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# Test: Plane Waves In Good Conductor And Dielectrics

## 24 Questions MCQ Test Electromagnetic Theory | Test: Plane Waves In Good Conductor And Dielectrics

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This mock test of Test: Plane Waves In Good Conductor And Dielectrics for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam. This contains 24 Multiple Choice Questions for Electrical Engineering (EE) Test: Plane Waves In Good Conductor And Dielectrics (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Plane Waves In Good Conductor And Dielectrics quiz give you a good mix of easy questions and tough questions. Electrical Engineering (EE) students definitely take this Test: Plane Waves In Good Conductor And Dielectrics exercise for a better result in the exam. You can find other Test: Plane Waves In Good Conductor And Dielectrics extra questions, long questions & short questions for Electrical Engineering (EE) on EduRev as well by searching above.
QUESTION: 1

### For conductors, the loss tangent will be

Solution:

Explanation: In conductors, the conductivity will be more. Thus the loss tangent σ/ωε will be maximum.

QUESTION: 2

### In metals, the total permittivity is

Solution:

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.

QUESTION: 3

### The total permeability in a conductor is

Solution:

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.

QUESTION: 4

Calculate the phase constant of a conductor with attenuation constant given by 0.04 units.

Solution:

Explanation: 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.

QUESTION: 5

Calculate the attenuation constant of a conductor of conductivity 200 units, frequency 1M radian/s in air.

Solution:

Explanation: The attenuation constant of a conductor is given by α = √(ωμσ/2). On substituting ω = 106, σ = 200 and μ = 4π x 10-7, we get α = 11.2 units.

QUESTION: 6

The skin depth of a conductor with attenuation constant of 7 neper/m is

Solution:

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.

QUESTION: 7

The expression for velocity of a wave in the conductor is

Solution:

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

QUESTION: 8

In conductors, the E and H vary by a phase difference of

Solution:

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.

QUESTION: 9

EM waves do not travel inside metals. State True/False.

Solution:

Explanation: 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.

QUESTION: 10

The propagation constant of the wave in a conductor with air as medium is

Solution:

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.

QUESTION: 11

An example for electromagnetic wave propagation is

Solution:

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.

QUESTION: 12

The phase shift in the electric and magnetic fields in an EM wave is given by which parameter?

Solution:

Explanation: The intrinsic impedance in a conductor is given by η = √(ωμ/2σ) x (1+j). The phase shift is represented by the 1+j term. In polar form it indicates 45 degree phase shift.

QUESTION: 13

The loss tangent of a perfect dielectric will be

Solution:

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

QUESTION: 14

In pure dielectrics, the parameter that is zero is

Solution:

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.

QUESTION: 15

The total permittivity of a dielectric transformer oil (relative permittivity is 2.2) will be (in order 10-11

Solution:

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.

QUESTION: 16

The permeability of a dielectric material in air medium will be

Solution:

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.

QUESTION: 17

The attenuation in a good dielectric will be non- zero. State True/False.

Solution:

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.

QUESTION: 18

Calculate the phase constant of a dielectric with frequency 6 x 106 in air.

Solution:

Explanation: The phase constant of a dielectric is given by β = ω√(με). On substituting for ω = 6 x 106 , μ = 4π x 10-7, ε = 8.854 x 10-12 in air medium, we get the phase constant as 0.02 units.

QUESTION: 19

The frequency in rad/sec of a wave with velocity of that of light and phase constant of 20 units is (in GHz)

Solution:

Explanation: The velocity of a wave is given by V = ω/β. To get ω, put v = 3 x 108 and β = 20. Thus ω = vβ = 3 x 108 x 20 = 60 x 108 = 6 GHz.

QUESTION: 20

The relation between the speed of light, permeability and permittivity is

Solution:

Explanation: The standard relation between speed of light, permeability and permittivity is given by c = 1/√(με). The value in air medium is 3 x 108 m/s.

QUESTION: 21

The phase constant of a wave with wavelength 2 units is

Solution:

Explanation: The phase constant is given by β = 2π/λ. On substituting λ = 2 units, we get β = 2π/2 = π = 3.14 units.

QUESTION: 22

The expression for intrinsic impedance is given by

Solution:

Explanation: 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.

QUESTION: 23

The electric and magnetic field components in the electromagnetic wave propagation are in phase. State True/False.

Solution: