectively. The transmission coefficient is(a) 1/3(b) 2/3(c) 1(d) 3I got this question in unit test.This intriguing question originated from Power, Power Loss and Return Loss in chapter EM Wave Propagation of Electromagnetic Theory

Answer»

Right answer is (B) 2/3

For explanation I WOULD SAY: The ratio of the reflected to the incident voltage is the REFLECTION COEFFICIENT. It is given by R = 5/15 = 1/3. To get the transmission coefficient, T = 1 – R = 1 – 1/3 = 2/3.

60.

The power transmitted by a wave with incident power of 16 units is(Given that the reflection coefficient is 0.5)(a) 12(b) 8(c) 16(d) 4The question was posed to me in quiz.The origin of the question is Power, Power Loss and Return Loss topic in chapter EM Wave Propagation of Electromagnetic Theory

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The correct answer is (a) 12

The best EXPLANATION: The fraction of the TRANSMITTED to the incident POWER is given by the reflection coefficient. Thus Pref = (1-R^2) Pinc. On substituting the given data, we GET Pref = (1- 0.5^2) x 16 = 12 units. In other words, it is the remaining power after reflection.

61.

The power reflected by a wave with incident power of 16 units is(Given that the reflection coefficient is 0.5)(a) 2(b) 8(c) 6(d) 4I had been asked this question in an interview.I'm obligated to ask this question of Power, Power Loss and Return Loss topic in division EM Wave Propagation of Electromagnetic Theory

Answer»

Correct option is (d) 4

Explanation: The fraction of the REFLECTED to the incident power is given by the reflection COEFFICIENT. Thus Pref = R^2xPinc. On substituting the given data, we get Pref = 0.5^2 X 16 = 4 units.

62.

The power of a wave of with voltage of 140V and a characteristic impedance of 50 ohm is(a) 1.96(b) 19.6(c) 196(d) 19600The question was asked in an online interview.My doubt is from Power, Power Loss and Return Loss topic in chapter EM Wave Propagation of Electromagnetic Theory

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The correct answer is (C) 196

Best EXPLANATION: The power of a wave is given by P = V^2/2Zo, where V is the generator VOLTAGE and ZO is the characteristic impedance. on substituting the given data, we get P = 140^2/(2×50) = 196 units.

63.

The power of the electromagnetic wave with electric and magnetic field intensities given by 12 and 15 respectively is(a) 180(b) 90(c) 45(d) 120The question was posed to me in unit test.My doubt is from Power, Power Loss and Return Loss in portion EM Wave Propagation of Electromagnetic Theory

Answer» RIGHT choice is (b) 90

Easy EXPLANATION: The Poynting VECTOR gives the power of an EM wave. Thus P = EH/2. On substituting for E = 12 and H = 15, we get P = 12 X 15/2 = 90 UNITS.
64.

The expression for intrinsic impedance is given by(a) √(με)(b) (με)(c) √(μ/ε)(d) (μ/ε)This question was addressed to me by my college professor while I was bunking the class.My question comes from Plane Waves in Dielectrics in section EM Wave Propagation of Electromagnetic Theory

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The correct option is (c) √(μ/ε)

The EXPLANATION is: The intrinsic impedance is given by the ratio of square ROOT of the PERMITTIVITY to the permeability. Thus η = √(μ/ε) is the intrinsic impedance. In free space or air medium, the intrinsic impedance will be 120π or 377 OHMS.

65.

The electric and magnetic field components in the electromagnetic wave propagation are in phase. State True/False.(a) True(b) FalseI had been asked this question during an internship interview.My question is based upon Plane Waves in Dielectrics in division EM Wave Propagation of Electromagnetic Theory

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The CORRECT answer is (a) True

The EXPLANATION is: In DIELECTRICS, the electric and magnetic FIELDS will be in phase or the phase difference between them is zero. This is due to the large attenuation which leads to increase in phase shift.

66.

The phase constant of a wave with wavelength 2 units is(a) 6.28(b) 3.14(c) 0.5(d) 2This question was posed to me in examination.I'd like to ask this question from Plane Waves in Dielectrics topic in section EM Wave Propagation of Electromagnetic Theory

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The correct ANSWER is (b) 3.14

To explain: The PHASE constant is GIVEN by β = 2π/λ. On SUBSTITUTING λ = 2 units, we get β = 2π/2 = π = 3.14 units.

67.

The relation between the speed of light, permeability and permittivity is(a) C = 1/√(με)(b) C = με(c) C = μ/ε(d) C = 1/μεI got this question during an interview.Question is taken from Plane Waves in Dielectrics topic in chapter EM Wave Propagation of Electromagnetic Theory

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Right choice is (a) C = 1/√(με)

For EXPLANATION I would say: The STANDARD relation between SPEED of light, PERMEABILITY and PERMITTIVITY is given by c = 1/√(με). The value in air medium is 3 x 10^8 m/s.

68.

The frequency in rad/sec of a wave with velocity of that of light and phase constant of 20 units is (in GHz)(a) 6(b) 60(c) 600(d) 0.6I got this question in an interview for internship.My question comes from Plane Waves in Dielectrics topic in portion EM Wave Propagation of Electromagnetic Theory

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Right CHOICE is (a) 6

The explanation: The velocity of a wave is GIVEN by V = ω/β. To get ω, put v = 3 X 10^8 and β = 20. THUS ω = vβ = 3 x 10^8^ x 20 = 60 x 10^8 = 6 GHz.

69.

Calculate the phase constant of a dielectric with frequency 6 x 10^6 in air.(a) 2(b) 0.2(c) 0.02(d) 0.002The question was asked by my college director while I was bunking the class.The question is from Plane Waves in Dielectrics topic in chapter EM Wave Propagation of Electromagnetic Theory

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Right ANSWER is (c) 0.02

For explanation I WOULD say: The PHASE constant of a dielectric is GIVEN by β = ω√(με). On substituting for ω = 6 x 10^6 , μ = 4π x 10^-7, ε = 8.854 x 10^-12 in air medium, we get the phase constant as 0.02 units.

70.

The attenuation in a good dielectric will be non- zero. State True/False.(a) True(b) FalseThis question was posed to me in final exam.My question comes from Plane Waves in Dielectrics topic in section EM Wave Propagation of Electromagnetic Theory

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Right CHOICE is (a) True

Easiest EXPLANATION: GOOD dielectrics attenuate the electromagnetic waves than any other material. Thus the attenuation CONSTANT of the dielectric will be non-zero, positive and large.

71.

The permeability of a dielectric material in air medium will be(a) Absolute permeability(b) Relative permeability(c) Product of absolute and relative permeability(d) UnityThis question was addressed to me in homework.Asked question is from Plane Waves in Dielectrics in division EM Wave Propagation of Electromagnetic Theory

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Right CHOICE is (a) Absolute permeability

Easiest EXPLANATION: The total permeability is the PRODUCT of the absolute and the relative permeability. In air medium, the relative permeability will be unity. Thus the total permeability is equal to the absolute permeability GIVEN by 4π X 10^-7 units.

72.

In pure dielectrics, the parameter that is zero is(a) Attenuation(b) Propagation(c) Conductivity(d) ResistivityThis question was posed to me during an internship interview.This is a very interesting question from Plane Waves in Dielectrics topic in chapter EM Wave Propagation of Electromagnetic Theory

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The CORRECT option is (c) Conductivity

Explanation: There are no free charge carriers available in a DIELECTRIC. In other words, the charge carriers are PRESENT in the valence BAND, which is very difficult to start to conduct. Thus conduction is low in dielectrics. For pure dielectrics, the conductivity is assumed to be zero.

73.

The total permittivity of a dielectric transformer oil (relative permittivity is 2.2) will be (in order 10^-11)(a) 1.94(b) 19.4(c) 0.194(d) 194This question was addressed to me during an online exam.The origin of the question is Plane Waves in Dielectrics topic in division EM Wave Propagation of Electromagnetic Theory

Answer» CORRECT answer is (a) 1.94

Easy explanation: The TOTAL permittivity is the product of the absolute and the relative permittivity. The absolute permittivity is 8.854 x 10^-12 and the relative permittivity(in this CASE for transformer OIL) is 2.2. Thus the total permittivity is 8.854 x 10^-12 x 2.2 = 1.94 x 10^-11 UNITS.
74.

The loss tangent of a perfect dielectric will be(a) Zero(b) Unity(c) Maximum(d) MinimumI got this question in an online interview.Asked question is from Plane Waves in Dielectrics topic in section EM Wave Propagation of Electromagnetic Theory

Answer»

The CORRECT OPTION is (d) Minimum

The EXPLANATION: Dielectrics have poor conductivity. The loss tangent σ/ωε will be LOW in dielectrics. For perfect dielectrics, the loss tangent will be minimum.

75.

An example for electromagnetic wave propagation is(a) refrigerator(b) electric fan(c) mobile transponder(d) relays in actuatorsThe question was asked in an interview for internship.Asked question is from Plane Waves in Good Conductor in portion EM Wave Propagation of Electromagnetic Theory

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Right option is (c) MOBILE transponder

The best EXPLANATION: The refrigerator, electric fan and RELAYS are electrical devices. They do not use ELECTROMAGNETIC energy as medium of energy transfer. The mobile transponder is an antenna, which uses the EM waves for communication with the SATELLITES.

76.

In conductors, the E and H vary by a phase difference of(a) 0(b) 30(c) 45(d) 60I have been asked this question at a job interview.Query is from Plane Waves in Good Conductor in division EM Wave Propagation of Electromagnetic Theory

Answer» CORRECT answer is (c) 45

Best explanation: The electric and MAGNETIC component, E and H respectively have a phase difference of 45 degrees. This is due to the WAVE propagation in conductors in the AIR medium.
77.

The propagation constant of the wave in a conductor with air as medium is(a) √(ωμσ)(b) ωμσ(c) √(ω/μσ)(d) ω/μσI had been asked this question during an online exam.Asked question is from Plane Waves in Good Conductor in chapter EM Wave Propagation of Electromagnetic Theory

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Right answer is (a) √(ωμσ)

The BEST explanation: The propagation constant is the sum of the attenuation constant and the PHASE constant. In CONDUCTORS, the attenuation and phase constant both are same and it is GIVEN by √(ωμσ/2). Their sum will be √(ωμσ), is the propagation constant.

78.

EM waves do not travel inside metals. State True/False.(a) True(b) FalseThis question was addressed to me during an interview for a job.I want to ask this question from Plane Waves in Good Conductor in division EM Wave Propagation of Electromagnetic Theory

Answer» CORRECT answer is (a) True

To elaborate: The CONDUCTORS or METALS do not support EM wave propagation onto them due the skin effect. This is the reason why mobile phones cannot be USED INSIDE lifts.
79.

The expression for velocity of a wave in the conductor is(a) V = √(2ω/μσ)(b) V = √(2ωμσ)(c) V = (2ω/μσ)(d) V = (2ωμσ)The question was posed to me by my college director while I was bunking the class.My doubt stems from Plane Waves in Good Conductor topic in division EM Wave Propagation of Electromagnetic Theory

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Right option is (a) V = √(2ω/μσ)

Best explanation: The velocity is the ratio of the frequency to the phase CONSTANT. In conductors, the phase constant is given by √(ωμσ/2). On SUBSTITUTING for β,ω in v, we GET v = √(2ω/μσ) UNITS.

80.

The skin depth of a conductor with attenuation constant of 7 neper/m is(a) 14(b) 49(c) 7(d) 1/7The question was posed to me in unit test.Question is taken from Plane Waves in Good Conductor in chapter EM Wave Propagation of Electromagnetic Theory

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Correct option is (d) 1/7

Explanation: The SKIN depth is the measure of the depth upto which an EM wave can penetrate through the conductor surface. It is the reciprocal of the ATTENUATION constant. On SUBSTITUTING for α = 7, we get δ = 1/α = 1/7 UNITS.

81.

Calculate the phase constant of a conductor with attenuation constant given by 0.04 units.(a) 0.02(b) 0.08(c) 0.0016(d) 0.04This question was addressed to me during an interview.My doubt is from Plane Waves in Good Conductor in section EM Wave Propagation of Electromagnetic Theory

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Correct OPTION is (d) 0.04

The explanation is: The phase CONSTANT and the attenuation constant are both the same in the CASE of conductors. Given that the attenuation constant is 0.04, implies that the phase constant is also 0.04.

82.

The total permeability in a conductor is(a) Absolute permeability(b) Relative permeability(c) Product of absolute and relative permeability(d) UnityI got this question in homework.My query is from Plane Waves in Good Conductor in portion EM Wave Propagation of Electromagnetic Theory

Answer»

Right answer is (c) Product of absolute and relative permeability

The BEST explanation: The total permeability is the product of the absolute and the relative permeability. For METALS or conductors, the relative permittivity is not UNITY. Thus the permittivity is the product of absolute and relative permeability.

83.

Calculate the attenuation constant of a conductor of conductivity 200 units, frequency 1M radian/s in air.(a) 11.2(b) 1.12(c) 56.23(d) 5.62I got this question by my school principal while I was bunking the class.This is a very interesting question from Plane Waves in Good Conductor topic in division EM Wave Propagation of Electromagnetic Theory

Answer» CORRECT choice is (a) 11.2

Explanation: The ATTENUATION constant of a conductor is given by α = √(ωμσ/2). On SUBSTITUTING ω = 10^6, σ = 200 and μ = 4π x 10^-7, we get α = 11.2 UNITS.
84.

In metals, the total permittivity is(a) Absolute permittivity(b) Relative permittivity(c) Product of absolute and relative permittivity(d) UnityThis question was posed to me in homework.I'd like to ask this question from Plane Waves in Good Conductor topic in portion EM Wave Propagation of Electromagnetic Theory

Answer»

Correct CHOICE is (a) Absolute permittivity

Easiest explanation: The total permittivity is the product of the absolute and the relative permittivity. For METALS or CONDUCTORS, the relative permittivity is unity. Thus the permittivity is SIMPLY the absolute permittivity.

85.

For conductors, the loss tangent will be(a) Zero(b) Unity(c) Maximum(d) MinimumThe question was asked in an online quiz.The above asked question is from Plane Waves in Good Conductor topic in portion EM Wave Propagation of Electromagnetic Theory

Answer» RIGHT option is (c) Maximum

Explanation: In conductors, the CONDUCTIVITY will be more. THUS the loss TANGENT σ/ωε will be maximum.
86.

The velocity of a wave travelling in the air medium without transmission lines or waveguides(wireless) is(a) 6 x 10^8(b) 3 x 10^8(c) 1.5 x 10^8(d) 9 x 10^8This question was addressed to me during an online exam.This question is from Plane Waves in Free Space topic in section EM Wave Propagation of Electromagnetic Theory

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Correct CHOICE is (b) 3 X 10^8

Explanation: In free space or air medium, the velocity of the wave propagating will be same as that of the LIGHT. Thus the velocity is the speed of light, V = C. It is given by 3 x 10^8m/s.

87.

In free space, the ratio of frequency to the velocity of light gives the phase constant. State True/False.(a) True(b) FalseThe question was asked in an interview for job.The question is from Plane Waves in Free Space topic in division EM Wave Propagation of Electromagnetic Theory

Answer»
88.

In free space, the condition that holds good is(a) Minimum attenuation and propagation(b) Minimum attenuation and maximum propagation(c) Maximum attenuation and minimum propagation(d) Maximum attenuation and propagationThis question was addressed to me in examination.Question is taken from Plane Waves in Free Space topic in portion EM Wave Propagation of Electromagnetic Theory

Answer»

Right option is (b) Minimum ATTENUATION and maximum propagation

Best EXPLANATION: The free SPACE does not have any BARRIER for attenuation. Thus it enables minimum attenuation and maximum propagation. This technique is employed in line of sight COMMUNICATION.

89.

The intrinsic impedance of free space is(a) 489(b) 265(c) 192(d) 377The question was asked in homework.My enquiry is from Plane Waves in Free Space topic in portion EM Wave Propagation of Electromagnetic Theory

Answer»

Right ANSWER is (d) 377

To EXPLAIN: The INTRINSIC impedance is the SQUARE root of ratio of the permeability to the permittivity. In free SPACE, the permeability and the permittivity is same as the absolute permeability and permittivity respectively. This is due to unity permeability and permittivity in free space. Thus η = √(μ/ε), where absolute permeability is given by 4π x 10^-7 and absolute permittivity is given by 8.854 x 10^-12. The intrinsic impedance will be 377 ohms.

90.

The conductivity in free space medium is(a) Infinity(b) Unity(c) Zero(d) NegativeI got this question in exam.My query is from Plane Waves in Free Space in chapter EM Wave Propagation of Electromagnetic Theory

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Right answer is (c) Zero

For EXPLANATION I WOULD say: As the charge carriers are not available in free space, the CONDUCTIVITY will be very low. For ideal cases, the conductivity can be taken as zero.

91.

Zero permeability/permittivity implies which state?(a) No ions are allowed in the medium(b) No current is generated in the medium(c) No magnetic or electric energy is permitted in the medium(d) No resistivityI got this question during an online exam.Origin of the question is Plane Waves in Free Space in portion EM Wave Propagation of Electromagnetic Theory

Answer»

Right answer is (c) No magnetic or ELECTRIC energy is permitted in the medium

Best EXPLANATION: The ZERO PERMITTIVITY in an electric field refers to the ability of the field/medium to permit electric charges in it. SIMILARLY, zero permeability in a magnetic field refers to the ability of the field/medium to permit the magnetic energy into the field.

92.

In free space, which parameter will be unity?(a) Permittivity(b) Absolute permittivity(c) Relative permittivity(d) PermeabilityThe question was posed to me in a national level competition.The query is from Plane Waves in Free Space topic in portion EM Wave Propagation of Electromagnetic Theory

Answer»

Right ANSWER is (C) Relative permittivity

Explanation: The relative permittivity is a constant for a PARTICULAR material. It is unity for free space or air. The absolute permittivity is a constant given by 8.854 x 10^-12 C/m2.

93.

Which parameter is unity in air medium?(a) Permittivity(b) Absolute permeability(c) Relative permeability(d) PermeabilityThe question was posed to me during an internship interview.This intriguing question originated from Plane Waves in Free Space topic in division EM Wave Propagation of Electromagnetic Theory

Answer»

Right ANSWER is (c) Relative permeability

To explain: In free space or air medium, the relative permeability is also UNITY, like relative PERMITTIVITY. The absolute permeability is GIVEN by 4π x 10^-7 units.

94.

In free space, the charge carriers will be(a) 0(b) 1(c) 100(d) InfinityThis question was posed to me during an internship interview.This interesting question is from Plane Waves in Free Space topic in chapter EM Wave Propagation of Electromagnetic Theory

Answer»

Correct option is (a) 0

To EXPLAIN I would say: Free space is not a CONDUCTOR. Thus the CHARGE carrier in free space is ASSUMED to be zero. But the free space consists of particles or ions that GET ionized during conduction.

95.

In waveguides, which of the following conditions will be true?(a) V > c(b) V < c(c) V = c(d) V >> cI had been asked this question during an internship interview.The above asked question is from Dielectric vs Conductor Wave Propagation in chapter EM Wave Propagation of Electromagnetic Theory

Answer»

Right choice is (a) V > c

To elaborate: In waveguides, the PHASE velocity will always be GREATER than the SPEED of light. This enables the wave to PROPAGATE through the waveguide. Thus V > c is the required condition.

96.

In perfect conductors, the phase shift between the electric field and magnetic field will be(a) 0(b) 30(c) 45(d) 90I had been asked this question by my college professor while I was bunking the class.This intriguing question originated from Dielectric vs Conductor Wave Propagation in section EM Wave Propagation of Electromagnetic Theory

Answer»
97.

The expression for phase constant is given by(a) Phase constant β = ωμε(b) Phase constant ω = με(c) Phase constant β = ω√(με)(d) Phase constant β = 1/ωμεI got this question in an online interview.Question is from Dielectric vs Conductor Wave Propagation topic in division EM Wave Propagation of Electromagnetic Theory

Answer»

Right option is (C) PHASE CONSTANT β = ω√(με)

For explanation: The phase constant is represented as β. It is a complex QUANTITY REPRESENTING the constant angle of the wave propagated. It is given byβ = ω√(με).

98.

For dielectrics, which two components will be in phase?(a) E and wave direction(b) H and wave direction(c) Wave direction and E x H(d) E and HThis question was posed to me in examination.I'd like to ask this question from Dielectric vs Conductor Wave Propagation in section EM Wave Propagation of Electromagnetic Theory

Answer»

The correct choice is (d) E and H

The BEST I can EXPLAIN: In DIELECTRICS, the electric and magnetic components E and H will be in phase with each other. This is due the variation in the permittivities and the permeabilities of the dielectric SURFACES. The phase difference between E and H will be 0.

99.

Calculate the velocity of wave propagation in a conductor with frequency 5 x 10^8 rad/s and phase constant of 3 x 10^8 units.(a) 3/5(b) 15(c) 5/3(d) 8I have been asked this question in homework.Asked question is from Dielectric vs Conductor Wave Propagation topic in chapter EM Wave Propagation of Electromagnetic Theory

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100.

For metals, the conductivity will be(a) 0(b) 1(c) -1(d) InfinityI had been asked this question in a job interview.My doubt is from Dielectric vs Conductor Wave Propagation topic in section EM Wave Propagation of Electromagnetic Theory

Answer»

Correct option is (d) Infinity

Best EXPLANATION: Metals are PURE conductors. Examples are iron, copper etc. Their CONDUCTIVITY will be very high. THUS the metal conductivity will be infinity. Practically the conductivity of conductors will be maximum.