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Multiple choice question for engineering

Set 1

1. Find the correct relation between current density and magnetization.
b) J = Div(M)
c) J = Curl(M)
d) M = Curl(J)

Answer: c [Reason:] The curl of the magnetization gives the magnetic field intensity theoretically. From Maxwell equation, we can correlate that with the current density (Ampere law)

2. 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 [Reason:] 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.

3. In air, the tangential component of flux density is continuous at the boundary. State True/False.
a) True
b) False

Answer: a [Reason:] 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.

4. 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 [Reason:] Unlike the electric fields, the magnetic flux density has normal component same in both the mediums. This gives Bn1 = Bn2.

5. 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 [Reason:] The normal component and tangential components of the magnetic flux density will be same. This holds good for any medium.

6. The line integral of the magnetic field intensity is the
a) Current density
b) Current
c) Magnetic flux density
d) Magnetic moment

Answer: b [Reason:] 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.

7. 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 [Reason:] The magnetization is the product of the susceptibility and the magnetic field intensity. Thus M = 6 x 13 = 78 units.

8. 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 [Reason:] 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.

9. 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 [Reason:] 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.

10. 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 [Reason:] 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.

Set 2

1. The distance vector is obtained in
a) Cartesian coordinate system
b) Spherical coordinate system
c) Circular coordinate system
d) Space coordinate system

Answer: d [Reason:] Vector formed by connecting two points in space is distance vector. Thus, it is obtained in space coordinate system.

2. The divergence of distance vector is
a) 0
b) 3
c) 2
d) 1

Answer: b [Reason:] The distance vector of any coordinates is generally, r = xi + yj + zk. The divergence of r is 1 + 1 + 1 = 3.

3. Find a vector normal to a plane consisting of points p1(0,1,0), p2(1,0,1) and p3(0,0,1)
a) –j – k
b) –i – j
c) –i – k
d) –i – j – k

Answer: a [Reason:] Distance vector from p1 and p2 is a = i – j + k. Distance vector from p1 and p3 is b = –j + k. The vector normal to these points is a X b = -j – k.

4. The unit vector to the points p1(0,1,0), p2(1,0,1), p3(0,0,1) is
a) (-j – k)/1.414
b) (-i – k)/1.414
c) (-i – j)/1.414
d) (-i – j – k)/1.414

Answer: a [Reason:] The cross product of p1, p2, p3 is a X b = -j – k and its magnitude is 1.414. The unit normal vector is given by, (-j – k)/1.414.

5. The polar form of Cartesian coordinates is
a) Circular coordinates
b) Spherical coordinates
c) Cartesian coordinates
d) Space coordinates

Answer: a [Reason:] The radius in the polar coordinates is the Pythagorean triplet-(r,x,y).Thus it is the circular coordinates.

6. The work-electric field relation is given by
a) Volume integral
b) Surface integral
c) Line integral
d) Relation impossible

Answer: c [Reason:] The work done is given by, W = -Q ∫E dl. Thus it is line integral.

7. The distance vector can be used to compute which of the following?
a) Dot product
b) Cross product
c) Unit normal vector
d) Area

Answer: c [Reason:] The distance vector is the distance between two points on space, thus the unit normal vector is computed using the distance vector.

8. Distance and position vectors rely on field strength. State True/False.
a) True
b) False

Answer: a [Reason:] Position or distance of a vector is dependent on the field strength.

9. Find the projection of A on B. Given A = 10j + 3k and B = 4j + 5k.
a) 6
b) 6.25
c) 6.5
d) 6.75

Answer: b [Reason:] Projection of A on B = (A . B)/|B|. Thus the answer is 40/6.4= 6.25.

10. The vector product of two vectors is given by area of the parallelogram. State True/False.
a) True
b) False

Answer: a [Reason:] The vector product of two vectors is A X B = AB sin θ. n, where n is the unit normal vector to the plane given by A and B. Their magnitude is given by |A X B|, which is the area of parallelogram.

Set 3

1. Identify the advantage of using method of images.
a) Easy approach
b) Boundaries are replaced by charges
c) Boundaries are replaced by images
d) Calculation using Poisson and Laplace equation

Answer: a [Reason:] Electrostatic boundary value problems are difficult if Poisson and Laplace equation is solved directly. But method of images helps us to solve problems without the equations. This is done by replacing boundary surfaces with appropriate image charges.

2. Calculate the electric field intensity of a line charge of length 2m and potential 24V.
a) 24
b) 12
c) 0.083
d) 12.67

Answer: b [Reason:] The electric field intensity is given by the ratio of potential to distance or length. E = V/d = 24/2 = 12 V/m.

3. Calculate potential of a metal plate of charge 28C and capacitance 12 mF.
a) 3.33 kohm
b) 2.33 kohm
c) 3.33 Mohm
d) 2.33 Mohm

Answer: b [Reason:] Potential is given by V = Q/C. Put Q = 28C and C = 12 mF. We get V = 28/12 x 10-3 = 2.333 x 103 ohm.

4. Find the dissipation factor when series resistance is 5 ohm and capacitive resistance is 10 unit.
a) 2
b) 0.5
c) 1
d) 0

Answer: b [Reason:] The dissipation factor is nothing but the tangent of loss angle of loss tangent. Tan δ = Series resistance/Capacitive resistance = 5/10 = 0.5.

5. A material with zero resistivity is said to have
a) Zero conductance
b) Infinite conductance
c) Zero resistance
d) Infinite resistance

Answer: c [Reason:] Since resistivity is directly proportional to the resistance, when the resistivity is zero, resistance is also zero. Thus we get zero resistance. The option infinite conductance is also possible ideally, but it is not possible practically. As there is always some loss in the form of heat, thus infinite conductance is impossible, but a short circuit (zero resistance) is practically possible.

6. Find the energy stored by the capacitor 3F having a potential of 12V across it.
a) 432
b) 108
c) 216
d) 54

Answer: c [Reason:] The energy stored in a capacitor is given by, E = 0.5 CV2. E = 0.5 x 3 x 122 = 0.5 x 432 = 216 units.

7. By method of images, the problem can be easily calculated by replacing the boundary with which polygon?
a) Rectangle
b) Trapezoid
c) Square
d) Triangle

Answer: d [Reason:] When any field or potential needs to be calculated for either line charge or coaxial cable or concentric cylinder, the method of images uses a triangle which converts the three dimensional problem to one dimensional analysis. From this, the result can be calculated.

8. Calculate the electric field due to a surface charge of 20 units on a plate in air(in 1012 order)
a) 2.19
b) 1.12
c) 9.21
d) 2.91

Answer: b [Reason:] The electric field due to plate of charge will be E = ρs/2εo. Put ρs = 20, we get E = 20/(2 x 8.854 x 10-12) = 1.129 x 1012 units.

9. Find the electric field due to charge density of 1/18 and distance from a point P is 0.5 in air(in 109 order)
a) 0
b) 1
c) 2
d) 3

Answer: c [Reason:] The electric field for this case is given by, E = ρl/2πεd. Put ρl = 1/18 and d = 0.5. We get E = 2 x 109 units.

10. Find the total capacitances when two capacitors 2F and 5F are in series.
a) 5/7
b) 12/7
c) 2/5
d) 10/7

Answer: d [Reason:] Two capacitances in series gives C = C1C2/C1 + C2 = 2 x 5/2 + 5 = 10/7 farad.

Set 4

1. The del operator is called as
b) Curl
c) Divergence
d) Vector differential operator

Answer: d [Reason:] The Del operator is used to replace the differential terms, thus called vector differential operator in electromagnetics.

2. The relation between vector potential and field strength is given by
b) Divergence
c) Curl
d) Del operator

Answer: a [Reason:] The vector potential and field is given by, E = -Del (V).

3. The Laplacian operator is actually
c) Curl(Div V)
d) Div(Curl V)

Answer: b [Reason:] The Laplacian operator is the divergence of gradient of a vector, which is also called del2V operator.

4. The divergence of curl of a vector is zero. State True or False.
a) True
b) False

Answer: a [Reason:] The curl of a vector is the circular flow of flux. The divergence of circular flow is considered to be zero.

5. The curl of gradient of a vector is non-zero. State True or False.
a) True
b) False

Answer: b [Reason:] The differential flow of flux in a vector is a vector. The curl of this quantity will be zero.

6. Identify the correct vector identity.
a) i . i = j . j = k . k = 0
b) i X j = j X k = k X i = 1
c) Div (u X v) = v . Curl(u) – u . Curl(v)
d) i . j = j . k = k . i = 1

Answer: c [Reason:] By standard proof, Div (u X v) = v . Curl(u) – u . Curl (v).

7. A vector is said to be solenoidal when its
a) Divergence is zero
b) Divergence is unity
c) Curl is zero
d) Curl is unity

Answer: a [Reason:] When the divergence of a vector is zero, it is said to be solenoidal /divergent-free.

8. The magnetic field intensity is said to be
a) Divergent
b) Curl free
c) Solenoidal
d) Rotational

Answer: c [Reason:] By Maxwell’s equation, the magnetic field intensity is solenoidal due to the absence of magnetic monopoles.

9. A field has zero divergence and it has curls. The field is said to be
a) Divergent, rotational
b) Solenoidal, rotational
c) Solenoidal, irrotational
d) Divergent, irrotational

Answer: b [Reason:] Since the path is not divergent, it is solenoidal and the path has curl, thus rotational.

10. When a vector is irrotational, which condition holds good?
a) Stoke’s theorem gives non-zero value
b) Stoke’s theorem gives zero value
c) Divergence theorem is invalid
d) Divergence theorem is valid

Answer: b [Reason:] Stoke’ theorem is given by, ∫ A.dl = ∫ (Curl A). ds, when curl is zero(irrotational), the theorem gives zero value.

Set 5

1. The loss tangent of a perfect dielectric will be
a) Zero
b) Unity
c) Maximum
d) Minimum

Answer: d [Reason:] Dielectrics have poor conductivity. The loss tangent σ/ωε will be low in dielectrics. For perfect dielectrics, the loss tangent will be minimum.

2. In pure dielectrics, the parameter that is zero is
a) Attenuation
b) Propagation
c) Conductivity
d) Resistivity

Answer: c [Reason:] There are no free charge carriers available in a dielectric. In other words, the charge carriers are present in the valence band, which is very difficult to start to conduct. Thus conduction is low in dielectrics. For pure dielectrics, the conductivity is assumed to be zero.

3. The total permittivity of a dielectric transformer oil (relative permittivity is 2.2) will be (in order 10-11)
a) 1.94
b) 19.4
c) 0.194
d) 194

Answer: a [Reason:] The total permittivity is the product of the absolute and the relative permittivity. The absolute permittivity is 8.854 x 10-12 and the relative permittivity(in this case for transformer oil) is 2.2. Thus the total permittivity is 8.854 x 10-12 x 2.2 = 1.94 x 10-11 units.

4. The permeability of a dielectric material in air medium will be
a) Absolute permeability
b) Relative permeability
c) Product of absolute and relative permeability
d) Unity

Answer: a [Reason:] The total permeability is the product of the absolute and the relative permeability. In air medium, the relative permeability will be unity. Thus the total permeability is equal to the absolute permeability given by 4π x 10-7 units.

5. The attenuation in a good dielectric will be non- zero. State True/False.
a) True
b) False

Answer: a [Reason:] Good dielectrics attenuate the electromagnetic waves than any other material. Thus the attenuation constant of the dielectric will be non-zero, positive and large.

6. Calculate the phase constant of a dielectric with frequency 6 x 106 in air.
a) 2
b) 0.2
c) 0.02
d) 0.002

Answer: c [Reason:] The phase constant of a dielectric is given by β = ω√(με). On substituting for ω = 6 x 106 , μ = 4π x 10-7, ε = 8.854 x 10-12 in air medium, we get the phase constant as 0.02 units.

7. The frequency in rad/sec of a wave with velocity of that of light and phase constant of 20 units is (in GHz)
a) 6
b) 60
c) 600
d) 0.6

Answer: a [Reason:] The velocity of a wave is given by V = ω/β. To get ω, put v = 3 x 108 and β = 20. Thus ω = vβ = 3 x 108 x 20 = 60 x 108 = 6 GHz.

8. The relation between the speed of light, permeability and permittivity is
a) C = 1/√(με)
b) C = με
c) C = μ/ε
d) C = 1/με

Answer: a [Reason:] The standard relation between speed of light, permeability and permittivity is given by c = 1/√(με). The value in air medium is 3 x 108 m/s.

9. The phase constant of a wave with wavelength 2 units is
a) 6.28
b) 3.14
c) 0.5
d) 2

Answer: b [Reason:] The phase constant is given by β = 2π/λ. On substituting λ = 2 units, we get β = 2π/2 = π = 3.14 units.

10.The expression for intrinsic impedance is given by
a) √(με)
b) (με)
c) √(μ/ε)
d) (μ/ε)

Answer: c [Reason:] The intrinsic impedance is given by the ratio of square root of the permittivity to the permeability. Thus η = √(μ/ε) is the intrinsic impedance. In free space or air medium, the intrinsic impedance will be 120π or 377 ohms.

11.The electric and magnetic field components in the electromagnetic wave propagation are in phase. State True/False.
a) True
b) False