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# Multiple choice question for engineering

## Set 1

1. The loss tangent refers to the
a) Power due to propagation in conductor to that in dielectric
b) Power loss
c) Current loss
d) Charge loss

Answer: a [Reason:] The loss tangent is the tangent angle formed by the plot of conduction current density vs displacement current density. It is the ratio of Jc by Jd. It represents the loss of power due to propagation in a dielectric, when compared to that in a conductor.

2. Calculate the conduction current density when the resistivity of a material with an electric field of 5 units is 4.5 units.
a) 22.5
b) 4.5/5
c) 5/4.5
d) 9.5

Answer: c [Reason:] The conduction current density is the product of the conductivity and the electric field. The resistivity is the reciprocal of the conductivity. Thus the required formula is Jc = σ E = E/ρ = 5/4.5 units.

3. At high frequencies, which parameter is significant?
a) Conduction current
b) Displacement current
c) Attenuation constant
d) Phase constant

Answer: b [Reason:] The conduction current occurs in metals and is independent of the frequency. The attenuation and phase constant highly depend on the varying frequency. The displacement current occurs due to dielectrics and is significant only at very high frequencies.

4. Find the loss tangent of a material with conduction current density of 5 units and displacement current density of 10 units.
a) 2
b) 0.5
c) 5
d) 10

Answer: b [Reason:] The loss tangent is the ratio of Jc by Jd. On substituting for Jc = 5 and Jd = 10, the loss tangent, tan δ = 5/10 = 0.5. It is to be noted that it is tangent angle, so that the maxima and minima lies between 1 and -1 respectively.

5. The loss tangent is also referred to as
a) Attenuation
b) Propagation
c) Dissipation factor
d) Polarization

Answer: c [Reason:] The loss tangent is the measure of the loss of power due to propagation in a dielectric, when compared to that in a conductor. Hence it is also referred to as dissipation factor.

6. The loss tangent of a wave propagation with an intrinsic angle of 20 degree is
a) Tan 20
b) Tan 40
c) Tan 60
d) Tan 80

Answer: b [Reason:] The angle of the loss tangent δ is twice the intrinsic angle θn. Thus tan δ = tan 2θn = tan 2(20) = tan 40.

7. The expression for the loss tangent is given by
a) σ/ωε
b) ωε/σ
c) σ/ω
d) ω/ε

Answer: a [Reason:] The conduction current density is Jc = σ E and the displacement current density is Jd = jωεE. Its magnitude will be ωεE. Thus the loss tangent tan δ = Jc /Jd = σ/ωε is the required expression.

8. Find the loss angle in degrees when the loss tangent is 1.
a) 0
b) 30
c) 45
d) 90

Answer: c [Reason:] The loss tangent is tan δ, where δ is the loss angle. Given that loss tangent tan δ = 1. Thus we get δ = tan-1(1) = 450.

9. The complex permittivity is given by 2-j. Find the loss tangent.
a) 1/2
b) -1/2
c) 2
d) -2

Answer: a [Reason:] The loss tangent for a given complex permittivity of ε = ε’ – jε’’ is given by tan δ = ε’’/ ε’. Thus the loss tangent is 1/2.

10. The intrinsic angle of the wave with a loss angle of 60 is
a) 120
b) 60
c) 90
d) 30

Answer: d [Reason:] The angle of the loss tangent δ is twice the intrinsic angle θn. Thus tan δ = tan 2θn. We get θn = δ/2 = 60/2 = 30 degrees.

## Set 2

1. The transmission line is said to be lossless when the
a) Conductor is perfect and dielectric is lossless
b) Conductor is perfect and dielectric is lossy
c) Conductor is imperfect and dielectric is lossy
d) Conductor is imperfect and dielectric is lossless

Answer: a [Reason:] Lossless transmission line refers to a line with no losses or attenuation. This is possible only when the conductor is perfect and the dielectric is lossless.

2. The resistance of a lossless transmission line is
a) 0
b) 1
c) -1
d) Infinity

Answer: a [Reason:] The lossless transmission line will have minimal loss of power. Thus in the ideal cases, the resistance is assumed to be zero.

3. Which two parameters given below are zero in the lossless line?
a) L, C
b) C, G
c) G, α
d) R, L

Answer: c [Reason:] Due to the minimal losses in the lossless transmission line, the parameters of conductance and attenuation constant are assumed to be zero. Practically, a lossless line is not possible.

4. The characteristic impedance of the line having primary constants L and C as 35 milli henry and 70 micro farad respectively is
a) 500
b) 22.36
c) 125
d) 50

Answer: b [Reason:] The characteristic impedance of a lossless line is given by Zo = √(L/C). On substituting for L = 35 x 10-3 and C = 70 x 10-6, we get Zo = √(500) = 22. 36 ohm.

5. The attenuation constant is measured in the units of
a) Ohm
b) Neper
c) Decibel

Answer: b [Reason:] The attenuation is the loss of power due to the transmission. It is measured in the Neper units. 1 neper = 8.68 decibel/m.

6. The velocity of the transmission line with a frequency of 35 radian/s and phase constant of 68.5 is
a) 1.95
b) 2.36
c) 4.56
d) 3.48

Answer: a [Reason:] The velocity of the wave in a transmission line is the ratio of the frequency to the phase constant. Thus v = ω/β. On substituting for ω = 35 and β = 68.5, we get v = 68.5/35 = 1.95 units.

7. The condition that holds good in a distortionless transmission line is
a) R/L = G/C
b) RL = GC
c) L/R = C/G
d) RG/LC

Answer: a [Reason:] For a distortionless transmission line, the primary constants are related by R/L = G/C.

8. Given that R = 20 ohm, L = 40 mH, C = 40 μF, G = 0.02 mho. Find whether the line is distortionless or not.
a) Distorted line
b) Distortionless line
c) All of the mentioned
d) None of the mentioned

Answer: b [Reason:] For a distortionless line, the condition R/L = G/C must be true. On substituting for R = 20, L = 40 mH, C = 40 μF, G = 0.02, we get R/L = 500 and G/C = 500. Thus the line is distortionless.

9. For R= 20 ohm and G= 0.8 mho, the attenuation constant will be
a) 16
b) 4
c) 2
d) 32

Answer: a [Reason:] The attenuation constant is given by α = √(RG), where R and G is the resistance and conductance respectively. On substituting for R = 20 and G = 0.8, we get α = √(20 x 0.8) = 4 units.

10. The velocity of an electromagnetic wave with frequency 6MHz and a skin depth of 1.6 μm.
a) 3.75
b) 0.26
c) 9.6
d) 7.8

Answer: c [Reason:] The velocity of a wave is the product of the frequency and the skin depth. It is given v = f. δ. On substituting for f = 6 x 106 and δ = 1.6 x 10-6, we get v = 6 x 106 x 1.6 x 10-6 = 9.6 units.

11. The characteristic impedance of the transmission line with R= 45 ohm and G= 0.45 mho is
a) 100
b) 1
c) 0.1
d) 10

Answer: d [Reason:] The characteristic impedance in terms of R, G is Zo = √(R/G). On substituting for R = 45 and G = 0.45, we get Zo = 10 units.

12. The distortionless line is one in which the attenuation constant and phase constant are independent of the frequency. State true/false.
a) True
b) False

Answer: b [Reason:] In a distortionless line, the attenuation constant is independent of the frequency and the phase constant is linearly dependent of the frequency.

## Set 3

1. The magnetic moment of a field with current 12A and area 1.6 units is
a) 19.2
b) 12.9
c) 21.9
d) 91.2

Answer: a [Reason:] The magnetic moment is the product of current and the area of the conductor. It is given by M = IA, where I = 12 and A = 1.6.Thus we get, M = 12 x 1.6 = 19.2 units.

2. Find the torque of a loop with magnetic moment 12.5 and magnetic flux density 7.65 units is
a) 95.625
b) 65.925
c) 56.525
d) 65.235

Answer: a [Reason:] The torque is defined as the product of the magnetic moment and the magnetic flux density given by T = MB, where M = 12.5 and B = 7.65. Thus we get T = 12.5 x 7.65 = 95.625 units.

3. The magnetization is defined by the ratio of
a) Magnetic moment to area
b) Magnetic moment to volume
c) Magnetic flux density to area
d) Magnetic flux density to volume

Answer: b [Reason:] The magnetization refers to the amount of dipole formation in a given volume when it is subjected to a magnetic field. It is given by the ratio of the magnetic moment to the volume. Thus Pm = M/V.

4. Find the orbital dipole moment in a field of dipoles of radius 20cm and angular velocity of 2m/s(in 10-22 order)
a) 64
b) 76
c) 54
d) 78

Answer: a [Reason:] The orbital dipole moment is given by M = 0.5 x eVangx r2, where e = 1.6 x 10-19 is the charge of the electron, Vang = 2 and r = 0.2. On substituting, we get M = 0.5 x 1.6 x 10-19x 2 x 0.22= 64 x 10-22 units.

5. Find the orbital angular moment of a dipole with angular velocity of 1.6m/s and radius 35cm(in 10-31 order)
a) 1.78
b) 8.71
c) 7.18
d) 2.43

Answer: a [Reason:] The orbital angular moment is given by Ma = m x Vangx r2,where m = 9.1 x 10-31, Vang = 1.6 and r = 0.35. On substituting, we get, Ma = 9.1 x 10-31 x 1.6 x 0.352 = 1.78 x 10-31 units.

6. The ratio of the orbital dipole moment to the orbital angular moment is given by
a) e/m
b) –e/m
c) e/2m
d) –e/2m

Answer: d [Reason:] The orbital dipole moment is given by M = 0.5 x eVangx r2 and the orbital angular moment is given by Ma = m x Vangx r2. Their ratio M/Ma is given by –e/2m, the negative sign indicates the charge of electron.

7. Calculate the Larmer angular frequency for a magnetic flux density of 12.34 x 10-10.
a) 108.36
b) 810.63
c) 368.81
d) 183.36

Answer: a [Reason:] The Larmer angular frequency is the product of magnitude of the ratio of orbital dipole moment to orbital angular moment and the magnetic flux density. It is given by fL = B e/2m, where is the charge of electron and m is the mass of the electron. On substituting, we get fL = 12.34 x 10-10 x 1.6 x 10-19/(2 x 9.1 x 10-31) = 108.36 units.

8. The Bohr magneton is given by
a) eh/2m
b) eh/2πm
c) eh/4m
d) eh/4πm

Answer: d [Reason:] In atomic physics, the Bohr magneton (symbol μB) is a physical constant and the natural unit for expressing the magnetic moment of an electron caused by either its orbital or spin angular momentum. It is given by eh/4πm, where h is the Planck’s constant, e is the charge of the electron and m is the mass of the electron.

9. Find the magnetization of the field which has a magnetic moment 16 units in a volume of 1.2 units.
a) 16.67
b) 13.33
c) 15.56
d) 18.87

Answer: b [Reason:] The magnetization is the ratio of the magnetic moment to the volume. Thus M = m/v, where m = 16 and v = 1.2. We get M = 16/1.2 = 13.33 units.

10. Which of the following is true regarding magnetic lines of force?
a) Real
b) Imaginary
c) Does not exist
d) Parallel to field

Answer: b [Reason:] Magnetic Lines of Force is a an imaginary line representing the direction of magnetic field such that the tangent at any point is the direction of the field vector at that point.

## Set 4

1. Identify which of the following is the unit of magnetic flux density?
a) Weber
b) Weber/m
c) Tesla
d) Weber-1

Answer: c [Reason:] The unit of magnetic flux density is weber/m2. It is also called as tesla.

2. The divergence of H will be
a) 1
b) -1
c) ∞
d) 0

Answer: d [Reason:] We know that the divergence of B is zero. Also B = μH. Thus divergence of H is also zero.

3. Find the flux contained by the material when the flux density is 11.7 Tesla and the area is 2 units.
a) 23.4
b) 12.3
c) 32.4
d) 21.3

Answer: a [Reason:] The total flux is given by φ = ∫ B.ds, where ∫ds is the area. Thus φ = BA. We get φ = 11.7 x 2 = 23.4 units.

4. Find the current when the magnetic field intensity is given by 2L and L varies as 0->1.
a) 2
b) 1
c) 0.5
d) 0

Answer: b [Reason:] From Ampere law, we get ∫ H.dL = I. Put H = 2L and L = 0->1. On integrating H with respect to L, the current will be 1A.

5. Find the magnetic field intensity when the flux density is 8 x 10-6 Tesla in the medium of air.
a) 6.36
b) 3.66
c) 6.63
d) 3.36

Answer: a [Reason:] We how that, B = μH. To get H = B/μ, put B = 8 x 10-6 and μ = 4π x 10-7. Thus H = 8 x 10-6/ 4π x 10-7 = 6.36 units.

6. If ∫ H.dL = 0, then which statement will be true?

Answer: c [Reason:] The given condition shows that the magnetic field intensity will be the negative gradient of the magnetic vector potential.

7. Find the magnetic flux density of the material with magnetic vector potential A = y i + z j + x k.
a) i + j + k
b) –i – j – k
c) –i-j
d) –i-k

Answer: b [Reason:] The magnetic flux density is the curl of the magnetic vector potential. B = Curl(A). Thus Curl(A) = i(-1) – j(1) + k(-1) = -i – j – k. We get B = -i – j – k.

8. Find the magnetic flux density when a flux of 28 units is enclosed in an area of 15cm.
a) 178.33
b) 186.67
c) 192.67
d) 124.33

Answer: b [Reason:] The total flux is the product of the magnetic flux density and the area. Total flux = B x A. To get B, put flux/area. B = 28/0.15 = 186.67 units.

9. Find the magnetic flux density B when E is given by 3sin y i + 4cos z j + ex k.
a) ∫(4sin z i – ex j – 3cos y k)dt
b) -∫(4sin z i – ex j – 3cos y k)dt
c) ∫(4sin y i – ex j + 3cos y k)dt
d) -∫(4sin y i + ex j + 3cos y k)dt

Answer: b [Reason:] We know that Curl (E) = -dB/dt. The curl of E is (4sin z i – ex j – 3cos y k). To get B, integrate the -curl(E) with respect to time to get B = -∫(4sin z i – ex j – 3cos y k)dt.

10. Find current density J when B = 50 x 10-6 units and area dS is 4 units.
a) 9.94
b) 8.97
c) 7.92
d) 10.21

Answer: a [Reason:] To get H, H = B/μ = 50 x 10-6/ 4π x 10-7 = 39.78 units. Also H = ∫ J.dS, where H = 39.78 and ∫ dS = 4. Thus J = 39.78/4 = 9.94 units.

## Set 5

1. The H quantity is analogous to which component in the following?
a) B
b) D
c) E
d) V

Answer: c [Reason:] The H quantity refers to magnetic field intensity in the magnetic field. This is analogous to the electric field intensity E in the electric field.

2. The magnetic flux density is directly proportional to the magnetic field intensity. State True/False.
a) True
b) False

Answer: a [Reason:] The magnetic field intensity is directly proportional to the magnetic field intensity for a particular material (Permeability). It is given by B = μH.

3. Ampere law states that,
a) Divergence of H is same as the flux
b) Curl of D is same as the current
c) Divergence of E is zero
d) Curl of H is same as the current density

Answer: d [Reason:] Ampere circuital law or Ampere law states that the closed integral of the magnetic field intensity is same as the current enclosed by it. It is given by Curl(H) = J.

4. Given the magnetic field is 2.4 units. Find the flux density in air(in 10-6 order).
a) 2
b) 3
c) 4
d) 5

Answer: b [Reason:] We know that B = μH. On substituting μ = 4π x 10-7 and H = 2.4, we get B = 4π x 10-7 x 2.4 = 3 x 10-6 units.

5. Find the electric field when the magnetic field is given by 2sin t in air.
a) 8π x 10-7 cos t
b) 4π x 10-7 sin t
c) -8π x 10-7 cos t
d) -4π x 10-7 sin t

Answer: a [Reason:] Given H = 2sin t. We get B = μH = 4π x 10-7 x 2sin t = 8πx10-7sin t. To get E, integrate B with respect to time, we get 8πx10-7cos t.

6. Find the height of an infinitely long conductor from point P which is carrying current of 6.28A and field intensity is 0.5 units.
a) 0.5
b) 2
c) 6.28
d) 1

Answer: b [Reason:] The magnetic field intensity of an infinitely long conductor is given by H = I/2πh. Put I = 6.28 and H = 0.5, we get h = 1/0.5 = 2 units.

7. Find the magnetic field intensity due to a solenoid of length 12cm having 30 turns and current of 1.5A.
a) 250
b) 325
c) 175
d) 375

Answer: d [Reason:] The magnetic field intensity of a solenoid is given by H = NI/L = 30 X 1.5/0.12 = 375 units.

8. Find the magnetic field intensity at the radius of 6cm of a coaxial cable with inner and outer radii are 1.5cm and 4cm respectively. The current flowing is 2A.
a) 2.73
b) 3.5
c) 0
d) 1.25

Answer: c [Reason:] The inner radius is 1.5cm and the outer radius is 4cm. It is clear that the magnetic field intensity needs to be calculated outside of the conductor ie, r>4cm. This will lead to zero, since H outside the conductor will be zero.

9. Find the magnetic field intensity of a toroid of turns 40 and radius 20cm. The current carried by the toroid be 3.25A.
a) 103.45
b) 102
c) 105.7
d) 171

Answer: a [Reason:] The magnetic field intensity of a toroid is given by H = NI/2πrm. Put N = 40, I = 3.25 and rm = 0.2, we get H = 40 x 3.25/2π x 0.2 = 103.45 units.

10. The magnetic field intensity of an infinite sheet of charge with charge density 36.5 units in air will be
a) 18.25
b) 11.25
c) 73
d) 1/36.5