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# Multiple choice question for engineering

## Set 1

1. The electric field intensity is defined as
a) Force per unit charge
b) Force on a test charge
c) Force per unit charge on a test charge
d) Product of force and charge

Answer: c [Reason:] The electric field intensity is the force per unit charge on a test charge, i.e, q1 = 1C. E = F/Q = Q/(4∏εr2).

2. Find the force on a charge 2C in a field 1V/m.
a) 0
b) 1
c) 2
d) 3

Answer: c [Reason:] 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

Answer: b [Reason:] 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

Answer: c [Reason:] 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

Answer: a [Reason:] 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

Answer: a [Reason:] 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.

7. Electric field of an infinitely long conductor of charge density λ, is given by E = λ/(2πεh).aN. State True/False.
a) True
b) False

Answer: a [Reason:] 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.

8. Electric field intensity due to infinite sheet of charge σ is
a) Zero
b) Unity
c) σ/ε
d) σ/2ε

Answer: d [Reason:] 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.

9. For a test charge placed at infinity, the electric field will be
a) Unity
b) +∞
c) Zero
d) -∞

Answer: c [Reason:] 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

Answer: c [Reason:] 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.

## Set 2

1. Convert the point (3,4,5) from Cartesian to spherical coordinates
a) (7.07,45⁰,53⁰)
b) (0.707,45⁰,53⁰)
c) (7.07,54⁰,63⁰)
d) (0.707,54⁰,63⁰)

Answer: a [Reason:] 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.

2. Example of spherical system in the following is
a) Charge in space
b) Charge in box
c) Charge in dielectric
d) Uncharged system

Answer: a [Reason:] 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.

3. Spherical systems are employed in waveguides. State True/False
a) True
b) False

Answer: b [Reason:] There is no waveguide designed spherically to avoid absorption, rather than propagation.

4. Choose which of following condition is not required for a waveguide to exist.
a) The dimensions should be in accordance with desired frequency
b) Cut-off frequency should be minimum 6GHz
c) The shape should be spherical
d) No specific condition is required for waveguide design

Answer: c [Reason:] A waveguide need not be spherical, it has to be rectangular or circular, as it violates the propagation of the wave.

5. Find the spherical coordinates of A(2,3,-1)
a) (3.74, 105.5⁰, 56.13⁰)
b) (3.74, 105.5⁰, 56.31⁰)
c) (3.74, 106.5⁰, 56.13⁰)
d) (3.74, 106.5⁰, 56.31⁰)

Answer: b [Reason:] 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.

6. Find the Cartesian coordinates of B(4,25⁰,120⁰)
a) (0.845, 1.462, 3.625)
b) (-0.845, 1.462, 3.625)
c) (-8.45, 2.462, 6.325)
d) (8.45, 2.462, 6.325)

Answer: b [Reason:] 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

Answer: a [Reason:] On double integrating the differential volume, the area can be computed for a sphere.

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

Answer: a [Reason:] 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.

9. The scalar factor of spherical coordinates is
a) 1, r, r sin θ
b) 1, r, r
c) r, r, 1
d) r, 1, r

Answer: a [Reason:] The radius varies from unity to infinity, the plane angle from zero to 360 ⁰ and the z plane from (-∞, ∞) .

10. Transform the vector (4,-2,-4) at (1,2,3) into spherical coordinates.
a) 3.197i – 2.393j + 4.472k
b) -3.197i + 2.393j – 4.472k
c) 3.197i + 2.393j + 4.472k
d) -3.197i – 2.393j – 4.472k

Answer: b [Reason:] 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.

## Set 3

1. In conductors, which condition will be true?
a) σ/ωε > 1
b) σωε > 1
c) σ/ωε < 1
d) σωε < 1

Answer: a [Reason:] For conductors, the conductivity will be maximum. Thus the loss tangent is greater than unity. This is given by σ/ωε >1.

2. For metals, the conductivity will be
a) 0
b) 1
c) -1
d) Infinity

Answer: d [Reason:] 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.

3. In conductors, which two parameters are same?
a) Wavelength and phase constant
b) Phase and attenuation constant
c) Attenuation constant and skin depth
d) Skin depth and wavelength

Answer: b [Reason:] In conductors, which are considered to be lossy, the attenuation and the phase constant are the same. It is given by α=β= √(ωμσ/2).

4. Calculate the velocity of wave propagation in a conductor with frequency 5 x 108 rad/s and phase constant of 3 x 108 units.
a) 3/5
b) 15
c) 5/3
d) 8

Answer: c [Reason:] The velocity of wave propagation is the ratio of the frequency to the phase constant. It is given by V = ω/β. On substituting the given values, we get V = 5/3 units.

5. Calculate the wavelength of the wave with phase constant of 3.14 units.
a) 1
b) 2
c) 0.5
d) 4

Answer: b [Reason:] The wavelength is the ratio of 2π to the phase constant β. On substituting for β = 3.14, we get λ = 2π/β = 2π/3.14 = 2 units.

6. 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 H

Answer: d [Reason:] 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.

7. In perfect conductors, the phase shift between the electric field and magnetic field will be
a) 0
b) 30
c) 45
d) 90

Answer: c [Reason:] For perfect conductors, the electric and magnetic field E and H respectively vary by a phase of 45 degree. This is due to the polarisation phenomenon in the conductors, unlike dielectrics.

8. 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/ωμε

Answer: c [Reason:] The phase constant is represented as β. It is a complex quantity representing the constant angle of the wave propagated. It is given by β = ω√(με).

9. In waveguides, which of the following conditions will be true?
a) V > c
b) V < c
c) V = c
d) V >> c

Answer: a [Reason:] 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.

10. The attenuation constant in lossless dielectrics will be
a) 0
b) 1
c) -1
d) ∞

Answer: a [Reason:] In lossless dielectrics, the attenuation constant will not be same as the phase constant, unlike conductors. Also, due to the lossless behaviour, the attenuation will be nearly zero. Practically, zero attenuation is not possible.

## Set 4

1. For a dielectric, the condition to be satisfied is
a) σ/ωε > 1
b) σ/ωε < 1
c) σ = ωε
d) ωε = 1

Answer: b [Reason:] In a dielectric, the conductivity will be very less. Thus the loss tangent will be less than unity. This implies σ/ωε < 1 is true.

2. For a perfect dielectric, which parameter will be zero?
a) Conductivity
b) Frequency
c) Permittivity
d) Permeability

Answer: a [Reason:] The conductivity will be minimum for a dielectric. For a perfect dielectric, the conductivity will be zero.

3. Calculate the phase constant of a wave with frequency 12 rad/s and velocity 3×108 m/s(in 10-8 order)
a) 0.5
b) 72
c) 4
d) 36

Answer: c [Reason:] The phase constant is given by β = ω√(με), where ω is the frequency in rad/s and 1/√(με) is the velocity of wave. On substituting √(με) = 3×108 and ω = 12, we get β = 12/(3×108) = 4 x 10-8m/s.

4. For a lossless dielectric, the attenuation will be
a) 1
b) 0
c) -1
d) Infinity

Answer: b [Reason:] The attenuation is the loss of power of the wave during its propagation. In a lossless dielectric, the loss of power will not occur. Thus the attenuation will be zero.

5. Calculate the velocity of a wave with frequency 2 x109 rad/s and phase constant of 4 x 108 units.
a) 0.5
b) 5
c) 0.2
d) 2

Answer: b [Reason:] The velocity of a wave is the ratio of the frequency to the phase constant. Thus V = ω/β. On substituting the given values, we get V = 2 x109/ 4 x 108 = 5 units.

6. Which of the following is the correct relation between wavelength and the phase constant of a wave?
a) Phase constant = 2π/wavelength
b) Phase constant = 2π x wavelength
c) Phase constant = 1/(2π x wavelength)
d) Phase constant = wavelength/2π

Answer: a [Reason:] The phase constant is the ratio of 2π to the wavelength λ. Thus β = 2π/λ is the correct relation.

7. In lossy dielectric, the phase difference between the electric field E and the magnetic field H is
a) 90
b) 60
c) 45
d) 0

Answer: d [Reason:] In a lossy dielectric, the E and H component will be in phase. This implies that the phase difference between E and H will be 0.

8. The intrinsic impedance is the ratio of square root of
a) Permittivity to permeability
b) Permeability to permittivity
c) Phase constant to wavelength
d) Wavelength to phase constant

Answer: b [Reason:] The intrinsic impedance is the impedance of a particular material. It is the ratio of square root of the permeability to permittivity. For air, the intrinsic impedance is 377 ohm or 120π.

9. Calculate the skin depth of a material with attenuation constant of 2 units.
a) 2
b) 1
c) 0.5
d) 4

Answer: c [Reason:] The skin depth of a material is the reciprocal of the attenuation constant. Thus δ = 1/α. On substituting for α = 2, we get δ = ½ = 0.5 units.

10. Calculate the phase constant of a wave with skin depth of 2.5 units.
a) 5/2
b) 5
c) 2
d) 2/5

Answer: d [Reason:] The skin depth is the reciprocal of the phase constant and the attenuation constant too. Thus δ = 1/β. On substituting for δ = 2.5, we get β = 1/δ = 1/2.5 = 2/5 units.

11. An example for lossless propagation is
a) Dielectric waveguide propagation
b) Conductor propagation
c) Cavity resonator propagation
d) It is not possible

Answer: d [Reason:] There are many techniques employed to achieve zero attenuation or maximum propagation. But it is not achievable practically. Thus lossless propagation is not possible practically.

12. Skin depth phenomenon is found in which materials?
a) Insulators
b) Dielectrics
c) Conductors
d) Semiconductors

Answer: c [Reason:] Skin depth is found in pure conductors. It the property of the conductor to allow a small amount of electromagnetic energy into its skin, but not completely. This is the reason why EM waves cannot travel inside a good conductor.

## Set 5

1. The magnetic energy of a magnetic material is given by
a) BH/2
b) B/2H
c) H/2B
d) B/H

Answer: a [Reason:] The magnetic energy of a material is given by half of the product of the magnetic flux density and the magnetic field intensity. It is represented as BH/2. Since B = μH, we can also write as μH2 or B2/2μ.

2. The induced emf in a material opposes the flux producing it. This is
b) Ampere law
c) Lenz law
d) Curie law

Answer: c [Reason:] The induced emf in a material under the influence of a magnetic field will oppose the flux that produces it. This is indicated by a negative sign in the emf equation. This phenomenon is called Lenz law.

3. The energy in a magnetic material is due to which process?
a) Emf
b) Magnetization
c) Magnetostriction
d) Polarization

Answer: b [Reason:] The energy in a magnetic material is due to the formation of magnetic dipoles which are held together due to magnetic force. This gives energy to the material. Hence it is due to magnetization process.

4. The flux lines of two energised coils overlapping on each other will give
a) Series aiding
b) Shunt aiding
c) Series opposing
d) Shunt opposing

Answer: a [Reason:] Flux lines are the magnetic lines of force of a magnetic material. Since the flux is overlapping, the total flux of the two coils together will be high. Thus it is an aiding flux. Also this type of overlapping is possible only when the two coils are back to back or in series connection.

5. The resistance in a magnetic material is called as
a) Capacitance
b) Inductance
c) Reluctance
d) Magnetic resistance

Answer: c [Reason:] The reluctance of a magnetic material is the ability of the material to oppose the magnetic flux. It is the ratio of the magnetic motive force mmf to the flux.

6. Calculate the reluctance of the material with a mmf of 3.5 units and flux of 7units.
a) 32.5
b) 10.5
c) 0.5
d) 2

Answer: c [Reason:] The reluctance is defined as the ratio of the mmf and the flux. It is given by S = mmf/φ. On substituting mmf = 3.5 and φ = 7, we get S = 3.5/7 = 0.5 units.

7. Which of the following relations is correct?
a) NI = Sφ
b) NS = Iφ
c) Nφ = SI
d) NI = S/φ

Answer: a [Reason:] The reluctance is also defined by the ratio of the current element to the flux. In other words, mmf = NI. Thus S = NI/φ. We get the relation NI = Sφ.

8. Calculate the reluctance of a material with length 2π x 10-4 in air with area 0.5.
a) 1
b) 10
c) 100
d) 1000

Answer: d [Reason:] The reluctance is given by S = L/μ A, where L is the length, A is the area and μ is the permeability. On substituting L = 2π x 10-4, A = 0.5 and μ = 4π x 10-7, we get S = 103/(2×0.5) = 1000 units.

9. Ampere turn is equivalent to which element?
a) Sφ
b) S/φ
c) φ/S
d) S