Electric Fields in Material Space MCQs ( Electromagnetic Theory ) MCQs – New Electromagnetic Theory MCQs
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Electromagnetic Theory MCQs – Electric Fields in Material Space MCQs ( Electromagnetic Theory ) MCQs
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Conductors
1. Which of the following are conductors?
a) Ceramics
b) Plastics
c) Mercury
d) Rubber
Answer: c
Explanation: Normally, metals are said to be good conductors. Here mercury is the only metal (which is in liquid form). The other options are insulators.
2. Find the range of band gap energy for conductors.
a) >6 eV
b) 0.2-0.4 eV
c) 0.4-2 eV
d) 2-6 eV
Answer: b
Explanation: Conductors are materials with least band gap energy. The smallest range in this group is 0.2-0.4 eV.
3. Conduction in metals is due to
a) Electrons only
b) Electrons and holes
c) Holes only
d) Applied electric field
Answer: a
Explanation: Conduction in metals is only due to majority carriers, which are electrons. Electrons and holes are responsible for conduction in a semiconductor.
4. Find the band gap energy when a light of wavelength 1240nm is incident on it.
a) 1eV
b) 2eV
c) 3eV
d) 4eV
Answer: a
Explanation: The band gap energy in electron volt when wavelength is given is, Eg = 1.24(μm)/λ = 1.24 x 10-6/1240 x 10-9 = 1eV.
5. Alternating current measured in a transmission line will be
a) Peak value
b) Average value
c) RMS value
d) Zero
Answer: c
Explanation: The instantaneous current flowing in a transmission line, when measured using an ammeter, will give RMS current value. This value is 70.7% of the peak value. This is because, due to oscillations in AC, it is not possible to measure peak value. Hence to normalize, we consider current at any time in a line will be the RMS current.
6. The current in a metal at any frequency is due to
a) Conduction current
b) Displacement current
c) Both conduction and displacement current
d) Neither conduction nor displacement current
Answer: a
Explanation: At any frequency, the current through the metal will be due to conduction current. Only at high frequencies and when medium is air, the conduction is due to displacement current. Thus in general the current in metal is due to conduction current, which depends on the mobility of the carriers.
7. For conductors, the free electrons will exist at
a) Valence band
b) Middle of valence and conduction band
c) Will not exist
d) Conduction band
Answer: d
Explanation: In conductors, the free electrons exist in the conduction band. Since the band gap energy is very low, less energy is required to transport the free electrons to the conduction band, as they are readily available to conduct.
8. The current flowing through an insulating medium is called
a) Conduction
b) Convection
c) Radiation
d) Susceptibility
Answer: b
Explanation: A beam of electrons in a vacuum tube is called convection current. It occurs when current flows through an insulating medium like liquid, vacuum etc.
9. Find the conduction current density when conductivity of a material is 500 units and corresponding electric field is 2 units.
a) 500
b) 250
c) 1000
d) 2000
Answer: c
Explanation: The conduction current density is given by, J = σE
J = 500 X 2 = 1000 units.
10. Calculate the convection current when electron density of 200 units is travelling at a speed of 12m/s.
a) 16.67
b) 2400
c) 2880
d) 0.06
Answer: b
Explanation: The convection current density is given by, J = ρeV
J = 200 X 12= 2400 units.
Dielectrics
1. A dielectric is always an insulator. But an insulator is not necessarily a dielectric. State True/False.
a) True
b) False
Answer: a
Explanation: For a material to be dielectric, its permittivity should be very high. This is seen in insulators. For a material to be insulator, the condition is to have large band gap energy. However, this is not necessary for a dielectric.
2. Identify a good dielectric.
a) Iron
b) Ceramics
c) Plastic
d) Magnesium
Answer: b
Explanation: Iron and magnesium are metals. Hence they need not be considered. Both ceramics and plastic are insulators. But dielectric constant is more for ceramics always. Hence ceramics is the best dielectric.
3. A dielectric can be made a conductor by
a) Compression
b) Heating
c) Doping
d) Freezing
Answer: b
Explanation: On increasing the temperature, the free electrons in an insulator can be promoted from valence to conduction band. Gradually, it can act as a conductor through heating process. This condition is called dielectric breakdown, wherein the insulator loses its dielectric property and starts to conduct.
4. Find the dielectric constant for a material with electric susceptibility of 4.
a) 3
b) 5
c) 8
d) 16
Answer: b
Explanation: The electric susceptibility is given by χe = εr – 1. For a susceptibility of 4, the dielectric constant will be 5. It has no unit.
5. For a dielectric which of the following properties hold good?
a) They are superconductors at high temperatures
b) They are superconductors at low temperatures
c) They can never become a superconductor
d) They have very less dielectric breakdown voltage
Answer: b
Explanation: Superconductors are characterised by diamagnetism behaviour and zero resistivity, which true for a dielectric. They occur only at low temperature. Thus a dielectric can become a superconductor at low temperatures with very high dielectric breakdown voltage.
6. The magnetic field which destroys the superconductivity is called
a) Diamagnetic field
b) Ferromagnetic field
c) Ferrimagnetic field
d) Critical field
Answer: d
Explanation: Critical field is that strong magnetic field which can destroy the superconductivity of a material. The temperature at which this occurs is called transition temperature.
7. The magnetic susceptibility in a superconductor will be
a) Positive
b) Negative
c) Zero
d) Infinity
Answer: b
Explanation: Due to perfect diamagnetism in a superconductor, its magnetic susceptibility will be negative. This phenomenon is called Meissner effect.
8. The superconducting materials will be independent of which of the following?
a) Magnetic field
b) Electric field
c) Magnetization
d) Temperature
Answer: b
Explanation: Superconducting materials depends only on the applied magnetic field, resultant magnetization at the temperature considered. It is independent of the applied electric field and the corresponding polarization.
9. Find the mean free path of an electron travelling at a speed of 18m/s in 2 seconds.
a) 9
b) 36
c) 0.11
d) 4.5
Answer: b
Explanation: The mean free path is defined as the average distance travelled by an electron before collision takes place. It is given by, d = v x τc, where v is the velocity and τc is the collision time. Thus d = 18 x 2 = 36m.
10. Find the velocity of an electron when its kinetic energy is equal to one electron volt (in 105m/s).
Given charge of an electron e = 1.6 x 10-19 and mass of an electron m = 9.1 x 10-31.
a) 3.9
b) 4.9
c) 5.9
d) 6.9
Answer: c
Explanation: When the kinetic energy and one electron volt are equal, we can equate mv2/2 = eV. Put e and m in the equation to get velocity v = 5.9 x 105 m/s.
Displacement and Conduction Current
1. Find the conductivity of a material with conduction current density 100 units and electric field of 4 units.
a) 25
b) 400
c) 0.04
d) 1600
Answer: a
Explanation: The conduction current density is given by, Jc = σE. To get conductivity, σ = J/E = 100/4 = 25 units.
2. Calculate the displacement current density when the electric flux density is 20sin 0.5t.
a) 10sin 0.5t
b) 10cos 0.5t
c) 20sin 2t
d) 20cos 2t
Answer: b
Explanation: The displacement current density is given by, Jd = dD/dt.
Jd = d(20sin 0.5t)/dt = 20cos 0.5t (0.5) = 10cos 0.5t.
3. Find the magnitude of the displacement current density in air at a frequency of 18GHz in frequency domain. Take electric field E as 4 units.
a) 18
b) 72
c) 36
d) 4
Answer: d
Explanation: Jd = dD/dt = εdE/dt in time domain. For frequency domain, convert using Fourier transform, Jd = εjωE. The magnitude of
Jd = εωE = ε(2πf)E. On substituting, we get 4 ampere.
4. Calculate the frequency at which the conduction and displacement currents become equal with unity conductivity in a material of permittivity 2.
a) 18 GHz
b) 9 GHz
c) 36 GHz
d) 24 GHz
Answer: b
Explanation: When Jd = Jc , we get εωE = σE. Thus εo(2∏f) = σ. On substituting conductivity as one and permittivity as 2, we get f = 9GHz.
5. The ratio of conduction to displacement current density is referred to as
a) Attenuation constant
b) Propagation constant
c) Loss tangent
d) Dielectric constant
Answer: c
Explanation: Jc /Jd is a standard ratio, which is referred to as loss tangent given by σ /ε ω. The loss tangent is used to determine if the material is a conductor or dielectric.
6. If the loss tangent is very less, then the material will be a
a) Conductor
b) Lossless dielectric
c) Lossy dielectric
d) Insulator
Answer: b
Explanation: If loss tangent is less, then σ /ε ω <<1. This implies the conductivity is very poor and the material should be a dielectric. Since it is specifically mentioned very less, assuming the conductivity to be zero, the dielectric will be lossless (ideal).
7. In good conductors, the electric and magnetic fields will be
a) 45 in phase
b) 45 out of phase
c) 90 in phase
d) 90 out of phase
Answer: b
Explanation: The electric and magnetic fields will be out of phase by 45 in good conductors. This is because their intrinsic impedance is given by η = √(ωμ/σ) X (1+j). In polar form we get 45 out of phase.
8. In free space, which of the following will be zero?
a) Permittivity
b) Permeability
c) Conductivity
d) Resistivity
Answer: c
Explanation: In free space, ε = ε0 and μ = μ0. The relative permittivity and permeability will be unity. Since the free space will contain no charges in it, the conductivity will be zero.
9. If the intrinsic angle is 20, then find the loss tangent.
a) tan 20
b) tan 40
c) tan 60
d) tan 80
Answer: b
Explanation: The loss tangent is given by tan 2θn, where θn = 20. Thus the loss tangent will be tan 40.
10. The intrinsic impedance of free space is given by
a) 272 ohm
b) 412 ohm
c) 740 ohm
d) 377 ohm
Answer: d
Explanation: The intrinsic impedance is given by η = √(μo/εo) ohm. Here εo = 8.854 x 10-12 and μo = 4π x 10-7.
On substituting the values, we get η = 377 ohm.
Polarization
1. The best definition of polarization is
a) Orientation of dipoles in random direction
b) Electric dipole moment per unit volume
c) Orientation of dipole moments
d) Change in polarity of every dipole
Answer: b
Explanation: The polarization is defined mathematically as the electric dipole moment per unit volume. It is also referred to as the orientation of the dipoles in the direction of applied electric field.
2. Calculate the polarization vector of the material which has 100 dipoles per unit volume in a volume of 2 units.
a) 200
b) 50
c) 400
d) 0.02
Answer: a
Explanation: polarization vector P = N x p, where N = 100 and p = 2. On substituting we get P = 200 units.
3. Polarizability is defined as the
a) Product of dipole moment and electric field
b) Ratio of dipole moment to electric field
c) Ratio of electric field to dipole moment
d) Product of dielectric constant and dipole moment
Answer: b
Explanation: Polarizability is a constant that is defined as the ratio of elemental dipole moment to the electric field strength.
4. Calculate the energy stored per unit volume in a dielectric medium due to polarization when P = 9 units and E = 8 units.
a) 1.77
b) 2.25
c) 36
d) 144
Answer: c
Explanation: The energy stored per unit volume in a dielectric medium is given by, W = 0.5 X PE = 0.5 X 9 X 8 = 36 units.
5. Identify which type of polarization depends on temperature.
a) Electronic
b) Ionic
c) Orientational
d) Interfacial
Answer: c
Explanation: The electronic, ionic and interfacial polarization depends on the atoms which are independent with respect to temperature. Only the orientational polarization is dependent on the temperature and is inversely proportional to it.
6. Calculate the polarization vector in air when the susceptibility is 5 and electric field is 12 units.
a) 3
b) 2
c) 60
d) 2.4
Answer: c
Explanation: The polarization vector is given by, P = ε0 x χe x E, where χe = 5 and ε0 = 12. On substituting, we get P = 1 x 5 x 12 = 60 units.
7. In isotropic materials, which of the following quantities will be independent of the direction?
a) Permittivity
b) Permeability
c) polarization
d) Polarizability
Answer: a
Explanation: Isotropic materials are those with radiate or absorb energy uniformly in all directions (eg. Isotropic antenna). Thus it is independent of the direction.
8. The total polarization of a material is the
a) Product of all types of polarization
b) Sum of all types of polarization
c) Orientation directions of the dipoles
d) Total dipole moments in the material
Answer: b
Explanation: The total polarization of a material is given by the sum of electronic, ionic, orientational and interfacial polarization of the material.
9. In the given types of polarization, which type exists in the semiconductor?
a) Electronic
b) Ionic
c) Orientational
d) Interfacial or space charge
Answer: d
Explanation: The interfacial type of polarization occurs due to accumulation of charges at the interface in a multiphase material. This interface or junction is found in a semiconductor material.
10. Solids do not have which type of polarization?
a) Ionic
b) Orientational
c) Interfacial
d) Electronic
Answer: c
Explanation: Solids possess permanent dipole moments. Moreover they do not have junction like semiconductors. Thus, solids neglect the interfacial and space charge polarization. They possess only electronic, ionic and orientational polarizations.
Dielectric Strength and Constant
1. Which of the following is not an example of elemental solid dielectric?
a) Diamond
b) Sulphur
c) Silicon
d) Germanium
Answer: c
Explanation: Elemental solid dielectrics are the materials consisting of single type of atoms. Such materials have neither ions nor permanent dipoles and possess only electronic polarization. Its examples are diamond, sulphur and germanium.
2. Ionic non polar solid dielectrics contain more than one type of atoms but no permanent dipoles. State True/False
a) True
b) False
Answer: a
Explanation: In ionic crystals, the total polarization is electronic and ionic in nature. Thus, it implies that it contains more than one type of atom and no permanent dipoles.
3. Compute the refractive index when the dielectric constant is 256 in air.
a) 2562
b) 16
c) 256
d) 64
Answer: b
Explanation: By Maxwell relation, εr = n2, where εro is the dielectric constant at optical frequencies and n is the refractive index. For the given dielectric constant we get n = 16.
4. Dielectric property impacts the behaviour of a material in the presence of electric field. State True/False.
a) True
b) False
Answer: a
Explanation: Based on the dielectric property, a material can be classified as piezoelectric, ferroelectric, pyroelectric and anti-ferroelectric materials under the influence of electric field.
5. Curie-Weiss law is applicable to which of the following materials?
a) Piezoelectric
b) Ferroelectric
c) Pyroelectric
d) Anti-ferroelectric
Answer: b
Explanation: Curie-Weiss law is given by χe = εr -1 = C/(T-θ), where C is the curie constant and θ is the characteristic temperature which is usually a few degrees higher than the curie temperature for ferromagnetic materials.
6. Curie-Weiss law is used to calculate which one of the following?
a) Permittivity
b) Permeability
c) Electric susceptibility
d) Magnetic susceptibility
Answer: c
Explanation: Curie-Weiss law is given by χe = εr -1. Thus it is used to calculate the electric susceptibility of a material.
7. Calculate the loss tangent when the dielectric constant in AC field is given by 3 + 2j.
a) (2/3)
b) (3/2)
c) (-3/2)
d) (-2/3)
Answer: d
Explanation: The AC dielectric constant is given by εr = ε` – jε“, where ε` is the real part of AC dielectric and ε“ is the imaginary part of AC dielectric. The loss tangent is given by tan δ = ε“/ε` = -2/3.
8. When a dielectric loses its dielectric property, the phenomenon is called
a) Dielectric loss
b) Dielectric breakdown
c) Polarization
d) Magnetization
Answer: b
Explanation: Due to various treatments performed on the dielectric, in order to make it conduct, the dielectric reaches a state, where it loses its dielectric property and starts to conduct. This phenomenon is called as dielectric breakdown.
9. Choose the best definition of dielectric loss.
a) Absorption of electric energy by dielectric in an AC field
b) Dissipation of electric energy by dielectric in a static field
c) Dissipation of heat by dielectric
d) Product of loss tangent and relative permittivity
Answer: a
Explanation: In the scenario of an AC field, the absorption of electrical energy by a dielectric material is called as dielectric loss. This will result in dissipation of energy in the form of heat.
10. Compute the loss factor when the loss tangent is 0.88 and the real part of dielectric is 24.
a) 12.12
b) 12.21
c) 21.21
d) 21.12
Answer: d
Explanation: The loss factor is nothing but the imaginary part of AC dielectric. It is given by, ε“ = ε` tan δ. We get loss factor as 24 x 0.88 = 21.12.
Continuity Equation
1. Find the current when the charge is a time function given by q(t) = 3t + t2 at 2 seconds.
a) 3
b) 5
c) 7
d) 9
Answer: c
Explanation: The current is defined as the rate of change of charge in a circuit ie, I = dq/dt. On differentiating the charge with respect to time, we get 3 + 2t. At time t = 2s, I = 7A.
2. The continuity equation is a combination of which of the two laws?
a) Ohm’s law and Gauss law
b) Ampere law and Gauss law
c) Ohm’s law and Ampere law
d) Maxwell law and Ampere law
Answer: b
Explanation: I = ∫ J.ds is the integral form of Ohm’s law and Div (J) = dq/dt is the Gauss law analogous to D. Through these two equations, we get Div(J) = -dρ/dt. This is the continuity equation.
3. Calculate the charge density for the current density given 20sin x i + ycos z j at the origin.
a) 20t
b) 21t
c) 19t
d) -20t
Answer: b
Explanation: Using continuity equation, the problem can be solved. Div(J) =
– dρ/dt. Div(J) = 20cos x + cos z. At origin, we get 20cos 0 + cos 0 = 21. To get ρ, on integrating the Div(J) with respect to t, the charge density will be 21t.
4. Compute the conductivity when the current density is 12 units and the electric field is 20 units. Also identify the nature of the material.
a) 1.67, dielectric
b) 1.67, conductor
c) 0.6, dielectric
d) 0.6, conductor
Answer: c
Explanation: The current density is the product of conductivity and electric field intensity. J = σE. To get σ, put J = 12 and E = 20. σ = 12/20 = 0.6. Since the conductivity is less than unity, the material is a dielectric.
5. Find the electron density when convection current density is 120 units and the velocity is 5m/s.
a) 12
b) 600
c) 24
d) 720
Answer: c
Explanation: The convection current density is given by J = ρe x v. To get ρe, put J = 120 and v = 5. ρe = 120/5 = 24 units.
6. Calculate the electric field when the conductivity is 20 units, electron density is 2.4 units and the velocity is 10m/s. Assume the conduction and convection current densities are same.
a) 2.4
b) 4.8
c) 3.6
d) 1.2
Answer: d
Explanation: The conduction current density is given by J = σE and the convection current density is J = ρe v. When both are equal, ρe v = σE. To get E, put σ = 20, ρe = 2.4 and v = 10, E = 2.4 x 10/20 = 1.2 units.
7. Find the mobility of the electrons when the drift velocity is 23 units and the electric field is 11 units.
a) 1.1
b) 2.2
c) 3.2
d) 0.9
Answer: b
Explanation: The mobility is defined as the drift velocity per unit electric field. Thus μe = vd/E = 23/11 = 2.1 units.
8. Find the resistance of a cylinder of area 200 units and length 100m with conductivity of 12 units.
a) 1/24
b) 1/48
c) 1/12
d) 1/96
Answer: a
Explanation: The resistance is given by R = ρL/A = L/σA. Put L = 100, σ = 12 and A = 200, we get R = 100/(12 x 200) = 1/24 units.
9. Calculate the potential when a conductor of length 2m is having an electric field of 12.3units.
a) 26.4
b) 42.6
c) 64.2
d) 24.6
Answer: d
Explanation: The electric field is given by E = V/L. To get V, put E = 12.3 and L = 2.Thus we get V = E x L = 12.3 x 2 = 24.6 units.
10. On equating the generic form of current density equation and the point form of Ohm’s law, we can obtain V=IR. State True/False.
a) True
b) False
Answer: a
Explanation: The generic current density equation is J = I/A and the point form of Ohm’s law is J = σ E. On equating both and substituting E = V/L, we get V = IL/σ A = IR which is the Ohm’s law.
Boundary Conditions
1. The charge within a conductor will be
a) 1
b) -1
c) 0
d) ∞
Answer: c
Explanation: No charges exist in a conductor. An illustration for this statement is that, it is safer to stay inside a car rather than standing under a tree during lightning. Since the car has a metal body, no charges will be possessed by it to get ionized by the lightning.
2. For a conservative field which of the following equations holds good?
a) ∫ E.dl = 0
b) ∫ H.dl = 0
c) ∫ B.dl = 0
d) ∫ D.dl = 0
Answer: a
Explanation: A conservative field implies the work done in a closed path will be zero. This is given by ∫ E.dl = 0.
3. Find the electric field if the surface density at the boundary of air is 10-9.
a) 12π
b) 24π
c) 36π
d) 48π
Answer: c
Explanation: It is the conductor-free space boundary. At the boundary, E = ρ/εo. Put ρ = 10-9 and εo = 10-9/36π. We get E = 36π units.
4. Find the flux density at the boundary when the charge density is given by 24 units.
a) 12
b) 24
c) 48
d) 96
Answer: b
Explanation: At the boundary of a conductor- free space interface, the flux density is equal to the charge density. Thus D = ρv = 24 units.
5. Which component of the electric field intensity is always continuous at the boundary?
a) Tangential
b) Normal
c) Horizontal
d) Vertical
Answer: a
Explanation: At the boundary of the dielectric-dielectric, the tangential component of the electric field intensity is always continuous. We get Et1 = Et2.
6. The normal component of which quantity is always discontinuous at the boundary?
a) E
b) D
c) H
d) B
Answer: b
Explanation: The normal component of an electric flux density is always discontinuous at the boundary for a dielectric-dielectric boundary. We get Dn1 = Dn2, when we assume the free surface charge exists at the interface.
7. The electric flux density of a surface with permittivity of 2 is given by 12 units. What the flux density of the surface in air?
a) 24
b) 6
c) 1/6
d) 0
Answer: b
Explanation: The relation between electric field and permittivity is given by Dt1/Dt2 = ε1/ε2. Put Dt1 = 12, ε1 = 2 and ε2 =1, we get Dt2 = 12 x 1/ 2 = 6 units.
8. The electric field intensity of a surface with permittivity 3.5 is given by 18 units. What the field intensity of the surface in air?
a) 5.14
b) 0.194
c) 63
d) 29
Answer: c
Explanation: The relation between flux density and permittivity is given by En1/En2 = ε2/ ε1. Put En1 = 18, ε1 = 3.5 and ε2 = 1. We get En2 = 18 x 3.5 = 63 units.
9. A wave incident on a surface at an angle 60 degree is having field intensity of 6 units. The reflected wave is at an angle of 30 degree. Find the field intensity after reflection.
a) 9.4
b) 8.4
c) 10.4
d) 7.4
Answer: c
Explanation: By Snell’s law, the relation between incident and reflected waves is given by, E1 sin θ1 = E2 sin θ2. Thus 6 sin 60 = E2 sin 30. We get E2 = 6 x 1.732 = 10.4 units.
10. Find the permittivity of the surface when a wave incident at an angle 60 is reflected by the surface at 45 in air.
a) 1.41
b) 3.5
c) 2.2
d) 1.73
Answer: d
Explanation: From the relations of the boundary conditions of a dielectric-dielectric interface, we get tan θ1/tan θ2 = ε1/ε2. Thus tan 60/tan 45 = ε1/1. We get ε1 = tan 60 = 1.73.
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