Electromagnetic Engineering Module-2

1. The best definition of polarisation 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 polarisation 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 polarisation 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: Polarisation 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 polarisation 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 polarisation depends on temperature.
a) Electronic
b) Ionic
c) Orientational
d) Interfacial
Answer: c
Explanation: The electronic, ionic and interfacial polarisation depends on the atoms which are independent with respect to temperature. Only the orientational polarisation is dependent on the temperature and is inversely proportional to it.

6. Calculate the polarisation 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 polarisation 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) Polarisation
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 polarisation of a material is the
a) Product of all types of polarisation
b) Sum of all types of polarisation
c) Orientation directions of the dipoles
d) Total dipole moments in the material
Answer: b
Explanation: The total polarisation of a material is given by the sum of electronic, ionic, orientational and interfacial polarisation of the material.

9. In the given types of polarisation, which type exists in the semiconductor?
a) Electronic
b) Ionic
c) Orientational
d) Interfacial or space charge
Answer: d
Explanation: The interfacial type of polarisation 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 polarisation?
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 polarisation. They possess only electronic, ionic and orientational polarisations.

11. The given equation satisfies the Laplace equation.
V = x² + y² – z². State True/False.
a) True
b) False
Answer: a
Explanation: Grad (V) = 2xi + 2yj – 4zk. Div (Grad (V)) = Del²(V) = 2+2-4 = 0. It satisfies the Laplacian equation. Thus the statement is true.

12. In free space, the Poisson equation becomes
a) Maxwell equation
b) Ampere equation
c) Laplace equation
d) Steady state equation
Answer: c
Explanation: The Poisson equation is given by Del²(V) = -ρ/ε. In free space, the charges will be zero. Thus the equation becomes, Del²(V) = 0, which is the Laplace equation.

13. If Laplace equation satisfies, then which of the following statements will be true?
a) Potential will be zero
b) Current will be infinite
c) Resistance will be infinite
d) Voltage will be same
Answer: b
Explanation: Laplace equation satisfying implies the potential is not necessarily zero due to subsequent gradient and divergence operations following. Finally, if potential is assumed to be zero, then resistance is zero and current will be infinite.

14. Suppose the potential function is a step function. The equation that gets satisfied is
a) Laplace equation
b) Poisson equation
c) Maxwell equation
d) Ampere equation
Answer: a
Explanation: Step is a constant function. The Laplace equation Div(Grad(step)) will become zero. This is because gradient of a constant is zero and divergence of zero vector will be zero.

15. Calculate the charge density when a potential function x² + y² + z² is in air(in 10-9 order)
a) 1/6π
b) 6/2π
c) 12/6π
d) 10/8π
Answer: a
Explanation: The Poisson equation is given by Del²(V) = -ρ/ε. To find ρ, put ε = 8.854 x 10^-12 in air and Laplacian of the function is 2 + 2 + 2 = 6. Ρ = 6 x 10^-9/36π = 1/6π units.

16. The function V = e^xsin y + z does not satisfy Laplace equation. State True/False.
a) True
b) False
Answer: b
Explanation: Grad (V) = e^xsin y i + e^x cos y j + k. Div(Grad(V)) = e^xsin y – e^xsin y + 0= 0.Thus Laplacian equation Div(Grad(V)) = 0 is true.

17. Poisson equation can be derived from which of the following equations?
a) Point form of Gauss law
b) Integral form of Gauss law
c) Point form of Ampere law
d) Integral form of Ampere law
Answer: a
Explanation: The point of Gauss law is given by, Div (D)= ρv. On putting
D= ε E and E=- Grad (V) in Gauss law, we get Del² (V)= -ρ/ε, which is the Poisson equation.

18. Find the charge density from the function of flux density given by 12x – 7z.
a) 19
b) -5
c) 5
d) -19
Answer: c
Explanation: From point form of Gauss law, we get Div (D) = ρv
Div (D) = Div(12x – 7z) = 12-7 = 5, which the charge density ρv. Thus ρv = 5 units.

19. Find the electric field of a potential function given by 20 log x + y at the point (1,1,0).
a) -20 i – j
b) -i -20 j
c) i + j
d) (i + j)/20
Answer: a
Explanation: The electric field is given by E = -Grad(V). The gradient of the given function is 20i/x + j. At the point (1,1,0), we get 20i + j. The electric field E = -(20i + j) = -20i – j.

20. When a material has zero permittivity, the maximum potential that it can possess is
a) ∞
b) -∞
c) Unity
d) Zero
Answer: d
Explanation: Permittivity is zero, implies that the ability of the material to store electric charges is zero. Thus the electric field and potential of the material is also zero.

21. The lines of force are said to be
a) Real
b) Imaginary
c) Drawn to trace the direction
d) Not significant
Answer: c
Explanation: The lines drawn to trace the direction in which a positive test charge will experience force due to the main charge are called lines of force. They are not real but drawn for our interpretation.

22. Electric flux density in electric field is referred to as
a) Number of flux lines
b) Ratio of flux lines crossing a surface and the surface area
c) Direction of flux at a point
d) Flux lines per unit area
Answer: b
Explanation: Electric flux density is given by the ratio between number of flux lines crossing a surface normal to the lines and the surface area. The direction of D at a point is the direction of the flux lines at that point.

23. The electric flux density is the
a) Product of permittivity and electric field intensity
b) Product of number of flux lines and permittivity
c) Product of permeability and electric field intensity
d) Product of number of flux lines and permeability
Answer: a
Explanation: D= εE, where ε=εoεr is the permittivity of electric field and E is the electric field intensity. Thus electric flux density is the product of permittivity and electric field intensity.

24. Which of the following correctly states Gauss law?
a) Electric flux is equal to charge
b) Electric flux per unit volume is equal to charge
c) Electric field is equal to charge density
d) Electric flux per unit volume is equal to volume charge density
Answer: d
Explanation: The electric flux passing through any closed surface is equal to the total charge enclosed by that surface. In other words, electric flux per unit volume leaving a point (vanishing small volume), is equal to the volume charge density.

25. The Gaussian surface is
a) Real boundary
b) Imaginary surface
c) Tangential
d) Normal
Answer: b
Explanation: It is any physical or imaginary closed surface around a charge which satisfies the following condition: D is everywhere either normal or tangential to the surface so that D.ds becomes either Dds or 0 respectively.

26. Find the flux density of a sheet of charge density 25 units in air.
a) 25
b) 12.5
c) 6.25
d) 3.125
Answer: b
Explanation: Electric field intensity of infinite sheet of charge E = σ/2ε.
Thus D = εE = σ/2 = 25/2 = 12.5.

27. A uniform surface charge of σ = 2 μC/m², is situated at z = 2 plane. What is the value of flux density at P(1,1,1)m?
a) 10^-6
b) -10^-6
c) 10⁶
d) -10⁶
Answer: b
Explanation: The flux density of any field is independent of the position (point). D = σ/2 = 2 X 10^-6(-az)/2 = -10^-6.

28. Find the flux density of line charge of radius (cylinder is the Gaussian surface) 2m and charge density is 3.14 units?
a) 1
b) 0.75
c) 0.5
d) 0.25
Answer: d
Explanation: The electric field of a line charge is given by, E = λ/(2περ), where ρ is the radius of cylinder, which is the Gaussian surface and λ is the charge density. The density D = εE = λ/(2πρ) = 3.14/(2π X 2) = 1/4 = 0.25.

29. If the radius of a sphere is 1/(4π)m and the electric flux density is 16π units, the total flux is given by,
a) 2
b) 3
c) 4
d) 5
Answer: c
Explanation: Total flux leaving the entire surface is, ψ = 4πr²D from Gauss law. Ψ = 4π(1/16π²) X 16π = 4.

30. Find the electric field intensity of transformer oil (εr = 2 approx) with density 1/4π (in 10⁹ units)
a) 2.5
b) 3.5
c) 4.5
d) 5.5
Answer: c
Explanation: D = εE. E = (1/4π)/(2Xεo) = 4.5 X 10⁹ units.

31. Capacitor is a device used to__________
a) store electrical energy
b) vary the resistance
c) store magnetic energy
d) dissipate energy
Answer: a
Explanation: Capacitor is used to store the charge. It stores electrical energy between the plates.

32. Capacitor stores which type of energy?
a) kinetic energy
b) vibrational energy
c) potential energy
d) heat energy
Answer: c
Explanation: Capacitor store charge in between the plates. This charge is stationary so we can say capacitor store potential energy.

33. Capacitor blocks__________ after long time.
a) alternating current
b) direct current
c) both alternating and direct current
d) neither alternating nor direct current
Answer: b
Explanation: Capacitor blocks direct current at steady state and pass alternating current.

34. Why does capacitor block dc signal at steady state?
a) due to high frequency of dc signal
b) due to zero frequency of dc signal
c) capacitor doesnot pass any current at steady state
d) due to zero frequency of dc signal
Answer: d
Explanation: Frequency of dc signal is zero. So, Capacitive reactance X_C=1/2πfc becomes infinite and capacitor behaves as open circuit for dc signal. Hence, capacitor block dc signal.

35. If a parallel plate capacitor of plate area 2m² and plate separation 1m store the charge of 1.77*10^-11 C. What is the voltage across the capacitor?
a) 1V
b) 2V
c) 3V
d) 4V
Answer: a
Explanation: C=€₀A/d
On substituting values of d, A, we get C=2€₀.
Q=CV
V=1 V.

36. Which of the following is a passive device?
a) Transistor
b) Rectifier
c) Capacitor
d) Vaccuum Tubes
Answer: c
Explanation: Capacitor is a passive device as it consumes power rest all generate power so, they are active devices.

37. What is the value of capacitance of a capacitor which has a voltage of 4V and has 16C of charge?
a) 2F
b) 4F
c) 6F
d) 8F
Answer: b
Explanation: Q is directly proportional to V. The constant of proportionality in this case is C, that is, the capacitance. Hence Q=CV. From the relation, C=Q/V= 16/4=4F.

38. For which medium capacitance is high?
a) Air
b) Mica
c) Water
d) Metal
Answer: d
Explanation: Metals are assumed to have a high value of dielectric constant so they have high capacitance.

Electromagnetic Engineering Module-3

1. Biot Savart law in magnetic field is analogous to which law in electric field?
a) Gauss law
b) Faraday law
c) Coulomb’s law
d) Ampere law
Answer: c
Explanation: Biot Savart law states that the magnetic flux density H = I.dl sinθ/4πr², which is analogous to the electric field F = q1q2/4πεr², which is the Coulomb’s law.

2. Which of the following cannot be computed using the Biot Savart law?
a) Magnetic field intensity
b) Magnetic flux density
c) Electric field intensity
d) Permeability
Answer: c
Explanation: The Biot Savart law is used to calculate magnetic field intensity. Using which we can calculate flux density and permeability by the formula B = μH.

3. Find the magnetic field of a finite current element with 2A current and height 1/2π is
a) 1
b) 2
c) 1/2
d) 1/4
Answer: a
Explanation: The magnetic field due to a finite current element is given by H = I/2πh. Put I = 2 and h = 1/2π, we get H = 1 unit.

4. Calculate the magnetic field at a point on the centre of the circular conductor of radius 2m with current 8A.
a) 1
b) 2
c) 3
d) 4
Answer: b
Explanation: The magnetic field due to a point in the centre of the circular conductor is given by H = I/2a. Put I = 8A and a = 2m, we get H = 8/4 = 2 units.

5. The current element of the solenoid of turns 100, length 2m and current 0.5A is given by,
a) 100 dx
b) 200 dx
c) 25 dx
d) 50 dx
Answer: c
Explanation: The current element of the solenoid is given by NI dx/L. Put N = 100, I = 0.5 and L = 2 to get, I dx = 100 x 0.5 x dx/2 = 25 dx.

6. Find the magnetic field intensity at the centre O of a square of the sides equal to 5m and carrying 10A of current.
a) 1.2
b) 1
c) 1.6
d) 1.8
Answer: d
Explanation: The magnetic field is given by H = 4I/√2πω. Put I = 10 and ω = 5m. Thus H = 4 x 10/√2π(5) = 1.8 unit.

7. Find the magnetic flux density when a point from a finite current length element of current 0.5A and radius 100nm.
a) 0
b) 0.5
c) 1
d) 2
Answer: c
Explanation: The magnetic flux density is B = μH, where H is given by I/2πr. Put μ = 4π x 10^-7, I = 0.5 and r = 10^-7, we get B = 4π x 10^-7 x 0.5/2π x 10^-7 = 1 unit.

8. In a static magnetic field only magnetic dipoles exist. State True/False.
a) True
b) False
Answer: a
Explanation: From Gauss law for magnetic field, we get divergence of the magnetic flux density is always zero (ie, Div(B) = 0). This implies the non-existence of magnetic monopole.

9. The magnetic field intensity will be zero inside a conductor. State true/false.
a) True
b) False
Answer: b
Explanation: Electric field will be zero inside a conductor and magnetic field will be zero outside the conductor. In other words, the conductor boundary, E will be maximum and H will be minimum.

10. Find the magnetic field when a circular conductor of very high radius is subjected to a current of 12A and the point P is at the centre of the conductor.
a) 1
b) ∞
c) 0
d) -∞
Answer: c
Explanation: The magnetic field of a circular conductor with point on the centre is given by I/2a. If the radius is assumed to be infinite, then H = 12/2(∞) = 0.

11. Identify which of the following is the unit of magnetic flux density?
a) Weber
b) Weber/m
c) Tesla
d) Weber^-1
Answer: c
Explanation: The unit of magnetic flux density is weber/m². It is also called as tesla.

12. The divergence of H will be
a) 1
b) -1
c) ∞
d) 0
Answer: d
Explanation: We know that the divergence of B is zero. Also B = μH. Thus divergence of H is also zero.

13. 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
Explanation: 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.

14. 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
Explanation: 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.

15. 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
Explanation: 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.

16. If ∫ H.dL = 0, then which statement will be true?
a) E = -Grad(V)
b) B = -Grad(D)
c) H = -Grad(Vm)
d) D = -Grad(A)
Answer: c
Explanation: The given condition shows that the magnetic field intensity will be the negative gradient of the magnetic vector potential.

17. 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
Explanation: 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.

18. 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
Explanation: 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.

19. 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
Explanation: 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.

20. 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
Explanation: 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.

21. 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
Answer: c
Explanation: The electric force is given by F = qE, where q = 2C and E = 6 units. Thus we get F = 2 x 6 = 12 units.

22. 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
Answer: b
Explanation: 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.

23. 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
Answer: b
Explanation: 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.

24. 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
Answer: a
Explanation: 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.

25. 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
Answer: d
Explanation: 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.

26. 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
Answer: a
Explanation: 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.

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

28. 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
Answer: a
Explanation: 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.

29. 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
Answer: a
Explanation: 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.

30. The magnetic force impacts the energy of the field. State True/false.
a) True
b) False
Answer: a
Explanation: 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.

31. 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
Explanation: 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.

32. 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
Explanation: 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.

33. 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
Explanation: 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.

34. 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
Explanation: 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.

35. 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
Explanation: 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.

36. 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
Explanation: 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.

37. 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
Explanation: 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.

38. The Bohr magneton is given by
a) eh/2m
b) eh/2πm
c) eh/4m
d) eh/4πm
Answer: d
Explanation: 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.

39. 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
Explanation: 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.

40. 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
Explanation: 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.

41. Find the Lorentz force due to a conductor of length 2m carrying a current of 1.5A and magnetic flux density of 12 units.
a) 24
b) 36
c) 32
d) 45
Answer: b
Explanation: The Lorentz is given by the product of the current, differential length and the magnetic flux density. Put B = 12, I = 1.5 and L = 2, thus we get F = BIL = 12 x 1.5 x 2 = 36 units.

42. Calculate the flux density due to a circular conductor of radius 100nm and current 5A in air.
a) 10
b) 100
c) 0.1
d) 1
Answer: a
Explanation: The field intensity of this conductor is I/2πR and since B = μH, the flux density will be B = μI/2πR. Put I = 5 and R = 100 x 10^-9, thus we get B = 4π x 10^-7x 5/(2π x 100 x 10^-9) = 10 units.

43. The torque expression of a current carrying conductor is
a) T = BIA cos θ
b) T = BA cos θ
c) T = BIA sin θ
d) T = BA sin θ
Answer: c
Explanation: The torque is given by the product of the flux density, magnetic moment IA and the sine angle of the conductor held by the field. This gives T = BIA sin θ.

44. Find the current in a dipole with a moment of 16 units and area of 9 units.
a) 1.78
b) 2.78
c) 1.87
d) 2.34
Answer: a
Explanation: The dipole moment is given by M = IA. To get I, put M = 16 and A = 9, we get I = M/A = 16/9 = 1.78 units.

45. The expression for magnetization is given by(I-current, A-area, V-volume)
a) M = IAV
b) M = IA/V
c) M = V/IA
d) M = 1/IAV
Answer: b
Explanation: The magnetization is defined as the magnetic moment per unit volume and the magnetic moment is IA. Thus M = IA/V is the expression.

46. Find the permeability of a medium whose susceptibility is 100.
a) -100
b) 99
c) -99
d) 101
Answer: d
Explanation: The susceptibility is given by χ_m = μ_r-1. To get permeability, μr = χm + 1 = 100 + 1 = 101 units.

47. Calculate the magnetization of a material with susceptibility of 50 and field intensity of 0.25 units.
a) 12.5
b) 25
c) 75
d) 37.5
Answer: a
Explanation: The magnetization is the product of the susceptibility and the field intensity given by M = χ_mH. Put χm = 50 and H = 0.25, then M = 50 x 0.25 = 12.5 units.

48. Very small and positive susceptibility is found in
a) Ferromagnetic
b) Diamagnetic
c) Paramagnetic
d) Antiferromagnetic
Answer: c
Explanation: Paramagnetic materials are characterized by a small and positive susceptibility. The susceptibility and the temperature are directly related.

49. Which of the following materials is ferrimagnetic?
a) Fe
b) Sn
c) Fe₂O₃
d) FeCl
Answer: c
Explanation: Fe is iron and a ferromagnetic material. Sn and FeCl are not magnetic materials. The oxides of iron like ferric oxide Fe₂O₃ is said to be a ferrimagnetic material.

50. Identify the diamagnetic material.
a) Silicon
b) Germanium
c) Silver
d) Cobalt
Answer: b
Explanation: The diamagnetic materials are characterised by very small or negative susceptibility. Also the susceptibility is independent of the temperature. The material having these properties is germanium from the given options. Metals like gold and atoms with closed shells are also diamagnetic.

51. Find the correct relation between current density and magnetization.
a) J = Grad(M)
b) J = Div(M)
c) J = Curl(M)
d) M = Curl(J)
Answer: c
Explanation: The curl of the magnetization gives the magnetic field intensity theoretically. From Maxwell equation, we can correlate that with the current density (Ampere law)

52. The tangential component of the magnetic field intensity is continuous at the boundary of separation of two media. State True/False.
a) True
b) False
Answer: a
Explanation: For two medium of separation, the tangential component of the magnetic field intensity will be continuous. This is analogous to the fact that the tangential component of the electric field intensity is continuous at the boundary.

53. In air, the tangential component of flux density is continuous at the boundary. State True/False.
a) True
b) False
Answer: a
Explanation: Since the tangential component of the magnetic field intensity will be continuous and B = μH, in air, the tangential component of the flux density will also be continuous.

54. The flux density of medium 1 has a normal component of 2.4 units, then the normal component of the flux density in the medium 2 will be
a) 1.2
b) 4.8
c) 2.4
d) 0
Answer: c
Explanation: Unlike the electric fields, the magnetic flux density has normal component same in both the mediums. This gives Bn1 = Bn2.

55. The normal component of magnetic field intensity at the boundary of separation of the medium will be
a) Same
b) Different
c) Negative
d) Inverse
Answer: a
Explanation: The normal component and tangential components of the magnetic flux density will be same. This holds good for any medium.

56. The line integral of the magnetic field intensity is the
a) Current density
b) Current
c) Magnetic flux density
d) Magnetic moment
Answer: b
Explanation: The line integral of the magnetic field intensity is given by ∫H.dl. This is same as the current component. From this relation, the Ampere law can be deduced.

57. Find the magnetization of the material with susceptibility of 6 units and magnetic field intensity of 13 units.
a) 2.16
b) 6.2
c) 78
d) 1.3
Answer: c
Explanation: The magnetization is the product of the susceptibility and the magnetic field intensity. Thus M = 6 x 13 = 78 units.

58. Find the ratio of permeability of the two media when the wave is incident on the boundary at 45 degree and reflected by the boundary at 60 degree.
a) 1:1
b) √3:1
c) 1:√3
d) 1:√2
Answer: c
Explanation: From the magnetic boundary conditions, the ratio of permeability μ1/μ2 = tan θ1/tan θ2 and θ1 = 45, θ2 = 60. Thus we get μ1/μ2 = 1/√3. The ratio will be 1:√3.

59. Find the magnetic moment of a material with magnetization 5 units in a volume of 35 units.
a) 7
b) 1/7
c) 15
d) 175
Answer: d
Explanation: The magnetization is the ratio of the magnetic moment and the volume. To get moment, put M = 5 and V = 35, thus moment will be 5 x 35 = 175 units.

60. A boundary of separation between two magnetic materials is identified by which factor?
a) Change in the permeability
b) Change in permittivity
c) Change in magnetization
d) Conduction
Answer: a
Explanation: Two materials are differentiated by their permeability in case of magnetic and permittivity in case of electric. Thus at the boundary of separation, change in permeability is identified for magnetic materials.

61. The point form of Ampere law is given by
a) Curl(B) = I
b) Curl(D) = J
c) Curl(V) = I
d) Curl(H) = J
Answer: d
Explanation: Ampere law states that the line integral of H about any closed path is exactly equal to the direct current enclosed by that path. ∫ H.dl = I The point form will be Curl (H) = J.

62. The Ampere law is based on which theorem?
a) Green’s theorem
b) Gauss divergence theorem
c) Stoke’s theorem
d) Maxwell theorem
Answer: c
Explanation: The proof of the Ampere’s circuital law is obtained from Stoke’s theorem for H and J only.

63. Electric field will be maximum outside the conductor and magnetic field will be maximum inside the conductor. State True/False.
a) True
b) False
Answer: a
Explanation: At the conductor-free space boundary, electric field will be maximum and magnetic field will be minimum. This implies electric field is zero inside the conductor and increases as the radius increases and the magnetic field is zero outside the conductor and decreases as it approaches the conductor.

64. Find the magnetic flux density of a finite length conductor of radius 12cm and current 3A in air( in 10^-6 order)
a) 4
b) 5
c) 6
d) 7
Answer: b
Explanation: The magnetic field intensity is given by H = I/2πr, where I = 3A and r = 0.12. The magnetic flux density in air B = μ H, where μ = 4π x 10^-7.Thus B = 4π x 10^-7 x 3/2π x 0.12 = 5x 10^-6 units.

65. Calculate the magnetic field intensity due to a toroid of turns 50, current 2A and radius 159mm.
a) 50
b) 75
c) 100
d) 200
Answer: c
Explanation: The magnetic field intensity is given by H = NI/2πrm, where N = 50, I = 2A and rm = 1/2π. Thus H = 50 x 2/2π x 0.159 = 100 units.

66. Find the magnetic field intensity due to an infinite sheet of current 5A and charge density of 12j units in the positive y direction and the z component is above the sheet.
a) -6
b) 12k
c) 60
d) 6
Answer: d
Explanation: The magnetic field intensity when the normal component is above the sheet is Hx = 0.5 K, where K = 12. Thus we get H = 0.5 x 12 = 6 units.

67. Find the magnetic field intensity due to an infinite sheet of current 5A and charge density of 12j units in the positive y direction and the z component is below the sheet.
a) 6
b) 0
c) -6
d) 60k
Answer: c
Explanation: The magnetic intensity when the normal component is below the sheet is Hy = -0.5 K, where K = 12.Thus we get H = -0.5 x 12 = -6 units.

68. Find the current density on the conductor surface when a magnetic field H = 3cos x i + zcos x j A/m, for z>0 and zero, otherwise is applied to a perfectly conducting surface in xy plane.
a) cos x i
b) –cos x i
c) cos x j
d) –cos x j
Answer: b
Explanation: By Ampere law, Curl (H) = J. The curl of H will be i(-cos x) – j(0) + k(-z sin x) = -cos x i – zsin x k. In the xy plane, z = 0. Thus Curl(H) = J = -cos x i.

69. When the rotational path of the magnetic field intensity is zero, then the current in the path will be
a) 1
b) 0
c) ∞
d) 0.5
Answer: b
Explanation: By Ampere law, Curl(H) = J. The rotational path of H is zero, implies the curl of H is zero. This shows the current density J is also zero. The current is the product of the current density and area, which is also zero.

70. Find the magnetic field intensity when the current density is 0.5 units for an area up to 20 units.
a) 10
b) 5
c) 20
d) 40
Answer: a
Explanation: We know that ∫ H.dl = I. By Stoke’s law, we can write Curl(H) = J. In integral form, H = ∫ J.ds, where J = 0.5 and ds is defined by 20 units. Thus H = 0.5 x 20 = 10 units.

1. For time varying currents, the field or waves will be
a) Electrostatic
b) Magneto static
c) Electromagnetic
d) Electrical
Answer: c
Explanation: For stationary charges, the field is electrostatic. For steady currents, the field is magneto static. But for time varying currents, the field or waves will be electromagnetic.

2. According to Faraday’s law, EMF stands for
a) Electromagnetic field
b) Electromagnetic force
c) Electromagnetic friction
d) Electromotive force
Answer: d
Explanation: The force in any closed circuit due to the change in the flux linkage of the circuit is called as electromotive force EMF. This phenomenon is called as Faraday’s law.

3. Calculate the emf when the flux is given by 3sin t + 5cos t
a) 3cos t – 5sin t
b) -3cos t + 5sin t
c) -3sin t – 5cos t
d) 3cos t + 5sin t
Answer: b
Explanation: The electromotive force is given by Vemf = -dλ/dt. Thus Vemf = -dλ/dt = -(3cos t – 5sin t) = -3cos t + 5sin t.

4. The induced voltage will oppose the flux producing it. State True/False.
a) True
b) False
Answer: a
Explanation: According to Lenz law, the induced voltage acts in such a way that it opposes the flux producing it. This is indicated by a negative sign.

5. Calculate the emf when a coil of 100 turns is subjected to a flux rate of 0.3 tesla/sec.
a) 3
b) 30
c) -30
d) -300
Answer: c
Explanation: The induced emf is given by Vemf = -dλ/dt = -Ndψ/dt. Thus emf will be -100 x 0.3 = -30 units.

6. Find the displacement current when the flux density is given by t³ at 2 seconds.
a) 3
b) 6
c) 12
d) 27
Answer: c
Explanation: The displacement current is given by Jd = dD/dt. Thus Jd = 3t². At time t = 2, we get Jd = 3(2)²= 12A.

7. Find the force due to a current element of length 2cm and flux density of 12 tesla. The current through the element will be 5A.
a) 1 N
b) 1.2 N
c) 1.4 N
d) 1.6 N
Answer: b
Explanation: The force due to a current element is given by F = BI x L. Thus F = 12 x 5 x 0.02 = 1.2 units.

8. Which of the following statements is true?
a) E is the cross product of v and B
b) B is the cross product of v and E
c) E is the dot product of v and B
d) B is the dot product of v and E
Answer: a
Explanation: The electric field is the cross product of the velocity and the magnetic field intensity. This is given by Lorentz equation.

9. The time varying electric field E is conservative. State True/False.
a) True
b) False
Answer: b
Explanation: The time varying electric field E(t) is not a closed path. Thus the curl will be non-zero. This implies E(t) is not conservative and the statement is false.

10. When the conduction current density and displacement current density are same, the dissipation factor will be
a) Zero
b) Minimum
c) Maximum
d) Unity
Answer: d
Explanation: Dissipation factor refers to the tangent of loss angle. It is the ratio of conduction current density to displacement current density. When both are same, the loss tangent or the dissipation factor will be unity.

11. On what factors does the skin effect depend upon?
a. Cross section of the conductors.
b. Supply frequency.
c. Permeability of the conductor.
d. All of these.
ANSWER: d. All of these.

12. What are the line constants in a transmission line?
a. Resistance and series conductance only.
b. Series and shunt conductance.
c. Resistance, inductance and capacitance.
d. Resistance, inductance, capacitance and shunt conductance.
ANSWER: d. Resistance, inductance, capacitance and shunt conductance.

13. What is the cause of skin effect?
a. Supply frequency.
b. Self inductance of conductor.
c. High sensitive material in the centre.
d. Both (a) and (b)
ANSWER: d. Both (a) and (b)

14. The conductor carries more current on the surface in comparison to its core. What is this phenomenon called?
a. Corona
b. Ferranti effect.
c. Skin effect.
d. Proximity effect.
ANSWER: c. Skin effect.

15. By using which conductor is the skin effect reduced?
a. Bundled conductors.
b. Stranded conductors.
c. Hollow conductors.
d. Solid conductors.
ANSWER: b. Stranded conductors.

16. In which of the transmission systems is the skin effect observed?
a. Cable carrying dc current.
b. DC transmission line only.
c. AC transmission line only.
d. DC as well as AC transmission lines.
ANSWER: c. AC transmission line only.

17. Skin effect in a conductor is proportional to
a. (Diameter of conductor)^1/2
b. Diameter of conductor.
c. (Diameter of conductor)²
d. (Diameter of conductor)⁴
ANSWER: c. (Diameter of conductor)²

18. Skin effect is negligible for what supply frequency and for what diameter?
a. < 50 Hz and < 1 cm.
b. < 50 Hz and > 1 cm.
c. > 50 Hz and < 1 cm.
d. > 50 Hz and > 1 cm.
ANSWER: a. < 50 Hz and < 1 cm.

19. The skin effect is a phenomenon observed in
a) Insulators
b) Dielectrics
c) Conductors
d) Semiconductors
View Answer
Answer: c
Explanation: The skin of the conductor allows a certain amount of electromagnetic power to pass through it. This phenomenon is called the skin effect. This is the reason why, electromagnetic waves cannot travel inside a conductor.

20. The skin depth is measured in
a) Meter
b) Millimetre
c) Centimetre
d) Micrometer
Answer: d
Explanation: The depth to which the electromagnetic waves pass through the conductor is very small. It is measured in μm.

21. The skin depth is calculated from the amplitude of the wave. State true/false
a) True
b) False
Answer: a
Explanation: The skin depth is the measure of the depth to which the amplitude of an EM wave will reduce to 36.8% of its initial value. Thus it can be calculated if the initial amplitude is known.

21. The attenuation constant is 0.5 units. The skin depth will be
a) 0.5
b) 0.25
c) 2
d) 4
Answer: c
Explanation: The skin depth is the reciprocal of the attenuation constant. Thus δ = 1/α. On substituting for α = 0.5, we get δ = 1/0.5 = 2 units.

22. Calculate the skin depth of a conductor, having a conductivity of 200 units. The wave frequency is 10 GHz in air.
a) 355.8
b) 3.558
c) 35.58
d) 0.3558
Answer: a
Explanation: The skin depth is calculated by δ = 1/√(πfμσ), where f is the frequency, μ is the permeability and σ is the conductivity. For the given data, f = 10 x 10⁹, μ = 4π x 10^-7 in air and σ = 200, we get δ = 355.8 μm.

23. The effective skin resistance of a material with conductivity 120 and skin depth of 2μm is
a) 4.16 kilo ohm
b) 4.16 mega ohm
c) 41.6 kilo ohm
d) 41.6 mega ohm
Answer: a
Explanation: The effective skin resistance is given by R_s = 1/δσ, where δ is the skin depth and σ is the conductivity. For the given data, δ = 2 x 10^-6 and σ = 120, we get Rs = 1/(120x2x10^-6) = 4.16 kilo ohm.

24. The skin depth is used to find which parameter?
a) DC resistance
b) AC resistance
c) Permittivity
d) Potential
Answer: b
Explanation: Since the skin depth varies for different frequencies, it can be used to calculate the varying AC resistance for a material.

25. The relation between the skin depth and frequency is given by
a) Skin depth α f
b) Skin depth α 1/f
c) Skin depth α √f
d) Skin depth α 1/√f
Answer: d
Explanation: The skin depth is given by δ = 1/√(πfμσ). Thus the relation between the skin depth and the frequency is, Skin depth α 1/√f.

26. A perfect dielectric acts as a
a) Perfect transmitter
b) Perfect reflector
c) Bad transmitter
d) Bad reflector
Answer: a
Explanation: A perfect dielectric acts as a perfect transmitter. In other words, a wave incident on a perfect dielectric will transmit completely through it.

27. A perfect conductor acts as a
a) Perfect transmitter
b) Perfect reflector
c) Bad transmitter
d) Bad reflector
Answer: b
Explanation: A perfect conductor acts as a perfect reflector. In other words, a wave incident on a perfect conductor will be totally reflected back into the same medium. There will be no skin effect.

28. The resultant electric field of two components in the x and y direction having amplitudes 6 and 8 respectively is
a) 100
b) 36
c) 64
d) 10
Answer: d
Explanation: The resultant electric field of two components is given by E = √(Ex² + Ey²). For the given data, the electric field will be E = √(6²+8²) = 10 units.

29. The skin depth of the wave having a frequency of 3MHz and a velocity of 12 m/s is
a) 2
b) 3
c) 4
d) 6
Answer: c
Explanation: The velocity of a wave is the product of the frequency and the skin depth. Thus v = f.δ. To get δ, put v = 12 and f = 3MHz, we get δ = 12/(3×10⁶) = 4 μm.

30. Unit of Poynting Vector is _____________
a) Watt
b) Watt/s
c) Watt/m
d) Watt/m²
Answer: d
Explanation: Poynting Vector represents the energy moving out per unit area per unit time. Thus, it’s unit is W/m². It is a vector quantity.

31. The energy transported by the fields per unit time per unit are is called __________
a) Poynting Energy
b) Electro-magnetic energy
c) Poynting vector
d) Flux density
Answer: c
Explanation: The work done on the charges by an electromagnetic field is equal to the decrease in energy stored in the field less the energy that flowed out through the surface. This energy, that is transported by the fields, per unit are per unit time is called Poynting vector.

32. The direction of Poynting vector is perpendicular to the direction of propagation of wave.
a) True
b) False
Answer: b
Explanation: The Poynting vector is proportional to the cross product of Electric and magnetic field, E X B. Therefore, its direction is perpendicular to Electric and Magnetic waves, i.e., in the direction of propagation of wave.

33. According to the Poynting theorem, the energy flow per unit time out of any closed surface is ___________
a) Integral of S over the length of the surface
b) Integral of S over the are of the surface
c) Differential of S over the length of the surface
d) Differential of S over the are of the surface
Answer: b
Explanation: According to the pointing theorem, the energy flow per unit time out of any closed surface is the integral of S over the surface i.e., P = ∫S.da.

34. The correct expression for the Poynting vector is __________
a) S = E X B
b) S = E X B/2
c) S = E X B/μ_o
d) S = E X B/2μ_o
Answer: c
Explanation: Poynting vector can be defined as the rate at which the energy is carried out of the volume across the bounding surface. It is always in the direction of the propagation of wave, as it is perpendicular to both electric and magnetic field.

35. In free space, E(z,t) = 10³sin(ωt – βz) y^. Obtain H(z,t).
a) 1.23 sin (ωt – βz) (-y^)
b) 1.23 sin(ωt – βz) (-x^)
c) 2.65 sin (ωt – βz) (-y^)
d) 2.65 sin(ωt – βz) (-x^)
Answer: d
Explanation: In the following case, the wave is propagating in z direction. Therefore, direction of S is in z direction.
We know, S = E X B. As E is in +ve y direction, B should be in negative x direction,
Also we know, ∣E∣/∣H∣ = 376.6
Therefore, ∣H∣ = 10³/376.6
∣H∣ = 2.65
Hence, H(z,t) = 2.65 sin(wt – az) – x^.

36. Earth receives 2 cal/min/cm² of solar energy. What is the value of \mid H\mid?
a) 1.98 A/m
b) 2.45 A/m
c) 3.75 A/m
d) 4.13 A/m
Answer: a
Explanation: We know, S = E X H. Here E and H would be in perpendicular direction.
Therefore, S = ∣E∣∣H∣
Now, given S = 2 cal/min/cm²
= 1400 J/sm²
We know, ∣E∣/∣H∣ = 376.6
Therefore, ∣E∣=376.6∣H∣
Hence, 1400 = 376.6 ∣H∣2
∣H∣2= 3.71
∣H∣ = 1.98 A/m.

37. In an electromagnetic wave, the electric field of amplitude 4 V/m is oscillating. The Energy density of the wave is ___________
a) 1.41 X 10^-10 J/m³
b) 2.41 X 10^-10 J/m³
c) 3.41 X 10^-10 J/m³
d) 4.41 X 10^-10 J/m³
Answer: a
Explanation: We know, Energy density = ε E²
Here, E_max = 4 V/m and ε = 8.85 X 10^-12 C²/Nm²
Therefore, Energy Density = 8.85 X 10^-12 X 4 X 4
= 1.41 X 10^-10 J/m³.

38. The amplitude of the electric field in a parallel beam of light of intensity 3 W/m² is __________
a) 23.4 N/C
b) 34.5 N/C
c) 47.53 N/C
d) 51.45 N/C
Answer: c
Explanation: We know, Intensity, I = εE²c/2
Now, E_max = 2Iε0c−−−√
Here, I = 3 W/m², ε = 8.85 X 10^-12 C²/Nm² and c = 3 X 10⁸ m/s
We get, E_max = 47.53 N/c.

39. What will be the direction of Poynting vector?

a) x direction
b) y direction
c) z direction
d) -z direction
Answer: d
Explanation: As we can see, the electric field is in +ve Y direction while the magnetic field is in the +ve X direction. Also, S = E X B/μ. Hence, S is in the negative Z direction.

40. Poynting vector gives the energy flow per unit area per unit time through a cross-sectional area along the direction of propagation of the wave.
a) True
b) False
Answer: b
Explanation: The Poynting vector is in the direction of the propagation of the wave. However, the cross-sectional area through which the energy is flowing out is perpendicular to the direction of propagation of wave.

41. The magnitude of average value of S at a point is called as __________
a) Amplitude of radiation
b) Frequency of radiation
c) Intensity of radiation
d) Momentum flow
Answer: c
Explanation: As the frequencies of electromagnetic waves are very high, the time variation of the Poynting vector is so rapid that we have to deal with the average value. The magnitude of average value of S at a point is called the intensity of the radiation at that point.

42. The radiation pressure is given by __________
a) S
b) S_av
c) S/c
d) S_av/c
Answer: d
Explanation: The momentum is transferred per unit surface area per unit time. This momentum transfer is responsible for radiation pressure which is given by S_av/c.

43. For time varying currents, the field or waves will be
a) Electrostatic
b) Magneto static
c) Electromagnetic
d) Electrical
Answer: c
Explanation: For stationary charges, the field is electrostatic. For steady currents, the field is magneto static. But for time varying currents, the field or waves will be electromagnetic.

44. According to Faraday’s law, EMF stands for
a) Electromagnetic field
b) Electromagnetic force
c) Electromagnetic friction
d) Electromotive force
Answer: d
Explanation: The force in any closed circuit due to the change in the flux linkage of the circuit is called as electromotive force EMF. This phenomenon is called as Faraday’s law.

45. Calculate the emf when the flux is given by 3sin t + 5cos t
a) 3cos t – 5sin t
b) -3cos t + 5sin t
c) -3sin t – 5cos t
d) 3cos t + 5sin t
Answer: b
Explanation: The electromotive force is given by Vemf = -dλ/dt. Thus Vemf = -dλ/dt = -(3cos t – 5sin t) = -3cos t + 5sin t.

46. The induced voltage will oppose the flux producing it. State True/False.
a) True
b) False
Answer: a
Explanation: According to Lenz law, the induced voltage acts in such a way that it opposes the flux producing it. This is indicated by a negative sign.

47. Calculate the emf when a coil of 100 turns is subjected to a flux rate of 0.3 tesla/sec.
a) 3
b) 30
c) -30
d) -300
Answer: c
Explanation: The induced emf is given by Vemf = -dλ/dt = -Ndψ/dt. Thus emf will be -100 x 0.3 = -30 units.

48. Find the displacement current when the flux density is given by t³ at 2 seconds.
a) 3
b) 6
c) 12
d) 27
Answer: c
Explanation: The displacement current is given by Jd = dD/dt. Thus Jd = 3t². At time t = 2, we get Jd = 3(2)²= 12A.

49. Find the force due to a current element of length 2cm and flux density of 12 tesla. The current through the element will be 5A.
a) 1 N
b) 1.2 N
c) 1.4 N
d) 1.6 N
Answer: b
Explanation: The force due to a current element is given by F = BI x L. Thus F = 12 x 5 x 0.02 = 1.2 units.

50. Which of the following statements is true?
a) E is the cross product of v and B
b) B is the cross product of v and E
c) E is the dot product of v and B
d) B is the dot product of v and E
Answer: a
Explanation: The electric field is the cross product of the velocity and the magnetic field intensity. This is given by Lorentz equation.

51. The time varying electric field E is conservative. State True/False.
a) True
b) False
Answer: b
Explanation: The time varying electric field E(t) is not a closed path. Thus the curl will be non-zero. This implies E(t) is not conservative and the statement is false.

52. When the conduction current density and displacement current density are same, the dissipation factor will be
a) Zero
b) Minimum
c) Maximum
d) Unity
Answer: d
Explanation: Dissipation factor refers to the tangent of loss angle. It is the ratio of conduction current density to displacement current density. When both are same, the loss tangent or the dissipation factor will be unity.

1. Which of the following parameters is not a primary parameter?
a) Resistance
b) Attenuation constant
c) Capacitance
d) Conductance
Answer: b
Explanation: The primary parameters of a transmission line are the resistance, inductance, capacitance and conductance. The attenuation, phase and propagation constant are secondary parameters. Thus the odd one out is the attenuation constant.

2. The networks in which the R, L, C parameters are individually concentrated or lumped at discrete points in the circuit are called
a) Lumped
b) Distributed
c) Parallel
d) Paired
Answer: a
Explanation: The networks in which the R, L, C parameters are individually concentrated or lumped at discrete points in the circuit are called lumped networks. These networks can be identified definitely as representing a particular parameter. An example is the filters.

3. The lines having R, L, C distributed along the circuit are called
a) Lumped
b) Distributed
c) Parallel
d) Paired
Answer: b
Explanation: In distributed lines, the primary parameters are distributed along the circuit with each elemental length having its own values and the concentration of the individual parameters is not possible. An example is the transmission of power.

4. Which primary parameter is uniformly distributed along the length of the conductor?
a) G
b) C
c) L
d) R
Answer: d
Explanation: The resistance is a primary parameter that is uniformly distributed along the length of the conductor. It depends on the cross section area and the length of the conductor.

5. The primary parameter that is associated with the magnetic flux linkage is
a) R
b) L
c) C
d) G
Answer: b
Explanation: When the conductors carry current, the conductor will be surrounded and linked by magnetic flux. The flux linkages per ampere of current gives rise to the effect of inductance. It is denoted by L.

6. The primary parameter that is associated with the electric charges is
a) G
b) R
c) C
d) L
Answer: c
Explanation: Conductors separated by insulating dielectrics in order to store electric charges, gives rise to the capacitance effect. The capacitance is distributed in the whole conductor length.

7. The leakage current in the transmission lines is referred to as the
a) Resistance
b) Radiation
c) Conductance
d) Polarisation
Answer: c
Explanation: The dielectrics or insulators of the open wire line may not be perfect and a leakage current will flow. This leakage conductance exists between the conductors.

8. Find the receiving impedance of a transmission line having a voltage of 24V and a conduction current of 1.2A is
a) 25.2
b) 22.8
c) 28.8
d) 20
Answer: d
Explanation: By Ohm’s law, the impedance is the ratio of the voltage to the current. On substituting for V = 24 and I = 1.2, we get Z = V/I = 24/1.2 = 20 units.

9. The characteristic impedance of a transmission line with impedance and admittance of 16 and 9 respectively is
a) 25
b) 1.33
c) 7
d) 0.75
Answer: b
Explanation: The characteristic impedance is given by Zo = √(Z/Y), where Z is the impedance and Y is the admittance. On substituting for Z = 16 and Y = 9, we get the characteristic impedance as √(16/9) = 1.33 units.

10. The propagation constant of a transmission line with impedance and admittance of 9 and 16 respectively is
a) 25
b) 144
c) 12
d) 7
Answer: c
Explanation: The propagation constant is given by γ = √(ZY), where Z is given by 9 and Y is 16. On substituting the given values, the propagation constant will be γ = √(ZY) = √(9 x 16) = 12 units.

11. Find the characteristic impedance expression in terms of the inductance and capacitance parameters.
a) Zo = √(LC)
b) Zo = LC
c) Zo = √(L/C)
d) Zo = L/C
Answer: c
Explanation: The characteristic impedance is given by the square root of the ratio of the inductance to the capacitance. Thus Zo = √(L/C) is the required expression.

12. When a transmission line has a load impedance same as that of the characteristic impedance, the line is said to be
a) Parallel
b) Perpendicular
c) Polarized
d) Matched
Answer: d
Explanation: When a transmission line load impedance is same as that of the characteristic impedance, the line is said to be matched. In such cases, full transmission of power will occur, with minimal losses.

13. The wavelength of a line with a phase constant of 6.28 units is
a) 2
b) 1
c) 0.5
d) 3.14
Answer: b
Explanation: The wavelength and the phase constant are related by λ = 2π/β, where β is given as 6.28. On substituting for β, we get λ = 2π/6.28 = 1 unit.

14. The wavelength of a wave with a frequency of 6 GHz in air is
a) 50
b) 5
c) 0.5
d) 0.05
Answer: d
Explanation: The wavelength is given by the ratio of the velocity to the frequency of the wave. In air medium, the velocity can be assumed as the speed of light. On substituting for v and f, we get λ = v/f = 3×10⁸/6×10⁹ = 0.05 units.

15. The phase constant of a wave with a wavelength of 2 units is given by
a) 2
b) 3.14
c) 6.28
d) 1
Answer: b
Explanation: The phase constant is given by β = 2π/λ. On substituting for λ = 2, we get β = 2π/2 = 3.14 units.

16. The frequency of a wave travelling in a transmission line with velocity 4 x 108 and wavelength 3 units is
a) 0.75 GHz
b) 0.133 GHz
c) 7.5 GHz
d) 1.33 GHz
Answer: b
Explanation: The frequency and wavelength relation is given by f = v/λ. On substituting for v and λ, we get f = 4 x 10⁸/3 = 0.133 GHz.

17. The velocity and phase constant relation is given by
a) V = ω/β
b) V = ωβ
c) V = β/ω
d) Vωβ = 1
Answer: a
Explanation: The velocity of a wave is the ratio of the frequency in radian/second to the phase constant. It is given by V = ω/β.

18. Find the phase constant of a wave travelling with a velocity of 1.2 x 108 and a frequency of 7.5 giga radian/sec
a) 62.5
b) 26.5
c) 56.2
d) 52.6
Answer: a
Explanation: The phase constant is given by β = ω/v, from the definition of phase constant and velocity. On substituting for ω = 7.5 x 10⁹ and v = 1.2 x 10⁸, we get the phase constant β = 7.5 x 10⁹/1.2 x 10⁸ = 62.5 units.

19. The electrical length in a transmission line refers to the
a) Product of attenuation constant and length
b) Ratio of attenuation constant and length
c) Product of phase constant and length
d) Ratio of phase constant and length
Answer: a
Explanation: The electrical length in a transmission line refers to the product of the attenuation constant α and the length of the line l. It is given by αl.

20. The unit of attenuation constant is
a) Decibel
b) Bel
c) Neper
d) No unit
Answer: c
Explanation: Attenuation constant is the measure of the power loss of the wave during its transmission. It is expressed in terms of neper and 1 neper= 8.686 decibel/m.

21. The attenuation constant causes phase distortion and the phase constant causes frequency distortion. State True/False.
a) True
b) False
Answer: b
Explanation: There are always some distortions, even in the perfect transmission line. This is due to the variation of the secondary parameters. The attenuation constant causes the frequency distortion, whereas the phase constant causes the phase distortion.

22. The propagation constant of a wave with attenuation and phase constant given by 2 and 3 respectively is
a) 2 – 3j
b) 3 – 2j
c) 2 + 3j
d) 3 + 2j
Answer: c
Explanation: The propagation constant is given by γ = α + jβ. Given that α = 2 and β = 3. Thus we get the propagation constant as γ = 2 + 3j.

23. The velocity of wave in the air medium is
a) 1 x 10⁸
b) 1.5 x 10⁸
c) 3 x 10⁸
d) 1 x 10⁹
Answer: c
Explanation: The light is travelling at its fastest speed in air medium. Thus the velocity of a wave in the air medium is assumed to have the speed of light. It is given by c = 3 x 10⁸.

24. Identify the secondary parameter from the options given below.
a) Resistance
b) Conductance
c) Phase constant
d) Capacitance
Answer: c
Explanation: Primary parameters are directly observed from the circuit characteristics. Secondary parameters are derived or calculated from the primary parameters. R, L, C, G are primary parameters, whereas α, β, γ, Zo are secondary parameters.

25. Which of the following parameters does not exist in the transmission line equation?
a) R
b) Zo
c) ZL
d) Propagation constant
Answer: a
Explanation: The transmission line equation consists of secondary parameters only, which are derived from the primary parameters. The propagation constant, load impedance and the characteristic impedance are related in the transmission line equation.

26. For an infinite transmission line, the characteristic impedance is given by 50 ohm. Find the input impedance.
a) 25
b) 100
c) 2500
d) 50
Answer: d
Explanation: From the transmission line equation, the infinite line will have an input impedance same as that of the characteristic impedance. Thus Zin = Zo for l->∞. This shows that the line will be matched. The input impedance for the given case is 50 ohm.

27. The best transmission length for effective transmission of power is
a) L = λ/4
b) L = λ/8
c) L = λ/2
d) L = ∞
Answer: b
Explanation: Maximum transmission of power will occur, when the transmission line is matched. This implies that the input and characteristic impedances are the same. This condition is possible for l = λ/8 and l = ∞. Since l = ∞ is not feasible, the best option is l = λ/8.

28. When the length of the transmission line is same as that of the wavelength, then which condition holds good?
a) Zin = Zo
b) Z = Zo
c) ZL = Zo
d) Zin = ZL
Answer: d
Explanation: When the transmission line has a length same as that of the wavelength of the wave propagating through it, the input impedance will be same as the load impedance. This is the case where the wave is not amplified. The transmission line acts as a buffer.

29. The input impedance of a half wave transmission line with a load impedance of 12.5 ohm is
a) 25
b) 50
c) 6.25
d) 12.5
Answer: d
Explanation: For a half wave transmission line L = λ/2, the input and the load impedances will be the same. Thus for the given data, the input impedance will be 12.5 ohm.

30. The condition for a quarter wave transformer is
a) Zo² = Zin ZL
b) Zo = Zin ZL
c) ZL² = Zin Zo
d) Zo = Zin
Answer: a
Explanation: The quarter wave transformer represents L = λ/4. In this case, the characteristic impedance is the geometric mean of the input and load impedances. Thus Zo² = Zin ZL is the required condition.

31. Find the characteristic impedance of a quarter wave with input and load impedances given by 50 and 25 respectively.
a) 50
b) 25
c) 75
d) 35.35
Answer: d
Explanation: For a quarter line wave, the characteristic impedance is the geometric mean of input and load impedances. Thus Zo² = Zin ZL. On substituting for Zin = 50 and ZL = 25, we get Zo² = 50 x 25. The characteristic impedance will be 35.35 ohm.

32. Find the load impedance in a quarter line transformer with characteristic impedance of 75 ohm and input impedance of 200 ohm.
a) 28.125
b) 12.285
c) 52.185
d) 85.128
Answer: a
Explanation: For a quarter line wave, the characteristic impedance is the geometric mean of input and load impedances. Thus Zo² = Zin ZL. On substituting for Zo = 75 and Zin = 200, we get ZL = Zo²/Zin = 75²/200 = 28.125 ohm.

33. The reflection coefficient of a perfectly matched transmission line is
a) 1
b) -1
c) 0
d) ∞
Answer: c
Explanation: In a perfectly matched line, maximum power transfer will occur. Losses will be minimal. This implies unity transmission coefficient and zero reflection coefficient.

34. The purpose of the transmission line equation is to
a) Find primary parameters
b) Find secondary parameters
c) Find the reflection cofficient
d) Impedance matching
Answer: d
Explanation: The transmission line equation is useful in finding the length of the line which gives maximum power transfer. Thus it is useful for impedance matching.

35. The quarter wave transformer can be considered as a
a) Impedance inverter
b) Impedance doubler
c) Impedance tripler
d) Impedance quadrupler
Answer: a
Explanation: A quarter wave transformer may be considered as an impedance inverter as it can transform a low impedance into a high impedance and vice-versa.

36. Which transmission line is called as one to one transformer?
a) L = λ
b) L = λ/2
c) L = λ/4
d) L = λ/8
Answer: b
Explanation: The half wave transformer line repeats its terminating impedance. In other words, when l = λ/2, Zin = ZL. Thus it is considered to be one to one transformer.

37. The reflection coefficient of a wave with transmission coefficient 0.35 is
a) 1.35
b) 0.65
c) 0.35
d) 0.7
Answer: b
Explanation: The reflection coefficient is the reverse of the transmission coefficient. Thus T + R = 1. On substituting for T = 0.35, the reflection coefficient R will be 1 – 0.35 = 0.65 (no unit).

38. The incident wave amplitude is 24 units. Find the reflected wave amplitude if the reflection coefficient is 0.6.
a) 14.4
b) 16.6
c) 13.3
d) 11.1
Answer: a
Explanation: The reflection coefficient is the ratio of the reflected amplitude to the incident amplitude. Thus R = Er/Ei. On substituting for Ei = 24 and R = 0.6, we get Er = R Ei = 0.6 X 24 = 14.4 units.

39. Find the reflection coefficient of the wave passing through two media having intrinsic impedances of 4 and 9 respectively.
a) 0.5
b) 1
c) 0.38
d) 0.1
Answer: c
Explanation: The reflection coefficient in terms of intrinsic impedances is given by R = η2 – η1/η2 + η1. On substituting for η1 = 4 and η2 = 9, we get R = 9 – 4/9 + 4 = 5/13 = 0.38.

40. The reflection coefficient of a wave travelling through two media having permittivities 4 and 9 respectively is
a) 0
b) 0.5
c) 0.25
d) 0.2
Answer: d
Explanation: The reflection coefficient in terms of permittivity is given by R = √ε2 – √ε1/√ε2 + √ε1. On substituting for ε1 = 4 and ε2 = 9, we get R = 3 – 2/3 + 2 = 1/5 = 0.2.

41. Calculate the transmission coefficient, when the incident and transmitted amplitudes are 10 and 7 respectively.
a) 17
b) 3
c) 10/7
d) 0.7
Answer: d
Explanation: The transmission coefficient is defined as the ratio of the transmitted amplitude to the incident amplitude. Thus T = Et/Ei. On substituting for Ei = 10 and Et = 7, we get T = 7/10 = 0.7.

42. The transmission coefficient in a wave travelling through two media having intrinsic impedances of 5.5 and 1.33 is
a) 0.389
b) 0.55
c) 0.133
d) 0.42
Answer: a
Explanation: The transmission coefficient is terms of the intrinsic impedance is given by T = 2η2/η1 + η2. On substituting for η1 = 5.5 and η2 = 1.33, we get T = 2 x 1.33/1.33 + 5.5 = 2.66/6.83 = 0.389.

43. The transmission coefficient in a wave travelling through two media having permittivities 4 and 1 is
a) 1/4
b) 3/2
c) 3/4
d) 2/3
Answer: d
Explanation: The transmission coefficient in terms of the permittivity is given by T= 2√ε2/√ε1 + √ε2. On substituting for ε1 = 4 and ε2 = 1, we get T= 2(1)/1 + 2 = 2/3.

44. The reflection coefficient of a transmission line having characteristic and load impedances as 50 and 30 ohm respectively is
a) 1/4
b) 1/8
c) 1/2
d) 3/4
Answer: a
Explanation: The reflection coefficient of a transmission line is given by, R = ZL – Zo/ZL + Zo, where ZL and Zo is the load and characteristic impedances respectively. On substituting ZL = 30 and Zo = 50, the reflection coefficient R = 50 – 30/50 + 30 = 20/80 = 1/4.

45. In matched line, the transmission coefficient is
a) 0
b) 1
c) -1
d) Infinity
Answer: b
Explanation: In matched line, the maximum power is transferred from transmitter to receiver. Such reflection coefficient will be zero and transmission coefficient is unity.

46. Find the power reflected in a transmission line, when the reflection coefficient and input power are 0.45 and 18V respectively.
a) 3.645
b) 6.453
c) 4.563
d) 5.463
Answer: a
Explanation: The ratio of the reflected to incident amplitudes gives the reflection. Similarly, the ratio of reflected to incident power gives square of the reflection coefficient. Thus Prefl = R²Pinc. On substituting for R = 0.45 and Pinc = 18, we get Prefl = 0.45² x 18 = 3.645 units.

47. The transmitted power in a transmission line, when the reflection coefficient and the incident power are 0.6 and 24V respectively, is
a) 15.36
b) 51.63
c) 15.63
d) 51.36
Answer: a
Explanation: The transmitted power in terms of the reflection coefficient and the incident power is Ptr = (1-R²)Pinc, where R = 0.6 and Pinc = 24, from the given data. Thus Ptr = (1- 0.6²) x 24 = 15.36 units.

48. The reflection coefficient of a short circuit transmission line is -1. State True/False.
a) True
b) False
Answer: a
Explanation: For a short circuit line, the losses are maximum due to heavy current flow. This leads to less transmission and more attenuation. Thus the reflection coefficient is negative. R = -1 for short circuit lines.

49. Standing waves occurs due to
a) Impedance match
b) Impedance mismatch
c) Reflection
d) Transmission
Answer: b
Explanation: Impedance mismatches result in standing waves along the transmission line. It shows the variation of the wave amplitudes due to mismatching.

50. Standing wave ratio is defined as the
a) Ratio of voltage maxima to voltage minima
b) Ratio of current maxima to current minima
c) Product of voltage maxima and voltage minima
d) Product of current maxima and current minima
Answer: a
Explanation: SWR is defined as the ratio of the partial standing wave’s amplitude at an antinode (maximum) to the amplitude at a node (minimum) along the line. It is given by S = V_MAX/V_MIN.

51. Given that the reflection coefficient is 0.6. Find the SWR.
a) 2
b) 4
c) 6
d) 8
Answer: b
Explanation: The relation between reflection coefficient and SWR is given by S = 1 + R/1 – R. On substituting for R = 0.6, we get S = 1 + 0.6/1 – 0.6 = 1.6/0.4 = 4.

52. The maxima and minima voltage of the standing wave are 6 and 2 respectively. The standing wave ratio is
a) 2
b) 3
c) 1/2
d) 4
Answer: b
Explanation: The ratio of voltage maxima to voltage minima is given by the standing wave ratio SWR. Thus S = V_MAX/V_MIN. On substituting the given data, we get S = 6/2 = 3.

53. Find the standing wave ratio, when a load impedance of 250 ohm is connected to a 75 ohm line.
a) 0.3
b) 75
c) 250
d) 3.33
Answer: d
Explanation: The standing wave ratio is the ratio of the load impedance to the characteristic impedance. Thus S = ZL/Zo. On substituting for ZL = 250 and Zo = 75, we get S = 250/75 = 3.33.

54. Find the reflection coefficient of the wave with SWR of 3.5.
a) 0.55
b) 0.23
c) 0.48
d) 0.68
Answer: a
Explanation: The reflection coefficient in terms of the SWR is given by R = S – 1/S + 1. On substituting for S = 3.5, we get 3.5 – 1/3.5 + 1 = 0.55.

55. The range of the standing wave ratio is
a) 0 < S < 1
b) -1 < S < 1
c) 1 < S < ∞
d) 0 < S < ∞
Answer: c
Explanation: The standing wave ratio is given by S = 1 – R/1 + R. Thus the minimum value of S is 1. It can extend upto infinity for long lines. Thus the range is 1 < S < ∞.

56. For matched line, the standing wave ratio will be
a) 0
b) ∞
c) -1
d) 1
Answer: d
Explanation: In a matched line, maximum transmission occurs. The reflection will be zero. The standing wave ratio S = 1 – R/1 + R. For R = 0, the SWR is unity for matched line.

57. The maximum impedance of a 50 ohm transmission line with SWR of 3 is
a) 50/3
b) 3/50
c) 150
d) 450
Answer: c
Explanation: The maximum impedance is given by the product of the characteristic impedance and the SWR. Thus Z_max = S Zo. On substituting for S = 3 and Zo = 50, we get Z_MAX = 3 X 50 = 150 units.

58. The minimum impedance of a 75 ohm transmission line with a SWR of 2.5 is
a) 100
b) 50
c) 25
d) 30
Answer: d
Explanation: The minimum impedance in terms of SWR is given by Z_MIN = Zo/S. Substituting the given data for S = 2.5 and Zo = 75, we get Z_min = 75/2.5 = 30.

59. The standing wave ratio of short circuited and open circuited lines will be
a) 0
b) 1
c) -1
d) ∞
Answer: d
Explanation: The transmission line will reflect high power when it is short or circuited. This will lead to high reflection coefficient. Thus the standing wave ratio will be infinity for these extreme cases.

60. The current reflection coefficient of a line with voltage reflection coefficient of 0.65 is given by
a) 0
b) 0.65
c) -0.65
d) 0.35
Answer: c
Explanation: The current reflection coefficient at any point on the line is the negative of the voltage reflection coefficient at that point, i.e, -R. Given that the voltage reflection coefficient is 0.65, thus the current reflection coefficient is -0.65.

61. The Smith chart is a polar chart which plots
a) R vs Z
b) R vs Znorm
c) T vs Z
d) T vs Znorm
Answer: b
Explanation: The Smith chart is a frequency domain plot. It is the polar chart of the reflection coefficient R with respect to the normalised impedance Znorm.

62. The Smith chart is graphical technique used in the scenario of transmission lines. State true/false.
a) True
b) False
Answer: a
Explanation: The Smith chart is used for calculating the reflection coefficient and standing wave ratio for normalised load impedance of a transmission line.

63. The Smith chart consists of the
a) Constant R and variable X circles
b) Variable R and constant X circles
c) Constant R and constant X circles
d) Variable R and variable X circles
Answer: c
Explanation: The Smith chart consists of the constant resistance circles and the constant reactance circles. The impedances are plotted using these circles. Also stub matching can be done using the Smith chart.

64. The circles in the Smith chart pass through which point?
a) (0,1)
b) (0,-1)
c) (-1,0)
d) (1,0)
Answer: d
Explanation: All the constant resistance and reactance circles in the Smith chart pass through the (1,0) point. This is the midpoint of the Smith Chart. The resistance is unity and reactance is zero at this point.

65. Moving towards the clockwise direction in the Smith chart implies moving
a) Towards generator
b) Towards load
c) Towards stub
d) Towards waveguide
Answer: a
Explanation: On moving towards the clockwise direction in the Smith chart, we are traversing towards the generator. This is used to calculate the normalised load impedance.

66. The centre of the point having a normalised resistance of 1.2 ohm and reactance of 1.5 ohm is
a) (0.54,0)
b) (0.45,0)
c) (0.36,0)
d) (0.78,0)
Answer: a
Explanation: The centre of a point in Smith chart is given by C = (r/1+r, 0). On substituting for r = 1.2, we get centre as (1.2/1+1.2,0) = (0.54,0).

67. The normalised load impedance of the transmission line 50 ohm with a load of 30 ohm is
a) 30
b) 150
c) 5/3
d) 3/5
Answer: d
Explanation: The normalised impedance is calculated by dividing the impedance with the characteristic impedance. Given that the load impedance is 30 ohm, the normalised load impedance of the 50 ohm transmission line is 30/50 = 3/5 ohm.

68. The radius of the point having a normalised resistance of 1 ohm is
a) 1
b) 0.2
c) 0.5
d) 0.25
Answer: c
Explanation: The radius of the point with a radius r is given by R = 1/r+1. On substituting for r = 1, we get R = 1/1 + 1 = ½ = 0.5.

69. The best stub selection for the transmission line will be
a) Series open
b) Series short
c) Shunt open
d) Shunt short
Answer: d
Explanation: Normally series stubs are not preferred as modification of the stub parameters requires changing the whole stub setup. Shunt stubs enable modification with ease. Open circuited stubs are not preferred as it will radiate power like an antenna, which is undesirable. Hence shorted stubs are used.

70. The centre and radius of a line with normalised load impedance of 1 + 0.5j is
a) (1,2) and 2
b) (2,1) and 2
c) (1,2) and 1
d) (2,1) and 1
Answer: a
Explanation: The centre and radius of a line are (1, 1/x) and 1/x, where x is the reactance. Here x = 0.5, from the given data. Thus C = (1,2) and R = 2.

71. The open circuit impedance of the transmission line is given by
a) Z_OC = j Zo tan βl
b) Z_OC = – j Zo tan βl
c) Z_OC = j Zo cot βl
d) Z_OC = -j Zo cot βl
Answer: d
Explanation: The open circuit in a transmission line refers to the load side open circuited. In this case, the load impedance will be infinite. Thus the transmission line equation will be Z_OC = -j Zo cot βl.

72. The short circuit impedance of the transmission line is given by
a) Z_SC = j Zo tan βl
b) Z_SC = -j Zo tan βl
c) Z_SC = j Zo cot βl
d) Z_SC = -j Zo cot βl
Answer: a
Explanation: The short circuit in a transmission line refers to the load side shorted. In this case, the load impedance will be zero. Thus the transmission line equation will be Z_SC = j Zo tan βl.

73. In a shorted line, the reflection coefficient will be
a) 0
b) 1
c) -1
d) ∞
Answer: c
Explanation: The shorted line will absorb more power than any other line. Thus the reflection coefficient is considered to be negative.

74. The open circuit line will have a reflection coefficient of
a) 0
b) 1
c) -1
d) ∞
Answer: b
Explanation: An open circuit line has infinite output impedance. Any wave incident at the output will be completely reflected. Thus the reflection coefficient is unity.

75. The standing wave ratio in short and open circuit transmission lines will be
a) 0
b) -1
c) 1
d) ∞
Answer: d
Explanation: The reflection coefficient is 1 and -1 in open and shorted lines respectively. This value of reflection coefficient will yield infinite standing wave ratio.

76. The characteristic impedance of a line having open and short impedances of 20 and 5 respectively is
a) 20
b) 100
c) 25
d) 10
Answer: d
Explanation: The characteristic impedance is the geometric mean of the short and open circuit impedance. It is given by Zo² = Zsc Zoc. On substituting Zoc = 20 and Zsc = 5, we get Zo² = 20 X 5 = 100. Thus Zo = 10 ohm.

77. The short circuit impedance is given by 18 ohm and the characteristic impedance is 50 ohm. Find the open circuit impedance.
a) 138.8
b) 188.3
c) 388.1
d) 838.1
Answer: a
Explanation: The relation between characteristic impedance, open and short impedance is given by Zo² = Zsc Zoc. For the given values Zo = 50 and Zsc = 18, we get Zoc = 50²/18 = 138.8 units.

78. For maximum power transfer theorem to be applied to the transmission line, the reflection coefficient has to be
a) 1
b) -1
c) 0
d) ∞
Answer: c
Explanation: Maximum power transfer between the load and source is possible, only when both are matched. This will lead to no reflections. Thus the reflection coefficient will be zero.

79. Find the transmission coefficient of a 75 ohm line with load impedance of 40 ohm.
a) 0.69
b) 0.96
c) 0.31
d) 0.13
Answer: a
Explanation: The transmission coefficient in terms of the load impedance is given by T = Z_L/Z₀. On substituting for Z_L = 40 and Zo = 75, we get T = 40/75 = 0.69.

80. The standing waves for open circuit voltage and short circuit current are the same. State true/false.
a) True
b) False
Answer: a
Explanation: The open circuit voltage and short circuit current will be same for a transmission line. The phase difference is λ/8.

81. The standing waves for open circuit current and short circuit voltage are the same. State true/false.
a) true
b) false
Answer: a
Explanation: The open circuit current and short circuit voltage will be same for a transmission line. The phase difference is λ/8.

82. The standing wave ratio for the maximum power transfer in a transmission line is
a) 1:2
b) 2:1
c) -1:1
d) 1:1
Answer: d
Explanation: The load and the source has to be matched for maximum power transfer. This is indicated by the ratio of 1:1.

1. Micro strip can be fabricated using:
a) Photo lithographic process
b) Electrochemical process
c) Mechanical methods
d) None of the mentioned
Answer: a
Explanation: Microstrip lines are planar transmission lines primarily because it can be fabricated by photolithographic processes and is easily miniaturized and integrated with both passive and active microwave devices.

2. The mode of propagation in a microstrip line is:
a) Quasi TEM mode
b) TEM mode
c) TM mode
d) TE mode
Answer: a
Explanation: The exact fields of a microstrip line constitute a hybrid TM-TE wave. In most practical applications, the dielectric substrate is very thin and so the fields are generally quasi-TEM in nature.

3. Microstrip line can support a pure TEM wave.
a) True
b) False
c) Microstrip supports only TM mode
d) Microstrip supports only TE mode
Answer: b
Explanation: The modeling of electric and magnetic fields of a microstrip line constitute a hybrid TM-TE model. Because of the presence of the very thin dielectric substrate, fields are quasi-TEM in nature. They do not support a pure TEM wave.

4. The effective di electric constant of a microstrip line is:
a) Equal to one
b) Equal to the permittivity of the material
c) Cannot be predicted
d) Lies between 1 and the relative permittivity of the micro strip line
Answer: d
Explanation: The effective dielectric constant of a microstrip line is given by (∈_r + 1)/2 + (∈_r-1)/2 * 1/ (√1+12d/w). Along with the relative permittivity, the effective permittivity also depends on the effective width and thickness of the microstrip line.

5. Effective dielectric constant of a microstrip is given by:
a) (∈_r + 1)/2 + (∈_r-1)/2 * 1/ (√1+12d/w)
b) (∈_r+1)/2 + (∈_r-1)/2
c) (∈_r+1)/2 (1/√1+12d/w)
d) (∈_r + 1)/2-(∈_r-1)/2
Answer: a
Explanation: The effective dielectric constant of a microstrip line is (∈_r + 1)/2 + (∈_r-1)/2 * 1/ (√1+12d/w). This relation clearly shows that the effective permittivity is a function of various parameters of a microstrip line, the relative permittivity, effective width and the thickness of the substrate.

6. The effective dielectric constant of a micro strip line is 2.4, then the phase velocity in the micro strip line is given by:
a) 1.5*10⁸ m/s
b) 1.936*10⁸ m/s
c) 3*10⁸ m/s
d) None of the mentioned
Answer: b
Explanation: The phase velocity in a microstrip line is given by C/√∈_r. substituting the value of relative permittivity and the speed of light in vacuum, the phase velocity is 1.936*10⁸ m/s.

7. The effective di electric constant of a micro strip line with relative permittivity being equal to 2.6, with a width of 5mm and thickness equal to 8mm is given by:
a) 2.6
b) 1.97
c) 1
d) 2.43
Answer: b
Explanation: The effective dielectric constant of a microstrip line is given by (∈_r + 1)/2 + (∈_r-1)/2 * 1/ (√1+12d/w). Substituting the given values of relative permittivity, effective width, and thickness, the effective dielectric constant is 1.97.

8. If the wave number of an EM wave is 301/m in air , then the propagation constant β on a micro strip line with effective di electric constant 2.8 is:
a) 602
b) 503.669
c) 150
d) 200
Answer: b
Explanation: The propagation constant β of a microstrip line is given by k₀√∈_e. ∈_e is the effective dielectric constant. Substituting the relevant values, the effective dielectric constant is 503.669.

9. For most of the micro strip substrates:
a) Conductor loss is more significant than di electric loss
b) Di electric loss is more significant than conductor loss
c) Conductor loss is not significant
d) Di-electric loss is less significant
Answer: a
Explanation: Surface resistivity of the conductor (microstrip line) contributes to the conductor loss of a microstrip line. Hence, conductor loss is more significant in a microstrip line than dielectric loss.

10. The wave number in air for EM wave propagating on a micro strip line operating at 10GHz is given by:
a) 200
b) 211
c) 312
d) 209
Answer: d
Explanation: The wave number in air is given by the relation 2πf/C. Substituting the given value of frequency and ‘C’, the wave number obtained is 209.

11. The effective dielectric constant ∈_r for a microstrip line:
a) Varies with frequency
b) Independent of frequency
c) It is a constant for a certain material
d) Depends on the material used to make microstrip
Answer: b
Explanation: The effective dielectric constant of a microstrip line is given by (∈_r + 1)/2 + (∈_r-1)/2 * 1/ (√1+12d/w). The equation clearly indicates that the effective dielectric constant is independent of the frequency of operation, but depends only on the design parameters of a microstrip line.

12. With an increase in the operating frequency of a micro strip line, the effective di electric constant of a micro strip line:
a) Increases
b) Decreases
c) Independent of frequency
d) Depends on the material of the substrate used as the microstrip line
Answer: c
Explanation: As the relation between effective permittivity and the other parameters of a microstrip line indicate, effective dielectric constant is not a frequency dependent parameter and hence remains constant irrespective of the operation of frequency.

13. What is the use of the RFID Module?
a) Object Identification
b) To provide 3G Connectivity
c) To measure temperature
d) To measure Wi-Fi strength
Answer: a
Explanation: The RFID Module is primarily used for object identification and tracking. It’s abbreviation stands for Radio Frequency Identification Module. It’s working is mostly wireless and uses electromagnetic fields.

14. What is the role of the MISO pin in the RFID Module?
a) Master In Slave Out
b) Manage Internal Slave Output
c) Master Internal Search Optimization
d) Manage Input Slave Op
Answer: a
Explanation: The RFID Module’s MISO pin acts as Master In Slave Out while the SPI Interface is activated, and as a serial clock when I2C Interface is enabled. However, when UART Mode is activated, it assumes the role of the Serial Data Output Pin.

15. Is there an SDA pin on the RFID Module?
a) Yes
b) No
Answer: a
Explanation: The RFID Module has an SDA/SS/Rx Pin which, when the SPI Interface is enabled acts as a signal pin, when the I2C Interface is enabled acts as a Serial Data Pin, and in case of UART Mode Activation acts as a Serial Data Input pin.

16. What will happen if we supply a voltage of 25V to the Vcc of the RFID Module?
a) Damage is caused
b) Module will shut down
c) Module will not respond for the time the voltage is applied
d) Module will function normally
Answer: a
Explanation: The RFID Modules are mostly built to work on a voltage of approximately 3.3V. Any voltage lower than that and the board will not be able to power on, but however any voltage significantly above that and the board may suffer permanent damage.

17. Which frequency does the RFID Module operate in?
a) 12.98 MHz
b) 14.67 MHz
c) 19.56 MHz
d) 13.56 MHz
Answer: d
Explanation: The RFID Module operates in the 13.56 MHz frequency since it is the universal unregulated frequency for all scientific and medical research purposes. It was set as the standard international frequency to be followed by all, since it does not interfere significantly with the environment.

18. What is the maximum data rate of the RFID Module?
a) 11 Mbps
b) 1 Kbps
c) 10 Mbps
d) 11 Gbps
Answer: c
Explanation: The maximum data rate of any given device is the ability of that given device to transfer as many bits of information from one location to another in unit time. Here, the maximum data rate of the RFID Module is 10 Mbps, which means that the module can transfer a maximum of 10 Megabytes of data in 1 second.

19. What is the maximum read range of the RFID Module?
a) 2 cm
b) 1 cm
c) 10 cm
d) 5 cm
Answer: d
Explanation: The maximum read range of an RFID Module is the maximum distance in space that the receiver module can read the signal sent by the transmitter module. In order for proper working of the module during operation, it is advised that both the modules are kept within the read range for accurate performance.

20. Is there an interrupt pin on the RFID Module?
a) Yes
b) No
Answer: a
Explanation: There is an interrupt pin on the RFID Module which keeps the module alert to incoming connection requests from any RFID transmitter that approaches its maximum read range so that it can throw a subsequent software interrupt in order for the new transmitter to function in sync with the receiver.

21. What is the minimum working current internally in the RFID Module?
a) 12 mA
b) 13 mA
c) 10 mA
d) 1 mA
Answer: b
Explanation: The internal working current inside the RFID Module ranges between 13 mA and 26 mA approximately. If there is a power surge in any portion of the module, it can mean that there is a faulty component which can be dangerous for the entire module as a whole.

22. Till what voltage are the logic pins on the RFID Module resistant to?
a) 5V
b) 3.3V
c) 2V
d) 12V
Answer: a
Explanation: The logic pins on the RFID Module are resistant up to 5V despite having an operational range of 2.5V to 3.3V. This means that one does not need to connect any logic level converter circuit in between the RFID Module and any other Arduino Microcontrollers which work on 5V.