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Multiple choice question for engineering

Set 1

1. Find the electric force when the charge of 2C is subjected to an electric field of 6 units.
a) 6
b) 3
c) 12
d) 24

View Answer

Answer: c [Reason:] The electric force is given by F = qE, where q = 2C and E = 6 units. Thus we get F = 2 x 6 = 12 units.

2. Find the magnetic force when a charge 3.5C with flux density of 4 units is having a velocity of 2m/s.
a) 14
b) 28
c) 7
d) 32

View Answer

Answer: b [Reason:] The magnetic force is given by F = q(v x B), where q = 3.5C, v = 2m/s and B = 4 units. Thus we get F = 3.5(2 x 4) = 28 units.

3. Find the electric field when the velocity of the field is 12m/s and the flux density is 8.75 units.
a) 510
b) 105
c) 150
d) 165

View Answer

Answer: b [Reason:] The electric field intensity is the product of the velocity and the magnetic flux density ie, E = v x B = 12 x 8.75 = 105 units.

4. Find the Lorentz force of a charge 2.5C having an electric field of 5 units and magnetic field of 7.25 units with a velocity 1.5m/s.
a) 39.68
b) 68.39
c) 86.93
d) 93.68

View Answer

Answer: a [Reason:] The Lorentz force is given by F = qE + q(v x B), it is the sum of electric and magnetic force. On substituting q = 2.5, E = 5, v = 1.5 and B = 7.25, F = 2.5(5) + 2.5(1.5 x 7.25) = 39.68 units.

5. The force on a conductor of length 12cm having current 8A and flux density 3.75 units at an angle of 300 is
a) 1.6
b) 2
c) 1.4
d) 1.8

View Answer

Answer: d [Reason:] The force on a conductor is given by F = BIL sin θ, where B = 3.75, I = 8, L = 0.12 and θ = 300. We get F = 3.75 x 8 x 0.12 sin 30 = 1.8 units.

6. The force per unit length of two conductors carrying equal currents of 5A separated by a distance of 20cm in air(in 10-6 order)
a) 25
b) 35
c) 40
d) 50

View Answer

Answer: a [Reason:] The force per unit length of two conductors is given by F = μ I1xI2/2πD, where I1 = I2 = 5 and D = 0.2. Thus F = 4π x 10-7 x 52/ 2π x 0.2 = 25 x 10-6 units.

7. When currents are moving in the same direction in two conductors, then the force will be
a) Attractive
b) Repulsive
c) Retracting
d) Opposing

View Answer

Answer: a [Reason:] When two conductors are having currents moving in the same direction then the forces of the two conductors will be moving towards each other or attractive.

8. Find the flux density due to a conductor of length 6m and carrying a current of 3A(in 10-7 order)
a) 1
b) 10
c) 100
d) 0.1

View Answer

Answer: a [Reason:] The flux density is B = μH, where H = I/2πR. Put I = 3 and R = 6, we get B = 4π x 10-7 x 3/2π x 6 = 1 x 10-7 units.

9. Find the maximum force of the conductor having length 60cm, current 2.75A and flux density of 9 units.
a) 14.85
b) 18.54
c) 84.25
d) 7.256

View Answer

Answer: a [Reason:] The force on a conductor is given by F = BIL sin θ, where B = 3.75, I = 8, L = 0.12 and θ = 90 for maximum force. We get F = BIL= 9 x 2.75 x 0.6 sin 90 = 14.85 units.

10. The magnetic force impacts the energy of the field. State True/false.
a) True
b) False

View Answer

Answer: a [Reason:] The magnetic force depends on the flux density of a material and the flux density is in turn dependent on the energy of the material. It can be shown that F = q(v x B) and E = 0.5 x B2/μ. It is clear that B and F are related.

Set 2

1. In a waveguide, which of the following condition is true always?
a) phase velocity = c
b) group velocity = c
c) phase velocity > c
d) phase velocity < c

View Answer

Answer: c [Reason:] The phase velocity is always greater than the speed of light in waveguides. This implies the group velocity is small.

2. The term cos θ is given by 2.5. Find the phase velocity.
a) 3
b) 5
c) 7.5
d) 2.5

View Answer

Answer: c [Reason:] The phase velocity is given by Vp = c cos θ. On substituting for cos θ = 2.5 and the speed of light, we get the phase velocity as 7.5 x 108 m/s.

3. The cut off wavelength and the guided wavelength are given by 0.5 and 2 units respectively. Find the wavelength of the wave.
a) 0.48
b) 0.32
c) 0.45
d) 0.54

View Answer

Answer: a [Reason:] The cut off wavelength and the guided wavelength are related as (1/λ)2 = (1/λc)2 + (1/λg)2. On substituting for λc = 0.5 and λg = 2, we get λ = 0.48 units.

4. The cut off wavelength of the rectangular waveguide in dominant mode with dimensions 6 cm x 4 cm is
a) 12cm
b) 6cm
c) 4cm
d) 2cm

View Answer

Answer: a [Reason:] The cut off wavelength in the dominant mode is given by λc = 2a/m, where a is the broad wall dimension. On substituting for m = 1 and a = 6cm, we get the cut off wavelength as 12cm.

5. The product of the phase and the group velocities is given by the
a) Speed of light
b) Speed of light/2
c) 2 x Speed of light
d) (speed of light)/4

View Answer

Answer: d [Reason:] The product of the phase and the group velocities is given by the square of the speed of the light. Thus Vp x Vg = c2 is the relation.

6. The phase velocity of a wave having a group velocity of 6 x 106 is (in order of 108 m/s)
a) 2.4
b) 3
c) 15
d) 150

View Answer

Answer: d [Reason:] We know that the phase and the group velocities are given by Vp x Vg = c2. On substituting for Vg = 6 x 106 and the speed of light, we get Vp = 150 x 108 m/s.

7. The group velocity of a wave with a phase velocity of 60 x 109 is (in 106 order)
a) 1.5
b) 2
c) 2.5
d) 3

View Answer

Answer: a [Reason:] We know that the phase and the group velocities are given by Vp x Vg = c2. On substituting for Vp = 60 x 109 and the speed of light, we get Vg = 1.5 x 106 m/s.

8. The phase velocity of a wave having a phase constant of 4 units and a frequency of 2.5 x 109 radian/sec is (in 108 order)
a) 3.25
b) 3.75
c) 6.25
d) 6.75

View Answer

Answer: c [Reason:] The phase velocity and the phase constant are related by Vp = ω/βg. On substituting for ω = 2.5 x 109 and β = 4, we get the phase velocity as 6.25 units.

9. The guided wavelength and the phase constant are related by
a) 2π/βg = λg
b) 1/βg = λg
c) 1/2πβg = λg
d) βg = λg

View Answer

Answer: a [Reason:] The guided wavelength and the phase constant are related by 2π/βg = λg, where βg is the guided phase constant and λg is the guided wavelength.

10. The phase velocity refers to a group of waves and the group velocity refers to a single wave. State true/false.
a) True
b) False

View Answer

Answer: b [Reason:] The phase velocity refers to a single wave and the group velocity refers to a group of waves.

11. The phase and group velocities does not depend on which of the following?
a) Frequency
b) Wavelength
c) Phase constant
d) Attenuation constant

View Answer

Answer: d [Reason:] The phase and the group velocities are directly related by the frequency, wavelength and the phase constant. It is independent of the attenuation constant.

12. The distance between two successive points in a waveguide is the
a) Guided wavelength
b) 2 x guided wavelength
c) Guided wavelength/2
d) (guided wavelength)/4

View Answer

Answer: c [Reason:] The distance between two successive points in a waveguide is equal to half of the guided wavelength.

Set 3

1. In free space, the charge carriers will be
a) 0
b) 1
c) 100
d) Infinity

View Answer

Answer: a [Reason:] 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.

2. In free space, which parameter will be unity?
a) Permittivity
b) Absolute permittivity
c) Relative permittivity
d) Permeability

View Answer

Answer: c [Reason:] 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.

3. Which parameter is unity in air medium?
a) Permittivity
b) Absolute permeability
c) Relative permeability
d) Permeability

View Answer

Answer: c [Reason:] 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.

4. The conductivity in free space medium is
a) Infinity
b) Unity
c) Zero
d) Negative

View Answer

Answer: c [Reason:] 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.

5. 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 resistivity

View Answer

Answer: c [Reason:] 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.

6. The intrinsic impedance of free space is
a) 489
b) 265
c) 192
d) 377

View Answer

Answer: d [Reason:] 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.

7. 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 propagation

View Answer

Answer: b [Reason:] 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.

8. In free space, the ratio of frequency to the velocity of light gives the phase constant. State True/False.
a) True
b) False

View Answer

Answer: a [Reason:] The phase constant is given by the ratio of the frequency in radian/sec to the velocity of the wave propagating. In free space, the velocity is considered to be the velocity of light. Thus the statement is true.

9. The velocity of a wave travelling in the air medium without transmission lines or waveguides(wireless) is
a) 6 x 108
b) 3 x 108
c) 1.5 x 108
d) 9 x 108

View Answer

Answer: b [Reason:] 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 108m/s.

10. The vectors of the electromagnetic wave propagation can be expressed in
a) Dot product
b) Cross product
c) Unit vector
d) Perpendicular vector

View Answer

Answer: b [Reason:] In an EM wave, the electric and the magnetic fields will be perpendicular to each other and with the direction of the propagation. Thus it can be expressed in cross product where iE x iH = iw. Here iE is the electric vector component, iH is the magnetic vector component and iw is the vector of the wave propagating.

Set 4

1. For conductors, the loss tangent will be
a) Zero
b) Unity
c) Maximum
d) Minimum

View Answer

Answer: c [Reason:] In conductors, the conductivity will be more. Thus the loss tangent σ/ωε will be maximum.

2. In metals, the total permittivity is
a) Absolute permittivity
b) Relative permittivity
c) Product of absolute and relative permittivity
d) Unity

View Answer

Answer: a [Reason:] 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.

3. The total permeability in a conductor is
a) Absolute permeability
b) Relative permeability
c) Product of absolute and relative permeability
d) Unity

View Answer

Answer: c [Reason:] 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.

4. 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) 0.04

View Answer

Answer: d [Reason:] 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.

5. 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) 5.62

View Answer

Answer: a [Reason:] 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.

6. The skin depth of a conductor with attenuation constant of 7 neper/m is
a) 14
b) 49
c) 7
d) 1/7

View Answer

Answer: d [Reason:] 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.

7. 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)

View Answer

Answer: a [Reason:] 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.

8. In conductors, the E and H vary by a phase difference of
a) 0
b) 30
c) 45
d) 60

View Answer

Answer: c [Reason:] 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.

9. EM waves do not travel inside metals. State True/False.
a) True
b) False

View Answer

Answer: a [Reason:] 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.

10. The propagation constant of the wave in a conductor with air as medium is
a) √(ωμσ)
b) ωμσ
c) √(ω/μσ)
d) ω/μσ

View Answer

Answer: a [Reason:] 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.

11. An example for electromagnetic wave propagation is
a) refrigerator
b) electric fan
c) mobile transponder
d) relays in actuators

View Answer

Answer: c [Reason:] 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.

12. The phase shift in the electric and magnetic fields in an EM wave is given by which parameter?
a) phase constant
b) attenuation constant
c) propagation constant
d) intrinsic impedance

View Answer

Answer: d [Reason:] 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.

Set 5

1. The total power of a wave with average power 15 units in a surface density of 0.5 units is
a) 15
b) 30
c) 7.5
d) 0.75

View Answer

Answer: c [Reason:] The total power is given by the surface integral of the average power. Thus ∫Pavg ds is the total power. On substituting for Pavg = 15 and ∫ds = 0.5, we get total power as 7.5 units.

2. The power of a wave with electric field intensity of 3 units in air is
a) 0.01
b) 0.03
c) 0.05
d) 0.07

View Answer

Answer: a [Reason:] The Poynting vector gives the power of a wave. It is given as P = E2/2η. On substituting for E = 3 and η = 377 in air, the power is P = 32/(2×377) = 0.01 units.

3. Find the power of an EM wave, given that the cross product of the E and H component is 2 + 3j.
a) 2
b) 1
c) 4
d) 8

View Answer

Answer: b [Reason:] The Poynting power vector for complex quantity of E x H is P = 0.5 x Re(E x H). In the given data, Re(E x H) = 2, thus we get P = 0.5 x 2 = 1 unit.

4. The power in a electromagnetic wave with electric field and magnetic field intensities 12 and 8 respectively is
a) 96
b) 12
c) 8
d) 48

View Answer

Answer: d [Reason:] The Poynting vector is given by P = 0.5 EH. Given that E = 12 and H = 8, we get P = 0.5 x 12 x 8 = 48 units.

5. The power in a wave given that H component is 0.82 units in air.
a) 126.74
b) 621.47
c) 216.47
d) 745.62

View Answer

Answer: a [Reason:] The power of a wave is given by P = ηH2/2. In air medium, η = 377 and given that H = 0.82. We get power P = 377 x 0.822/2 = 126.74 units.

6. Find the power of a wave given that the RMS value of E and H are 6 and 4.5 respectively.
a) 24
b) 27
c) 29
d) 32

View Answer

Answer: b [Reason:] The power is the product of the RMS electric field and the RMS magnetic field. Thus P = Erms X Hrms. On substituting Erms = 6 and Hrms = 4.5, the power is P = 6 x 4.5 = 27 units.

7. The electric and magnetic fields vary with time in which of the following fields?
a) DC
b) AC
c) Static
d) It does not vary with time

View Answer

Answer: b [Reason:] The electric and magnetic fields vary with time in oscillating fields. It is certain that such fields are AC fields.

8. The power per unit velocity of a wave with electric field as 8 units and density 10 units is
a) 40
b) 20
c) 80
d) 160

View Answer

Answer: c [Reason:] The power per unit velocity P/v is given by the product of electric field and the density. Thus P/v = E.d = 8 x 10 = 80 units.

9. The power of a wave in a cylindrical waveguide of radius 2m with electric field 12 units is
a) 2.39
b) 3.92
c) 9.23
d) 9.32

View Answer

Answer: a [Reason:] The power of a wave is given by ∫P ds, where P = E2/2η and ∫ds = πr2. On substituting for E = 12, η = 377 in air and r = 2, we get P = 2.39 units.

10. The work done in the power transmission with E and H given by 50 and 65 respectively. The velocity of propagation is 20m/s.
a) 162.5
b) 621.5
c) 562.1
d) 261.5

View Answer

Answer: a [Reason:] The work done is given by W = EH/v, where E = 50, H = 65 and v = 20. On substituting, we get W = 50 x 65/20 = 162.5 units.

11. The Poynting vector is the power component that is calculated by the
a) Product of E and H
b) Ratio of E and H
c) Dot product of E and H
d) Cross product of E and H

View Answer

Answer: d [Reason:] The Poynting vector P is the cross product of the electric field E and the magnetic field H. It is given by P = E X H. It is used to calculate the power in wave propagation in transmission lines, waveguides and antenna.

12. The maximum power transceived by the antenna is in the region of
a) Aperture
b) Effective aperture
c) Maxima lobe
d) Minima lobe

View Answer

Answer: b [Reason:] An antenna is a device that transmits and receives power. The transmitting and receiving is in the region called aperture. Maximum or efficient power transmission occurs in the region called effective aperture.

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