Electromagnetic Theory Pune University MCQs
Electromagnetic Theory Pune University MCQs
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Dot and Cross Product”.
1. When two vectors are perpendicular, their
a) Dot product is zero
b) Cross product is zero
c) Both are zero
d) Both are not necessarily zero
Answer: a
Explanation: Dot product of two perpendicular vectors is given by A.B = |a||b|cos 90, which is zero. Thus, dot product is zero and vectors are perpendicular.
2. The cross product of the vectors 3i + 4j – 5k and –i + j – 2k is,
a) 3i – 11j + 7k
b) -3i + 11j + 7k
c) -3i – 11j – 7k
d) -3i + 11j – 7k
Answer: b
Explanation: Cross product of two vectors is, A X B = i – j + k. Using the formula, the answer can be calculated.
3. Which of the following are not vector functions in Electromagnetics?
a) Gradient
b) Divergence
c) Curl
d) There is no non- vector functions in Electromagnetics
Answer: d
Explanation: Since all the coordinates in electromagnetic are space coordinates, direction and magnitude both are important. Thus all functions are vector only.
4. The work done of vectors force F and distance d, separated by angle θ can be calculated using,
a) Cross product
b) Dot product
c) Addition of two vectors
d) Cannot be calculated
Answer: b
Explanation: Force is a vector quantity, whereas distance is scalar. Work is defined as the product of force and distance, which is given by dot product.
5. Find whether the vectors are parallel, and
a) Parallel
b) Collinearly parallel
c) Not parallel
d) Data insufficient
Answer: c
Explanation: Two vectors are parallel when their cross product is zero. Since their cross product is 4i + 2j – 6k , the vectors are not parallel.
6. Lorentz force is based on,
a) Dot product
b) Cross product
c) Both dot and cross product
d) Independent of both
Answer: b
Explanation: Lorentz force is given by, F = q .Thus cross product is the answer.
7. Electromagnetic forces are defined by
a) Fleming’s right hand rule
b) Fleming’s left hand rule
c) Faraday’s law
d) Ampere law
Answer: b
Explanation: The three left hand fingers denote electric field, magnetic field and wave propagation in free space, analogous to force, magnetic field and current respectively in any conductor.
8. The dot product of two vectors is a scalar. The cross product of two vectors is a vector. State True/False.
a) True
b) False
Answer: a
Explanation: Dot product is an algebraic operation that takes two equal length sequences and returns a scalar. Cross product is a binary operation that calculates area of two vectors, thus vector quantity.
9. Which of the Pythagorean Theorem is valid in Electromagnetics?
a) |dot product| + |dot product| = 1
b) |cross product| – |cross product| = 1
c) |dot product| 2 + |cross product| 2 = 1
d) |dot product| + |cross product| = 0
Answer: c
Explanation: Option |dot product| 2 + |cross product| 2 = 1 gives |cos| 2 + |sin| 2 = 1, which is the right answer.
Answer: c
Explanation: Cross product of dot product of two vectors is a vector value.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Position and Distance Vectors”.
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
Explanation: 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
Explanation: 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, p2 and p3
a) –j – k
b) –i – j
c) –i – k
d) –i – j – k
Answer: a
Explanation: 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, p2, p3 is
a) /1.414
b) /1.414
c) /1.414
d) /1.414
Answer: a
Explanation: 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, /1.414.
5. The polar form of Cartesian coordinates is
a) Circular coordinates
b) Spherical coordinates
c) Cartesian coordinates
d) Space coordinates
Answer: a
Explanation: The radius in the polar coordinates is the Pythagorean triplet-.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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: Projection of A on B = /|B|. Thus the answer is 40/6.4= 6.25.
Answer: a
Explanation: 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.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Vector Properties”.
1. The del operator is called as
a) Gradient
b) Curl
c) Divergence
d) Vector differential operator
Answer: d
Explanation: 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
a) Gradient
b) Divergence
c) Curl
d) Del operator
Answer: a
Explanation: The vector potential and field is given by, E = -Del .
3. The Laplacian operator is actually
a) Grad
b) Div
c) Curl
d) Div
Answer: b
Explanation: The Laplacian operator is the divergence of gradient of a vector, which is also called del 2 V operator.
4. The divergence of curl of a vector is zero. State True or False.
a) True
b) False
Answer: a
Explanation: 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
Explanation: 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 = v . Curl – u . Curl
d) i . j = j . k = k . i = 1
Answer: c
Explanation: By standard proof, Div = v . Curl – u . Curl .
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
Explanation: 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
Explanation: 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
Explanation: Since the path is not divergent, it is solenoidal and the path has curl, thus rotational.
Answer: b
Explanation: Stoke’ theorem is given by, ∫ A.dl = ∫ . ds, when curl is zero, the theorem gives zero value.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Cylindrical Coordinate System”.
1. The Cartesian system is also called as
a) Circular coordinate system
b) Rectangular coordinate system
c) Spherical coordinate system
d) Space coordinate system
Answer: b
Explanation: The other name for Cartesian is rectangular system, which is given by .
2. The volume of a parallelepiped in Cartesian is
a) dV = dx dy dz
b) dV = dx dy
c) dV = dy dz
d) dV = dx dz
Answer: a
Explanation: The volume of a parallelepiped is given by product of differential length, breadth and height.
3. A charge is placed in a square container. The position of the charge with respect to the origin can be found by
a) Spherical system
b) Circular system
c) Cartesian system
d) Space coordinate system
Answer: c
Explanation: Since the container possesses dimensions of a square , it can be found by Cartesian system.
4. The scalar factor of Cartesian system is unity. State True/False.
a) True
b) False
Answer: a
Explanation: The range of Cartesian system is one to infinity. Thus the minimum scalar value of the system is unity.
5. The angular separation between the vectors A = 4i + 3j + 5k and B = i – 2j + 2k is
a) 65.8
b) 66.8
c) 67.8
d) 68.8
Answer: c
Explanation: The dot product the vector is 8. Angle of separation is cos θ = 8/ = 0.377 and θ = cos -1 = 67.8.
6. The Cartesian coordinates can be related to cylindrical coordinates and spherical coordinates. State True/False.
a) True
b) False
Answer: a
Explanation: All the coordinate systems are inter-convertible and all the vector operations are applicable to it.
7. Transform the vector A = 3i – 2j – 4k at P to cylindrical coordinates
a) -3.6j – 4k
b) -3.6j + 4k
c) 3.6j – 4k
d) 3.6j + 4k
Answer: a
Explanation: Convert the Cartesian form to cylindrical form by formula and substitute the points to get -3.6j – 4k.
8. The spherical equivalent of the vector B = yi + j located at is given by
a)
b)
c)
d)
Answer: d
Explanation: Substitute the points in the vector and convert the Cartesian to cylindrical form to get radius as 7, plane angle1 as 64.62 and plane angle2 as -71.57.
9. Which of the following criteria is used to choose a coordinate system?
a) Distance
b) Intensity
c) Magnitude
d) Geometry
Answer: d
Explanation: The coordinate system is chosen based on the geometry of the given problem. From a point charge +Q, the electric field spreads in all 360 degrees. The calculation of electric field in this case will be spherical system.
Answer: a
Explanation: The order of vector transformation and point substitution will not affect the result, only when the vector is a constant.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Spherical Coordinate System”.
1. The cylindrical coordinate system is also referred to as
a) Cartesian system
b) Circular system
c) Spherical system
d) Space system
Answer: b
Explanation: The cylindrical coordinates is also called as circular system and is used for systems with circular dimensions.
2. Transform the vector B=yi+j located at point into cylindrical coordinates.
a)
b)
c)
d)
Answer: a
Explanation: ρ = √(x 2 +y 2 ) = √40 = 6.325
Φ = tan -1 = tan -1 = -71.57
z = 3.
3. Cylindrical systems have the following scalar values respectively
a) 1, ρ ,1
b) 1, 1, 1
c) 0,1,0
d) 1,0,0
Answer: a
Explanation: The range of radius is one to unity, that of plane angle is one to 360 degree and that of z plane is one to infinity. Thus the minimum scalar factor has to be 1, ρ , 1.
4. A charge located at point p ⁰ is said to be in which coordinate system?
a) Cartesian system
b) Cylindrical system
c) Spherical system
d) Space system
Answer: b
Explanation: The cylindrical system is of the form , which relates the point given in the question.
5. Cylindrical system is employed in waveguides. State True/False.
a) True
b) False
Answer: a
Explanation: Cylindrical systems are employed in circular waveguides, whereas Cartesian systems are employed in rectangular waveguides.
6. The pressure inside a piston cylinder is a variable of
a) Radius
b) Plane angle
c) Z plane distance
d) Constant, not a variable
Answer: c
Explanation: Pressure varies up and down in a cylinder due to suction. Thus it is dependent on the z plane distance of the cylinder.
7. Charges filled inside a cylindrical will possess flux in which direction?
a) Upwards
b) Downwards
c) Laterally outwards
d) Inwards
Answer: c
Explanation: The flux due to the charges will act outside the cylinder. Since the cylinder possesses curved surfaces, it will flow laterally outwards.
8. Rectangular waveguides dominate the circular waveguides. Find the reason.
a) Low cut-off frequency
b) Easy to design
c) More wave propagation
d) The statement is false
Answer: b
Explanation: Due to linear design, the desired dimensions can be easily constructed using rectangular waveguides than circular ones.
9. Transform the spherical system B = i + j + k into cylindrical form at
a) 2.467i + j + 1.167k
b) 2.467i – j + 1.167k
c) 2.467i – j – 1.167k
d) 2.467i + j – 1.167k
Answer: a
Explanation: The equivalent cylindrical form is given by,
B = i + j + k
At , r = √(5 2 +-2 2 ) = √29
sin θ = 5/√29 and cos θ = -2/√29
Thus, B = 2.467i + j + 1.167k.
Answer: d
Explanation: ρ = √(x 2 +y 2 ) = √13 = 3.61
Φ = tan -1 = 56.31
z = 1
Thus, A = .
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Spherical Coordinate System”.
1. Convert the point from Cartesian to spherical coordinates
a) ⁰⁰
b) ⁰⁰
c) ⁰⁰
d) ⁰⁰
Answer: a
Explanation: r = √(x 2 +y 2 +z 2 ) = √50 = 7.07
Θ = cos -1 = cos -1 = 45⁰
Φ = tan -1 = tan -1 = 53⁰.
2. Example of spherical system in the following is
a) Charge in space
b) Charge in box
c) Charge in dielectric
d) Uncharged system
Answer: a
Explanation: From a point charge +Q, the electric field spreads in all 360 degrees. The calculation of electric field in this case will be spherical system. Thus it is charge in the space.
3. Spherical systems are employed in waveguides. State True/False
a) True
b) False
Answer: b
Explanation: There is no waveguide designed spherically to avoid absorption, rather than propagation.
4. Choose which of following condition is not required for a waveguide to exist.
a) The dimensions should be in accordance with desired frequency
b) Cut-off frequency should be minimum 6GHz
c) The shape should be spherical
d) No specific condition is required for waveguide design
Answer: c
Explanation: A waveguide need not be spherical, it has to be rectangular or circular, as it violates the propagation of the wave.
5. Find the spherical coordinates of A
a) ⁰⁰
b) ⁰⁰
c) ⁰⁰
d) ⁰⁰
Answer: b
Explanation: r = √(x 2 +y 2 +z 2 ) = √14 = 3.74
Θ = cos -1 = cos -1 = 105.5⁰
Φ = tan -1 = tan -1 = 56.31⁰.
6. Find the Cartesian coordinates of B⁰⁰
a)
b)
c)
d)
Answer: b
Explanation: x = r sin θ cos φ = 4 sin25⁰ cos 120⁰ = -0.845
y = r sin θ sin φ = 4 sin 25⁰ sin 120⁰ = 1.462
z = r cos θ = 4 cos 25⁰ = 3.625.
7. The area of sphere can be computed from the sphere volume. State True/False.
a) True
b) False
Answer: a
Explanation: On double integrating the differential volume, the area can be computed for a sphere.
8. Given B= i+ j+k in spherical coordinates. Find Cartesian points at
a) -2i + j
b) 2i + k
c) i + 2j
d) –i – 2k
Answer: a
Explanation: r = √(x 2 +y 2 +z 2 ) = √25 = 5
Θ = cos -1 = 1
Φ = tan -1 = tan -1
Thus, B = -2i + j.
9. The scalar factor of spherical coordinates is
a) 1, r, r sin θ
b) 1, r, r
c) r, r, 1
d) r, 1, r
Answer: a
Explanation: The radius varies from unity to infinity, the plane angle from zero to 360 ⁰ and the z plane from .
Answer: b
Explanation: r = √(x 2 +y 2 +z 2 ) = 3.74
Θ = cos -1 = cos -1 = 36.7⁰
Φ = tan -1 = tan -1 = 63.4⁰
A = i + j + k
On substituting r, θ, φ, A = -3.197i + 2.393j – 4.472k.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Gradient”.
1. Gradient of a function is a constant. State True/False.
a) True
b) False
Answer: b
Explanation: Gradient of any scalar function may be defined as a vector. The vector’s magnitude and direction are those of the maximum space rate of change of φ.
2. The mathematical perception of the gradient is said to be
a) Tangent
b) Chord
c) Slope
d) Arc
Answer: c
Explanation: The gradient is the rate of change of space of flux in electromagnetics. This is analogous to the slope in mathematics.
3. Divergence of gradient of a vector function is equivalent to
a) Laplacian operation
b) Curl operation
c) Double gradient operation
d) Null vector
Answer: a
Explanation: Div = 2 V, which is the Laplacian operation. A function is said to be harmonic in nature, when its Laplacian tends to zero.
4. The gradient of xi + yj + zk is
a) 0
b) 1
c) 2
d) 3
Answer: d
Explanation: Grad = 1 + 1 + 1 = 3. In other words, the gradient of any position vector is 3.
5. Find the gradient of t = x 2 y+ e z at the point p
a) i + 10j + 0.135k
b) 10i + j + 0.135k
c) i + 0.135j + 10k
d) 10i + 0.135j + k
Answer: b
Explanation: Grad = 2xy i + x 2 j + e z k. On substituting p, we get 10i + j + 0.135k.
6. Curl of gradient of a vector is
a) Unity
b) Zero
c) Null vector
d) Depends on the constants of the vector
Answer: c
Explanation: Gradient of any function leads to a vector. Similarly curl of that vector gives another vector, which is always zero for all constants of the vector. A zero value in vector is always termed as null vector.
7. Find the gradient of the function given by, x 2 + y 2 + z 2 at
a) i + j + k
b) 2i + 2j + 2k
c) 2xi + 2yj + 2zk
d) 4xi + 2yj + 4zk
Answer: b
Explanation: Grad(x 2 +y 2 +z 2 ) = 2xi + 2yj + 2zk. Put x=1, y=1, z=1, the gradient will be 2i + 2j + 2k.
8. The gradient can be replaced by which of the following?
a) Maxwell equation
b) Volume integral
c) Differential equation
d) Surface integral
Answer: c
Explanation: Since gradient is the maximum space rate of change of flux, it can be replaced by differential equations.
9. When gradient of a function is zero, the function lies parallel to the x-axis. State True/False.
a) True
b) False
Answer: a
Explanation: Gradient of a function is zero implies slope is zero. When slope is zero, the function will be parallel to x-axis or y value is constant.
Answer: a
Explanation: Grad gives partial differentiation of sin x+ cos y with respect to x and partial differentiation of sin x + cos y with respect to y and similarly with respect to z. This gives cos x i – sin y j + 0 k = cos x i – sin y j.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Divergence”.
1. The divergence of a vector is a scalar. State True/False.
a) True
b) False
Answer: a
Explanation: Divergence can be computed only for a vector. Since it is the measure of outward flow of flux from a small closed surface as the volume shrinks to zero, the result will be directionless .
2. The divergence concept can be illustrated using Pascal’s law. State True/False.
a) True
b) False
Answer: a
Explanation: Consider the illustration of Pascal’s law, wherein a ball is pricked with holes all over its body. After water is filled in it and pressure is applied on it, the water flows out the holes uniformly. This is analogous to the flux flowing outside a closed surface as the volume reduces.
3. Compute the divergence of the vector xi + yj + zk.
a) 0
b) 1
c) 2
d) 3
Answer: d
Explanation: The vector given is a position vector. The divergence of any position vector is always 3.
4. Find the divergence of the vector yi + zj + xk.
a) -1
b) 0
c) 1
d) 3
Answer: b
Explanation: Div = Dx + Dy + Dz, which is zero. Here D refers to partial differentiation.
5. Given D = e -x sin y i – e -x cos y j
Find divergence of D.
a) 3
b) 2
c) 1
d) 0
Answer: d
Explanation: Div = Dx(e -x sin y) + Dy(-e -x cos y ) = -e -x sin y + e -x sin y = 0.
6. Find the divergence of the vector F= xe -x i + y j – xz k
a) (1 + e -x )
b) (1 + e -x )
c)
d)
Answer: a
Explanation: Div = Dx(xe -x ) + Dy+Dz = -xe -x + e -x + 1 – x =
e -x + = (1 + e -x ).
7. Determine the divergence of F = 30 i + 2xy j + 5xz 2 k at and state the nature of the field.
a) 1, solenoidal
b) 0, solenoidal
c) 1, divergent
d) 0, divergent
Answer: b
Explanation: Div = Dx + Dy + Dz(5xz 2 ) = 0 + 2x + 10xz = 2x + 10xz
Divergence at will give zero. As the divergence is zero, field is solenoidal.
Alternate/Shortcut: Without calculation, we can easily choose option “0, solenoidal”, as by theory when the divergence is zero, the vector is solenoidal. “0, solenoidal” is the only one which is satisfying this condition.
8. Find whether the vector is solenoidal, E = yz i + xz j + xy k
a) Yes, solenoidal
b) No, non-solenoidal
c) Solenoidal with negative divergence
d) Variable divergence
Answer: a
Explanation: Div = Dx + Dy + Dz = 0. The divergence is zero, thus vector is divergentless or solenoidal.
9. Find the divergence of the field, P = x 2 yz i + xz k
a) xyz + 2x
b) 2xyz + x
c) xyz + 2z
d) 2xyz + z
Answer: b
Explanation: Div = Dx(x 2 yz) + Dy + Dz = 2xyz + x, which is 2xyz + x. For different values of x, y, z the divergence of the field varies.
Answer: c
Explanation: Since the vector field does not diverge , the divergence is zero. Also, the path does not possess any curls, so the field is irrotational.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Line Integral”.
1. An electric field is given as E = 6y 2 z i + 12xyz j + 6xy 2 k. An incremental path is given by dl = -3 i + 5 j – 2 k mm. The work done in moving a 2mC charge along the path if the location of the path is at p is
a) 0.64
b) 0.72
c) 0.78
d) 0.80
Answer: b
Explanation: W = -Q E.dl
W = -2 X 10 -3 X (6y 2 z i + 12xyz j + 6xy 2 k) .
At p, W = -2 X 10 -3 = 0.72 J.
2. The integral form of potential and field relation is given by line integral. State True/False
a) True
b) False
Answer: a
Explanation: Vab = -∫ E.dl is the relation between potential and field. It is clear that it is given by line integral.
3. If V = 2x 2 y – 5z, find its electric field at point
a) 47.905
b) 57.905
c) 67.905
d) 77.905
Answer: b
Explanation: E = -Grad = -4xy i – 2×2 j + 5k
At , E = 48 i – 32 j + 5 k, |E| = √3353 = 57.905 units.
4. Find the potential between two points p and q with E = 40xy i + 20x 2 j + 2 k
a) 104
b) 105
c) 106
d) 107
Answer: c
Explanation: V = -∫ E.dl = -∫ (40xy dx + 20x 2 dy + 2 dz) , from q to p.
On integrating, we get 106 volts.
5. Find the potential between a and b. Given E = (-6y/x 2 )i + j + 5 k.
a) -8.014
b) -8.114
c) -8.214
d) -8.314
Answer: c
Explanation: V = -∫ E.dl = -∫ dx + dy + 5 dz, from b to a.
On integrating, we get -8.214 volts.
6. The potential of a uniformly charged line with density λ is given by,
λ/ ln. State True/False.
a) True
b) False
Answer: a
Explanation: The electric field intensity is given by, E = λ/
Vab = -∫ E.dr = -∫ λ/. On integrating from b to a, we get λ/ ln.
7. A field in which a test charge around any closed surface in static path is zero is called
a) Solenoidal
b) Rotational
c) Irrotational
d) Conservative
Answer: d
Explanation: Work done in moving a charge in a closed path is zero. It is expressed as, ∫ E.dl = 0. The field having this property is called conservative or lamellar field.
8. The potential in a lamellar field is
a) 1
b) 0
c) -1
d) ∞
Answer: b
Explanation: Work done in a lamellar field is zero. ∫ E.dl = 0,thus ∑V = 0. The potential will be zero.
9. Line integral is used to calculate
a) Force
b) Area
c) Volume
d) Length
Answer: d
Explanation: Length is a linear quantity, whereas area is two dimensional and volume is three dimensional. Thus single or line integral can be used to find length in general.
Answer: a
Explanation: dw = ei dt = Li di, W = L∫ i.di
Energy E = 0.5LI 2 = 0.5 X 0.1 X 2 2 = 0.2 Joule.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Surface Integral”.
1. Gauss law for electric field uses surface integral. State True/False
a) True
b) False
Answer: a
Explanation: Gauss law states that the electric flux passing through any closed surface is equal to the total charge enclosed by the surface. Thus the charge is defined as a surface integral.
2. Surface integral is used to compute
a) Surface
b) Area
c) Volume
d) density
Answer: b
Explanation: Surface integral is used to compute area, which is the product of two quantities length and breadth. Thus it is two dimensional integral.
3. Coulomb’s law can be derived from Gauss law. State True/ False
a) True
b) False
Answer: a
Explanation: Gauss law, Q = ∫∫D.ds
By considering area of a sphere, ds = r 2 sin θ dθ dφ.
On integrating, we get Q = 4πr 2 D and D = εE, where E = F/Q.
Thus, we get Coulomb’s law F = Q1 x Q2/4∏εR 2 .
4. Evaluate Gauss law for D = 5r 2 /4 i in spherical coordinates with r = 4m and θ = π/2.
a) 600
b) 599.8
c) 588.9
d) 577.8
Answer: c
Explanation: ∫∫ ( 5r 2 /4) . (r 2 sin θ dθ dφ), which is the integral to be evaluated.
Put r = 4m and substitute θ = 0→ π/4 and φ = 0→ 2π, the integral evaluates to 588.9.
5. Compute the Gauss law for D= 10ρ 3 /4 i, in cylindrical coordinates with ρ= 4m, z=0 and z=5.
a) 6100 π
b) 6200 π
c) 6300 π
d) 6400 π
Answer: d
Explanation: ∫∫ D.ds = ∫∫ (10ρ 3 /4)., which is the integral to be evaluated. Put ρ = 4m, z = 0→5 and φ = 0→2π, the integral evaluates to 6400π.
6. Compute divergence theorem for D= 5r 2 /4 i in spherical coordinates between r=1 and r=2.
a) 80π
b) 5π
c) 75π
d) 85π
Answer: c
Explanation: ∫∫ ( 5r 2 /4) . (r 2 sin θ dθ dφ), which is the integral to be evaluated. Since it is double integral, we need to keep only two variables and one constant compulsorily. Evaluate it as two integrals keeping r = 1 for the first integral and r = 2 for the second integral, with φ = 0→2π and θ = 0→ π. The first integral value is 80π, whereas second integral gives -5π. On summing both integrals, we get 75π.
7. Find the value of divergence theorem for A = xy 2 i + y 3 j + y 2 z k for a cuboid given by 0<x<1, 0<y<1 and 0<z<1.
a) 1
b) 4/3
c) 5/3
d) 2
Answer: c
Explanation: A cuboid has six faces. ∫∫A.ds = ∫∫Ax=0 dy dz + ∫∫Ax=1 dy dz + ∫∫Ay=0 dx dz + ∫∫Ay=1 dx dz + ∫∫Az=0 dy dx + ∫∫Az=1 dy dx. Substituting A and integrating we get + 1 + = 5/3.
8. The ultimate result of the divergence theorem evaluates which one of the following?
a) Field intensity
b) Field density
c) Potential
d) Charge and flux
Answer: d
Explanation: Gauss law states that the electric flux passing through any closed surface is equal to the total charge enclosed by the surface. Thus, it is given by, ψ = ∫∫ D.ds= Q, where the divergence theorem computes the charge and flux, which are both the same.
9. Find the value of divergence theorem for the field D = 2xy i + x 2 j for the rectangular parallelepiped given by x = 0 and 1, y = 0 and 2, z = 0 and 3.
a) 10
b) 12
c) 14
d) 16
Answer: b
Explanation: While evaluating surface integral, there has to be two variables and one constant compulsorily. ∫∫D.ds = ∫∫Dx=0 dy dz + ∫∫Dx=1 dy dz + ∫∫Dy=0 dx dz + ∫∫Dy=2 dx dz + ∫∫Dz=0 dy dx + ∫∫Dz=3 dy dx. Put D in equation, the integral value we get is 12.
Answer: c
Explanation: By Gauss law, ψ = ∫∫ D.ds, where ds = dydz i at the x-plane. Put x = 3 and integrate at -1<y<2 and 0<z<4, we get 12 X 3 = 36.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Volume Integral”.
1. The divergence theorem converts
a) Line to surface integral
b) Surface to volume integral
c) Volume to line integral
d) Surface to line integral
Answer: b
Explanation: The divergence theorem is given by, ∫∫ D.ds = ∫∫∫ Div dv. It is clear that it converts surface integral to volume integral.
2. The triple integral is used to compute volume. State True/False
a) True
b) False
Answer: a
Explanation: The triple integral, as the name suggests integrates the function/quantity three times. This gives volume which is the product of three independent quantities.
3. The volume integral is three dimensional. State True/False
a) True
b) False
Answer: a
Explanation: Volume integral integrates the independent quantities by three times. Thus it is said to be three dimensional integral or triple integral.
4. Find the charged enclosed by a sphere of charge density ρ and radius a.
a) ρ (4πa 2 )
b) ρ(4πa 3 /3)
c) ρ(2πa 2 )
d) ρ(2πa 3 /3)
Answer: b
Explanation: The charge enclosed by the sphere is Q = ∫∫∫ ρ dv.
Where, dv = r 2 sin θ dr dθ dφ and on integrating with r = 0->a, φ = 0->2π and θ = 0->π, we get Q = ρ(4πa 3 /3).
5. Evaluate Gauss law for D = 5r 2 /4 i in spherical coordinates with r = 4m and θ = π/2 as volume integral.
a) 600
b) 588.9
c) 577.8
d) 599.7
Answer: b
Explanation: ∫∫ D.ds = ∫∫∫ Div dv, where RHS needs to be computed.
The divergence of D given is, Div = 5r and dv = r 2 sin θ dr dθ dφ. On integrating, r = 0->4, φ = 0->2π and θ = 0->π/4, we get Q = 588.9.
6. Compute divergence theorem for D = 5r 2 /4 i in spherical coordinates between r = 1 and r = 2 in volume integral.
a) 80 π
b) 5 π
c) 75 π
d) 85 π
Answer: c
Explanation: D.ds = ∫∫∫ Div dv, where RHS needs to be computed.
The divergence of D given is, Div = 5r and dv = r 2 sin θ dr dθ dφ. On integrating, r = 1->2, φ = 0->2π and θ = 0->π, we get Q = 75 π.
7. Compute the Gauss law for D = 10ρ 3 /4 i, in cylindrical coordinates with ρ = 4m, z = 0 and z = 5, hence find charge using volume integral.
a) 6100 π
b) 6200 π
c) 6300 π
d) 6400 π
Answer: d
Explanation: Q = D.ds = ∫∫∫ Div dv, where RHS needs to be computed.
The divergence of D given is, Div = 10 ρ 2 and dv = ρ dρ dφ dz. On integrating, ρ = 0->4, φ = 0->2π and z = 0->5, we get Q = 6400 π.
8. Using volume integral, which quantity can be calculated?
a) area of cube
b) area of cuboid
c) volume of cube
d) distance of vector
Answer: c
Explanation: The volume integral gives the volume of a vector in a region. Thus volume of a cube can be computed.
9. Compute the charge enclosed by a cube of 2m each edge centered at the origin and with the edges parallel to the axes. Given D = 10y 3 /3 j.
a) 20
b) 70/3
c) 80/3
d) 30
Answer: c
Explanation: Div = 10y 2
∫∫∫Div dv = ∫∫∫ 10y 2 dx dy dz. On integrating, x = -1->1, y = -1->1 and z = -1->1, we get Q = 80/3.
Answer: b
Explanation: Div = 2y
∫∫∫Div dv = ∫∫∫ 2y dx dy dz. On integrating, x = 0->1, y = 0->2 and z = 0->3, we get Q = 12.
This set of Electromagnetic Theory Interview Questions and Answers focuses on “Laplacian Operator”.
1. The point form of Gauss law is given by, Div = ρv
State True/False.
a) True
b) False
Answer: a
Explanation: The integral form of Gauss law is ∫∫∫ ρv dv = V. Thus differential or point form will be Div = ρv.
2. If a function is said to be harmonic, then
a) Curl = 0
b) Div = 0
c) Div = 0
d) Grad = 0
Answer: c
Explanation: Though option Curl = 0 & Div = 0 are also correct, for harmonic fields, the Laplacian of electric potential is zero. Now, Laplacian refers to Div, which is zero for harmonic fields.
3. The Poisson equation cannot be determined from Laplace equation. State True/False.
a) True
b) False
Answer: b
Explanation: The Poisson equation is a general case for Laplace equation. If volume charge density exists for a field, then 2V= -ρv/ε, which is called Poisson equation.
4. Given the potential V = 25 sin θ, in free space, determine whether V satisfies Laplace’s equation.
a) Yes
b) No
c) Data sufficient
d) Potential is not defined
Answer: a
Explanation: 2 V = 0
2 V = 2 , which is not equal to zero. Thus the field does not satisfy Laplace equation.
5. If a potential V is 2V at x = 1mm and is zero at x=0 and volume charge density is -106εo, constant throughout the free space region between x = 0 and x = 1mm. Calculate V at x = 0.5mm.
a) 0.875
b) 0.675
c) 0.475
d) 0.275
Answer: d
Explanation: Del 2 = -ρv/εo= +106
On integrating twice with respect to x, V = 106. (x 2 /2) + C1x + C2.
Substitute the boundary conditions, x = 0, V = 0 and x = 1mm, V = 2V in V,
C1 = 1500 and C2 = 0. At x = 0.5mm, we get, V = 0.875V.
6. Find the Laplace equation value of the following potential field
V = x 2 – y 2 + z 2
a) 0
b) 2
c) 4
d) 6
Answer: b
Explanation: V = 2x – 2y + 2z
2 V = 2 – 2 + 2= 2, which is non zero value. Thus it doesn’t satisfy Laplace equation.
7. Find the Laplace equation value of the following potential field
V = ρ cosφ + z
a) 0
b) 1
c) 2
d) 3
Answer: a
Explanation: 2 = – + 0
= 0, this satisfies Laplace equation. The value is 0.
8. Find the Laplace equation value of the following potential field
V = r cos θ + φ
a) 3
b) 2
c) 1
d) 0
Answer: d
Explanation: 2 = – + 0
= 0, this satisfies Laplace equation. This value is 0.
9. The Laplacian operator cannot be used in which one the following?
a) Two dimensional heat equation
b) Two dimensional wave equation
c) Poisson equation
d) Maxwell equation
Answer: d
Explanation: Poisson equation, two-dimensional heat and wave equations are general cases of Laplacian equation. Maxwell equation uses only divergence and curl, which is first order differential equation, whereas Laplacian operator is second order differential equation. Thus Maxwell equation will not employ Laplacian operator.
Answer: d
Explanation: A field satisfying the Laplace equation is termed as harmonic field.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Stoke’s Theorem”.
1. Find the value of Stoke’s theorem for y i + z j + x k.
a) i + j
b) j + k
c) i + j + k
d) –i – j – k
Answer: d
Explanation: The curl of y i + z j + x k is i – j + k =
-i –j –k. Since the curl is zero, the value of Stoke’s theorem is zero. The function is said to be irrotational.
2. The Stoke’s theorem uses which of the following operation?
a) Divergence
b) Gradient
c) Curl
d) Laplacian
Answer: c
Explanation: ∫A.dl = ∫∫ Curl .ds is the expression for Stoke’s theorem. It is clear that the theorem uses curl operation.
3. Which of the following theorem convert line integral to surface integral?
a) Gauss divergence and Stoke’s theorem
b) Stoke’s theorem only
c) Green’ s theorem only
d) Stoke’s and Green’s theorem
Answer: d
Explanation: The Stoke’s theorem is given by ∫A.dl = ∫∫ Curl .ds. Green’s theorem is given by, ∫ F dx + G dy = ∫∫ dx dy. It is clear that both the theorems convert line to surface integral.
4. Find the value of Stoke’s theorem for A = x i + y j + z k. The state of the function will be
a) Solenoidal
b) Divergent
c) Rotational
d) Curl free
Answer: d
Explanation: Since curl is required, we need not bother about divergence property. The curl of the function will be i – j + k = 0. The curl is zero, thus the function is said to be irrotational or curl free.
5. The Stoke’s theorem can be used to find which of the following?
a) Area enclosed by a function in the given region
b) Volume enclosed by a function in the given region
c) Linear distance
d) Curl of the function
Answer: a
Explanation: It states that the line integral of a function gives the surface area of the function enclosed by the given region. This is computed using the double integral of the curl of the function.
6. The energy stored in an inductor 2H and current 4A is
a) 4
b) 8
c) 12
d) 16
Answer: d
Explanation: From Stoke’s theorem, we can calculate energy stored in an inductor as 0.5Li 2 . E = 0.5 X 2 X 4 2 = 16 units.
7. The voltage of a capacitor 12F with a rating of 2J energy is
a) 0.57
b) 5.7
c) 57
d) 570
Answer: a
Explanation: We can compute the energy stored in a capacitor from Stoke’s theorem as 0.5Cv 2 . Thus given energy is 0.5 X 12 X v 2 . We get v = 0.57 volts.
8. Find the power, given energy E = 2J and current density J = x 2 varies from x = 0 and x = 1.
a) 1/3
b) 2/3
c) 1
d) 4/3
Answer: b
Explanation: From Stoke’s theorem, we can calculate P = E X I = ∫ E. J ds
= 2∫ x 2 dx as x = 0->1. We get P = 2/3 units.
9. The conductivity of a material with current density 1 unit and electric field 200 μV is
a) 2000
b) 3000
c) 4000
d) 5000
Answer: d
Explanation: The current density is given by, J = σE. To find conductivity, σ = J/E = 1/200 X 10 -6 = 5000.
Answer: c
Explanation: Resistance calculated from Ohm’s law and Stoke’s theorem will be R = ρL/A. To get resistivity, ρ = RA/L = 200 X 20/10 = 400.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Green’s Theorem”.
1. Mathematically, the functions in Green’s theorem will be
a) Continuous derivatives
b) Discrete derivatives
c) Continuous partial derivatives
d) Discrete partial derivatives
Answer: c
Explanation: The Green’s theorem states that if L and M are functions of in an open region containing D and having continuous partial derivatives then,
∫ = ∫∫dx dy, with path taken anticlockwise.
2. Find the value of Green’s theorem for F = x 2 and G = y 2 is
a) 0
b) 1
c) 2
d) 3
Answer: a
Explanation: ∫∫dx dy = ∫∫dx dy = 0. The value of Green’s theorem gives zero for the functions given.
3. Which of the following is not an application of Green’s theorem?
a) Solving two dimensional flow integrals
b) Area surveying
c) Volume of plane figures
d) Centroid of plane figures
Answer: c
Explanation: In physics, Green’s theorem is used to find the two dimensional flow integrals. In plane geometry, it is used to find the area and centroid of plane figures.
4. The path traversal in calculating the Green’s theorem is
a) Clockwise
b) Anticlockwise
c) Inwards
d) Outwards
Answer: b
Explanation: The Green’s theorem calculates the area traversed by the functions in the region in the anticlockwise direction. This converts the line integral to surface integral.
5. Calculate the Green’s value for the functions F = y 2 and G = x 2 for the region x = 1 and y = 2 from origin.
a) 0
b) 2
c) -2
d) 1
Answer: c
Explanation: ∫∫dx dy = ∫∫dx dy. On integrating for x = 0->1 and y = 0->2, we get Green’s value as -2.
6. If two functions A and B are discrete, their Green’s value for a region of circle of radius a in the positive quadrant is
a) ∞
b) -∞
c) 0
d) Does not exist
Answer: d
Explanation: Green’s theorem is valid only for continuous functions. Since the given functions are discrete, the theorem is invalid or does not exist.
7. Applications of Green’s theorem are meant to be in
a) One dimensional
b) Two dimensional
c) Three dimensional
d) Four dimensional
Answer: b
Explanation: Since Green’s theorem converts line integral to surface integral, we get the value as two dimensional. In other words the functions are variable with respect to x,y, which is two dimensional.
8. The Green’s theorem can be related to which of the following theorems mathematically?
a) Gauss divergence theorem
b) Stoke’s theorem
c) Euler’s theorem
d) Leibnitz’s theorem
Answer: b
Explanation: The Green’s theorem is a special case of the Kelvin- Stokes theorem, when applied to a region in the x-y plane. It is a widely used theorem in mathematics and physics.
9. The Shoelace formula is a shortcut for the Green’s theorem. State True/False.
a) True
b) False
Answer: a
Explanation: The Shoelace theorem is used to find the area of polygon using cross multiples. This can be verified by dividing the polygon into triangles. It is a special case of Green’s theorem.
Answer: d
Explanation: dM/dx = cos x and dL/dy = -sin y
∫∫dx dy = ∫∫ dx dy. On integrating with x = 0->90 and y = 0->90, we get area of right angled triangle as -180 units . Since area cannot be negative, we take 180 units.
This set of Electromagnetic Theory Questions and Answers for Freshers focuses on “Gauss Divergence Theorem”.
1. Gauss theorem uses which of the following operations?
a) Gradient
b) Curl
c) Divergence
d) Laplacian
Answer: c
Explanation: The Gauss divergence theorem uses divergence operator to convert surface to volume integral. It is used to calculate the volume of the function enclosing the region given.
2. Evaluate the surface integral ∫∫ . dS, where S is the sphere given by x 2 + y 2 + z 2 = 9.
a) 120π
b) 180π
c) 240π
d) 300π
Answer: b
Explanation: We could parameterise surface and find surface integral, but it is wise to use divergence theorem to get faster results. The divergence theorem is given by ∫∫ F.dS = ∫∫∫ Div .dV
Div = 3 + 2 = 5. Now the volume integral will be ∫∫∫ 5.dV, where dV is the volume of the sphere 4πr 3 /3 and r = 3units.Thus we get 180π.
3. The Gauss divergence theorem converts
a) line to surface integral
b) line to volume integral
c) surface to line integral
d) surface to volume integral
Answer: d
Explanation: The divergence theorem for a function F is given by ∫∫ F.dS = ∫∫∫ Div .dV. Thus it converts surface to volume integral.
4. The divergence theorem for a surface consisting of a sphere is computed in which coordinate system?
a) Cartesian
b) Cylindrical
c) Spherical
d) Depends on the function
Answer: d
Explanation: Seeing the surface as sphere, we would immediately choose spherical system, but it is wrong. The divergence operation is performed in that coordinate system in which the function belongs to. It is independent of the surface region.
5. Find the Gauss value for a position vector in Cartesian system from the origin to one unit in three dimensions.
a) 0
b) 3
c) -3
d) 1
Answer: b
Explanation: The position vector in Cartesian system is given by R = x i + y j + z k. Div = 1 + 1 + 1 = 3. By divergence theorem, ∫∫∫3.dV, where V is a cube with x = 0->1, y = 0->1 and z = 0->1. On integrating, we get 3 units.
6. The divergence theorem value for the function x 2 + y 2 + z 2 at a distance of one unit from the origin is
a) 0
b) 1
c) 2
d) 3
Answer: d
Explanation: Div = 2x + 2y + 2z. The triple integral of the divergence of the function is ∫∫∫dx dy dz, where x = 0->1, y = 0->1 and z = 0->1. On integrating, we get 3 units.
7. If a function is described by F = (3x + z, y 2 − sin x 2 z, xz + ye x5 ), then the divergence theorem value in the region 0<x<1, 0<y<3 and 0<z<2 will be
a) 13
b) 26
c) 39
d) 51
Answer: c
Explanation: Div = 3 + 2y + x. By divergence theorem, the triple integral of Div F in the region is ∫∫∫ dx dy dz. On integrating from x = 0->1, y = 0->3 and z = 0->2, we get 39 units.
8. Find the divergence theorem value for the function given by (e z , sin x, y 2 )
a) 1
b) 0
c) -1
d) 2
Answer: b
Explanation: Since the divergence of the function is zero, the triple integral leads to zero. The Gauss theorem gives zero value.
9. For a function given by F = 4x i + 7y j +z k, the divergence theorem evaluates to which of the values given, if the surface considered is a cone of radius 1/2π m and height 4π 2 m.
a) 1
b) 2
c) 3
d) 4
Answer: b
Explanation: Div = 4 + 7 + 1 = 12. The divergence theorem gives ∫∫∫.dV, where dV is the volume of the cone πr 3 h/3, where r = 1/2π m and h = 4π 2 m. On substituting the radius and height in the triple integral, we get 2 units.
Answer: a
Explanation: The divergence theorem is given by, ∫∫ F.dS = ∫∫∫ Div .dV, for a function F. If the function is solenoidal, its divergence will be zero. Thus the theorem computes to zero.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Coulomb law”.
1. Coulomb is the unit of which quantity?
a) Field strength
b) Charge
c) Permittivity
d) Force
Answer: b
Explanation: The standard unit of charge is Coulomb. One coulomb is defined as the 1 Newton of force applied on 1 unit of electric field.
2. Coulomb law is employed in
a) Electrostatics
b) Magnetostatics
c) Electromagnetics
d) Maxwell theory
Answer: a
Explanation: Coulomb law is applied to static charges. It states that force between any two point charges is proportional to the product of the charges and inversely proportional to square of the distance between them. Thus it is employed in electrostatics.
3. Find the force between 2C and -1C separated by a distance 1m in air.
a) 18 X 10 6
b) -18 X 10 6
c) 18 X 10 -6
d) -18 X 10 -6
Answer: b
Explanation: F = q1q2/(4∏εor 2 ) = -2 X 9/(10 -9 X 12) = -18 X 10 9 .
4. Two charges 1C and -4C exists in air. What is the direction of force?
a) Away from 1C
b) Away from -4C
c) From 1C to -4C
d) From -4C to 1C
Answer: c
Explanation: Since the charges are unlike, the force will be attractive. Thus the force directs from 1C to -4C.
5. Find the force of interaction between 60 stat coulomb and 37.5 stat coulomb spaced 7.5cm apart in transformer oil in 10 -4 N,
a) 8.15
b) 5.18
c) 1.518
d) 1.815
Answer: d
Explanation: 1 stat coulomb = 1/(3 X 10 9 ) C
F = (1.998 X 1.2488 X 10 -16 )/(4∏ X 8.854 X 10 -12 X 2.2 X (7.5 X 10 -2 ) 2 ) = 1.815 X 10 -4 N.
6. Find the force between two charges when they are brought in contact and separated by 4cm apart, charges are 2nC and -1nC, in μN.
a) 1.44
b) 2.44
c) 1.404
d) 2.404
Answer: c
Explanation: Before the charges are brought into contact, F = 11.234 μN.
After charges are brought into contact and then separated, charge on each sphere is, /2 = 0.5nC
On calculating the force with q1 = q2 = 0.5nC, F = 1.404μN.
7. The Coulomb law is an implication of which law?
a) Ampere law
b) Gauss law
c) Biot Savart law
d) Lenz law
Answer: b
Explanation: The Coulomb law can be formulated from the Gauss law, using the divergence theorem. Thus it is an implication of Gauss law.
8. Two small diameter 10gm dielectric balls can slide freely on a vertical channel. Each carry a negative charge of 1μC. Find the separation between the balls if the lower ball is restrained from moving.
a) 0.5
b) 0.4
c) 0.3
d) 0.2
Answer: c
Explanation: F = mg = 10 X 10 -3 X 9.81 = 9.81 X 10 -2 N.
On calculating r by substituting charges, we get r = 0.3m.
9. A charge of 2 X 10 -7 C is acted upon by a force of 0.1N. Determine the distance to the other charge of 4.5 X 10 -7 C, both the charges are in vacuum.
a) 0.03
b) 0.05
c) 0.07
d) 0.09
Answer: d
Explanation: F = q1q2/(4∏εor 2 ) , substituting q1, q2 and F, r2 = q1q2/ =
We get r = 0.09m.
Answer: b
Explanation: The force of two charges with respect with each other is given by F1 and F2. Thus F1 + F2 = 0 and F1 = -F2.
This set of Electromagnetic Theory Interview Questions and Answers for freshers focuses on “Electric Field Intensity”.
1. The electric field intensity is defined as
a) Force per unit charge
b) Force on a test charge
c) Force per unit charge on a test charge
d) Product of force and charge
Answer: c
Explanation: The electric field intensity is the force per unit charge on a test charge, i.e, q1 = 1C. E = F/Q = Q/.
2. Find the force on a charge 2C in a field 1V/m.
a) 0
b) 1
c) 2
d) 3
Answer: c
Explanation: Force is the product of charge and electric field.
F = q X E = 2 X 1 = 2 N.
3. Find the electric field intensity of two charges 2C and -1C separated by a distance 1m in air.
a) 18 X 10 9
b) 9 X 10 9
c) 36 X 10 9
d) -18 X 10 9
Answer: b
Explanation: F = q1q2/ = -2 X 9/(10 -9 X 12) = -18 X 10 9
E = F/q = 18 X 10 9 /2 = 9 X 10 9 .
4. What is the electric field intensity at a distance of 20cm from a charge 2 X 10 -6 C in vacuum?
a) 250,000
b) 350,000
c) 450,000
d) 550,000
Answer: c
Explanation: E = Q/
= (2 X 10 -6 )/(4∏ X εo X 0.2 2 ) = 450,000 V/m.
5. Determine the charge that produces an electric field strength of 40 V/cm at a distance of 30cm in vacuum(in 10 -8 C)
a) 4
b) 2
c) 8
d) 6
Answer: a
Explanation: E = Q/ (4∏εor 2 )
Q = (4000 X 0.3 2 )/ (9 X 10 9 ) = 4 X 10 -8 C.
6. The field intensity of a charge defines the impact of the charge on a test charge placed at a distance. It is maximum at d = 0cm and minimises as d increases. State True/False
a) True
b) False
Answer: a
Explanation: If a test charge +q is situated at a distance r from Q, the test charge will experience a repulsive force directed radially outward from Q. Since electric field is inversely proportional to distance, thus the statement is true.
7. Electric field of an infinitely long conductor of charge density λ, is given by E = λ/.aN. State True/False.
a) True
b) False
Answer: a
Explanation: The electric field intensity of an infinitely long conductor is given by, E = λ/.i + j
For an infinitely long conductor, α = 0. E = λ/. = λ/.aN.
8. Electric field intensity due to infinite sheet of charge σ is
a) Zero
b) Unity
c) σ/ε
d) σ/2ε
Answer: d
Explanation: E = σ/2ε., where α = h/(√(h 2 +a 2 ))
Here, h is the distance of the sheet from point P and a is the radius of the sheet. For infinite sheet, α = 90. Thus E = σ/2ε.
9. For a test charge placed at infinity, the electric field will be
a) Unity
b) +∞
c) Zero
d) -∞
Answer: c
Explanation: E = Q/ (4∏εor 2 )
When distance d is infinity, the electric field will be zero, E= 0.
Answer: c
Explanation: In an electromagnetic wave, the electric field and magnetic field will be perpendicular to each other. Both of these fields will be perpendicular to the wave propagation.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Electric Field Density”.
1. 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.
2. 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.
3. 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.
4. 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 , is equal to the volume charge density.
5. 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.
6. 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.
7. A uniform surface charge of σ = 2 μC/m 2 , is situated at z = 2 plane. What is the value of flux density at Pm?
a) 10 -6
b) -10 -6
c) 10 6
d) -10 6
Answer: b
Explanation: The flux density of any field is independent of the position . D = σ/2 = 2 X 10 -6 /2 = -10 -6 .
8. Find the flux density of line charge of radius 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/, where ρ is the radius of cylinder, which is the Gaussian surface and λ is the charge density. The density D = εE = λ/ = 3.14/ = 1/4 = 0.25.
9. If the radius of a sphere is 1/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 = 4πr 2 D from Gauss law. Ψ = 4π(1/16π 2 ) X 16π = 4.
Answer: c
Explanation: D = εE. E = / = 4.5 X 10 9 units.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Electric Potential”.
1. Potential difference is the work done in moving a unit positive charge from one point to another in an electric field. State True/False.
a) True
b) False
Answer: a
Explanation: The electric potential is the ratio of work done to the charge. Also it is the work done in moving a unit positive charge from infinity to a point in an electric field.
2. A point charge 2nC is located at origin. What is the potential at ?
a) 12
b) 14
c) 16
d) 18
Answer: d
Explanation: V = Q/, where r = 1m
V = (2 X 10 -9 )/ = 18 volts.
3. Six equal point charges Q = 10nC are located at 2,3,4,5,6,7m. Find the potential at origin.
a) 140.35
b) 141.35
c) 142.35
d) 143.35
Answer: d
Explanation: V = ∑Q/r = (10 X 10 -9 /4πεo)
= 143.35 volts.
4. A point charge 0.4nC is located at . Find the potential differences between m and m due to the charge.
a) 2.5
b) 2.6
c) 2.7
d) 2.8
Answer: c
Explanation: Vab = + , where rA and rB are position vectors rA = 1m and rB = 4m. Thus Vab = 2.7 volts.
5. Find the potential of V = 60sin θ/r 2 at P
a) 5.774
b) 6.774
c) 7.774
d) 8.774
Answer: a
Explanation: V = 60sin θ/r 2 , put r = 3m, θ = 60 and φ = 25, V = 60 sin 60/3 2 = 5.774 volts.
6. Given E = 40xyi + 20x 2 j + 2k. Calculate the potential between two points and .
a) 105
b) 106
c) 107
d) 108
Answer: b
Explanation: V = -∫ E.dl = -∫ (40xy dx + 20x 2 dy + 2 dz), from to , we get Vpq on integrating from Q to P. Vpq = 106 volts.
7. The potential difference in an open circuit is
a) Zero
b) Unity
c) Infinity
d) Circuit does not exist open
Answer: c
Explanation: In an open circuit no current exists due to non-existence of loops. Also voltage/potential will be infinity in an open circuit.
8. The potential taken between two points across a resistor will be
a) Positive
b) Negative
c) Zero
d) Infinity
Answer: b
Explanation: The resistor will absorb power and dissipate it in the form of heat energy. The potential between two points across a resistor will be negative.
9. What is the potential difference between 10sinθcosφ/r 2 at A and B?
a) 2.386
b) 3.386
c) 4.386
d) 5.386
Answer: c
Explanation: Potential at A, Va = 10sin30cos20/1 2 = 4.6985 and Potential at B, Vb = 10sin90cos60/4 2 = 0.3125. Potential difference between A and B is, Vab = 4.6985 – 0.3125 = 4.386 volts.
Answer: b
Explanation: In any ac circuit, the voltage measured will not be exact maximum. In order to normalise, we assume the instantaneous voltage at any point be 70.7% of the peak value, which is called the root mean square voltage.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Gauss Law”.
1. Divergence theorem is based on
a) Gauss law
b) Stoke’s law
c) Ampere law
d) Lenz law
Answer: a
Explanation: The divergence theorem relates surface integral and volume integral. Div = ρv, which is Gauss’s law.
2. The Gaussian surface for a line charge will be
a) Sphere
b) Cylinder
c) Cube
d) Cuboid
Answer: b
Explanation: A line charge can be visualized as a rod of electric charges. The three dimensional imaginary enclosed surface of a rod can be a cylinder.
3. The Gaussian surface for a point charge will be
a) Cube
b) Cylinder
c) Sphere
d) Cuboid
Answer: c
Explanation: A point charge is single dimensional. The three dimensional imaginary enclosed surface of a point charge will be sphere.
4. A circular disc of radius 5m with a surface charge density ρs = 10sinφ is enclosed by surface. What is the net flux crossing the surface?
a) 3
b) 2
c) 1
d) 0
Answer: d
Explanation: Q = ∫ ρsds = ∫∫ 10sinφ rdrdφ, on integrating with r = 0->5 and φ = 0->2π, we get Q = ψ = 0.
5. The total charge of a surface with densities 1,2,…,10 is
a) 11
b) 33
c) 55
d) 77
Answer: c
Explanation: Q = ∫∫D.ds. Since the data is discrete, the total charge will be summation of 1,2,…,10,i.e, 1+2+…+10 = 10/2 = 55.
6. The work done by a charge of 10μC with a potential 4.386 is
a) 32.86
b) 43.86
c) 54.68
d) 65.68
Answer: b
Explanation: By Gauss law principles, W = Q X V = 10 X 10 -6 X 4.386 = 43.86 X 10 -6 joule.
7. The potential of a coaxial cylinder with charge density 1 unit , inner radius 1m and outer cylinder 2m is (in 10 9 )
a) 12.74
b) 13.47
c) 12.47
d) 13.74
Answer: c
Explanation: The potential of a coaxial cylinder will be ρl ln/2πε, where ρl = 1, b = 2m and a = 1m. We get V = 12.47 X 10 9 volts.
8. Find the potential due to a charged ring of density 2 units with radius 2m and the point at which potential is measured is at a distance of 1m from the ring.
a) 18π
b) 24π
c) 36π
d) 72π
Answer: d
Explanation: The potential due to a charged ring is given by λa/2εr, where a = 2m and r = 1m. We get V = 72π volts.
9. Gauss law cannot be used to find which of the following quantity?
a) Electric field intensity
b) Electric flux density
c) Charge
d) Permittivity
Answer: d
Explanation: Permittivity is constant for a particular material. It cannot be determined from Gauss law, whereas the remaining options can be computed from Gauss law.
Answer: b
Explanation: The divergence of magnetic flux density is always zero. This is called Gauss law for magnetic fields. It implies the non-existence of magnetic monopoles in any magnetic field.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Applications of Gauss Law”.
1. Gauss law can be used to compute which of the following?
a) Permittivity
b) Permeability
c) Radius of Gaussian surface
d) Electric potential
Answer: c
Explanation: Gauss law relates the electric flux density and the charge density. Thus it can be used to compute radius of the Gaussian surface. Permittivity and permeability are constants for a particular material.
2. Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ = -3 at R = 5m. Find the flux density at R = 1m.
a) 0
b) 1
c) 2
d) 3
Answer: a
Explanation: Since 1m does not enclose any cylinder , the charge density and charge becomes zero according to Gauss law. Thus flux density is also zero.
3. Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ = -3 at R = 5m. Find the flux density at R = 3m.
a) 3
b) 10/3
c) 11/3
d) 4
Answer: b
Explanation: The radius is 3m, hence it will enclose one Gaussian cylinder of R = 2m.
By Gauss law, ψ = Q
D = σ, D = σ, Thus D = 10/3 units.
4. Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ =-3 at R = 5m. Find the flux density at R = 4.5m.
a) 4/4.5
b) 3/4.5
c) 2/4.5
d) 1/4.5
Answer: c
Explanation: The Gaussian cylinder of R = 4.5m encloses sum of charges of two cylinders .
By Gauss law, ψ = Q
D = σ, D = Q1 + Q2 = σ1 + σ2, here σ1 = 5 and σ2 = -2. We get D = 2/4.5 units.
5. Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ = -3 at R = 5m. Find the flux density at R = 6m.
a) 17/6
b) -17/6
c) 13/6
d) -13/6
Answer: d
Explanation: The radius R = 6m encloses all the three Gaussian cylinders.
By Gauss law, ψ = Q
D = σ, D = Q1 + Q2 + Q3 = σ1 + σ2 + σ3, here σ1 = 5, σ2 = -2 and σ3 = -3. We get D = -13/6 units.
6. Gauss law can be evaluated in which coordinate system?
a) Cartesian
b) Cylinder
c) Spherical
d) Depends on the Gaussian surface
Answer: d
Explanation: The Gauss law exists for all materials. Depending on the Gaussian surface of the material, we take the coordinate systems accordingly. Suppose if the material is a coaxial cable, the Gaussian surface is in the form of cylinder. Thus we take Cylinder/Circular coordinate system.
7. Gauss law cannot be expressed in which of the following forms?
a) Differential
b) Integral
c) Point
d) Stokes theorem
Answer: d
Explanation: Gauss law can be expressed in differential or point form as,
Div = ρv and in integral form as ∫∫ D.ds = Q = ψ . It is not possible to express it using Stoke’s theorem.
8. The tangential component of electric field intensity is always continuous at the interface. State True/False.
a) True
b) False
Answer: a
Explanation: Consider a dielectric-dielectric boundary, the electric field intensity in both the surfaces will be Et1 = Et2, which implies that the tangential component of electric field intensity is always continuous at the boundary.
9. The normal component of the electric flux density is always discontinuous at the interface. State True/False.
a) True
b) False
Answer: a
Explanation: In a dielectric-dielectric boundary, if a free surface charge density exists at the interface, then the normal components of the electric flux density are discontinuous at the boundary, which means Dn1 = Dn2.
Answer: c
Explanation: From Gauss law, we can compute the electric flux density. This in turn can be used to find electric field intensity. We know that F = qE. Hence force can be computed. This gives the Coulomb’s law.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Relation of E,D,V”.
1. The electric flux density and electric field intensity have which of the following relation?
a) Linear
b) Nonlinear
c) Inversely linear
d) Inversely nonlinear
Answer: a
Explanation: The electric flux density is directly proportional to electric field intensity. The proportionality constant is permittivity. D=ε E. It is clear that both are in linear relationship.
2. The electric field intensity is the negative gradient of the electric potential. State True/False.
a) True
b) False
Answer: a
Explanation: V = -∫E.dl is the integral form. On differentiating both sides, we get E = -Grad . Thus the electric field intensity is the negative gradient of the electric potential.
3. Find the electric potential for an electric field 3units at a distance of 2m.
a) 9
b) 4
c) 6
d) 3/2
Answer: c
Explanation: The electric field intensity is the ratio of electric potential to the distance. E = V/d. To get V = E X d = 3 X 2 = 6units.
4. Find the potential at a point for the function V = 2x 2 y + 5z.
a) 96
b) 66
c) 30
d) -66
Answer: b
Explanation: The electric potential for the function V = 2x 2 y + 5z at the point is given by V = 22 + 5 = 96-30 = 66 units.
5. Find the electric flux density surrounding a material with field intensity of 2xyz placed in transformer oil at the point P is
(in 10 -10 units)
a) 2.1
b) 2.33
c) 2.5
d) 2.77
Answer: c
Explanation: D = εE, where ε = εo εr. The flux density is given by,
D = 8.854 X 10 -12 X 2.2 X 2 = 2.33 X 10 -10 units.
6. If potential V = 20/(x 2 + y 2 ). The electric field intensity for V is
40/(x 2 + y 2 ) 2 . State True/False.
a) True
b) False
Answer: a
Explanation: E = -Grad = -Grad(20/(x 2 + y 2 )) = -(-40x i /(x 2 + y 2 ) 2 – 40/(x 2 + y 2 ) 2 ) = 40/(x 2 + y 2 ) 2 . Thus the statement is true.
7. Find the potential of the function V = 60cos θ/r at the point P.
a) 20
b) 10
c) 30
d) 60
Answer: b
Explanation: Given V = 60cos θ/r. For r = 3m and θ = 60, we get V = 60cos 60/3 = 20cos 60 = 10 units.
8. Find the work done moving a charge 2C having potential V = 24volts is
a) 96
b) 24
c) 36
d) 48
Answer: d
Explanation: The work done is the product of charge and potential.
W = Q X V = 2 X 24 = 48 units.
9. If the potential is given by, V = 10sin θ cosφ/r, find the density at the point P
(in 10 -12 units)
a) 13.25
b) 22.13
c) 26.31
d) 31.52
Answer: b
Explanation: Since V is given find out E.E = -Grad = – Grad. From E, we can easily compute D. D = εE = 8.854 X 10 -12 X 5/2 = 22.13 units.
Answer: a
Explanation: Find E from V, E = -Grad . We get E at A as 59.97i – 71.98j -20k. Thus D = εE = 8.854 X 10 -12 X
= nC/m 2 .
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Real Time Applications”.
1. Calculate the capacitance of a material in air with area 20 units and distance between plates is 5m.
a) 35.36pF
b) 3.536pF
c) 35.36nF
d) 3.536nF
Answer: a
Explanation: The capacitance of any material is given by, C = εA/d, where ε = εoεr is the permittivity in air and the material respectively. Thus C = 1 X 8.854 X 10 -12 X 20/5 = 35.36pF.
2. The resistance of a material with conductivity 2millimho/m 2 , length 10m and area 50m is
a) 500
b) 200
c) 100
d) 1000
Answer: c
Explanation: The resistance is given by, R = ρL/A, where ρ is the resistivity, the inverse of conductivity. R = 10/ = 100 ohm.
3. Find the inductance of a coil with permeability 3.5, turns 100 and length 2m. Assume the area to be thrice the length.
a) 131.94mH
b) 94.131mH
c) 131.94H
d) 94.131H
Answer: a
Explanation: The inductance is given by L = μ N 2 A/l, where μ= μoμr is the permeability of air and the material respectively. N = 100 and Area = 3 X 2 = 6. L = 4π X 10 -7 X 100 2 X 6/2 = 131.94mH.
4. Find the current density of a material with resistivity 20 units and electric field intensity 2000 units.
a) 400
b) 300
c) 200
d) 100
Answer: d
Explanation: The current density is given by J = σ E, where σ is the conductivity. Thus resistivity ρ = 1/σ. J = E/ρ = 2000/20 = 100 units.
5. Find the current in a conductor with resistance 2 ohm, electric field 2 units and distance 100cm.
a) 1A
b) 10mA
c) 10A
d) 100mA
Answer: a
Explanation: We know that E = V/d. To get potential, V = E X d = 2 X 1 = 2 volts. From Ohm’s law, V = IR and current I = V/R = 2/2 = 1A.
6. In electric fields, D= ε E. The correct expression which is analogous in magnetic fields will be
a) H = μ B
b) B = μ H
c) A = μ B
d) H = μ A
Answer: b
Explanation: In electric fields, the flux density is a product of permittivity and field intensity. Similarly, for magnetic fields, the magnetic flux density is the product of permeability and magnetic field intensity, given by B= μ H.
7. Find the force on a conductor of length 12m and magnetic flux density 20 units when a current of 0.5A is flowing through it.
a) 60
b) 120
c) 180
d) 200
Answer: b
Explanation: The force on a conductor is given by F = BIL, where B = 20, I = 0.5 and L = 12. Force F = 20 X 0.5 x 12 = 120 N.
8. From the formula F = qE, can prove that work done is a product of force and displacement. State True/False
a) True
b) False
Answer: a
Explanation: We know that F = qE = qV/d and W = qV. Thus it is clear that qV = W and qV = Fd. On equating both, we get W = Fd, which is the required proof.
9. Calculate the power of a material with electric field 100 units at a distance of 10cm with a current of 2A flowing through it.
a) 10
b) 20
c) 40
d) 80
Answer: b
Explanation: Power is defined as the product of voltage and current.
P = V X I, where V = E X d. Thus P = E X d X I = 100 X 0.1 X 2 = 20 units.
Answer: c
Explanation: Power is given by, P= V X I, where I = J X A is the current.
Thus power P = V X J X A = 20 X 15 X 100 = 30,000 joule = 30kJ.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Electric Dipole”.
1. Choose the best definition of a dipole.
a) A pair of equal and like charges located at the origin
b) A pair of unequal and like charges located at the origin
c) A pair of equal and unlike charges separated by a small distance
d) A pair of unequal and unlike charges separated by a small distance
Answer: c
Explanation: An electric dipole generally refers to two equal and unlike charges separated by a small distance. It can be anywhere, not necessarily at origin.
2. The potential due to a dipole at a point P from it is the
a) Sum of potentials at the charges
b) Difference of potentials at the charges
c) Multiplication of potentials at the charges
d) Ratio of potentials at the charges
Answer: b
Explanation: The total potential at the point P due to the dipole is given by the difference of the potentials of the individual charges.
V = V1 + , since both the charges are unlike. Thus V = V1 – V2.
3. Calculate the dipole moment of a dipole with equal charges 2C and -2C separated by a distance of 2cm.
a) 0.02
b) 0.04
c) 0.06
d) 0.08
Answer: b
Explanation: The dipole moment of charge 2C and distance 2cm will be,
M = Q x d. Thus, M = 2 x 0.02 = 0.04 C-m.
4. Find the angle at which the potential due a dipole is measured, when the distance from one charge is 12cm and that due to other is 11cm, separated to each other by a distance of 2cm.
a) 15
b) 30
c) 45
d) 60
Answer: d
Explanation: Here, the two charges are separated by d = 2cm.
The distance from one charge will be R1 = 11cm. The distance from another charge will be R2 = 12cm. If R1 and R2 is assumed to be parallel, then R2 – R1 = d cos θ. We get 1 = 2cos θ and cos θ = 0.5. Then θ =
cos -1 = 60.
5. Find the potential due the dipole when the angle subtended by the two charges at the point P is perpendicular.
a) 0
b) Unity
c) ∞
d) -∞
Answer: a
Explanation: The potential due the dipole is given by, V = m cos θ/(4πεr 2 ). When the angle becomes perpendicular . The potential becomes zero since cos 90 will become zero.
6. For two charges 3C and -3C separated by 1cm and are located at distances 5cm and 7cm respectively from the point P, then the distance between their midpoint and the point P will be
a) 5.91
b) 12.6
c) 2
d) 9
Answer: a
Explanation: For a distant point P, the R1 and R2 will approximately be equal.
R1 = R2 = r, where r is the distance between P and the midpoint of the two charges. Thus they are in geometric progression, R1R2=r 2
Now, r2 = 5 x 7 = 35. We get r = 5.91cm.
7. Calculate the distance between two charges of 4C forming a dipole, with a dipole moment of 6 units.
a) 1
b) 1.5
c) 2
d) 2.5
Answer: b
Explanation: The dipole moment is given by, M = Q x d. To get d, we rearrange the formula d = M/Q = 6/4 = 1.5units.
8. The potential due to the dipole on the midpoint of the two charges will be
a) 0
b) Unity
c) ∞
d) -∞
Answer: c
Explanation: The potential due a dipole at a point P will be V = m cos θ/(4πεr 2 ).
Now it is given that potential on the midpoint, which means P is on midpoint, then the distance from midpoint and P will be zero. When r = 0 is put in the above equation, we get V = ∞. This shows that the potential of a dipole at its midpoint will be maximum/infinity.
9. Dipoles in any electric field undergo
a) Magnetism
b) Electromagnetism
c) Magnetisation
d) Polarisation
Answer: d
Explanation: Dipoles in any pure electric field will undergo polarisation. It is the process of alignment of dipole moments in accordance with the electric field applied.
Answer: b
Explanation: Dipole moment implicates the strength of the dipole in the electric field. They are then used to compute the polarisation patterns based on the applied field. Once the polarisation is determined we can find its susceptibility. Though all options seem to be correct, the apt answer is to calculate polarisation, provided applied field is known.
This set of Electromagnetic Theory Questions and Answers for Experienced people focuses on “Electrostatic Energy”.
1. The electrostatic energy in an electric field does not depend on which of the following?
a) Magnitude of charges
b) Permittivity
c) Applied electric field
d) Flux lines
Answer: c
Explanation: The energy in an electric field directly magnitude of charges. Thus electric field and flux density are also dependent. But the applied field affects only the polarisation and it is independent of the energy in the field.
2. Calculate the energy in an electric field with flux density 6 units and field intensity of 4 units.
a) 12
b) 24
c) 36
d) 48
Answer: a
Explanation: The energy in an electric field is given by, W = 0.5 x D x E, where D = 6 and E = 4. We get W = 0.5 x 6 x 4 = 12 units.
3. Calculate the energy in an electric field with permittivity of 56 and field intensity of 36π
a) 3.16
b) 5.16
c) 7.16
d) 9.16
Answer: a
Explanation: The energy in an electric field is given by, W = 0.5 x D x E. Since D = εE, we get W = 0.5 x ε x E 2 . On substituting the data, we get 3.16 microjoule.
4. Equipotential surface is a
a) Real surface
b) Complex surface
c) Imaginary surface
d) Not existing surface
Answer: c
Explanation: Equipotential surface is an imaginary surface in an electric field of a given charge distribution in which all the points on the surface are at the same electric potential.
5. The work done in moving a test charge from one point to another in an equipotential surface is zero. State True/False.
a) True
b) False
Answer: a
Explanation: Since the electric potential in the equipotential surface is the same, the work done will be zero.
6. When curl of a path is zero, the field is said to be conservative. State True/False.
a) True
b) False
Answer: a
Explanation: By Stoke’s theorem, when curl of a path becomes zero, then
∫ E.dl = 0. In other words the work done in a closed path will always be zero. Fields having this property is called conservative or lamellar fields.
7. If the electric potential is given, which of the following cannot be calculated?
a) Electrostatic energy
b) Electric field intensity
c) Electric flux density
d) Permittivity
Answer: a
Explanation: Using potential, we can calculate electric field directly by gradient operation. From E, the flux density D can also be calculated. Thus it is not possible to calculate energy directly from potential.
8. Superconductors exhibit which of the following properties?
a) Ferromagnetism
b) Polarisation
c) Diamagnetism
d) Ferrimagnetism
Answer: c
Explanation: Since superconductors have very good conductivity at low temperatures , they have nearly zero resistivity and exhibit perfect diamagnetism.
9. Debye is the unit used to measure
a) Permittivity
b) Electric dipole moment
c) Magnetic dipole moment
d) Susceptibility
Answer: b
Explanation: Debye is the standard unit for measurement of electric dipole moment. 1 Debye = 3.336 x 10 -30 Coulomb-meter.
Answer: c
Explanation: Ceramic materials are generally brittle. Since these materials are used in capacitors, they have higher dielectric constant than polymer. With respect to energy, they possess high electrostatic energy due to very high dielectric constant .
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Electrostatic Properties”.
1. The permittivity is also called
a) Electrostatic energy
b) Dielectric constant
c) Dipole moment
d) Susceptibility
Answer: b
Explanation: The term permittivity or dielectric constant is the measurement of electrostatic energy stored within it and therefore depends on the material.
2. Dielectric constant will be high in
a) Conductors
b) Semiconductors
c) Insulators
d) Superconductors
Answer: c
Explanation: Materials that have very less conductivity like ceramics, plastics have higher dielectric constants. Due to their low conductivity, the dielectric materials are said to be good insulators.
3. Under the influence of electric field, the dielectric materials will get charged instantaneously. State True/False.
a) True
b) False
Answer: a
Explanation: The dielectrics have the ability of storing energy easily when an electric field is applied as their permittivity is relatively higher than any other materials.
4. Insulators perform which of the following functions?
a) Conduction
b) Convection
c) Provide electrical insulation
d) Allows current leakage at interfaces
Answer: c
Explanation: Insulators is a non-conducting material which prevents the leakage of electric current in unwanted directions. Thus it is used to provide electrical insulation.
5. Which of the following properties distinguish a material as conductor, insulator and semiconductor?
a) Free electron charges
b) Fermi level after doping
c) Energy band gap
d) Electron density
Answer: c
Explanation: The only parameter that classifies the material as conductor or insulator or semiconductor is the band gap energy. It is the energy required to make the electrons conduct. This is low of conductors, average for semiconductors and very high for insulators. This means it requires very high energy to make an insulator conduct.
6. Semiconductors possess which type of bonding?
a) Metallic
b) Covalent
c) Ionic
d) Magnetic
Answer: b
Explanation: Conductors exhibit metallic bonding. Insulators exhibit ionic bonding and semiconductors exhibit covalent bonding due to sharing of atoms.
7. Find the susceptibility of a material whose dielectric constant is 2.26.
a) 1.26
b) 3.26
c) 5.1
d) 1
Answer: a
Explanation: Electric susceptibility is the measure of ability of the material to get polarised. It is given by, χe = εr – 1.Thus we get 1.26.
8. The bound charge density and free charge density are 12 and 6 units respectively. Calculate the susceptibility.
a) 1
b) 0
c) 2
d) 72
Answer: c
Explanation: The electric susceptibility is given by, χe = Bound free density/Free charge density. χe = 12/6 = 2. It has no unit.
9. The susceptibility of free space is
a) 1
b) 0
c) 2
d) ∞
Answer: b
Explanation: For free space/air, the relative permittivity is unity i.e, εr = 1. Thus χe = εr – 1 = 0. The susceptibility will become zero in air.
Answer: b
Explanation: The electric flux density of a field is the sum of εE and polarisation P. It gives D = εE + P. When electric field becomes zero, it is clear that D = P.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Conductors”.
1. Which of the following are conductors?
a) Ceramics
b) Plastics
c) Mercury
d) Rubber
Answer: c
Explanation: Normally, metals are said to be good conductors. Here mercury is the only metal . The other options are insulators.
2. Find the range of band gap energy for conductors.
a) >6 eV
b) 0.2-0.4 eV
c) 0.4-2 eV
d) 2-6 eV
Answer: b
Explanation: Conductors are materials with least band gap energy. The smallest range in this group is 0.2-0.4 eV.
3. Conduction in metals is due to
a) Electrons only
b) Electrons and holes
c) Holes only
d) Applied electric field
Answer: a
Explanation: Conduction in metals is only due to majority carriers, which are electrons. Electrons and holes are responsible for conduction in a semiconductor.
4. Find the band gap energy when a light of wavelength 1240nm is incident on it.
a) 1eV
b) 2eV
c) 3eV
d) 4eV
Answer: a
Explanation: The band gap energy in electron volt when wavelength is given is, Eg = 1.24/λ = 1.24 x 10 -6 /1240 x 10 -9 = 1eV.
5. Alternating current measured in a transmission line will be
a) Peak value
b) Average value
c) RMS value
d) Zero
Answer: c
Explanation: The instantaneous current flowing in a transmission line, when measured using an ammeter, will give RMS current value. This value is 70.7% of the peak value. This is because, due to oscillations in AC, it is not possible to measure peak value. Hence to normalise, we consider current at any time in a line will be the RMS current.
6. The current in a metal at any frequency is due to
a) Conduction current
b) Displacement current
c) Both conduction and displacement current
d) Neither conduction nor displacement current
Answer: a
Explanation: At any frequency, the current through the metal will be due to conduction current. Only at high frequencies and when medium is air, the conduction is due to displacement current. Thus in general the current in metal is due to conduction current, which depends on the mobility of the carriers.
7. For conductors, the free electrons will exist at
a) Valence band
b) Middle of valence and conduction band
c) Will not exist
d) Conduction band
Answer: d
Explanation: In conductors, the free electrons exist in the conduction band. Since the band gap energy is very low, less energy is required to transport the free electrons to the conduction band, as they are readily available to conduct.
8. The current flowing through an insulating medium is called
a) Conduction
b) Convection
c) Radiation
d) Susceptibility
Answer: b
Explanation: A beam of electrons in a vacuum tube is called convection current. It occurs when current flows through an insulating medium like liquid, vacuum etc.
9. Find the conduction current density when conductivity of a material is 500 units and corresponding electric field is 2 units.
a) 500
b) 250
c) 1000
d) 2000
Answer: c
Explanation: The conduction current density is given by, J = σE
J = 500 X 2 = 1000 units.
Answer: b
Explanation: The convection current density is given by, J = ρeV
J = 200 X 12= 2400 units.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Dielectrics”.
1. A dielectric is always an insulator. But an insulator is not necessarily a dielectric. State True/False.
a) True
b) False
Answer: a
Explanation: For a material to be dielectric, its permittivity should be very high. This is seen in insulators. For a material to be insulator, the condition is to have large band gap energy. However, this is not necessary for a dielectric.
2. Identify a good dielectric.
a) Iron
b) Ceramics
c) Plastic
d) Magnesium
Answer: b
Explanation: Iron and magnesium are metals. Hence they need not be considered. Both ceramics and plastic are insulators. But dielectric constant is more for ceramics always. Hence ceramics is the best dielectric.
3. A dielectric can be made a conductor by
a) Compression
b) Heating
c) Doping
d) Freezing
Answer: b
Explanation: On increasing the temperature, the free electrons in an insulator can be promoted from valence to conduction band. Gradually, it can act as a conductor through heating process. This condition is called dielectric breakdown, wherein the insulator loses its dielectric property and starts to conduct.
4. Find the dielectric constant for a material with electric susceptibility of 4.
a) 3
b) 5
c) 8
d) 16
Answer: b
Explanation: The electric susceptibility is given by χe = εr – 1. For a susceptibility of 4, the dielectric constant will be 5. It has no unit.
5. For a dielectric which of the following properties hold good?
a) They are superconductors at high temperatures
b) They are superconductors at low temperatures
c) They can never become a superconductor
d) They have very less dielectric breakdown voltage
Answer: b
Explanation: Superconductors are characterised by diamagnetism behaviour and zero resistivity, which true for a dielectric. They occur only at low temperature. Thus a dielectric can become a superconductor at low temperatures with very high dielectric breakdown voltage.
6. The magnetic field which destroys the superconductivity is called
a) Diamagnetic field
b) Ferromagnetic field
c) Ferrimagnetic field
d) Critical field
Answer: d
Explanation: Critical field is that strong magnetic field which can destroy the superconductivity of a material. The temperature at which this occurs is called transition temperature.
7. The magnetic susceptibility in a superconductor will be
a) Positive
b) Negative
c) Zero
d) Infinity
Answer: b
Explanation: Due to perfect diamagnetism in a superconductor, its magnetic susceptibility will be negative. This phenomenon is called Meissner effect.
8. The superconducting materials will be independent of which of the following?
a) Magnetic field
b) Electric field
c) Magnetization
d) Temperature
Answer: b
Explanation: Superconducting materials depends only on the applied magnetic field, resultant magnetization at the temperature considered. It is independent of the applied electric field and the corresponding polarization.
9. Find the mean free path of an electron travelling at a speed of 18m/s in 2 seconds.
a) 9
b) 36
c) 0.11
d) 4.5
Answer: b
Explanation: The mean free path is defined as the average distance travelled by an electron before collision takes place. It is given by, d = v x τc, where v is the velocity and τc is the collision time. Thus d = 18 x 2 = 36m.
Answer: c
Explanation: When the kinetic energy and one electron volt are equal, we can equate mv 2 /2 = eV. Put e and m in the equation to get velocity v = 5.9 x 10 5 m/s.
This set of Electromagnetic Theory Interview Questions and Answers for Experienced people focuses on “Displacement and Conduction Current”.
1. Find the conductivity of a material with conduction current density 100 units and electric field of 4 units.
a) 25
b) 400
c) 0.04
d) 1600
Answer: a
Explanation: The conduction current density is given by, Jc = σE. To get conductivity, σ = J/E = 100/4 = 25 units.
2. Calculate the displacement current density when the electric flux density is 20sin 0.5t.
a) 10sin 0.5t
b) 10cos 0.5t
c) 20sin 2t
d) 20cos 2t
Answer: b
Explanation: The displacement current density is given by, Jd = dD/dt.
Jd = d/dt = 20cos 0.5t = 10cos 0.5t.
3. Find the magnitude of the displacement current density in air at a frequency of 18GHz in frequency domain. Take electric field E as 4 units.
a) 18
b) 72
c) 36
d) 4
Answer: d
Explanation: Jd = dD/dt = εdE/dt in time domain. For frequency domain, convert using Fourier transform, Jd = εjωE. The magnitude of
Jd = εωE = εE. On substituting, we get 4 ampere.
4. Calculate the frequency at which the conduction and displacement currents become equal with unity conductivity in a material of permittivity 2.
a) 18 GHz
b) 9 GHz
c) 36 GHz
d) 24 GHz
Answer: b
Explanation: When Jd = Jc , we get εωE = σE. Thus εo = σ. On substituting conductivity as one and permittivity as 2, we get f = 9GHz.
5. The ratio of conduction to displacement current density is referred to as
a) Attenuation constant
b) Propagation constant
c) Loss tangent
d) Dielectric constant
Answer: c
Explanation: Jc /Jd is a standard ratio, which is referred to as loss tangent given by σ /ε ω. The loss tangent is used to determine if the material is a conductor or dielectric.
6. If the loss tangent is very less, then the material will be a
a) Conductor
b) Lossless dielectric
c) Lossy dielectric
d) Insulator
Answer: b
Explanation: If loss tangent is less, then σ /ε ω <<1. This implies the conductivity is very poor and the material should be a dielectric. Since it is specifically mentioned very less, assuming the conductivity to be zero, the dielectric will be lossless .
7. In good conductors, the electric and magnetic fields will be
a) 45 in phase
b) 45 out of phase
c) 90 in phase
d) 90 out of phase
Answer: b
Explanation: The electric and magnetic fields will be out of phase by 45 in good conductors. This is because their intrinsic impedance is given by η = √ X . In polar form we get 45 out of phase.
8. In free space, which of the following will be zero?
a) Permittivity
b) Permeability
c) Conductivity
d) Resistivity
Answer: c
Explanation: In free space, ε = ε0 and μ = μ0. The relative permittivity and permeability will be unity. Since the free space will contain no charges in it, the conductivity will be zero.
9. If the intrinsic angle is 20, then find the loss tangent.
a) tan 20
b) tan 40
c) tan 60
d) tan 80
Answer: b
Explanation: The loss tangent is given by tan 2θn, where θn = 20. Thus the loss tangent will be tan 40.
Answer: d
Explanation: The intrinsic impedance is given by η = √ ohm. Here εo = 8.854 x 10 -12 and μo = 4π x 10 -7 .
On substituting the values, we get η = 377 ohm.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Polarization”.
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 . 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.
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.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Continuity Equation”.
1. Find the current when the charge is a time function given by q = 3t + t 2 at 2 seconds.
a) 3
b) 5
c) 7
d) 9
Answer: c
Explanation: The current is defined as the rate of change of charge in a circuit ie, I = dq/dt. On differentiating the charge with respect to time, we get 3 + 2t. At time t = 2s, I = 7A.
2. The continuity equation is a combination of which of the two laws?
a) Ohm’s law and Gauss law
b) Ampere law and Gauss law
c) Ohm’s law and Ampere law
d) Maxwell law and Ampere law
Answer: b
Explanation: I = ∫ J.ds is the integral form of Ohm’s law and Div = dq/dt is the Gauss law analogous to D. Through these two equations, we get Div = -dρ/dt. This is the continuity equation.
3. Calculate the charge density for the current density given 20sin x i + ycos z j at the origin.
a) 20t
b) 21t
c) 19t
d) -20t
Answer: b
Explanation: Using continuity equation, the problem can be solved. Div =
– dρ/dt. Div = 20cos x + cos z. At origin, we get 20cos 0 + cos 0 = 21. To get ρ, on integrating the Div with respect to t, the charge density will be 21t.
4. Compute the conductivity when the current density is 12 units and the electric field is 20 units. Also identify the nature of the material.
a) 1.67, dielectric
b) 1.67, conductor
c) 0.6, dielectric
d) 0.6, conductor
Answer: c
Explanation: The current density is the product of conductivity and electric field intensity. J = σE. To get σ, put J = 12 and E = 20. σ = 12/20 = 0.6. Since the conductivity is less than unity, the material is a dielectric.
5. Find the electron density when convection current density is 120 units and the velocity is 5m/s.
a) 12
b) 600
c) 24
d) 720
Answer: c
Explanation: The convection current density is given by J = ρe x v. To get ρe, put J = 120 and v = 5. ρe = 120/5 = 24 units.
6. Calculate the electric field when the conductivity is 20 units, electron density is 2.4 units and the velocity is 10m/s. Assume the conduction and convection current densities are same.
a) 2.4
b) 4.8
c) 3.6
d) 1.2
Answer: d
Explanation: The conduction current density is given by J = σE and the convection current density is J = ρe v. When both are equal, ρe v = σE. To get E, put σ = 20, ρe = 2.4 and v = 10, E = 2.4 x 10/20 = 1.2 units.
7. Find the mobility of the electrons when the drift velocity is 23 units and the electric field is 11 units.
a) 1.1
b) 2.2
c) 3.2
d) 0.9
Answer: b
Explanation: The mobility is defined as the drift velocity per unit electric field. Thus μe = vd/E = 23/11 = 2.1 units.
8. Find the resistance of a cylinder of area 200 units and length 100m with conductivity of 12 units.
a) 1/24
b) 1/48
c) 1/12
d) 1/96
Answer: a
Explanation: The resistance is given by R = ρL/A = L/σA. Put L = 100, σ = 12 and A = 200, we get R = 100/ = 1/24 units.
9. Calculate the potential when a conductor of length 2m is having an electric field of 12.3units.
a) 26.4
b) 42.6
c) 64.2
d) 24.6
Answer: d
Explanation: The electric field is given by E = V/L. To get V, put E = 12.3 and L = 2.Thus we get V = E x L = 12.3 x 2 = 24.6 units.
Answer: a
Explanation: The generic current density equation is J = I/A and the point form of Ohm’s law is J = σ E. On equating both and substituting E = V/L, we get V = IL/σ A = IR which is the Ohm’s law.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Boundary Conditions”.
1. The charge within a conductor will be
a) 1
b) -1
c) 0
d) ∞
Answer: c
Explanation: No charges exist in a conductor. An illustration for this statement is that, it is safer to stay inside a car rather than standing under a tree during lightning. Since the car has a metal body, no charges will be possessed by it to get ionized by the lightning.
2. For a conservative field which of the following equations holds good?
a) ∫ E.dl = 0
b) ∫ H.dl = 0
c) ∫ B.dl = 0
d) ∫ D.dl = 0
Answer: a
Explanation: A conservative field implies the work done in a closed path will be zero. This is given by ∫ E.dl = 0.
3. Find the electric field if the surface density at the boundary of air is 10 -9 .
a) 12π
b) 24π
c) 36π
d) 48π
Answer: c
Explanation: It is the conductor-free space boundary. At the boundary, E = ρ/εo. Put ρ = 10 -9 and εo = 10 -9 /36π. We get E = 36π units.
4. Find the flux density at the boundary when the charge density is given by 24 units.
a) 12
b) 24
c) 48
d) 96
Answer: b
Explanation: At the boundary of a conductor- free space interface, the flux density is equal to the charge density. Thus D = ρv = 24 units.
5. Which component of the electric field intensity is always continuous at the boundary?
a) Tangential
b) Normal
c) Horizontal
d) Vertical
Answer: a
Explanation: At the boundary of the dielectric-dielectric, the tangential component of the electric field intensity is always continuous. We get Et1 = Et2.
6. The normal component of which quantity is always discontinuous at the boundary?
a) E
b) D
c) H
d) B
Answer: b
Explanation: The normal component of an electric flux density is always discontinuous at the boundary for a dielectric-dielectric boundary. We get Dn1 = Dn2, when we assume the free surface charge exists at the interface.
7. The electric flux density of a surface with permittivity of 2 is given by 12 units. What the flux density of the surface in air?
a) 24
b) 6
c) 1/6
d) 0
Answer: b
Explanation: The relation between electric field and permittivity is given by Dt1/Dt2 = ε1/ε2. Put Dt1 = 12, ε1 = 2 and ε2 =1, we get Dt2 = 12 x 1/ 2 = 6 units.
8. The electric field intensity of a surface with permittivity 3.5 is given by 18 units. What the field intensity of the surface in air?
a) 5.14
b) 0.194
c) 63
d) 29
Answer: c
Explanation: The relation between flux density and permittivity is given by En1/En2 = ε2/ ε1. Put En1 = 18, ε1 = 3.5 and ε2 = 1. We get En2 = 18 x 3.5 = 63 units.
9. A wave incident on a surface at an angle 60 degree is having field intensity of 6 units. The reflected wave is at an angle of 30 degree. Find the field intensity after reflection.
a) 9.4
b) 8.4
c) 10.4
d) 7.4
Answer: c
Explanation: By Snell’s law, the relation between incident and reflected waves is given by, E1 sin θ1 = E2 sin θ2. Thus 6 sin 60 = E2 sin 30. We get E2 = 6 x 1.732 = 10.4 units.
Answer: d
Explanation: From the relations of the boundary conditions of a dielectric-dielectric interface, we get tan θ1/tan θ2 = ε1/ε2. Thus tan 60/tan 45 = ε1/1. We get ε1 = tan 60 = 1.73.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Poisson and Laplace equation”.
1. The given equation satisfies the Laplace equation.
V = x 2 + y 2 – z 2 . State True/False.
a) True
b) False
Answer: a
Explanation: Grad = 2xi + 2yj – 4zk. Div ) = Del 2 = 2+2-4 = 0. It satisfies the Laplacian equation. Thus the statement is true.
2. 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 Del 2 = -ρ/ε. In free space, the charges will be zero. Thus the equation becomes, Del 2 = 0, which is the Laplace equation.
3. 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.
4. 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) will become zero. This is because gradient of a constant is zero and divergence of zero vector will be zero.
5. Calculate the charge density when a potential function x 2 + y 2 + z 2 is in air
a) 1/6π
b) 6/2π
c) 12/6π
d) 10/8π
Answer: a
Explanation: The Poisson equation is given by Del 2 = -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.
6. The function V = e x sin y + z does not satisfy Laplace equation. State True/False.
a) True
b) False
Answer: b
Explanation: Grad = e x sin y i + e x cos y j + k. Div) = e x sin y – e x sin y + 0= 0.Thus Laplacian equation Div) = 0 is true.
7. 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 = ρv. On putting
D= ε E and E=- Grad in Gauss law, we get Del 2 = -ρ/ε, which is the Poisson equation.
8. 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 = ρv
Div = Div = 12-7 = 5, which the charge density ρv. Thus ρv = 5 units.
9. Find the electric field of a potential function given by 20 log x + y at the point .
a) -20 i – j
b) -i -20 j
c) i + j
d) /20
Answer: a
Explanation: The electric field is given by E = -Grad. The gradient of the given function is 20i/x + j. At the point , we get 20i + j. The electric field E = - = -20i – j.
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.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Resistance and Capacitance”.
1. Find the resistivity of a material having resistance 20kohm, area 2 units and length of 12m.
a) 6666.6
b) 3333.3
c) 1200
d) 2000
Answer: b
Explanation: The resistance of a material is given by R = ρL/A. To get ρ, put R = 20 x 10 3 , A = 2 and L = 12. We get ρ = 3333.3 units.
2. A resistor value of colour code orange violet orange will be
a) 37 kohm
b) 37 Mohm
c) 48 kohm
d) 48 Mohm
Answer: a
Explanation: Orange refers to number 3. Violet refers to number 7. The third colour code orange refers to 103. Thus the resistor value will be 37 kilo ohm.
3. A infinite resistance is considered as a/an
a) Closed path
b) Open path
c) Not defined
d) Ammeter with zero reading
Answer:b
Explanation: When there exists infinite resistance in a path, the current flowing will ideally be zero. This is possible only for an open path/circuit.
4. Find the time constant in a series R-L circuit when the resistance is 4 ohm and the inductance is 2 H.
a) 0.25
b) 0.2
c) 2
d) 0.5
Answer: d
Explanation: The time constant for an R-L series circuit will be τ = L/R. Put R = 4 and L = 2. We get τ = 2/4 = 0.5 second.
5. Find the time constant for a R-C circuit for resistance R = 24 kohm and C = 16 microfarad.
a) 1.5 millisecond
b) 0.6 nanosecond
c) 384 millisecond
d) 384 microsecond
Answer: c
Explanation: The time constant for R-C circuit is τ = RC. Put R = 24 kilo ohm and C = 16 micro farad. We get τ = 24 x 10 3 x 16 x 10 -6 = 0.384 = 384 millisecond.
6. Find the capacitance when charge is 20 C has a voltage of 1.2V.
a) 32.67
b) 16.67
c) 6.67
d) 12.33
Answer: b
Explanation: Capacitance is related to Q and V as C = Q/V. Put C = 20C and V = 1.2V, we get Q = 20/1.2 = 16.67 farad.
7. Calculate the capacitance of two parallel plates of area 2 units separated by a distance of 0.2m in air
a) 8.84
b) 88.4
c) 884.1
d) 0.884
Answer: b
Explanation: Capacitance is given by, C = εo A/d. Put A = 2, d = 0.2, εo = 8.854 x 10 -12 , we get C = 8.841 x 10 -11 = 88. 41 pF.
8. Compute the capacitance between two concentric shells of inner radius 2m and the outer radius is infinitely large.
a) 0.111 nF
b) 0.222 nF
c) 4.5 nF
d) 5.4 nF
Answer: b
Explanation: The concentric shell with infinite outer radius is considered to be an isolated sphere. The capacitance C = 4πε/. If b->∞, then C = 4πεa. Put a = 2m, we get C = 4π x 8.854 x 10 -12 x 2 = 0.222 nF.
9. The capacitance of a material refers to
a) Ability of the material to store magnetic field
b) Ability of the material to store electromagnetic field
c) Ability of the material to store electric field
d) Potential between two charged plates
Answer: c
Explanation: The capacitance of a material is a measure of the ability of the material to store electric field. It is the ratio of charge stored to the voltage across the parallel plates.
Answer: a
Explanation: Capacitance between coaxial cylinders of inner radius 1.25cm and outer radius 1.25 + 2.13 = 3.38cm will be C = 2πεL/ ln. Put b = 3.38, a = 1.25 and L = 1000m, we get C = 1.957 x 10 -7 = 195.7 nF.
This set of Electromagnetic Theory Quiz focuses on “Method of Images”.
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
Explanation: 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
Explanation: 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
Explanation: 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 10 3 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
Explanation: 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
Explanation: 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 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
Explanation: The energy stored in a capacitor is given by, E = 0.5 CV 2 .
E = 0.5 x 3 x 12 2 = 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
Explanation: 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 10 12 order)
a) 2.19
b) 1.12
c) 9.21
d) 2.91
Answer: b
Explanation: 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 10 12 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 10 9 order)
a) 0
b) 1
c) 2
d) 3
Answer: c
Explanation: 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 10 9 units.
Answer: d
Explanation: Two capacitances in series gives C = C1C2/C1 + C2 = 2 x 5/2 + 5 = 10/7 farad.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Biot Savart Law”.
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/4πr 2 , which is analogous to the electric field F = q1q2/4πεr 2 , 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 = 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 = 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.
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.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Faraday Law, EMF and Lenz Law”.
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.
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 t 3 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 = 3t 2 . At time t = 2, we get Jd = 3 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 is not a closed path. Thus the curl will be non-zero. This implies E is not conservative and the statement is false.
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.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Ampere Law”.
1. The point form of Ampere law is given by
a) Curl = I
b) Curl = J
c) Curl = I
d) Curl = 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 = J.
2. 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.
3. 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.
4. 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.
5. 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.
6. 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.
7. 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.
8. 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 = J. The curl of H will be i – j + k = -cos x i – zsin x k. In the xy plane, z = 0. Thus Curl = J = -cos x i.
9. 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 = 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.
Answer: a
Explanation: We know that ∫ H.dl = I. By Stoke’s law, we can write Curl = 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.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Maxwell Law”.
1. The divergence of which quantity will be zero?
a) E
b) D
c) H
d) B
Answer: d
Explanation: The divergence of the magnetic flux density is always zero. This is because of the non existence of magnetic monopoles in a magnetic field.
2. Find the charge density when the electric flux density is given by 2x i + 3y j + 4z k.
a) 10
b) 9
c) 24
d) 0
Answer: b
Explanation: The charge density is the divergence of the electric flux density by Maxwell’s equation. Thus ρ = Div and Div = 2 + 3 + 4 = 9. We get ρ = 9 units.
3. Find the Maxwell equation derived from Faraday’s law.
a) Div = J
b) Div = I
c) Curl = -dB/dt
d) Curl = -dH/dt
Answer: c
Explanation: From the Faraday’s law and Lenz law, using Stoke’s theorem, we get Curl = -dB/dt. This is the Maxwell’s first law of electromagnetics.
4. Find the Maxwell law derived from Ampere law.
a) Div = H
b) Div = J
c) Curl = J
d) Curl = D
Answer: c
Explanation: From the current density definition and Ohm’s law, the Ampere circuital law Curl = J can be derived. This is Maxwell’s second law of electromagnetics.
5. The Faraday’s law states about which type of EMF?
a) Transformer EMF
b) Back EMF
c) Generator EMF
d) Secondary EMF
Answer: a
Explanation: The stationary loop in a varying magnetic field results in an induced emf due to the change in the flux linkage of the loop. This emf is called as induced or transformer EMF.
6. In which of the following forms can Maxwell’s equation not be represented?
a) Static
b) Differential
c) Integral
d) Harmonic
Answer: a
Explanation: Maxwell equations can be represented in differential/point form and integral form alternatively. Sometimes, it can be represented by time varying fields called harmonic form.
7. The charge build up in the capacitor is due to which quantity?
a) Conduction current
b) Displacement current
c) Convection current
d) Direct current
Answer: b
Explanation: The charge in the capacitor is due to displacement current. It is the current in the presence of the dielectric placed between two parallel metal plates.
8. In metals which of the following equation will hold good?
a) Curl = J
b) Curl = dD/dt
c) Curl = D
d) Curl = dB/dt
Answer: a
Explanation: Generally, the Curl is the sum of two currents- conduction and displacement. In case of metals, it constitutes conduction J and in case of dielectrics, it constitutes the displacement current dD/dt.
9. Find the flux enclosed by a material of flux density 12 units in an area of 80cm.
a) 9.6
b) 12/80
c) 80/12
d) 12/0.8
Answer: a
Explanation: The total flux in a material is the product of the flux density and the area. It is given by flux = 12 x 0.8= 9.6 units.
Answer: c
Explanation: The electric flux density from Maxwell’s equation is given by D = ∫ ρ dv. On substituting ρ = 16 and ∫dv = 1, we get D = 16 units.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Magnetic Field Intensity”.
1. The H quantity is analogous to which component in the following?
a) B
b) D
c) E
d) V
Answer: c
Explanation: The H quantity refers to magnetic field intensity in the magnetic field. This is analogous to the electric field intensity E in the electric field.
2. The magnetic flux density is directly proportional to the magnetic field intensity. State True/False.
a) True
b) False
Answer: a
Explanation: The magnetic field intensity is directly proportional to the magnetic field intensity for a particular material . It is given by B = μH.
3. Ampere law states that,
a) Divergence of H is same as the flux
b) Curl of D is same as the current
c) Divergence of E is zero
d) Curl of H is same as the current density
Answer: d
Explanation: Ampere circuital law or Ampere law states that the closed integral of the magnetic field intensity is same as the current enclosed by it. It is given by Curl = J.
4. Given the magnetic field is 2.4 units. Find the flux density in air(in 10 -6 order).
a) 2
b) 3
c) 4
d) 5
Answer: b
Explanation: We know that B = μH. On substituting μ = 4π x 10 -7 and H = 2.4, we get B = 4π x 10 -7 x 2.4 = 3 x 10 -6 units.
5. Find the electric field when the magnetic field is given by 2sin t in air.
a) 8π x 10 -7 cos t
b) 4π x 10 -7 sin t
c) -8π x 10 -7 cos t
d) -4π x 10 -7 sin t
Answer: a
Explanation: Given H = 2sin t. We get B = μH = 4π x 10 -7 x 2sin t = 8πx10 -7 sin t.
To get E, integrate B with respect to time, we get 8πx10 -7 cos t.
6. Find the height of an infinitely long conductor from point P which is carrying current of 6.28A and field intensity is 0.5 units.
a) 0.5
b) 2
c) 6.28
d) 1
Answer: b
Explanation: The magnetic field intensity of an infinitely long conductor is given by H = I/2πh. Put I = 6.28 and H = 0.5, we get h = 1/0.5 = 2 units.
7. Find the magnetic field intensity due to a solenoid of length 12cm having 30 turns and current of 1.5A.
a) 250
b) 325
c) 175
d) 375
Answer: d
Explanation: The magnetic field intensity of a solenoid is given by H = NI/L = 30 X 1.5/0.12 = 375 units.
8. Find the magnetic field intensity at the radius of 6cm of a coaxial cable with inner and outer radii are 1.5cm and 4cm respectively. The current flowing is 2A.
a) 2.73
b) 3.5
c) 0
d) 1.25
Answer: c
Explanation: The inner radius is 1.5cm and the outer radius is 4cm. It is clear that the magnetic field intensity needs to be calculated outside of the conductor ie, r>4cm. This will lead to zero, since H outside the conductor will be zero.
9. Find the magnetic field intensity of a toroid of turns 40 and radius 20cm. The current carried by the toroid be 3.25A.
a) 103.45
b) 102
c) 105.7
d) 171
Answer: a
Explanation: The magnetic field intensity of a toroid is given by H = NI/2πrm. Put N = 40, I = 3.25 and rm = 0.2, we get H = 40 x 3.25/2π x 0.2 = 103.45 units.
Answer: a
Explanation: The magnetic field intensity of an infinite sheet of charge is given by H = 0.5 K, for the point above the sheet and –0.5 K, for the point below the sheet. Here k is the charge density. Thus H = 0.5 x 36.5 = 18.25 units.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Magnetic Field Density”.
1. Identify which of the following is the unit of magnetic flux density?
a) Weber
b) Weber/m
c) Tesla
d) Weber -1
Answer: c
Explanation: The unit of magnetic flux density is weber/m 2 . It is also called as tesla.
2. 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.
3. Find the flux contained by the material when the flux density is 11.7 Tesla and the area is 2 units.
a) 23.4
b) 12.3
c) 32.4
d) 21.3
Answer: a
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.
4. Find the current when the magnetic field intensity is given by 2L and L varies as 0->1.
a) 2
b) 1
c) 0.5
d) 0
Answer: b
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.
5. Find the magnetic field intensity when the flux density is 8 x 10 -6 Tesla in the medium of air.
a) 6.36
b) 3.66
c) 6.63
d) 3.36
Answer: a
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.
6. If ∫ H.dL = 0, then which statement will be true?
a) E = -Grad
b) B = -Grad
c) H = -Grad
d) D = -Grad
Answer: c
Explanation: The given condition shows that the magnetic field intensity will be the negative gradient of the magnetic vector potential.
7. Find the magnetic flux density of the material with magnetic vector potential A = y i + z j + x k.
a) i + j + k
b) –i – j – k
c) –i-j
d) –i-k
Answer: b
Explanation: The magnetic flux density is the curl of the magnetic vector potential. B = Curl. Thus Curl = i – j + k = -i – j – k. We get B = -i – j – k.
8. Find the magnetic flux density when a flux of 28 units is enclosed in an area of 15cm.
a) 178.33
b) 186.67
c) 192.67
d) 124.33
Answer: b
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.
9. Find the magnetic flux density B when E is given by 3sin y i + 4cos z j + ex k.
a) ∫dt
b) -∫dt
c) ∫dt
d) -∫dt
Answer: b
Explanation: We know that Curl = -dB/dt. The curl of E is . To get B, integrate the -curl with respect to time to get B = -∫dt.
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.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Magnetic Vector Potential”.
1. The magnetic vector potential is a scalar quantity.
a) True
b) False
Answer: b
Explanation: The magnetic vector potential could be learnt as a scalar. But it is actually a vector quantity, which means it has both magnitude and direction.
2. Find the magnetic field intensity when the magnetic vector potential x i + 2y j + 3z k.
a) 6
b) -6
c) 0
d) 1
Answer: b
Explanation: The magnetic field intensity is given by H = -Grad. The gradient of Vm is 1 + 2 + 3 = 6. Thus H = -6 units.
3. The value of ∫ H.dL will be
a) J
b) I
c) B
d) H
Answer: b
Explanation: By Stoke’s theorem, ∫ H.dL = ∫ Curl.dS and from Ampere’s law, Curl = J. Thus ∫ H.dL = ∫ J.dS which is nothing but current I.
4. Given the vector potential is 16 – 12sin y j. Find the field intensity at the origin.
a) 28
b) 16
c) 12
d) 4
Answer: c
Explanation: The field intensity is given by H = – Grad. The gradient is given by 0 – 12cos y. At the origin, the gradient will be -12 cos 0 = -12. Thus the field intensity will be 12 units.
5. Find the vector potential when the field intensity 60x 2 varies from to .
a) 120
b) -20
c) -180
d) 60
Answer: b
Explanation: The field intensity H = -Grad. To get V, integrate H with respect to the variable. Thus V = -∫H.dl = -∫60x 2 dx = -20x 3 as x = 0->1 to get -20.
6. Find the flux density B when the potential is given by x i + y j + z k in air.
a) 12π x 10 -7
b) -12π x 10 -7
c) 6π x 10 -7
d) -6π x 10 -7
Answer: b
Explanation: The field intensity H = -Grad. Since the given potential is a position vector, the gradient will be 3 and H = -3. Thus the flux density B = μH = 4π x 10 -7 x = -12π x 10 -7 units.
7. The Laplacian of the magnetic vector potential will be
a) –μ J
b) – μ I
c) –μ B
d) –μ H
Answer: a
Explanation: The Laplacian of the magnetic vector potential is given by Del 2 = -μ J, where μ is the permeability and J is the current density.
8. The magnetic vector potential for a line current will be inversely proportional to
a) dL
b) I
c) J
d) R
Answer: d
Explanation: The magnetic vector potential for the line integral will be A = ∫ μIdL/4πR. It is clear that the potential is inversely proportional to the distance or radius R.
9. The current element of the magnetic vector potential for a surface current will be
a) J dS
b) I dL
c) K dS
d) J dV
Answer: c
Explanation: The magnetic vector potential for the surface integral is given by A = ∫ μKdS/4πR. It is clear that the current element is K dS.
Answer: a
Explanation: The magnetic flux density B can be expressed as the space derivative of the magnetic vector potential A. Thus B = Curl.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Magnetostatic Energy”.
1. Find the induced EMF in an inductor of 2mH and the current rate is 2000 units.
a) 4
b) -4
c) 1
d) -1
Answer: b
Explanation: The induced emf is given by e = -Ldi/dt. Put L = 2 x 10 -3 and di/dt = 2000 in the equation. We get e = -2 x 10 -3 x 2000 = -4 units.
2. Find the work done in an inductor of 4H when a current 8A is passed through it?
a) 256
b) 128
c) 64
d) 512
Answer: b
Explanation: The work done in the inductor will be W = 0.5 x LI 2 . On substituting L = 4 and I = 8, we get, W = 0.5 x 4 x 8 2 = 128 units.
3. Find the inductance of a material with 100 turns, area 12 units and current of 2A in air.
a) 0.75mH
b) 7.5mH
c) 75mH
d) 753mH
Answer: a
Explanation: The inductance of any material is given by L = μ N 2 A/I. On substituting N = 100, A = 0.12 and I = 2, we get L = 4π x 10 -7 x 100 2 x 0.12/2 = 0.75 units.
4. Calculate the magnetic energy when the magnetic intensity in air is given as 14.2 units(in 10 -4 order)
a) 1.26
b) 2.61
c) 6.12
d) 1.62
Answer: a
Explanation: The magnetic energy is given by E = 0.5 μ H 2 . Put H = 14.2 and in air μ = 4π x 10 -7 , we get E = 0.5 x 4π x 10 -7 x 14.2 2 = 1.26 x 10 -4 units.
5. Calculate the magnetic energy when the magnetic flux density is given by 32 units(in 10 8 order)
a) 4.07
b) 7.4
c) 0.47
d) 7.04
Answer: a
Explanation: The magnetic energy is given by E = 0.5 μ H 2 and we know that μH = B. On substituting we get a formula E = 0.5 B 2 /μ. Put B = 32 and in air μ = 4π x 10 -7 , we get E = 0.5 x 32 2 /4π x 10 -7 = 4.07 x 10 8 units.
6. Calculate the energy when the magnetic intensity and magnetic flux density are 15 and 65 respectively.
a) 755
b) 487.5
c) 922
d) 645
Answer: b
Explanation: The magnetic energy can also be written as E = 0.5 μH 2 = 0.5 BH, since B = μH. On substituting B = 65 and H = 15 we get E = 0.5 x 65 x 15 = 487.5 units.
7. Find the inductance when the energy is given by 2 units with a current of 16A.
a) 15.6mH
b) 16.5mH
c) 16.8mH
d) 15.8mH
Answer: a
Explanation: The energy stored in an inductor is given by E = 0.5 LI 2 . To get L, put E = 2 and I = 16 and thus L = 2E/I 2 = 2 x 2/16 2 = 15.6mH.
8. Find the power of an inductor of 5H and current 4.5A after 2 seconds.
a) 25.31
b) 50.62
c) 102.4
d) 204.8
Answer: a
Explanation: The energy stored in an inductor is given by E = 0.5 LI 2 . Thus, put L = 5 and I = 4.5 and we get E = 0.5 x 5 x 4.5 2 = 50.625 units To get power P = E/t = 50.625/2 = 25.31 units.
9. Find the turns in an solenoid of inductance 23.4mH , current 2A and area 15cm.
a) 900
b) 400
c) 498
d) 658
Answer: c
Explanation: The inductance of any material is given by L = μ N 2 A/I.
Put L = 23.4 x 10 -3 , I = 2 and A = 0.15, we get N as 498 turns.
Answer: a
Explanation: The inductance is directly proportional to square of the turns. Since the energy is directly proportional to the inductance, we can say both are dependent on each other.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Magnetostatic Properties”.
1. The magnetostatics highly relies on which property?
a) Resistance
b) Capacitance
c) Inductance
d) Moment
Answer: c
Explanation: The magnetostatics highly relies on the inductance of the magnetic materials, which decides its behavior in the influence of magnetic field.
2. The inductance is the measure of
a) Electric charges stored by the material
b) Emf generated by energising the coil
c) Magnetic field stored by the material
d) Magnetization of dipoles
Answer: b
Explanation: The inductance is a property of an electric conductor/coil which measures the amount of emf generated by passing current through the coil.
3. Find the total flux in a coil of magnetic flux density 12 units and area 7 units.
a) 0.84
b) 0.96
c) 8.4
d) 9.6
Answer: a
Explanation: The total flux in a coil is defined by φ = BA, where B = 12 and A = 0.07. On substituting these values, we get φ = 12 x 0.07 = 0.84 units.
4. Find the energy of a coil of inductance 18mH and current passing through it 1.25A.(in 10 -3 order)
a) 14.06
b) 61
c) 46.1
d) 28.12
Answer: a
Explanation: The magnetic energy possessed by a coil is given by E = 0.5 x LI 2 . Put L = 18 x 10 -3 and I = 1.25, thus we get E = 0.5 x 18 x 10 -3 x 1.25 2 = 14.06 x 10 -3 units.
5. Using Maxwell equation which of the following cannot be calculated directly?
a) B
b) D
c) A
d) H
Answer: c
Explanation: The Maxwell equations can be used to compute E,H,D,B and J directly. It is not possible to find the magnetic vector potential A directly.
6. Which of the following relation will hold good?
a) D = μ H
b) B = ε E
c) E = ε D
d) B = μ H
Answer: d
Explanation: The magnetic flux density is the product the permeability and the magnetic field intensity. This statement is always true for any material .
7. The permeability and permittivity of air or free space is unity. State true/false.
a) True
b) False
Answer: b
Explanation: The permeability and permittivity of free space or air is always unity. This implies that the air is always ready to store electric or magnetic charges subjected to it.
8. Choose the best relation.
a) A = -Div
b) V = Curl
c) H = -Grad
d) V = Div
Answer: c
Explanation: For any magnetic field, the magnetic field intensity will be the negative gradient of the potential of the field. This is given by H = -Grad.
9. Find the magnetic field when the magnetic vector potential is a unit vector.
a) 1
b) -1
c) 0
d) 2
Answer: c
Explanation: We know that H = -Grad, where is a unit vector. The gradient of a constant/unit vector will be zero. Thus the magnetic field intensity will be zero.
Answer: d
Explanation: The electromagnetic wave experiences Lorentz force which is the combination of the electrostatic force and magneto static force. It is given by F = QE + Q.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Magnetic Force and Lorentz Force”.
1. Find the electric force when the charge of 2C is subjected to an electric field of 6 units.
a) 6
b) 3
c) 12
d) 24
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.
2. Find the magnetic force when a charge 3.5C with flux density of 4 units is having a velocity of 2m/s.
a) 14
b) 28
c) 7
d) 32
Answer: b
Explanation: The magnetic force is given by F = q, where q = 3.5C, v = 2m/s and B = 4 units. Thus we get F = 3.5 = 28 units.
3. Find the electric field when the velocity of the field is 12m/s and the flux density is 8.75 units.
a) 510
b) 105
c) 150
d) 165
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.
4. Find the Lorentz force of a charge 2.5C having an electric field of 5 units and magnetic field of 7.25 units with a velocity 1.5m/s.
a) 39.68
b) 68.39
c) 86.93
d) 93.68
Answer: a
Explanation: The Lorentz force is given by F = qE + q, 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 + 2.5 = 39.68 units.
5. The force on a conductor of length 12cm having current 8A and flux density 3.75 units at an angle of 300 is
a) 1.6
b) 2
c) 1.4
d) 1.8
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.
6. The force per unit length of two conductors carrying equal currents of 5A separated by a distance of 20cm in air(in 10 -6 order)
a) 25
b) 35
c) 40
d) 50
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.
7. When currents are moving in the same direction in two conductors, then the force will be
a) Attractive
b) Repulsive
c) Retracting
d) Opposing
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.
8. Find the flux density due to a conductor of length 6m and carrying a current of 3A(in 10 -7 order)
a) 1
b) 10
c) 100
d) 0.1
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.
9. Find the maximum force of the conductor having length 60cm, current 2.75A and flux density of 9 units.
a) 14.85
b) 18.54
c) 84.25
d) 7.256
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.
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 and E = 0.5 x B 2 /μ. It is clear that B and F are related.
This set of Electromagnetic Theory MCQs focuses on “Magnetic Torque”.
1. Find the force that exists in an electromagnetic wave.
a) Electrostatic force
b) Magnetostatic force
c) Lorentz force
d) Electromotive force
Answer: c
Explanation: In an electromagnetic wave, the force of the electric and magnetic field both coexist. This is given by F = qE + q. It is called Lorentz force.
2. In an field having a force of 12N and distance 20cm, the torque will be
a) 0.24
b) 2.4
c) 24
d) 12/20
Answer: b
Explanation: The torque is defined as the product of the force and distance in a field. Thus T = F x d = 12 x 0.2 = 2.4 units.
3. Find the torque in a conductor having current 2A, flux density 50 units, length 15cm and distance of 8m.
a) 120
b) 240
c) 800
d) 350
Answer: a
Explanation: The torque on a conductor is given by T = BILd, where L x d is the area of the conductor. Thus the torque will be, T = 50 x 2 x 0.15 x 8 = 120 units.
4. The distance of the conductor when the area and length of the conductor is 24m2 and 13.56m.
a) 1.76
b) 2.67
c) 1.52
d) 2.15
Answer: a
Explanation: We know that the surface integral is the area component which is the product of two dimensions given by length and distance in a conductor. Thus A = L x d. To get d, d = A/L = 24/13.56 = 1.76 units.
5. The torque on a conductor with flux density 23 units, current 1.6A and area 6.75 units will be
a) 248.4
b) 192.6
c) 175.4
d) 256.9
Answer: a
Explanation: The maximum torque on a conductor will be at perpendicular angle ie, at 90. The torque will be given as T = BIA, where B = 23, I = 1.6 and A = 6.75.Thus we get, T = 23 x 1.6 x 6.75 = 248.4 units.
6. Consider the conductor to be a coil of turns 60 and the flux density to be 13.5 units, current 0.12A and area 16units. The torque will be
a) 1555.2
b) 1222.5
c) 525.1
d) 255.6
Answer: a
Explanation: For a single turn or loop, the torque will be BIA. For N turns, the torque will be T = NBIA, where N = 60, B = 13.5, I = 0.12 and A = 16. Thus T = 60 x 13.5 x 0.12 x 16 = 1555.2 units.
7. The torque of a conductor is defined only in the case when
a) The field is perpendicular to the loop
b) The plane of the loop is parallel to the field
c) The plane of the loop is perpendicular to the current direction
d) The field and the current direction are same
Answer: b
Explanation: The torque of a conductor is given by T = NBIA. This equation of the conductor is valid only when the plane of the loop is parallel to the magnetic field applied to it.
8. Find the angle at which the torque is minimum.
a) 30
b) 45
c) 60
d) 90
Answer: d
Explanation: The torque of a conductor loop is given by T = BIA cos θ. The torque is minimum refers to zero torque. This is possible only when the angle is 90 or perpendicular.
9. The magnetic moment and torque are related as follows
a) T = BM
b) B = TM
c) M = TB
d) T = M
Answer: a
Explanation: The torque is defined as the product of the magnetic flux density and the magnetic moment. It is given by T = BM, where M = IA is the magnetic moment.
Answer: a
Explanation: The magnetic moment is given by the ratio of the torque and the magnetic flux density. Thus M = T/B, where T = 20 and B = 51 units. We get M = 20/51 = 0.39 units.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Magnetic Dipole”.
1. The magnetic moment of a field with current 12A and area 1.6 units is
a) 19.2
b) 12.9
c) 21.9
d) 91.2
Answer: a
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.
2. Find the torque of a loop with magnetic moment 12.5 and magnetic flux density 7.65 units is
a) 95.625
b) 65.925
c) 56.525
d) 65.235
Answer: a
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.
3. The magnetization is defined by the ratio of
a) Magnetic moment to area
b) Magnetic moment to volume
c) Magnetic flux density to area
d) Magnetic flux density to volume
Answer: b
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.
4. Find the orbital dipole moment in a field of dipoles of radius 20cm and angular velocity of 2m/s(in 10 -22 order)
a) 64
b) 76
c) 54
d) 78
Answer: a
Explanation: The orbital dipole moment is given by M = 0.5 x eVangx r 2 , 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 -19 x 2 x 0.2 2 = 64 x 10 -22 units.
5. Find the orbital angular moment of a dipole with angular velocity of 1.6m/s and radius 35cm
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 r 2 ,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 0.35 2 = 1.78 x 10 -31 units.
6. The ratio of the orbital dipole moment to the orbital angular moment is given by
a) e/m
b) –e/m
c) e/2m
d) –e/2m
Answer: d
Explanation: The orbital dipole moment is given by M = 0.5 x eVangx r 2 and the orbital angular moment is given by Ma = m x Vangx r 2 . Their ratio M/Ma is given by –e/2m, the negative sign indicates the charge of electron.
7. Calculate the Larmer angular frequency for a magnetic flux density of 12.34 x 10 -10 .
a) 108.36
b) 810.63
c) 368.81
d) 183.36
Answer: a
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.
8. 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 is a physical constant and the natural unit for expressing the magnetic moment of an electron caused by either its orbital or spin angular momentum. It is given by eh/4πm, where h is the Planck’s constant, e is the charge of the electron and m is the mass of the electron.
9. Find the magnetization of the field which has a magnetic moment 16 units in a volume of 1.2 units.
a) 16.67
b) 13.33
c) 15.56
d) 18.87
Answer: b
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.
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.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Magnetic Materials”.
1. The presence of parallel alignment of magnetic dipole moment is given by which materials?
a) Diamagnetic
b) Ferromagnetic
c) Paramagnetic
d) Ferromagnetic
Answer: b
Explanation: The ferromagnetic materials are characterized by parallel alignment of magnetic dipole moments. Their susceptibility is very large.
2. The magnetic materials follow which law?
a) Faraday’s law
b) Ampere law
c) Lenz law
d) Curie Weiss law
Answer: d
Explanation: Generally, the ferromagnetic, paramagnetic and diamagnetic materials follow the Curie Weiss law, which relates the magnetization and the applied field.
3. Find the internal field when the applied field is 12 units, molecular field constant is 0.1 units and the magnetization is 74 units.
a) 86
b) 62
c) 752
d) 19.4
Answer: d
Explanation: From Curie law, the internal field of a magnetic material is given by H = Ho + χ M, where χ is the molecular field constant. Put χ = 0.1, M = 74 and Ho = 12, we get H = 12 + 74 = 19.4 units.
4. In which materials the magnetic anisotropy is followed?
a) Diamagnetic
b) Paramagnetic
c) Ferromagnetic
d) Ferromagnetic
Answer: c
Explanation: In materials like iron, the magnetic properties depend on the direction in which they are measured. This is magnetic anisotropy. The material iron is a ferromagnetic material type.
5. Piezoelectric effect is analogous to which phenomenon?
a) Electrostriction
b) Magnetostriction
c) Anisotropy
d) Magnetization
Answer: b
Explanation: The piezoelectric effect is the mechanical strain caused on a material like quartz when subjected to an electric field. The same is observed in a ferromagnetic material called magnetostriction.
6. The converse of magnetostriction is called the
a) Magnetization
b) Magnetic anisotropy
c) Villari effect
d) Curie effect
Answer: c
Explanation: When a strain is applied, the change in magnetic field is observed. This is the converse of the magnetostriction phenomenon and is called Villari effect.
7. The materials having very small susceptibility at all temperatures are
a) Antiferromagnetic
b) Diamagnetic
c) Ferromagnetic
d) Paramagnetic
Answer: a
Explanation: In antiferromagnetic materials, the susceptibility will decrease with increase in temperature. They have relatively small susceptibility at all temperatures.
8. Find the susceptibility when the curie constant is 0.2 and the difference in critical temperature and paramagnetic curie temperature is 0.01.
a) 2
b) 20
c) 0.02
d) 200
Answer: b
Explanation: The susceptibility in magnetic materials is given by χm = C/, where C is the curie constant, T is the critical temperature and θ is the paramagnetic curie temperature. Put C = 0.2 and T-θ = 0.01, thus we get susceptibility as 0.2/0.01 = 20.
9. The susceptibility is independent of temperature in which material?
a) Paramagnetic
b) Ferromagnetic
c) Diamagnetic
d) Ferromagnetic
Answer: c
Explanation: In the diamagnetic materials, the susceptibility is very small and negative. Thus the susceptibility will be independent of the temperature. The atoms of solids having closed shells and metals like gold have this property.
Answer: a
Explanation: The ferromagnetic materials are iron, nickel, cobalt which are highly attracted by magnetic field. Thus their susceptibility is also very high and nearing infinity. Also ferrimagnetics have infinite susceptibility.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Magnetization”.
1. 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.
2. 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 -7 x 5/(2π x 100 x 10 -9 ) = 10 units.
3. 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.
4. 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.
5. The expression for magnetization is given by
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.
6. 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.
7. 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 m H. Put χm = 50 and H = 0.25, then M = 50 x 0.25 = 12.5 units.
8. 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.
9. Which of the following materials is ferrimagnetic?
a) Fe
b) Sn
c) Fe 2 O 3
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 Fe 2 O 3 is said to be a ferrimagnetic material.
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.
This set of Electromagnetic Theory online test focuses on “Magnetic Boundary Conditions”.
1. Find the correct relation between current density and magnetization.
a) J = Grad
b) J = Div
c) J = Curl
d) M = Curl
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
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
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.
3. 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.
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
Explanation: 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
Explanation: 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
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.
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
Explanation: 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
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.
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
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.
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.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Inductance”.
1. Calculate the emf of a coil with turns 100 and flux rate 5 units.
a) 20
b) -20
c) 500
d) -500
Answer: d
Explanation: The emf is the product of the turns of the coil and the flux rate. Thus e = -N dφ/dt, where the negative sign indicates that the emf induced is opposing the flux. Thus e = -100 x 5 = -500 units.
2. The equivalent inductances of two coils 2H and 5H in series aiding flux with mutual inductance of 3H is
a) 10
b) 30
c) 1
d) 13
Answer: d
Explanation: The equivalent inductance of two coils in series is given by L = L1 + L2 + 2M, where L1 and L2 are the self inductances and M is the mutual inductance. Thus L = 2 + 5 + 2 = 13H.
3. The expression for the inductance in terms of turns, flux and current is given by
a) L = N dφ/di
b) L = -N dφ/di
c) L = Niφ
d) L = Nφ/i
Answer: a
Explanation: We know that e = -N dφ/dt and also e = -L di/dt. On equating both we get, L = Ndφ/di is the expression for inductance.
4. The equivalent inductance of two coils with series opposing flux having inductances 7H and 2H with a mutual inductance of 1H.
a) 10
b) 7
c) 11
d) 13
Answer: b
Explanation: The equivalent inductance of two coils in series with opposing flux is L = L1 + L2 – 2M, where L1 and L2 are the self inductances and M is the mutual inductance. Thus L = 7 + 2 – 2 = 7H.
5. A coil is said to be loosely coupled with which of the following conditions?
a) K>1
b) K<1
c) K>0.5
d) K<0.5
Answer: d
Explanation: k is the coefficient of coupling. It lies between 0 and 1. For loosely coupled coil, the coefficient of coupling will be very less. Thus the condition K<0.5 is true.
6. With unity coupling, the mutual inductance will be
a) L1 x L2
b) L1/L2
c) √
d) L2/L1
Answer: c
Explanation: The expression for mutual inductance is given by M = k √, where k is the coefficient of coupling. For unity coupling, k = 1, then M = √.
7. The inductance is proportional to the ratio of flux to current. State True/False.
a) True
b) False
Answer: a
Explanation: The expression is given by L = Ndφ/di. It can be seen that L is proportional to the ratio of flux to current. Thus the statement is true.
8. Calculate the mutual inductance of two tightly coupled coils with inductances 49H and 9H.
a) 21
b) 58
c) 40
d) 49/9
Answer: a
Explanation: For tightly coupled coils, the coefficient of coupling is unity. Then the mutual inductance will be M = √= √ = 21 units.
9. Find the inductance of a coil with turns 50, flux 3 units and a current of 0.5A
a) 150
b) 300
c) 450
d) 75
Answer: b
Explanation: The self inductance of a coil is given by L = Nφ/I, where N = 50, φ = 3 and I = 0.5. Thus L = 50 x 3/0.5 = 300 units.
Answer: a
Explanation: The inductance of a coaxial cable with inner radius a and outer radius b, from a distance d, is a standard formula derived from the definition of the inductance. This is given by L = μd ln/2π.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Magnetic Energy and Circuits”.
1. The magnetic energy of a magnetic material is given by
a) BH/2
b) B/2H
c) H/2B
d) B/H
Answer: a
Explanation: The magnetic energy of a material is given by half of the product of the magnetic flux density and the magnetic field intensity. It is represented as BH/2. Since B = μH, we can also write as μH 2 or B 2 /2μ.
2. The induced emf in a material opposes the flux producing it. This is
a) Faraday law
b) Ampere law
c) Lenz law
d) Curie law
Answer: c
Explanation: The induced emf in a material under the influence of a magnetic field will oppose the flux that produces it. This is indicated by a negative sign in the emf equation. This phenomenon is called Lenz law.
3. The energy in a magnetic material is due to which process?
a) Emf
b) Magnetization
c) Magnetostriction
d) Polarization
Answer: b
Explanation: The energy in a magnetic material is due to the formation of magnetic dipoles which are held together due to magnetic force. This gives energy to the material. Hence it is due to magnetization process.
4. The flux lines of two energised coils overlapping on each other will give
a) Series aiding
b) Shunt aiding
c) Series opposing
d) Shunt opposing
Answer: a
Explanation: Flux lines are the magnetic lines of force of a magnetic material. Since the flux is overlapping, the total flux of the two coils together will be high. Thus it is an aiding flux. Also this type of overlapping is possible only when the two coils are back to back or in series connection.
5. The resistance in a magnetic material is called as
a) Capacitance
b) Inductance
c) Reluctance
d) Magnetic resistance
Answer: c
Explanation: The reluctance of a magnetic material is the ability of the material to oppose the magnetic flux. It is the ratio of the magnetic motive force mmf to the flux.
6. Calculate the reluctance of the material with a mmf of 3.5 units and flux of 7units.
a) 32.5
b) 10.5
c) 0.5
d) 2
Answer: c
Explanation: The reluctance is defined as the ratio of the mmf and the flux. It is given by S = mmf/φ. On substituting mmf = 3.5 and φ = 7, we get S = 3.5/7 = 0.5 units.
7. Which of the following relations is correct?
a) NI = Sφ
b) NS = Iφ
c) Nφ = SI
d) NI = S/φ
Answer: a
Explanation: The reluctance is also defined by the ratio of the current element to the flux. In other words, mmf = NI. Thus S = NI/φ. We get the relation NI = Sφ.
8. Calculate the reluctance of a material with length 2π x 10 -4 in air with area 0.5.
a) 1
b) 10
c) 100
d) 1000
Answer: d
Explanation: The reluctance is given by S = L/μ A, where L is the length, A is the area and μ is the permeability. On substituting L = 2π x 10 -4 , A = 0.5 and μ = 4π x 10 -7 , we get S = 10 3 / = 1000 units.
9. Ampere turn is equivalent to which element?
a) Sφ
b) S/φ
c) φ/S
d) S
Answer: a
Explanation: Ampere turn refers to the current element, which is the product of the turns and the current. It is given by NI. From the definition of reluctance, S = NI/φ. Thus NI = Sφ is the best equivalent.
Answer: d
Explanation: The line integral of H is given by ∫H. dl. From Ampere law it can be related to the current density and hence the current element NI for a coil of N turns. Thus, ∫H. dl = NI.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Maxwell Law 1”.
1. The first Maxwell law is based on which law?
a) Ampere law
b) Faraday law
c) Lenz law
d) Faraday and Lenz law
Answer: d
Explanation: The first Maxwell equation states that Curl = -dB/dt. It is based on the emf concept. Thus it is derived from the Faraday and Lenz law.
2. The benefit of Maxwell equation is that
a) Any parameter can be calculated
b) Antenna can be designed
c) Polarisation of the wave can be calculated
d) Transmission line constants can be found
Answer: a
Explanation: The Maxwell equation relates the parameters E, D, H, B. When one parameter is known the other parameters can be easily calculated. In other words, it is used to relate an electric field parameter with its equivalent magnetic field.
3. The correct sequence to find H, when D is given is
a) D-E-B-H
b) D-B-E-H
c) It cannot be computed from the data given
d) D-H
Answer: a
Explanation: There is no direct relation between D and H, so the option D-H is not possible. Using the formula D = εE, the parameter E can be computed from D. By Maxwell equation, Curl = -dB/dt, the parameter B can be calculated. Using the formula B = μH, the parameter H can be calculated. Thus the sequence is D-E-B-H.
4. The curl of the electric field intensity is
a) Conservative
b) Rotational
c) Divergent
d) Static
Answer: b
Explanation: The curl of electric field intensity is Curl. From Maxwell law, the curl of E is a non-zero value. Thus E will be rotational.
5. Which of the following identities is always zero for static fields?
a) Grad
b) Curl
c) Div
d) Curl
Answer: d
Explanation: The curl of gradient of a vector is always zero. This is because the gradient of V is E and the curl of E is zero for static fields.
6. Find the Maxwell first law value for the electric field intensity is given by A sin wt az
a) 0
b) 1
c) -1
d) A
Answer: a
Explanation: The value of Maxwell first equation is Curl. The curl of E is zero. Thus for the given field, the value of Maxwell equation is zero. Thus the field is irrotational.
7. Find the electric field applied on a system with electrons having a velocity 5m/s subjected to a magnetic flux of 3.6 units.
a) 15
b) 18
c) 1.38
d) 0.72
Answer: b
Explanation: The electric field intensity is the product of the velocity and the magnetic flux density. Thus E = v x B, on substituting v = 5 and B = 3.6, we get E = 5 x 3.6 = 18 units.
8. Which of the following relations holds good?
a) Bq = ILE
b) E = ILBq
c) Eq = ILB
d) B = ILEq
Answer: c
Explanation: The force of a electrostatic field in given by F = Eq. The force on a conductor is given by F = BIL. In the case when a charge exists on a conductor, both the forces can be equated. Thus Eq = BIL is true.
9. When the Maxwell equation is expressed in frequency domain, then which substitution is possible?
a) d/dt = w/j
b) d/dt = j/w
c) d/dt = jw
d) Expression in frequency domain is not possible
Answer: c
Explanation: The conversion of time to frequency domain in Maxwell equation is given by the Fourier Transform. Differentiation in time gives jw in frequency domain. Thus d/dt = jw in frequency domain.
10. Calculate the emf of a material having a flux linkage of 2t 2 at time t = 1second.
a) 2
b) 4
c) 8
d) 16
Answer: b
Explanation: The emf of a material is given by Vemf = -dλ/dt. On substituting λ = 2t 2 , the emf is 4t. At t = 1 sec, the emf will be 4 units.
11. Calculate the emf of a material having flux density 5sin t in an area of 0.5 units.
a) 2.5 sin t
b) -2.5 cos t
c) -5 sin t
d) 5 cos t
Answer: d
Explanation: The emf can be written as Vemf = -d/dt. It can be written as Vemf = -B= -5sin t, since the integration and differentiation gets cancelled.
Answer: a
Explanation: Using Maxwell equation, from B we can calculate E by Curl = -dB /dt. From E, D can be calculated by D = εE. Thus the sequence is B->E->D.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Maxwell Law 2”.
1. Maxwell second equation is based on which law?
a) Ampere law
b) Faraday law
c) Lenz law
d) Coulomb law
Answer: a
Explanation: The second Maxwell equation is based on Ampere law. It states that the field intensity of a system is same as the current enclosed by it, i.e, Curl = J.
2. The Maxwell second equation that is valid in any conductor is
a) Curl = Jc
b) Curl = Jc
c) Curl = Jd
d) Curl = Jd
Answer: a
Explanation: For conductors, the conductivity parameter σ is significant and only the conduction current density exists. Thus the component J = Jc and Curl = Jc.
3. In dielectric medium, the Maxwell second equation becomes
a) Curl = Jd
b) Curl = Jc
c) Curl = Jd
d) Curl = Jd
Answer: a
Explanation: In dielectric medium conductivity σ will be zero. So the current density has only the displacement current density. Thus the Maxwell equation will be Curl = Jd.
4. Find the displacement current density of a material with flux density of 5sin t
a) 2.5cos t
b) 2.5sin t
c) 5cos t
d) 5sin t
Answer: c
Explanation: The displacement current density is the derivative of the flux density. Thus Jd = dD/dt. Put D = 5sin t in the equation, we get Jd = 5cos t units.
5. Find the conduction current density of a material with conductivity 200units and electric field 1.5 units.
a) 150
b) 30
c) 400/3
d) 300
Answer: d
Explanation: The conduction current density is given by Jc = σE, where σ = 200 and E = 1.5. Thus we get, Jc = 200 x 1.5 = 300 units.
6. Calculate the conduction density of a material with resistivity of 0.02 units and electric intensity of 12 units.
a) 300
b) 600
c) 50
d) 120
Answer: b
Explanation: The conduction density is given by Jc = σE, where σ is the inverse of resistivity and it is 1/0.02 = 50. Thus we get, Jc = 50 x 12 = 600 units.
7. In the conversion of line integral of H into surface integral, which theorem is used?
a) Green theorem
b) Gauss theorem
c) Stokes theorem
d) It cannot be converted
Answer: c
Explanation: To convert line integral to surface integral, i.e, in this case from line integral of H to surface integral of J, we use the Stokes theorem. Thus the Maxwell second equation can be written as ∫H.dl = ∫∫J.ds.
8. An implication of the continuity equation of conductors is given by
a) J = σ E
b) J = E/σ
c) J = σ/E
d) J = jwEσ
Answer: a
Explanation: The continuity equation indicates the current density in conductors. This is the product of the conductivity of the conductor and the electric field subjected to it. Thus J = σE is the implication of the continuity equation for conductors.
9. Find the equation of displacement current density in frequency domain.
a) Jd = jwεE
b) Jd = jwεH
c) Jd = wεE/j
d) Jd = jεE/w
Answer: a
Explanation: The displacement current density is Jd = dD/dt. Since D = εE and in frequency domain d/dt = jw, thus we get Jd = jwεE.
10. The total current density is given as 0.5i + j – 1.5k units. Find the curl of the magnetic field intensity.
a) 0.5i – 0.5j + 0.5k
b) 0.5i + j -1.5k
c) i – j + k
d) i + j – k
Answer: b
Explanation: By Maxwell second equation, the curl of H is same as the sum of conduction current density and displacement current density. Thus Curl = J = 0.5i + j – 1.5k units.
11. At dc field, the displacement current density will be
a) 0
b) 1
c) Jc
d) ∞
Answer: a
Explanation: The DC field refers to zero frequency. The conduction current is independent of the frequency, whereas the displacement current density is dependent on the frequency, i.e, Jd = jwεE. Thus at DC field, the displacement current density will be zero.
Answer: c
Explanation: Conduction density exists only for good conductors and displacement density is for dielectrics in any medium at high frequency. Thus both coexist when a conductor is placed in a dielectric medium.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Maxwell Law 3”.
1. The charge density of a electrostatic field is given by
a) Curl of E
b) Divergence of E
c) Curl of D
d) Divergence of D
Answer: d
Explanation: From the Gauss law for electric field, the volume charge density is the divergence of the electric flux density of the field. Thus Div = ρv.
2. In the medium of free space, the divergence of the electric flux density will be
a) 1
b) 0
c) -1
d) Infinity
Answer: b
Explanation: In free space or air, the charge density will be zero. In other words, the conduction is possible in mere air medium. By gauss law, since the charge density is same as the divergence of D, the Div in air/free space will be zero.
3. In a medium other than air, the electric flux density will be
a) Solenoidal
b) Curl free
c) Irrotational
d) Divergent
Answer: d
Explanation: In any medium other than the air, the conduction is possible, due to the charge carriers. Thus charge density is also non-zero. We can write from Gauss law that Div is non-zero. When the divergence is said to be non-zero, the field is not solenoidal or called as divergent field.
4. For a solenoidal field, the surface integral of D will be,
a) 0
b) 1
c) 2
d) 3
Answer: a
Explanation: For a solenoidal field, the divergence will be zero. By divergence theorem, the surface integral of D and the volume integral of Div is same. So as the Div is zero for a solenoidal field, the surface integral of D is also zero.
5. In a dipole, the Gauss theorem value will be
a) 1
b) 0
c) -1
d) 2
Answer: b
Explanation: The Gauss theorem for an electric field is given by Div= ρ. In a dipole only static charge exists and the divergence will be zero. Thus the Gauss theorem value for the dipole will be zero.
6. Find the electric flux density of a material whose charge density is given by 12 units in a volume region of 0.5 units.
a) 12
b) 24
c) 6
d) 48
Answer: c
Explanation: By Gauss law, Div = ρv. To get D, integrate the charge density given. Thus D = ∫ρv dv, where ρv = 12 and ∫dv = 0.5. We get, D = 12 x 0.5 = 6 units.
7. From the Gauss law for electric field, we can compute which of the following parameters?
a) B
b) H
c) E
d) A
Answer: c
Explanation: From the Gauss law for electric field, we can find the electric flux density directly. On substituting, D= ε E, the electric field intensity can be calculated.
8. The charge density of a system with the position vector as electric flux density is
a) 0
b) 1
c) 2
d) 3
Answer: d
Explanation: The divergence of the electric flux density is the charge density. For a position vector xi + yj + zk, the divergence will be 1 + 1 + 1 = 3. Thus by Gauss law, the charge density is also 3.
9. The sequence for finding E when charge density is given is
a) E-D-ρv
b) E-B-ρv
c) E-H-ρv
d) E-V-ρv
Answer: a
Explanation: From the given charge density ρv, we can compute the electric flux density by Gauss law. Since, D = εE, the electric field intensity can also be computed. Thus the sequence is E-D-ρv.
Answer: c
Explanation: The Gauss divergence theorem is given by ∫ D.ds = ∫Div.dv. From the theorem value, we can compute the charge density. Thus Gauss law employs the Gauss divergence theorem.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Maxwell Law 4”.
1. Which quantity is solenoidal in the electromagnetic theory?
a) Electric field intensity
b) Electric flux density
c) Magnetic field intensity
d) Magnetic flux density
Answer: d
Explanation: The divergence of the magnetic flux density is zero. This is the Maxwell fourth equation. As the divergence is zero, the quantity will be solenoidal or divergent less.
2. Which equation will be true, if the medium is considered to be air?
a) Curl = 0
b) Div = 0
c) Grad = 0
d) Div = 1
Answer: b
Explanation: From the Gauss law for magnetic field, the divergence of the magnetic flux density is zero. Also B = μH. Thus divergence of H is also zero, i.e, Div = 0 is true.
3. Find the sequence to find B when E is given.
a) E-D-H-B
b) B-E-D
c) H-B-E-D
d) V-E-B
Answer: a
Explanation: From E, D can be computed as D = εE. Using the Ampere law, H can be computed from D. Finally, B can be calculated from H by B = μH.
4. The Gauss law for magnetic field is valid in
a) Air
b) Conductor
c) Dielectric
d) All cases
Answer: d
Explanation: The Gauss law for magnetic field states that the divergence of B is always zero. This is valid for all cases like free space, dielectric medium etc.
5. The sequence for finding H from E is
a) E-B-H
b) E-V-H
c) E-D-H
d) E-A-H
Answer: a
Explanation: From E, we can compute B using the Maxwell first law. Using B, the parameter H can be found since B = μH. Thus the sequence is E-B-H is true.
6. The reason for non existence of magnetic monopoles is
a) The magnetic field cannot be split
b) Due to permeability
c) Due to magnetization
d) Due to magnetostriction
Answer: a
Explanation: Practically monopoles do not exist, due to the connection between north and south poles. But theoretically, they exist. The reason for their non- existence practically is that, the magnetic field confined to two poles cannot be split or confined to a single pole.
7. The non existence of the magnetic monopole is due to which operation?
a) Gradient
b) Divergence
c) Curl
d) Laplacian
Answer: b
Explanation: The Maxwell fourth law or the Gauss law for magnetic field states that the divergence of B is zero, implies the non existence of magnetic monopoles. Thus the operation involved is divergence.
8. Will dielectric breakdown lead to formation of magnetic monopole?
a) Yes
b) No
Answer: b
Explanation: When dielectric breakdown occurs, the material loses its dielectric property and becomes a conductor. When it is subjected to a magnetic field, north and south flux lines coexists, giving magnetic force. Thus there exists magnetic dipole. Suppose if the conductor is broken into very small pieces, still there exist a magnetic dipole in every broken part. In other words, when a piece is broken into half, there cannot exist a north pole in one half and a south pole in the other. Thus monopoles never exist.
9. Which equation will hold good for a magnetic material?
a) Line integral of H is zero
b) Surface integral of H is zero
c) Line integral of B is zero
d) Surface integral of B is zero
Answer: d
Explanation: We know that the divergence of B is zero. From Stokes theorem, the surface integral of B is equal to the volume integral of divergence of B. Thus surface integral of B is also zero.
Answer: a
Explanation: In any magnetic material or magnet, the dipoles exist. This is due to the magnetic lines of force joining the north to south poles. The interaction between these two poles together leads to dipole formation.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Maxwell Law in Time Static Fields”.
1. Calculate the emf in a material with flux linkage of 3.5t 2 at 2 seconds.
a) 3.5
b) -7
c) -14
d) 28
Answer: c
Explanation: The emf induced in a material with flux linkage is given by Vemf = -dλ/dt. On substituting λ= 3.5t 2 , we get emf = -7t. At time t = 2sec, the emf will be -14 units.
2. Find the emf induced in a coil of 60 turns with a flux rate of 3 units.
a) -60
b) -180
c) 60
d) 180
Answer: b
Explanation: The emf induced is the product of the turns and the flux rate. Thus Vemf = -Ndφ/dt. On substituting N = 60 and dφ/dt = 3, we get emf as -60 x 3 = -180 units.
3. Find the electric field intensity of a charge 2.5C with a force of 3N.
a) -7.5
b) 7.5
c) 2.5/3
d) 3/2.5
Answer: d
Explanation: The electric field intensity is the electric force per unit charge. It is given by E = F/q. On substituting F = 2.5 and q = 3, we get E = 3/2.5 units.
4. The electric field intensity of a field with velocity 10m/s and flux density of 2.8 units is
a) 0.28
b) 28
c) 280
d) 10/2.8
Answer: b
Explanation: The electric field is the product of the velocity and the magnetic flux density given by E = v x B. On substituting v = 10 and B = 2.8, we get E = 10 x 2.8 = 28 units.
5. The line integral of the electric field intensity is
a) Mmf
b) Emf
c) Electric potential
d) Magnetic potential
Answer: b
Explanation: From the Maxwell first law, the transformer emf is given by the line integral of the electric field intensity. Thus the emf is given by ∫ E.dl.
6. Which of the following relations is correct?
a) MMF = ∫ B.dl
b) MMF = ∫ H.dl
c) EMF = ∫ E.dl
d) EMF = ∫ D.dl
Answer: c
Explanation: The emf induced in a material is given by the line integral of the electric field intensity. Thus EMF = ∫ E.dl is the correct relation.
7. For static fields, the curl of E will be
a) Rotational
b) Irrotational
c) Solenoidal
d) Divergent
Answer: b
Explanation: For static fields, the charges will be constant and the field is constant. Thus curl of the electric field intensity will be zero. This implies the field is irrotational.
8. The line integral of which parameter is zero for static fields?
a) E
b) H
c) D
d) B
Answer: a
Explanation: The field is irrotational for static fields. Thus curl of E is zero. From Stokes theorem, the line integral of E is same as the surface integral of the curl of E. Since it is zero, the line integral of E will also be zero.
9. The magnitude of the conduction current density for a magnetic field intensity of a vector yi + zj + xk will be
a) 1.414
b) 1.732
c) -1.414
d) -1.732
Answer: b
Explanation: From the Ampere circuital law, the curl of H is the conduction current density. The curl of H = yi + zj + xk is –i – j – k. Thus conduction current density is –i – j – k. The magnitude will be √ = √3 = 1.732 units.
Answer: d
Explanation: The Gauss law for electric field states that the divergence of the electric flux density is the charge density. Thus Div = ρ. For D as a position vector, the divergence of the position vector D will be always 3. Thus the charge density is also 3.
This set of Electromagnetic Theory online quiz focuses on “Maxwell Law in Time Varying Fields”.
1. Find the curl of E when B is given as 15t.
a) 15
b) -15
c) 7.5
d) -7.5
Answer: b
Explanation: From Maxwell first law, we get Curl of E as the negative derivative of B with respect to time. Thus Curl = -dB/dt. On substituting B= 15t and differentiating, Curl = -15 units.
2. The charge build up in a capacitor is due to
a) Conduction current density
b) Displacement current density
c) Polarisation
d) Magnetization
Answer: b
Explanation: The capacitor consists of a dielectric placed between two conducting plates, subjected to a field. The current due to a dielectric is always due to the displacement current density.
3. The surface integral of which parameter is zero?
a) E
b) D
c) B
d) H
Answer: c
Explanation: The divergence of the magnetic flux density is always zero. By Stokes theorem, the surface integral of B is same as the volume integral of the divergence of B. Thus the surface integral of B is also zero.
4. Harmonic electromagnetic fields refer to fields varying sinusoidally with respect to time. State True/False.
a) True
b) False
Answer: a
Explanation: Fields that varying sinusoidally with respect to time are called as harmonic fields. An example for harmonic fields is A sin wt.
5. When electric potential is null, then the electric field intensity will be
a) 0
b) 1
c) dA/dt
d) –dA/dt
Answer: d
Explanation: The electric field intensity is given by E = -Grad- dA/dt, where V is the electric potential and A is the magnetic vector potential. When V is zero, then E = -dA/dt.
6. The gradient of the magnetic vector potential can be expressed as
a) –με dV/dt
b) +με dE/dt
c) –με dA/dt
d) +με dB/dt
Answer: a
Explanation: The gradient of A is the ratio of the negative gradient of electric potential to the speed of light c. We can write c = 1/√. Thus grad = -με dV/dt is the required expression.
7. Find the time constant of a capacitor with capacitance of 2 microfarad having an internal resistance of 4 megaohm.
a) 2
b) 0.5
c) 8
d) 0.25
Answer: c
Explanation: The time constant of capacitor is given by T = RC, where R = 4×10 6 and C = 2×10 -6 . Thus T = 4×10 6 x2x10 -6 = 8 seconds.
8. Which components exist in an electromagnetic wave?
a) Only E
b) Only H
c) Both E and H
d) Neither E or H
Answer: c
Explanation: In an electromagnetic wave, the electric and magnetic components coexist. They propagate perpendicular to each other and to the direction of propagation in space.
9. The propagation of the electromagnetic waves can be illustrated by
a) Faraday law
b) Ampere law
c) Flemming rule
d) Coulomb law
Answer: c
Explanation: By Flemming’s rule, when the thumb and the middle finger represent the inputs , then the fore finger represents the output . The EM propagation can be illustrated by this rule.
Answer: d
Explanation: The Gauss law, Faraday law and the Ampere law are directly used to find the parameters E, H, D, B. Thus it contributes to the Maxwell equations. The Curie Weiss law pertains to the property of any magnetic material. Thus it is not related to the Maxwell equation.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Loss Tangent”.
1. The loss tangent refers to the
a) Power due to propagation in conductor to that in dielectric
b) Power loss
c) Current loss
d) Charge loss
Answer: a
Explanation: The loss tangent is the tangent angle formed by the plot of conduction current density vs displacement current density. It is the ratio of Jc by Jd. It represents the loss of power due to propagation in a dielectric, when compared to that in a conductor.
2. Calculate the conduction current density when the resistivity of a material with an electric field of 5 units is 4.5 units.
a) 22.5
b) 4.5/5
c) 5/4.5
d) 9.5
Answer: c
Explanation: The conduction current density is the product of the conductivity and the electric field. The resistivity is the reciprocal of the conductivity. Thus the required formula is Jc = σ E = E/ρ = 5/4.5 units.
3. At high frequencies, which parameter is significant?
a) Conduction current
b) Displacement current
c) Attenuation constant
d) Phase constant
Answer: b
Explanation: The conduction current occurs in metals and is independent of the frequency. The attenuation and phase constant highly depend on the varying frequency. The displacement current occurs due to dielectrics and is significant only at very high frequencies.
4. Find the loss tangent of a material with conduction current density of 5 units and displacement current density of 10 units.
a) 2
b) 0.5
c) 5
d) 10
Answer: b
Explanation: The loss tangent is the ratio of Jc by Jd. On substituting for Jc = 5 and Jd = 10, the loss tangent, tan δ = 5/10 = 0.5. It is to be noted that it is tangent angle, so that the maxima and minima lies between 1 and -1 respectively.
5. The loss tangent is also referred to as
a) Attenuation
b) Propagation
c) Dissipation factor
d) Polarization
Answer: c
Explanation: The loss tangent is the measure of the loss of power due to propagation in a dielectric, when compared to that in a conductor. Hence it is also referred to as dissipation factor.
6. The loss tangent of a wave propagation with an intrinsic angle of 20 degree is
a) Tan 20
b) Tan 40
c) Tan 60
d) Tan 80
Answer: b
Explanation: The angle of the loss tangent δ is twice the intrinsic angle θn. Thus tan δ = tan 2θn = tan 2 = tan 40.
7. The expression for the loss tangent is given by
a) σ/ωε
b) ωε/σ
c) σ/ω
d) ω/ε
Answer: a
Explanation: The conduction current density is Jc = σ E and the displacement current density is Jd = jωεE. Its magnitude will be ωεE. Thus the loss tangent tan δ = Jc /Jd = σ/ωε is the required expression.
8. Find the loss angle in degrees when the loss tangent is 1.
a) 0
b) 30
c) 45
d) 90
Answer: c
Explanation: The loss tangent is tan δ, where δ is the loss angle. Given that loss tangent tan δ = 1. Thus we get δ = tan -1 = 450.
9. The complex permittivity is given by 2-j. Find the loss tangent.
a) 1/2
b) -1/2
c) 2
d) -2
Answer: a
Explanation: The loss tangent for a given complex permittivity of ε = ε’ – jε’’ is given by tan δ = ε’’/ ε’. Thus the loss tangent is 1/2.
Answer: d
Explanation: The angle of the loss tangent δ is twice the intrinsic angle θn. Thus tan δ = tan 2θn. We get θn = δ/2 = 60/2 = 30 degrees.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Lossy and Lossless Dielectrics”.
1. For a dielectric, the condition to be satisfied is
a) σ/ωε > 1
b) σ/ωε < 1
c) σ = ωε
d) ωε = 1
Answer: b
Explanation: In a dielectric, the conductivity will be very less. Thus the loss tangent will be less than unity. This implies σ/ωε < 1 is true.
2. For a perfect dielectric, which parameter will be zero?
a) Conductivity
b) Frequency
c) Permittivity
d) Permeability
Answer: a
Explanation: The conductivity will be minimum for a dielectric. For a perfect dielectric, the conductivity will be zero.
3. Calculate the phase constant of a wave with frequency 12 rad/s and velocity 3×10 8 m/s(in 10 -8 order)
a) 0.5
b) 72
c) 4
d) 36
Answer: c
Explanation: The phase constant is given by β = ω√, where ω is the frequency in rad/s and 1/√ is the velocity of wave. On substituting √ = 3×10 8 and ω = 12, we get β = 12/(3×10 8 ) = 4 x 10 -8 m/s.
4. For a lossless dielectric, the attenuation will be
a) 1
b) 0
c) -1
d) Infinity
Answer: b
Explanation: The attenuation is the loss of power of the wave during its propagation. In a lossless dielectric, the loss of power will not occur. Thus the attenuation will be zero.
5. Calculate the velocity of a wave with frequency 2 x10 9 rad/s and phase constant of 4 x 10 8 units.
a) 0.5
b) 5
c) 0.2
d) 2
Answer: b
Explanation: The velocity of a wave is the ratio of the frequency to the phase constant. Thus V = ω/β. On substituting the given values, we get V = 2 x10 9 / 4 x 10 8 = 5 units.
6. Which of the following is the correct relation between wavelength and the phase constant of a wave?
a) Phase constant = 2π/wavelength
b) Phase constant = 2π x wavelength
c) Phase constant = 1/
d) Phase constant = wavelength/2π
Answer: a
Explanation: The phase constant is the ratio of 2π to the wavelength λ. Thus β = 2π/λ is the correct relation.
7. In lossy dielectric, the phase difference between the electric field E and the magnetic field H is
a) 90
b) 60
c) 45
d) 0
Answer: d
Explanation: In a lossy dielectric, the E and H component will be in phase. This implies that the phase difference between E and H will be 0.
8. The intrinsic impedance is the ratio of square root of
a) Permittivity to permeability
b) Permeability to permittivity
c) Phase constant to wavelength
d) Wavelength to phase constant
Answer: b
Explanation: The intrinsic impedance is the impedance of a particular material. It is the ratio of square root of the permeability to permittivity. For air, the intrinsic impedance is 377 ohm or 120π.
9. Calculate the skin depth of a material with attenuation constant of 2 units.
a) 2
b) 1
c) 0.5
d) 4
Answer: c
Explanation: The skin depth of a material is the reciprocal of the attenuation constant. Thus δ = 1/α. On substituting for α = 2, we get δ = ½ = 0.5 units.
10. Calculate the phase constant of a wave with skin depth of 2.5 units.
a) 5/2
b) 5
c) 2
d) 2/5
Answer: d
Explanation: The skin depth is the reciprocal of the phase constant and the attenuation constant too. Thus δ = 1/β. On substituting for δ = 2.5, we get β = 1/δ = 1/2.5 = 2/5 units.
11. An example for lossless propagation is
a) Dielectric waveguide propagation
b) Conductor propagation
c) Cavity resonator propagation
d) It is not possible
Answer: d
Explanation: There are many techniques employed to achieve zero attenuation or maximum propagation. But it is not achievable practically. Thus lossless propagation is not possible practically.
Answer: c
Explanation: Skin depth is found in pure conductors. It the property of the conductor to allow a small amount of electromagnetic energy into its skin, but not completely. This is the reason why EM waves cannot travel inside a good conductor.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Dielectric and Conductor Wave Propagation”.
1. In conductors, which condition will be true?
a) σ/ωε > 1
b) σωε > 1
c) σ/ωε < 1
d) σωε < 1
Answer: a
Explanation: For conductors, the conductivity will be maximum. Thus the loss tangent is greater than unity. This is given by σ/ωε >1.
2. For metals, the conductivity will be
a) 0
b) 1
c) -1
d) Infinity
Answer: d
Explanation: Metals are pure conductors. Examples are iron, copper etc. Their conductivity will be very high. Thus the metal conductivity will be infinity. Practically the conductivity of conductors will be maximum.
3. In conductors, which two parameters are same?
a) Wavelength and phase constant
b) Phase and attenuation constant
c) Attenuation constant and skin depth
d) Skin depth and wavelength
Answer: b
Explanation: In conductors, which are considered to be lossy, the attenuation and the phase constant are the same. It is given by α=β= √.
4. Calculate the velocity of wave propagation in a conductor with frequency 5 x 10 8 rad/s and phase constant of 3 x 10 8 units.
a) 3/5
b) 15
c) 5/3
d) 8
Answer: c
Explanation: The velocity of wave propagation is the ratio of the frequency to the phase constant. It is given by V = ω/β. On substituting the given values, we get V = 5/3 units.
5. Calculate the wavelength of the wave with phase constant of 3.14 units.
a) 1
b) 2
c) 0.5
d) 4
Answer: b
Explanation: The wavelength is the ratio of 2π to the phase constant β. On substituting for β = 3.14, we get λ = 2π/β = 2π/3.14 = 2 units.
6. For dielectrics, which two components will be in phase?
a) E and wave direction
b) H and wave direction
c) Wave direction and E x H
d) E and H
Answer: d
Explanation: In dielectrics, the electric and magnetic components E and H will be in phase with each other. This is due the variation in the permittivities and the permeabilities of the dielectric surfaces. The phase difference between E and H will be 0.
7. In perfect conductors, the phase shift between the electric field and magnetic field will be
a) 0
b) 30
c) 45
d) 90
Answer: c
Explanation: For perfect conductors, the electric and magnetic field E and H respectively vary by a phase of 45 degree. This is due to the polarisation phenomenon in the conductors, unlike dielectrics.
8. The expression for phase constant is given by
a) Phase constant β = ωμε
b) Phase constant ω = με
c) Phase constant β = ω√
d) Phase constant β = 1/ωμε
Answer: c
Explanation: The phase constant is represented as β. It is a complex quantity representing the constant angle of the wave propagated. It is given by β = ω√.
9. In waveguides, which of the following conditions will be true?
a) V > c
b) V < c
c) V = c
d) V >> c
Answer: a
Explanation: In waveguides, the phase velocity will always be greater than the speed of light. This enables the wave to propagate through the waveguide. Thus V > c is the required condition.
Answer: a
Explanation: In lossless dielectrics, the attenuation constant will not be same as the phase constant, unlike conductors. Also, due to the lossless behaviour, the attenuation will be nearly zero. Practically, zero attenuation is not possible.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Plane Waves in Free Space”.
1. In free space, the charge carriers will be
a) 0
b) 1
c) 100
d) Infinity
Answer: a
Explanation: Free space is not a conductor. Thus the charge carrier in free space is assumed to be zero. But the free space consists of particles or ions that get ionized during conduction.
2. In free space, which parameter will be unity?
a) Permittivity
b) Absolute permittivity
c) Relative permittivity
d) Permeability
Answer: c
Explanation: The relative permittivity is a constant for a particular material. It is unity for free space or air. The absolute permittivity is a constant given by 8.854 x 10 -12 C/m2.
3. Which parameter is unity in air medium?
a) Permittivity
b) Absolute permeability
c) Relative permeability
d) Permeability
Answer: c
Explanation: In free space or air medium, the relative permeability is also unity, like relative permittivity. The absolute permeability is given by 4π x 10 -7 units.
4. The conductivity in free space medium is
a) Infinity
b) Unity
c) Zero
d) Negative
Answer: c
Explanation: As the charge carriers are not available in free space, the conductivity will be very low. For ideal cases, the conductivity can be taken as zero.
5. Zero permeability/permittivity implies which state?
a) No ions are allowed in the medium
b) No current is generated in the medium
c) No magnetic or electric energy is permitted in the medium
d) No resistivity
Answer: c
Explanation: The zero permittivity in an electric field refers to the ability of the field/medium to permit electric charges in it. Similarly, zero permeability in a magnetic field refers to the ability of the field/medium to permit the magnetic energy into the field.
6. The intrinsic impedance of free space is
a) 489
b) 265
c) 192
d) 377
Answer: d
Explanation: The intrinsic impedance is the square root of ratio of the permeability to the permittivity. In free space, the permeability and the permittivity is same as the absolute permeability and permittivity respectively. This is due to unity permeability and permittivity in free space. Thus η = √, where absolute permeability is given by 4π x 10 -7 and absolute permittivity is given by 8.854 x 10 -12 . The intrinsic impedance will be 377 ohms.
7. In free space, the condition that holds good is
a) Minimum attenuation and propagation
b) Minimum attenuation and maximum propagation
c) Maximum attenuation and minimum propagation
d) Maximum attenuation and propagation
Answer: b
Explanation: The free space does not have any barrier for attenuation. Thus it enables minimum attenuation and maximum propagation. This technique is employed in line of sight communication.
8. In free space, the ratio of frequency to the velocity of light gives the phase constant. State True/False.
a) True
b) False
Answer: a
Explanation: The phase constant is given by the ratio of the frequency in radian/sec to the velocity of the wave propagating. In free space, the velocity is considered to be the velocity of light. Thus the statement is true.
9. The velocity of a wave travelling in the air medium without transmission lines or waveguides is
a) 6 x 10 8
b) 3 x 10 8
c) 1.5 x 10 8
d) 9 x 10 8
Answer: b
Explanation: In free space or air medium, the velocity of the wave propagating will be same as that of the light. Thus the velocity is the speed of light, V = c. It is given by 3 x 10 8 m/s.
Answer: b
Explanation: In an EM wave, the electric and the magnetic fields will be perpendicular to each other and with the direction of the propagation. Thus it can be expressed in cross product where iE x iH = iw. Here iE is the electric vector component, iH is the magnetic vector component and iw is the vector of the wave propagating.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Planes Waves in Good Conductor”.
1. For conductors, the loss tangent will be
a) Zero
b) Unity
c) Maximum
d) Minimum
Answer: c
Explanation: In conductors, the conductivity will be more. Thus the loss tangent σ/ωε will be maximum.
2. In metals, the total permittivity is
a) Absolute permittivity
b) Relative permittivity
c) Product of absolute and relative permittivity
d) Unity
Answer: a
Explanation: The total permittivity is the product of the absolute and the relative permittivity. For metals or conductors, the relative permittivity is unity. Thus the permittivity is simply the absolute permittivity.
3. The total permeability in a conductor is
a) Absolute permeability
b) Relative permeability
c) Product of absolute and relative permeability
d) Unity
Answer: c
Explanation: The total permeability is the product of the absolute and the relative permeability. For metals or conductors, the relative permittivity is not unity. Thus the permittivity is the product of absolute and relative permeability.
4. Calculate the phase constant of a conductor with attenuation constant given by 0.04 units.
a) 0.02
b) 0.08
c) 0.0016
d) 0.04
Answer: d
Explanation: The phase constant and the attenuation constant are both the same in the case of conductors. Given that the attenuation constant is 0.04, implies that the phase constant is also 0.04.
5. Calculate the attenuation constant of a conductor of conductivity 200 units, frequency 1M radian/s in air.
a) 11.2
b) 1.12
c) 56.23
d) 5.62
Answer: a
Explanation: The attenuation constant of a conductor is given by α = √. On substituting ω = 10 6 , σ = 200 and μ = 4π x 10 -7 , we get α = 11.2 units.
6. The skin depth of a conductor with attenuation constant of 7 neper/m is
a) 14
b) 49
c) 7
d) 1/7
Answer: d
Explanation: The skin depth is the measure of the depth upto which an EM wave can penetrate through the conductor surface. It is the reciprocal of the attenuation constant. On substituting for α = 7, we get δ = 1/α = 1/7 units.
7. The expression for velocity of a wave in the conductor is
a) V = √
b) V = √
c) V =
d) V =
Answer: a
Explanation: The velocity is the ratio of the frequency to the phase constant. In conductors, the phase constant is given by √. On substituting for β,ω in v, we get v = √ units.
8. In conductors, the E and H vary by a phase difference of
a) 0
b) 30
c) 45
d) 60
Answer: c
Explanation: The electric and magnetic component, E and H respectively have a phase difference of 45 degrees. This is due to the wave propagation in conductors in the air medium.
9. EM waves do not travel inside metals. State True/False.
a) True
b) False
Answer: a
Explanation: The conductors or metals do not support EM wave propagation onto them due the skin effect. This is the reason why mobile phones cannot be used inside lifts.
10. The propagation constant of the wave in a conductor with air as medium is
a) √
b) ωμσ
c) √
d) ω/μσ
Answer: a
Explanation: The propagation constant is the sum of the attenuation constant and the phase constant. In conductors, the attenuation and phase constant both are same and it is given by √. Their sum will be √, is the propagation constant.
11. An example for electromagnetic wave propagation is
a) refrigerator
b) electric fan
c) mobile transponder
d) relays in actuators
Answer: c
Explanation: The refrigerator, electric fan and relays are electrical devices. They do not use electromagnetic energy as medium of energy transfer. The mobile transponder is an antenna, which uses the EM waves for communication with the satellites.
Answer: d
Explanation: The intrinsic impedance in a conductor is given by η = √ x . The phase shift is represented by the 1+j term. In polar form it indicates 45 degree phase shift.
This set of Electromagnetic Theory Question Bank focuses on “Plane Waves in Dielectrics”.
1. The loss tangent of a perfect dielectric will be
a) Zero
b) Unity
c) Maximum
d) Minimum
Answer: d
Explanation: 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
Explanation: 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 will be (in order 10 -11 )
a) 1.94
b) 19.4
c) 0.194
d) 194
Answer: a
Explanation: 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 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
Explanation: 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
Explanation: 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 10 6 in air.
a) 2
b) 0.2
c) 0.02
d) 0.002
Answer: c
Explanation: The phase constant of a dielectric is given by β = ω√. On substituting for ω = 6 x 10 6 , 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
a) 6
b) 60
c) 600
d) 0.6
Answer: a
Explanation: The velocity of a wave is given by V = ω/β. To get ω, put v = 3 x 10 8 and β = 20. Thus ω = vβ = 3 x 10 8 x 20 = 60 x 10 8 = 6 GHz.
8. The relation between the speed of light, permeability and permittivity is
a) C = 1/√
b) C = με
c) C = μ/ε
d) C = 1/με
Answer: a
Explanation: The standard relation between speed of light, permeability and permittivity is given by c = 1/√. The value in air medium is 3 x 10 8 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
Explanation: 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) √
b)
c) √
d)
Answer: c
Explanation: The intrinsic impedance is given by the ratio of square root of the permittivity to the permeability. Thus η = √ 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
Answer: a
Explanation: In dielectrics, the electric and magnetic fields will be in phase or the phase difference between them is zero. This is due to the large attenuation which leads to increase in phase shift.
Answer: b
Explanation: The skin depth is the reciprocal of the phase constant. On substituting for β = 12, we get δ = 1/β = 1/12 units.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Power, Power Loss and Return Loss”.
1. The power of the electromagnetic wave with electric and magnetic field intensities given by 12 and 15 respectively is
a) 180
b) 90
c) 45
d) 120
Answer: b
Explanation: The Poynting vector gives the power of an EM wave. Thus P = EH/2. On substituting for E = 12 and H = 15, we get P = 12 x 15/2 = 90 units.
2. The power of a wave of with voltage of 140V and a characteristic impedance of 50 ohm is
a) 1.96
b) 19.6
c) 196
d) 19600
Answer: c
Explanation: The power of a wave is given by P = V 2 /2Zo, where V is the generator voltage and Zo is the characteristic impedance. on substituting the given data, we get P = 140 2 / = 196 units.
3. The power reflected by a wave with incident power of 16 units is
a) 2
b) 8
c) 6
d) 4
Answer: d
Explanation: The fraction of the reflected to the incident power is given by the reflection coefficient. Thus Pref = R 2 xPinc. On substituting the given data, we get Pref = 0.5 2 x 16 = 4 units.
4. The power transmitted by a wave with incident power of 16 units is
a) 12
b) 8
c) 16
d) 4
Answer: a
Explanation: The fraction of the transmitted to the incident power is given by the reflection coefficient. Thus Pref = (1-R 2 ) Pinc. On substituting the given data, we get Pref = (1- 0.5 2 ) x 16 = 12 units. In other words, it is the remaining power after reflection.
5. The incident and the reflected voltage are given by 15 and 5 respectively. The transmission coefficient is
a) 1/3
b) 2/3
c) 1
d) 3
Answer: b
Explanation: The ratio of the reflected to the incident voltage is the reflection coefficient. It is given by R = 5/15 = 1/3. To get the transmission coefficient, T = 1 – R = 1 – 1/3 = 2/3.
6. The current reflection coefficient is given by -0.75. Find the voltage reflection coefficient.
a) -0.75
b) 0.25
c) -0.25
d) 0.75
Answer: d
Explanation: The voltage reflection coefficient is the negative of the current reflection coefficient. For a current reflection coefficient of -0.75, the voltage reflection coefficient will be 0.75.
7. The attenuation is given by 20 units. Find the power loss in decibels.
a) 13.01
b) 26.02
c) 52.04
d) 104.08
Answer: a
Explanation: The attenuation refers to the power loss. Thus the power loss is given by 20 units. The power loss in dB will be 10 log 20 = 13.01 decibel.
8. The reflection coefficient is 0.5. Find the return loss.
a) 12.12
b) -12.12
c) 6.02
d) -6.02
Answer: c
Explanation: The return loss is given by RL = -20log R, where is the reflection coefficient. It is given as 0.5. Thus the return loss will be RL = -20 log 0.5 = 6.02 decibel.
9. The radiation resistance of an antenna having a power of 120 units and antenna current of 5A is
a) 4.8
b) 9.6
c) 3.6
d) 1.8
Answer: a
Explanation: The power of an antenna is given by Prad = Ia 2 Rrad, where Ia is the antenna current and Rrad is the radiation resistance. On substituting the given data, we get Rrad = Prad/Ia 2 = 120/5 2 = 4.8 ohm.
10. The transmission coefficient is given by 0.65. Find the return loss of the wave.
a) 9.11
b) 1.99
c) 1.19
d) 9.91
Answer: a
Explanation: The transmission coefficient is the reverse of the reflection coefficient, i.e, T + R = 1. When T = 0.65, we get R = 0.35. Thus the return loss RL = -20log R = -20log 0.35 = 9.11 decibel.
11. The return loss is given as 12 decibel. Calculate the reflection coefficient.
a) 0.35
b) 0.55
c) 0.25
d) 0.75
Answer: c
Explanation: The return loss is given by RL = -20log R. The reflection coefficient can be calculated as R = 10 , by anti logarithm property. For the given return loss RL = 12, we get R = 10 = 0.25.
Answer: a
Explanation: The return loss is given by RL = -20log R. The reflection coefficient can be calculated as R = 10 , by anti logarithm property. For the given return loss RL = 6, we get R = 10 = 0.501. The transmission coefficient will be T = 1 – R = 1-0.501 = 0.498.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Refractive Index and Numerical Aperture”.
1. The expression for refractive index is given by
a) N = v/c
b) N = c/v
c) N = cv
d) N = 1/cv
Answer: b
Explanation: The refractive index is defined as the ratio of the velocity of light in a vacuum to its velocity in a specified medium. It is given by n = c/v. It is constant for a particular material.
2. Numerical aperture is expressed as the
a) NA = sin θa
b) NA = cos θa
c) NA = tan θa
d) NA = sec θa
Answer: a
Explanation: The numerical aperture is the measure of how much light the fiber can collect. It is the sine of the acceptance angle, the angle at which the light must be transmitted in order to get maximum reflection. Thus it is given by NA = sin θa.
3. For total internal reflection to occur, which condition must be satisfied?
a) N1 = N2
b) N1 > N2
c) N1 < N2
d) N1 x N2=1
Answer: b
Explanation: The refractive of the transmitting medium should be greater than that of the receiving medium. In other words, the light must flow from denser to rarer medium, for total internal reflection to occur.
4. Find the refractive index of a medium having a velocity of 1.5 x 10 8 .
a) 0.5
b) 5
c) 0.2
d) 2
Answer: d
Explanation: The refractive index is given by the ratio of the speed of light to the velocity in a particular medium. It is given by n = c/v. On substituting for v = 1.5 x 10 8 and c = 3 x 10 8 , we get n = 3/1.5 = 2. The quantity has no unit.
5. The refractive index of water will be
a) 1
b) 2.66
c) 5
d) 1.33
Answer: d
Explanation: The velocity of light in water as medium will be 2.25 x 10 8 . On substituting for the speed of light, we get refractive index as n = 3/2.25 = 1.33.
6. The refractive index of air is unity. State True/False.
a) True
b) False
Answer: a
Explanation: The velocity of light in the air medium and the speed of light are both the same. Since light travels at maximum velocity in air only. Thus the refractive index n = c/v will be unity.
7. The numerical aperture of a coaxial cable with core and cladding indices given by 2.33 and 1.4 respectively is
a) 3.73
b) 0.83
c) 3.46
d) 1.86
Answer: d
Explanation: The numerical aperture is given by NA = √(n1 2 – n2 2 ), where n1 and n2 are the refractive indices of core and cladding respectively. On substituting for n1 = 2.33 and n2 = 1.4, we get NA = √(2.33 2 -1.4 2 ) = 1.86.
8. Find the acceptance angle of a material which has a numerical aperture of 0.707 in air.
a) 30
b) 60
c) 45
d) 90
Answer: c
Explanation: The numerical aperture is given by NA = n sin θa, where n is the refractive index. It is unity in air. Thus NA = sin θa. To get θ= sin -1 , put NA = 0.707, thus θa = sin -1 = 45 degree.
9. The numerical aperture of a material with acceptance angle of 60 degree in water will be
a) 1.15
b) 2.15
c) 5.21
d) 1.52
Answer: a
Explanation: The numerical aperture is given by NA = n sin θa, where n is the refractive index. It is 1.33 for water medium. Given that the acceptance angle is 60, we get NA = 1.33 sin 60 = 1.15.
10. The core refractive index should be lesser than the cladding refractive index for a coaxial cable. State True/False
a) True
b) False
Answer: b
Explanation: The light should pass through the core region only, for effective transmission. When light passes through cladding, losses will occur, as cladding is meant for protection. Thus core refractive index must be greater than the cladding refractive index.
11. The refractive index is 2.33 and the critical angle is 350. Find the numerical aperture.
a) 2
b) 1.9
c) 2.33
d) 12
Answer: b
Explanation: The numerical aperture is given by NA = n cos θc, where θc is the critical angle and n is the refractive index. On substituting for n = 2.33 and θc = 35, we get NA = 2.33 cos 35 = 1.9.
Answer: c
Explanation: Silica is the most dominant optical fibre material. This is because of its hardness, flexibility, melting point. Also it is an easily available material.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Brewster Angle”.
1. Brewster angle is valid for which type of polarisation?
a) Perpendicular
b) Parallel
c) S polarised
d) P polarised
Answer: b
Explanation: The parallel polarisation of the electromagnetic waves is possible only when the transmission occurs at the Brewster angle.
2. The Brewster angle is expressed as
a) Tan -1
b) Tan -1
c) Tan -1
d) Tan
Answer: c
Explanation: The tangent of the Brewster angle is the ratio of the refractive indices of the second medium to that of the first medium. It is given by tan θb= n2/n1. Thus the Brewster angle will be θb = tan -1 .
3. The refractive index of a material with permittivity 16 is given by
a) 16
b) 256
c) 4
d) 8
Answer: c
Explanation: The refractive index is the square root of the permittivity. Thus n = √ε. Given that ε = 16, we get refractive index as n = 4. It has no unit.
4. The reflection coefficient in the wave propagation when it is transmitted with the Brewster angle is
a) 0
b) 1
c) -1
d) Infinity
Answer: a
Explanation: Brewster angle propagation refers to complete transmission. The wave transmitted at the Brewster angle will be completely transmitted without reflection. Thus the reflection coefficient will be zero.
5. The transmission coefficient of a wave propagating in the Brewster angle is
a) 0
b) 1
c) -1
d) Infinity
Answer: b
Explanation: The transmission coefficient is the reverse of the reflection coefficient. At Brewster angle, the reflection will be zero. Thus the transmission is T = 1-R. Since R = 0, T = 1. It is to be noted that T and R lies in the range of 0 to 1.
6. A circularly polarised wave transmitted at the Brewster angle will be received as linearly polarised wave. State True/False
a) True
b) False
Answer: a
Explanation: The Brewster angle is said to be the polarisation angle. When a circularly polarised wave is incident at the Brewster angle, the resultant wave will be linearly polarised.
7. An elliptically polarised wave transmitted at the Brewster angle will be received as an elliptically polarised wave. State True/False
a) True
b) False
Answer: b
Explanation: Any polarised wave transmitted at the Brewster angle will be linearly polarised. It can be a parallel, perpendicular, circular or elliptical polarisation. The resultant wave is always linearly polarised. This is the reason why the Brewster angle is called polarisation angle.
8. Find the Brewster angle of a wave transmitted from a medium of permittivity 4 to a medium of permittivity 2.
a) 35.26
b) 53.62
c) 26.35
d) 62.53
Answer: a
Explanation: The Brewster angle is given by θb = tan -1 , where n = √ε. Thus we can express the formula in terms of permittivity as θb = tan -1 √ . Here ε1 = 4 and ε2 = 2. Thus we get θb = tan -1 √ = tan -1 = 35.26 degree.
9. Find the ratio of refractive index of medium 2 to that of medium 1, when the Brewster angle is 60 degree.
a) 0.707
b) 1.5
c) 0.866
d) 1.732
Answer: d
Explanation: The tangent of the Brewster angle is the ratio of the medium 2 permittivity to the medium 1 permittivity. Thus tan θb = . Given that θb = 60 degree, the ratio n2/n1 will be tan 60 = 1.732.
Answer: a
Explanation: The Brewster angle is the angle of incidence at which complete transmission of the electromagnetic wave occurs.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Snell Law and Critical Angle”.
1. The Snell’s law can be derived from which type of incidence?
a) Incidence angle
b) Reflected angle
c) Refracted angle
d) Oblique incidence
Answer: d
Explanation: The oblique incidence refers to the interface between dielectric media. Consider a planar interface between two dielectric media. A plane wave is incident at an angle from medium 1 and reflected from medium 2. The interface plane defines the boundary between the media. This is the oblique medium.
2. The Snell’s law is given by
a) N1 sin θi = N2 sin θt
b) N2 sin θi = N1 sin θt
c) sin θi = sin θt
d) N1 cos θi = N2 cos θt
Answer: a
Explanation: The Snell law states that in an oblique medium, the product of the refractive index and sine of incidence angle in medium 1 is same as that of medium 2. Thus it is given by N1 sin θi = N2 sin θt.
3. Calculate the ratio of sine of incident angle to the sine of reflected angle when the refractive indices of medium 1 and 2 are given as 2.33 and 1.66 respectively.
a) 0.71
b) 1.4
c) 2
d) 3.99
Answer: a
Explanation: The Snell law is given by N1 sin θi = N2 sin θt. To get sin θi/sin θt, the ratio is N2/N1. On substituting for N1 = 2.33 and N2 = 1.66, we get 1.66/2.33 = 0.71.
4. Find the ratio of the refractive index of medium 1 to that of medium 2, when the incident and reflected angles are given by 300 and 450 respectively.
a) 0.5
b) 1
c) 2
d) 4
Answer: c
Explanation: The Snell law is given by N1 sin θi = N2 sin θt. For getting N1/N2, the ratio is sin θt/sin θi. On substituting for θi = 30 and θt = 45, we get sin 45/sin 30 = 2.
5. The refractive index of a medium with permittivity of 2 and permeability of 3 is given by
a) 3.56
b) 2.45
c) 3.21
d) 1.78
Answer: b
Explanation: The refractive index is given by n = c √, where c is the speed of light. Given that relative permittivity and relative permeability are 2 and 3 respectively. Thus n = 3 x 10 8 √(2 x 4π x 10 -7 x 3 x 8.854 x 10 -12 ) = 2.45.
6. The critical angle is defined as the angle of incidence at which the total internal reflection starts to occur. State True/False.
a) True
b) False
Answer: a
Explanation: The critical angle is the minimum angle of incidence which is required for the total internal reflection to occur. This is the angle that relates the refractive index with the angle of reflection in an oblique incidence medium.
7. The critical angle for two media of refractive indices of medium 1 and 2 given by 2 and 1 respectively is
a) 0
b) 30
c) 45
d) 60
Answer: b
Explanation: The sine of the critical angle is the ratio of refractive index of medium 2 to that in medium 1. Thus sin θc = n2/n1. To get θc, put n1 = 2 and n2 = 1. Thus we get θc = sin -1 = sin -1 = 30 degree.
8. The critical angle for two media with permittivities of 16 and 9 respectively is
a) 48.59
b) 54.34
c) 60
d) 45
Answer: a
Explanation: The sine of the critical angle is the ratio of refractive index of medium 2 to that in medium 1. Thus sin θc = n2/n1. Also n = √ε, thus sin θc = √ε2/√ε1. Put ε1 = 16 and ε2 = 9, we get θc = sin -1 = 48.59 degree.
9. The angle of incidence is equal to the angle of reflection for perfect reflection. State True/False.
a) True
b) False
Answer: a
Explanation: For complete wave reflection, the angle of incidence should be same as the angle of the reflection. In such cases, the reflection coefficient is unity and the transmission coefficient is zero.
10. The angle of incidence of a wave of a wave with angle of transmission 45 degree and the refractive indices of the two media given by 2 and 1.3 is
a) 41.68
b) 61.86
c) 12.23
d) 27.89
Answer: a
Explanation: The Snell law is given by N1 sin θi = N2 sin θt. To get θi, put N1 = 2, N2 = 1.3, θt = 45 degree. Thus we get θi = sin -1 /2 = 41.68 degree.
11. The angle at which the wave must be transmitted in air media if the angle of reflection is 45 degree is
a) 45
b) 30
c) 60
d) 90
Answer: a
Explanation: In air media, n1 = n2 = 1. Thus, sin θi=sin θt and the angle of incidence and the angle of reflection are same. Given that the reflection angle is 45, thus the angle of incidence is also 45 degree.
Answer: b
Explanation: From the definition of Snell law, sin θc = n2/n1. To get n2, put n1 = 1.732 and θc = 60. Thus we get sin 60 = n2/1.732 and n2 = 1.5.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Types of Polarization”.
1. When the phase angle between the Ex and Ey component is 0 0 or 180 0 , the polarisation is
a) Elliptical
b) Circular
c) Linear
d) Perpendicular
Answer: c
Explanation: The phase angle between the Ex and Ey component is 0 0 and 180 0 for linearly polarised wave. The wave is assumed to be propagating in the z direction.
2. The magnitude of the Ex and Ey components are same in which type of polarisation?
a) Linear
b) Circular
c) Elliptical
d) Perpendicular
Answer: b
Explanation: In circular polarisation, the magnitude of the Ex and Ey components are the same. This is a form of the elliptical polarisation in which the major and minor axis are the same.
3. When the Ex and Ey components of a wave are not same, the polarisation will be
a) Linear
b) Elliptical
c) Circular
d) Parallel
Answer: b
Explanation: In elliptical polarisation, the magnitude of Ex and Ey components are not same. This is due to the variation in the major and minor axes of the waves representing its magnitude.
4. Identify the polarisation of the wave given, Ex = Exo cos wt and Ey = Eyo sin wt. The phase difference is +90 0 .
a) Left hand circularly polarised
b) Right hand circularly polarised
c) Left hand elliptically polarised
d) Right hand elliptically polarised
Answer: c
Explanation: The magnitude of the Ex and Ey components are not same. Thus it is elliptical polarisation. For +90 phase difference, the polarisation is left handed. In other words, the rotation is in clockwise direction. Thus the polarisation is left hand elliptical.
5. Identify the polarisation of the wave given, Ex = 2 cos wt and Ey = sin wt. The phase difference is -90 0 .
a) Left hand circularly polarised
b) Right hand circularly polarised
c) Left hand elliptically polarised
d) Right hand elliptically polarised
Answer: d
Explanation: The magnitude of the Ex and Ey components are not same. Thus it is elliptical polarisation. For -90 phase difference, the polarisation is right handed. In other words, the rotation is in anti-clockwise direction. Thus the polarisation is right hand elliptical.
6. Identify the polarisation of the wave given, Ex = 2 cos wt and Ey = 2 sin wt. The phase difference is +90 0 .
a) Left hand circularly polarised
b) Right hand circularly polarised
c) Left hand elliptically polarised
d) Right hand elliptically polarised
Answer: a
Explanation: The magnitude of the Ex and Ey components are the same. Thus it is circular polarisation. For +90 phase difference, the polarisation is left handed. In other words, the rotation is in clockwise direction. Thus the polarisation is left hand circular.
7. Identify the polarisation of the wave given, Ex = cos wt and Ey = sin wt. The phase difference is -90 0 .
a) Left hand circularly polarised
b) Right hand circularly polarised
c) Left hand elliptically polarised
d) Right hand elliptically polarised
Answer: b
Explanation: The magnitude of the Ex and Ey components are the same. Thus it is circular polarisation. For -90 phase difference, the polarisation is right handed. In other words, the rotation is in anti-clockwise direction. Thus the polarisation is right hand circular.
8. For a non-zero Ex component and zero Ey component, the polarisation is
a) Parallel
b) Perpendicular
c) Elliptical
d) Circular
Answer: a
Explanation: When the Ex is non-zero and the Ey is zero, the polarisation is parallel. The parallel polarisation is classified under the linear polarisation type.
9. Identify the polarisation of the wave given that, Ex = 2 cos wt and Ey = cos wt.
a) Elliptical
b) Circular
c) Parallel
d) Linear
Answer: d
Explanation: The magnitude of the Ex and Ey components are not the same. Thus it cannot be circular polarisation. For a phase difference of 0, the polarisation is linear. In other words, the waves are in phase. Thus the polarisation is linear.
10. The Snell law is applicable for perpendicular polarisation and the Brewster law is applicable for parallel polarisation. State True/False.
a) True
b) False
Answer: a
Explanation: The Snell law is calculated from the oblique incidence media. Thus it is applicable for perpendicular polarisation. The Brewster law is applicable for perpendicular polarisation.
11. When the polarisation of the receiving antenna is unknown, to ensure that it receives atleast half the power, the transmitted wave should be
a) Linearly polarised
b) Elliptically polarised
c) Circularly polarised
d) Normally polarised
Answer: c
Explanation: The polarisation of the transmitting and receiving antenna has to be the same. This is the condition for maximum power transfer to occur. This is possible only when the polarisation is circular.
12. Identify the polarisation of the wave given that, Ex = 2 sin wt and Ey = 3 sin wt.
a) Linear
b) Elliptical
c) Circular
d) Parallel
Answer: a
Explanation: The magnitude of the Ex and Ey components are not the same. Thus it cannot be circular polarisation. For a phase difference of 0, the polarisation is linear. In other words, the waves are in phase. Thus the polarisation is linear.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “S and P Polarised Waves”.
1. The resultant electric field of a wave with Ex = 3 and Ey = 4 will be
a) 7
b) 1
c) 25
d) 5
Answer: d
Explanation: The resultant electric field of two electric components Ex and Ey is E = √(Ex 2 + Ey 2 ). On substituting for Ex = 3 and Ey = 4, we get E = 5 units.
2. In S polarisation, the electric field lies in the plane perpendicular to that of the interface. State True/False
a) True
b) False
Answer: a
Explanation: In the EM wave propagation, the electric and magnetic fields are perpendicular to each other. The S polarised wave is similar to the transverse magnetic wave, the electric field lies in the plane perpendicular to that of the interface.
3. In P polarisation, the electric field lies in the same plane as the interface. State True/False.
a) True
b) False
Answer: a
Explanation: In the EM wave propagation, the electric and magnetic fields are perpendicular to each other. The P polarised wave is similar to the transverse electric wave, the magnetic field lies in the plane perpendicular to that of the interface or the electric field lies in the same plane as the interface.
4. The group delay of a wave with phase constant 2.5 units and frequency of 1.2 radian/sec is
a) 3.7
b) 1.3
c) 3
d) 2.08
Answer: d
Explanation: The group delay is given by td = β/ω. On substituting for β = 2.5 and ω = 1.2, we get the group delay as td = 2.5/1.2 = 2.08 units.
5. The Brewster angle is valid for which type of polarisation?
a) S polarised
b) P polarised
c) Elliptical
d) Linear
Answer: b
Explanation: The Brewster angle is valid for perpendicular polarisation. The P polarised wave is also a type of perpendicular polarisation. In P polarisation, the electric field lies in the plane of the interface.
6. Find the reflection coefficient of a wave with an incident electric field of 5 units and reflected electric field of 2 units.
a) 2.5
b) 0.4
c) 0.8
d) 1.2
Answer: b
Explanation: The reflection coefficient is the ratio of the reflected electric field to the incident electric field. Thus τ = Er/Ei. On substituting for Ei = 5 and Er = 2, we get τ = 2/5 = 0.4.
7. The transmission coefficient of a wave with incident and transmitted electric field of 5 and 5 respectively is
a) 0
b) 1
c) 10
d) 5
Answer: b
Explanation: The transmission coefficient is the ratio of the transmitted electric field to the incident electric field. Thus T = Et/Ei. On substituting for Et = 5 and Ei = 5, we get T = 5/5 = 1. Simply, when the incident and transmitted field are same, no reflection occurs and the transmission is unity.
8. Find the relative permittivity of the medium having a refractive index of 1.6
a) 0.4
b) 2.56
c) 3.2
d) 4.8
Answer: b
Explanation: The refractive index is the square root of the relative permittivity. It is given by n = √εr. To get εr, put n = 1.6. We get εr = n 2 = 1.62 = 2.56.
9. Calculate the transmission coefficient of a wave with a reflection coefficient of 0.6
a) 0.6
b) 1
c) 0
d) 0.4
Answer: d
Explanation: The transmission coefficient is the reverse of the reflection coefficient. Thus T + τ = 1. On substituting for τ = 0.6, we get T = 0.4. It has no unit.
Answer: b
Explanation: The group delay expression is td = β/ω. To get β, put ω = 35 and td = 7.5. Thus we get β = td x ω = 7.5 x 35 = 262.5 units.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Transmission Line Primary Parameters”.
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 = √, where Z is the impedance and Y is the admittance. On substituting for Z = 16 and Y = 9, we get the characteristic impedance as √ = 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 = √, where Z is given by 9 and Y is 16. On substituting the given values, the propagation constant will be γ = √ = √ = 12 units.
11. Find the characteristic impedance expression in terms of the inductance and capacitance parameters.
a) Zo = √
b) Zo = LC
c) Zo = √
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 = √ is the required expression.
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.
This set of Electromagnetic Theory Questions and Answers for Entrance exams focuses on “Transmission Line Secondary Parameters”.
1. 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.
2. 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 = 3×10 8 /6×10 9 = 0.05 units.
3. 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.
4. 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 10 8 /3 = 0.133 GHz.
5. 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.
6. 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 10 9 and v = 1.2 x 10 8 , we get the phase constant β = 7.5 x 10 9 /1.2 x 10 8 = 62.5 units.
7. 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.
8. 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.
9. 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.
10. 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.
11. The velocity of wave in the air medium is
a) 1 x 10 8
b) 1.5 x 10 8
c) 3 x 10 8
d) 1 x 10 9
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 10 8 .
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.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Transmission Line Equations”.
1. 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.
2. 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.
3. 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.
4. 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.
5. 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.
6. The condition for a quarter wave transformer is
a) Zo 2 = Zin ZL
b) Zo = Zin ZL
c) ZL 2 = 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 Zo 2 = Zin ZL is the required condition.
7. 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 Zo 2 = Zin ZL. On substituting for Zin = 50 and ZL = 25, we get Zo 2 = 50 x 25. The characteristic impedance will be 35.35 ohm.
8. 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 Zo 2 = Zin ZL. On substituting for Zo = 75 and Zin = 200, we get ZL = Zo 2 /Zin = 75 2 /200 = 28.125 ohm.
9. 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.
10. 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.
11. 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.
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.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Input and Characteristic Impedances”.
1. The characteristic impedance of a quarter wave transformer with load and input impedances given by 30 and 75 respectively is
a) 47.43
b) 37.34
c) 73.23
d) 67.45
Answer: a
Explanation: In quarter wave transformer, the characteristic impedance will be the geometric mean of the input impedance and the load impedance. Thus Zo 2 = Z IN Z L . On substituting for Z IN = 75 and Z L = 30, we get the characteristic impedance as 47.43 units.
2. The input impedance of a quarter wave line 50 ohm and load impedance of 20 ohm is
a) 50
b) 20
c) 1000
d) 125
Answer: d
Explanation: The characteristic impedance will be the geometric mean of the input impedance and the load impedance. Thus Zo 2 = Zin ZL. On substituting for Zo = 50 and ZL = 20, we get the input impedance as 50 2 /20 = 125 ohm.
3. For a matched line, the input impedance will be equal to
a) Load impedance
b) Characteristic impedance
c) Output impedance
d) Zero
Answer: b
Explanation: A matched line refers to the input and characteristic impedance being the same. In such condition, maximum transmission will occur with minimal losses. The reflection will be very low.
4. The reflection coefficient lies in the range of
a) 0 < τ < 1
b) -1 < τ < 1
c) 1 < τ < ∞
d) 0 < τ < ∞
Answer: a
Explanation: The reflection coefficient lies in the range of 0 < τ < 1. For full transmission, the reflection will be zero. For no transmission, the reflection will be unity.
5. When the ratio of load voltage to input voltage is 5, the ratio of the characteristic impedance to the input impedance is
a) 1/5
b) 5
c) 10
d) 25
Answer: b
Explanation: From the transmission line equation, the ratio of the load voltage to the input voltage is same as the ratio of the characteristic impedance to the input impedance. Thus the required ratio is 5.
6. The power of the transmitter with a radiation resistance of 12 ohm and an antenna current of 3.5A is
a) 147
b) 741
c) 174
d) 471
Answer: a
Explanation: The power in a transmitter is given by Prad = Iant 2 Rrad. On substituting Irad = 3.5 and Rrad =12, we get Prad = 3.5 2 x 12 = 147 units.
7. The group delay of the wave with phase constant of 62.5 units and frequency of 4.5 radian/sec is
a) 13.88
b) 31.88
c) 88.13
d) 88.31
Answer: a
Explanation: The group delay is given by td = β/ω. Given that β = 62.5 and ω = 4.5, we get the group delay as td = 62.5/4.5 = 13.88 units.
8. The maximum impedance of a transmission line 50 ohm and the standing wave ratio of 2.5 is
a) 20
b) 125
c) 200
d) 75
Answer: b
Explanation: The maximum impedance of a line is given by Zmax = SZo. On substituting for S = 2.5 and Zo = 50, we get Zmax = 2.5 x 50 = 125 ohm.
9. The minimum impedance of a transmission line 75 ohm with a standing wave ratio of 4 is
a) 75
b) 300
c) 18.75
d) 150
Answer: c
Explanation: The minimum impedance of a line is given by Zmin = Zo/S. On substituting for Zo = 75 and S = 4, we get Zmin = 75/4 = 18.75 units.
10. The average power in an electromagnetic wave is given by
a) propagation constant
b) poynting vector
c) phase constant
d) attenuation constant
Answer: b
Explanation: The Poynting vector is the cross product of the electric field and magnetic field intensities. It gives the total power of an electromagnetic wave.
11. The characteristic impedance of a transmission line is normally chosen to be
a) 50
b) 75
c) 50 or 75
d) 100
Answer: c
Explanation: The characteristic impedance is always 50 ohm or 75 ohm for a transmission line. This is because of the GHz range of operation and the load impedences employed.
Answer: d
Explanation: The transmission line setup consists of antennae for transmitting and receiving power. It consists of waveguides and cavity resonator for guided transmission of electromagnetic waves. Thus oscillator is the odd one out.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Reflection and Transmission Coefficients”.
1. 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 .
2. 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.
3. 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.
4. 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.
5. 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.
6. 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.
7. 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 + 2 = 2/3.
8. 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.
9. 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.
10. 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 = R 2 Pinc. On substituting for R = 0.45 and Pinc = 18, we get Prefl = 0.45 2 x 18 = 3.645 units.
11. 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-R 2 )Pinc, where R = 0.6 and Pinc = 24, from the given data. Thus Ptr = (1- 0.6 2 ) x 24 = 15.36 units.
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.
This set of Electromagnetic Theory Questions and Answers for Campus interviews focuses on “Standing Waves and SWR”.
1. 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.
2. 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 to the amplitude at a node along the line. It is given by S = V MAX /V MIN .
3. 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.
4. 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.
5. 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.
6. 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.
7. 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 < ∞.
8. 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.
9. 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.
10. 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.
11. 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.
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.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Power, Power Loss and Return Loss”.
1. The power of the electromagnetic wave with electric and magnetic field intensities given by 12 and 15 respectively is
a) 180
b) 90
c) 45
d) 120
Answer: b
Explanation: The Poynting vector gives the power of an EM wave. Thus P = EH/2. On substituting for E = 12 and H = 15, we get P = 12 x 15/2 = 90 units.
2. The power of a wave of with voltage of 140V and a characteristic impedance of 50 ohm is
a) 1.96
b) 19.6
c) 196
d) 19600
Answer: c
Explanation: The power of a wave is given by P = V 2 /2Zo, where V is the generator voltage and Zo is the characteristic impedance. on substituting the given data, we get P = 140 2 / = 196 units.
3. The power reflected by a wave with incident power of 16 units is
a) 2
b) 8
c) 6
d) 4
Answer: d
Explanation: The fraction of the reflected to the incident power is given by the reflection coefficient. Thus Pref = R 2 xPinc. On substituting the given data, we get Pref = 0.5 2 x 16 = 4 units.
4. The power transmitted by a wave with incident power of 16 units is
a) 12
b) 8
c) 16
d) 4
Answer: a
Explanation: The fraction of the transmitted to the incident power is given by the reflection coefficient. Thus Pref = (1-R 2 ) Pinc. On substituting the given data, we get Pref = (1- 0.5 2 ) x 16 = 12 units. In other words, it is the remaining power after reflection.
5. The incident and the reflected voltage are given by 15 and 5 respectively. The transmission coefficient is
a) 1/3
b) 2/3
c) 1
d) 3
Answer: b
Explanation: The ratio of the reflected to the incident voltage is the reflection coefficient. It is given by R = 5/15 = 1/3. To get the transmission coefficient, T = 1 – R = 1 – 1/3 = 2/3.
6. The current reflection coefficient is given by -0.75. Find the voltage reflection coefficient.
a) -0.75
b) 0.25
c) -0.25
d) 0.75
Answer: d
Explanation: The voltage reflection coefficient is the negative of the current reflection coefficient. For a current reflection coefficient of -0.75, the voltage reflection coefficient will be 0.75.
7. The attenuation is given by 20 units. Find the power loss in decibels.
a) 13.01
b) 26.02
c) 52.04
d) 104.08
Answer: a
Explanation: The attenuation refers to the power loss. Thus the power loss is given by 20 units. The power loss in dB will be 10 log 20 = 13.01 decibel.
8. The reflection coefficient is 0.5. Find the return loss.
a) 12.12
b) -12.12
c) 6.02
d) -6.02
Answer: c
Explanation: The return loss is given by RL = -20log R, where is the reflection coefficient. It is given as 0.5. Thus the return loss will be RL = -20 log 0.5 = 6.02 decibel.
9. The radiation resistance of an antenna having a power of 120 units and antenna current of 5A is
a) 4.8
b) 9.6
c) 3.6
d) 1.8
Answer: a
Explanation: The power of an antenna is given by Prad = Ia 2 Rrad, where Ia is the antenna current and Rrad is the radiation resistance. On substituting the given data, we get Rrad = Prad/Ia 2 = 120/5 2 = 4.8 ohm.
10. The transmission coefficient is given by 0.65. Find the return loss of the wave.
a) 9.11
b) 1.99
c) 1.19
d) 9.91
Answer: a
Explanation: The transmission coefficient is the reverse of the reflection coefficient, i.e, T + R = 1. When T = 0.65, we get R = 0.35. Thus the return loss RL = -20log R = -20log 0.35 = 9.11 decibel.
11. The return loss is given as 12 decibel. Calculate the reflection coefficient.
a) 0.35
b) 0.55
c) 0.25
d) 0.75
Answer: c
Explanation: The return loss is given by RL = -20log R. The reflection coefficient can be calculated as R = 10 , by anti logarithm property. For the given return loss RL = 12, we get R = 10 = 0.25.
Answer: a
Explanation: The return loss is given by RL = -20log R. The reflection coefficient can be calculated as R = 10 , by anti logarithm property. For the given return loss RL = 6, we get R = 10 = 0.501. The transmission coefficient will be T = 1 – R = 1-0.501 = 0.498.
This set of Electromagnetic Theory Questions and Answers for Aptitude test focuses on “Intrinsic Impedance and Propagation Constant”.
1. The intrinsic impedance of a wave with electric and magnetic field of 10 and 8 respectively is
a) 1.6
b) 1.11
c) 1.25
d) 0.8
Answer: b
Explanation: The intrinsic impedance of a wave is given by η = √. on substituting for E = 10 and H = 8, we get η = 1.11 units.
2. The intrinsic impedance is defined as the ratio of the magnetic field to the electric field of the electromagnetic wave. State true/false
a) True
b) False
Answer: b
Explanation: The intrinsic impedance is defined as the ratio of the electric field intensity to the magnetic field intensity. It is denoted by η. It is a complex quantity.
3. The intrinsic impedance in free space is
a) 60π
b) 12π
c) 6π
d) 120π
Answer: d
Explanation: The intrinsic impedance of the free space is the ratio of the permittivity to the permeability. For air medium, the value is 120π or 377 ohm.
4. The propagation constant is a complex quantity. State true/false.
a) True
b) False
Answer: a
Explanation: The propagation constant is a complex quantity. It is given by γ = α + jβ, where α is the attenuation constant, a real value and β is the phase constant, a complex value.
5. Calculate the propagation constant of a wave with impedance and admittance given by 32 and 12 respectively.
a) 19.6
b) 17.6
c) 15.6
d) 13.6
Answer: a
Explanation: The propagation constant is given by γ = √, where Z is the impedance and Y is the admittance. On substituting for Z = 32 and Y = 12, we get γ = 19.6 units.
6. The intrinsic angle is 25 0 , find the loss angle.
a) 12.5
b) 50
c) 25
d) 75
Answer: b
Explanation: The loss angle is twice of the intrinsic angle. It is given by δ = 2θn. On substituting for θn = 25, the loss angle δ = 2 = 50 degree.
7. The standing wave ratio of the wave with maximum and minimum electric field intensities of 12 and 4 is
a) 12
b) 4
c) 3
d) 48
Answer: c
Explanation: The SWR is defined as the ratio of maximum electric field intensities to the minimum field intensities. It is given by S = E MAX /E MIN . Thus for the given data, S = 12/4 = 3.
8. The standing wave ratio of a wave travelling through two media having intrinsic impedances of 3 and 2 is
a) 2 ⁄ 3
b) 3 ⁄ 2
c) 1 ⁄ 5
d) 5
Answer: b
Explanation: The standing wave ratio is given by the ratio of the intrinsic impedance of medium 1 to the intrinsic impedance of medium 2. Thus S = η1/η2. On substituting for η1 = 3 and η2 = 2, we get S = 3/2.
9. The reflection coefficient of a wave travelling through two media having electric intrinsic impedances of 3 and 5 respectively is
a) 2
b) 8
c) 5 ⁄ 3
d) 1/4
Answer: d
Explanation: The reflection coefficient in terms for intrinsic impedance is R= η2-η1/η2+η1. On substituting the given data, we get R = 5 – 3/5 + 3 = 2/8 = ¼.
10. The reflection coefficient of a wave travelling through two media having magnetic intrinsic impedances of 2 and 1 respectively is
a) 1 ⁄ 2
b) 1 ⁄ 3
c) 1 ⁄ 4
d) 1 ⁄ 5
Answer: b
Explanation: The reflection coefficient in terms for magnetic intrinsic impedance is R = η1 – η2/η1 + η2. On substituting the given data, we get R = 2 – 1/2 + 1 = 1/3.
11. The transmission coefficient of a wave travelling through two media having electric intrinsic impedances of 3 and 2 respectively is
a) 2 ⁄ 2
b) 8 ⁄ 9
c) 4 ⁄ 5
d) 1 ⁄ 4
Answer: c
Explanation: The transmission coefficient in terms for intrinsic impedance is R = 2η2/η2 + η1. On substituting the given data, we get R = 2 x 2/2 + 3 = 4/5.
Answer: d
Explanation: The transmission coefficient in terms for intrinsic impedance is R = 2η1/η2 + η1. On substituting the given data, we get R = 2 x 3 ⁄ 4 + 3 = 6/7.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Skin Effect”.
1. The skin effect is a phenomenon observed in
a) Insulators
b) Dielectrics
c) Conductors
d) Semiconductors
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.
2. 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.
3. 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.
4. 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.
5. 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/√, where f is the frequency, μ is the permeability and σ is the conductivity. For the given data, f = 10 x 10 9 , μ = 4π x 10 -7 in air and σ = 200, we get δ = 355.8 μm.
6. 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.
7. 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.
8. 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/√. Thus the relation between the skin depth and the frequency is, Skin depth α 1/√f.
9. 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.
10. 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.
11. 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 = √(Ex 2 + Ey 2 ). For the given data, the electric field will be E = √(6 2 +8 2 ) = 10 units.
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/(3×10 6 ) = 4 μm.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Types of Transmission Lines”.
1. Identify which of the following is not a transmission line?
a) Telephone lines
b) Power transmission
c) Underground cables
d) Cavity resonators
Answer: d
Explanation: The types of transmission line are telephone lines, power transmission lines, underground cables, coaxial cables, fibre optic cable transmission etc. Cavity resonators are not transmission lines, they are components that aid maximum transmission.
2. The open wire transmission line consists of
a) Conductor
b) Dielectric
c) Both conductor and dielectric
d) Either conductor or dielectric
Answer: c
Explanation: The open wire is a common form of transmission line. The open wire consists of conductors. The conductors of such lines are considered to be parallel and separated by a dielectric.
3. Telephone lines and power lines are open wire transmission lines. State true/false.
a) True
b) False
Answer: a
Explanation: The open wire lines are long and used for distant wired communication. Such lines are telephone and power lines.
4. The cable transmission line consists of
a) Conductors
b) Insulators
c) Insulated conductors
d) Insulated conductors with dielectric
Answer: d
Explanation: The cable lines are underground transmission lines. The line consists of hundreds of individual paper insulated conductors twisted in pairs and combined inside a protective lead or plastic sheath, which is usually a solid dielectric.
5. The conductors lie perpendicularly with the dielectric in the cable line. State true/false.
a) True
b) False
Answer: b
Explanation: In all the transmission lines, the conductors are considered to be parallel with a solid dielectric.
6. The coaxial cable consists of
a) Conductors
b) Dielectric
c) Conductor with dielectric
d) Two conductors with dielectric
Answer: d
Explanation: The coaxial cable consists of a hollow conductor and the second conductor is located inside and coaxial with the tube. The dielectric may be solid or gaseous.
7. The coaxial cable are used in
a) Telephone cables transmission
b) Power transmission
c) Television signal transmission
d) Short wave transmission
Answer: c
Explanation: Practically, the coaxial cables are employed in the transmission of the television signals from the dish antenna to the transponder.
8. Identify which is not a type of waveguide.
a) Rectangular
b) Circular
c) Cylindrical
d) Cavity resonator
Answer: d
Explanation: Rectangular waveguide is a commonly used waveguide. Cylindrical and circular waveguides are the same. Cavity resonator is not a waveguide.
9. The range of frequencies handled by the waveguides is in
a) Hz
b) KHz
c) MHz
d) GHz
Answer: d
Explanation: The waveguides are operated in the GHz range. In particular, the waveguides are active above 6 GHz. The range goes upto several tens of GHz. Beyond this range, the transmission is handled by optic fibre cables.
10. The range of frequencies operated by the coaxial cables is in
a) Hz
b) kHz
c) MHz
d) GHz
Answer: c
Explanation: Coaxial cables are operated in the MHz range. The main application includes television cable line transmission.
11. The cut off frequency for waveguide operation is
a) 2 MHz
b) 6 GHz
c) 4 MHz
d) 6 MHz
Answer: b
Explanation: The waveguides should be operated above the cut off frequency of 6 GHz. This will lead to effective power transmission. At a frequency below this, will lead to attenuation.
Answer: a
Explanation: Though all the means of communication use the radio frequencies, the internet communication is the fastest. It involves email, voice message, video message etc. The telephone, television and radio use transmission lines for communication through radio frequencies, whereas the internet uses both wired and wireless means. Also it employs optic fibre, which uses light as the medium, since it is faster than the radio signals.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Lossless and Distortionless Line”.
1. The transmission line is said to be lossless when the
a) Conductor is perfect and dielectric is lossless
b) Conductor is perfect and dielectric is lossy
c) Conductor is imperfect and dielectric is lossy
d) Conductor is imperfect and dielectric is lossless
Answer: a
Explanation: Lossless transmission line refers to a line with no losses or attenuation. This is possible only when the conductor is perfect and the dielectric is lossless.
2. The resistance of a lossless transmission line is
a) 0
b) 1
c) -1
d) Infinity
Answer: a
Explanation: The lossless transmission line will have minimal loss of power. Thus in the ideal cases, the resistance is assumed to be zero.
3. Which two parameters given below are zero in the lossless line?
a) L, C
b) C, G
c) G, α
d) R, L
Answer: c
Explanation: Due to the minimal losses in the lossless transmission line, the parameters of conductance and attenuation constant are assumed to be zero. Practically, a lossless line is not possible.
4. The characteristic impedance of the line having primary constants L and C as 35 milli henry and 70 micro farad respectively is
a) 500
b) 22.36
c) 125
d) 50
Answer: b
Explanation: The characteristic impedance of a lossless line is given by Zo = √. On substituting for L = 35 x 10 -3 and C = 70 x 10 -6 , we get Zo = √ = 22. 36 ohm.
5. The attenuation constant is measured in the units of
a) Ohm
b) Neper
c) Decibel
d) Radian/sec
Answer: b
Explanation: The attenuation is the loss of power due to the transmission. It is measured in the Neper units. 1 neper = 8.68 decibel/m.
6. The velocity of the transmission line with a frequency of 35 radian/s and phase constant of 68.5 is
a) 1.95
b) 2.36
c) 4.56
d) 3.48
Answer: a
Explanation: The velocity of the wave in a transmission line is the ratio of the frequency to the phase constant. Thus v = ω/β. On substituting for ω = 35 and β = 68.5, we get v = 68.5/35 = 1.95 units.
7. The condition that holds good in a distortionless transmission line is
a) R/L = G/C
b) RL = GC
c) L/R = C/G
d) RG/LC
Answer: a
Explanation: For a distortionless transmission line, the primary constants are related by R/L = G/C.
8. Given that R = 20 ohm, L = 40 mH, C = 40 μF, G = 0.02 mho. Find whether the line is distortionless or not.
a) Distorted line
b) Distortionless line
c) All of the mentioned
d) None of the mentioned
Answer: b
Explanation: For a distortionless line, the condition R/L = G/C must be true. On substituting for R = 20, L = 40 mH, C = 40 μF, G = 0.02, we get R/L = 500 and G/C = 500. Thus the line is distortionless.
9. For R= 20 ohm and G= 0.8 mho, the attenuation constant will be
a) 16
b) 4
c) 2
d) 32
Answer: b
Explanation: The attenuation constant is given by α = √, where R and G is the resistance and conductance respectively. On substituting for R = 20 and G = 0.8, we get α = √ = 4 units.
10. The velocity of an electromagnetic wave with frequency 6MHz and a skin depth of 1.6 μm.
a) 3.75
b) 0.26
c) 9.6
d) 7.8
Answer: c
Explanation: The velocity of a wave is the product of the frequency and the skin depth. It is given v = f. δ. On substituting for f = 6 x 10 6 and δ = 1.6 x 10 -6 , we get v = 6 x 10 6 x 1.6 x 10 -6 = 9.6 units.
11. The characteristic impedance of the transmission line with R= 45 ohm and G= 0.45 mho is
a) 100
b) 1
c) 0.1
d) 10
Answer: d
Explanation: The characteristic impedance in terms of R, G is Zo = √. On substituting for R = 45 and G = 0.45, we get Zo = 10 units.
Answer: b
Explanation: In a distortionless line, the attenuation constant is independent of the frequency and the phase constant is linearly dependent of the frequency.
This set of Electromagnetic Theory Assessment Questions and Answers focuses on “Smith Chart”.
1. 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.
2. 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.
3. 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.
4. The circles in the Smith chart pass through which point?
a)
b)
c)
d)
Answer: d
Explanation: All the constant resistance and reactance circles in the Smith chart pass through the point. This is the midpoint of the Smith Chart. The resistance is unity and reactance is zero at this point.
5. 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.
6. The centre of the point having a normalised resistance of 1.2 ohm and reactance of 1.5 ohm is
a)
b)
c)
d)
Answer: a
Explanation: The centre of a point in Smith chart is given by C = . On substituting for r = 1.2, we get centre as = .
7. 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.
8. 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.
9. 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.
Answer: a
Explanation: The centre and radius of a line are and 1/x, where x is the reactance. Here x = 0.5, from the given data. Thus C = and R = 2.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Types of Waveguides”.
1. The phenomenon employed in the waveguide operation is
a) Reflection
b) Refraction
c) Total internal reflection
d) Adsorption
Answer: c
Explanation: The waveguides use total internal reflection phenomenon to transmit the waves passing through it. Thus the acceptance angle and critical angle are important for effective transmission.
2. The dominant mode in waveguide is the mode which has
a) Highest frequency
b) Highest wavelength
c) Lowest phase constant
d) Highest attenuation
Answer: b
Explanation: The dominant mode is the mode which has the minimum frequency or maximum wavelength available for propagation of the waves.
3. The modes are calculated from which parameter?
a) Frequency
b) Wavelength
c) Phase constant
d) V number
Answer: d
Explanation: The modes are calculated from the V number of the waveguides. It is given by M= V 2 /2.
4. The circular waveguides use which function in the frequency calculation?
a) Laplace function
b) Schottky function
c) Bessel function
d) Transfer function
Answer: c
Explanation: The circular or cylindrical waveguides use the Bessel function for the frequency calculation of a particular mode.
5. The scattering parameters are used to indicate the
a) Permittivity and permeability
b) Electric and magnetic field intensities
c) Reflection and transmission coefficients
d) Frequency and wavelength
Answer: c
Explanation: The scattering matrix consists of the transmission coefficients in the main diagonal and the reflection coefficients in the opposite diagonal.
6. Which of the following two parameter models cannot be used to represent a transmission line?
a) H parameter model
b) T parameter model
c) ABCD parameter model
d) S parameter model
Answer: a
Explanation: The T, ABCD and S parameter models are used in the transmission line modelling. The h parameter is not used for the same.
7. For the matched line, the parameters S 12 and S 21 are
a) 1
b) 0
c) -1
d) ∞
Answer: b
Explanation: The parameters S 12 and S 21 are the reflection coefficients. For a matched line, the reflection coefficients are zero. Thus the parameters S 12 and S 21 are also zero.
8. The waveguides are materials with characteristics of
a) Low bulk resistivity
b) High bulk resistivity
c) High conductivity
d) Low conductivity
Answer: a
Explanation: Generally, the waveguides are made of materials with low bulk resistivity like brass, copper, silver etc. But if the interior walls are properly plated, it is possible with poor conductivity materials too. It is even possible to make plastic waveguides.
9. The parameters S 11 and S 22 indicate the transmission coefficients. State true/false.
a) True
b) False
Answer: a
Explanation: In a scattering matrix, the parameters S 11 and S 22 indicate the transmission coefficients and the parameters S 21 and S 12 indicate the reflection coefficients.
Answer: a
Explanation: The waveguides aid in effective transmission of the electromagnetic power from the source antenna to the destination antenna.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Properties of Waveguides”.
1. The waveguide is employed in the transmission lines, when operated at the range of
a) Hz
b) KHz
c) MHz
d) GHz
Answer: d
Explanation: Waveguides are employed for effective transmission, when the lines carry electromagnetic waves in the GHz range.
2. The cut off frequency for a waveguide to operate is
a) 3 MHz
b) 3 GHz
c) 6 MHz
d) 6 GHz
Answer: d
Explanation: The cut off frequency of the waveguide is 6 GHz. This is the frequency at which the waveguide will start to operate.
3. In rectangular waveguides, the dimensions a and b represent the
a) Broad wall dimensions
b) Broad wall and side wall dimension respectively
c) Side wall and broad wall dimension respectively
d) Side wall dimensions
Answer: b
Explanation: In rectangular waveguide, the a parameter is the broad wall dimension of the waveguide and the b parameter is the side wall dimension of the waveguide. Always, a > b in a waveguide.
4. The Bessel function is denoted by
a) J n
b) J m
c) J n
d) J m
Answer: a
Explanation: The Bessel function is used in the circular waveguides. Normally J n = 0. Here n is the order of the Bessel function.
5. In a waveguide, always which condition holds good?
a) phase velocity = c
b) phase velocity greater than c
c) phase velocity lesser than c
d) group velocity = c
Answer: b
Explanation: In air medium, the phase velocity is assumed to be the speed of light. For waveguides, the phase velocity is always greater than the speed of the light.
6. The group wavelength is greater than the wavelength at any point. State true/false.
a) True
b) False
Answer: a
Explanation: In a waveguide, the phase velocity is greater than the velocity of light. Thus the group velocity will be less. This implies the group wavelength will be greater than the wavelength at any point.
7. Find the group wavelength of a wave, given that the group phase constant is 6.28 units.
a) 2
b) 3.14
c) 6.28
d) 1
Answer: d
Explanation: The group wavelength is given by λg = 2π/βg, where βg is the group wavelength of the wave. On substituting for βg = 6.28, we get group wavelength as unity.
8. The phase velocity of a wave with frequency of 15 radian/sec and group phase constant of 2 units is
a) 30
b) 15
c) 7.5
d) 2/15
Answer: c
Explanation: The phase velocity of a wave is given by Vp = ω/βg. on substituting for ω = 15 and βg = 2, we get phase velocity as 15/2 = 7.5 units.
9. The modes in a material having a V number of 20 is
a) 20
b) 400
c) 200
d) 100
Answer: c
Explanation: The relation between the modes and the V number is given by m = v 2 /2. Given that v = 20, we get m = 20 2 /2 = 200 modes.
Answer: b
Explanation: The relation between the modes and the V number is given by m = v 2 /2. Given that m = 50, we get v 2 = 2 x 50 = 100. The V number is 10.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Cut off Frequency and Wavelength”.
1. The real part of the propagation constant is the
a) Attenuation constant
b) Phase constant
c) Permittivity
d) Permeability
Answer: a
Explanation: The propagation constant is given by γ = α + jβ. Here the real part is the attenuation constant and the imaginary part is the phase constant.
2. The phase constant of a wave is given by
a) ω√
b) ω√
c) ω√
d) ω√
Answer: a
Explanation: The phase constant of a wave in a transmission line is given by β = ω√, where L and C are the specifications of the line.
3. The cut off frequency of the dominant mode in a TE wave in the line having a and b as 2.5 cm and 1 cm respectively is
a) 4.5 GHz
b) 5 GHz
c) 5.5 GHz
d) 6 GHz
Answer: d
Explanation: The dominant mode in TE is TE 10 . The cut off frequency will be mc/2a, where m = 1 and a = 0.025 are given. On substituting, we get the frequency as 1 x 3 x 10 8 /2 x 0.025 = 6 GHz.
4. The cut off frequency of the TE 01 mode will be
a) mc/2a
b) mc/2b
c) nc/2a
d) nc/2b
Answer: d
Explanation: The cut off frequency consists of modes m and n. For m = 0, the dimension b will be considered. Thus the frequency is nc/2b, where c is the speed of the light.
5. The condition which will satisfy the dimensions of the waveguide is
a) a = b
b) a > b
c) a < b
d) ab = 0
Answer: b
Explanation: The dimensions a and b represent the broad wall and the side wall dimensions respectively. The broad wall will be greater than the side wall. Thus the condition a>b is true.
6. The cut off wavelength of the TE 10 mode having a broad wall dimension of 5cm is
a) 0.1
b) 1
c) 10
d) 0.01
Answer: a
Explanation: The cut off wavelength of the waveguide is given by λc = 2a/m. on substituting for a = 0.05 and m = 1, we get λc = 2 x 0.05/1 = 0.1 units.
7. The broad wall dimension of a waveguide having a cut off frequency of 7.5 GHz is
a) 1
b) 2
c) 3
d) 4
Answer: b
Explanation: The cut off frequency and the broad wall dimension are related by fc = mc/2a. On substituting for m = 1 and fc = 7.5 GHz, we get a = 0.02 or 2 cm.
8. The sin θ in the waveguide refers to the ratio of the
a) Frequency to wavelength
b) Wavelength to frequency
c) Cut off frequency to frequency
d) Frequency to cut off frequency
Answer: c
Explanation: The ratio of the cut off frequency to the frequency at any point gives the sin θ in a waveguide.
9. Is the transmission of a frequency 5 GHz possible in waveguides?
a) Yes
b) No
Answer: a
Explanation: The cut off frequency for waveguide operation is 6 GHz. Thus a wave of 5 GHz is not possible for transmission in a waveguide.
10. The dimension for a waveguide in dominant mode with a cut off wavelength of 2 units is
a) 2
b) 4
c) 6
d) 8
Answer: b
Explanation: The cut off wavelength of a waveguide is given by λc = 2a/m. For the dominant mode, m = 1. Given that λc = 2, thus we get a = 4 units.
11. The waveguides are used in a transmission line for
a) Increasing transmission coefficient
b) Increasing reflection coefficient
c) Decreasing transmission coefficient
d) Decreasing reflection coefficient
Answer: a
Explanation: The waveguides are used to increase the transmission efficiency of the waves travelling through it.
Answer: c
Explanation: The attenuation coefficient of a wave with a resistance of R in a line of characteristic impedance Zo is α = R/2Z o . On substituting for R = 15 and Z o = 50, we get α = 15/ = 0.15 units.
This set of Electromagnetic Theory Problems focuses on “Transverse Electric Waves”.
1. In transverse electric waves, which of the following is true?
a) E is parallel to H
b) E is parallel to wave direction
c) E is transverse to wave direction
d) H is transverse to wave direction
Answer: c
Explanation: In TE waves, the electric field strength will be transverse to the wave direction. Thus the TE waves are also called H waves.
2. The dominant mode in rectangular waveguide is
a) TE 01
b) TE 10
c) TM 01
d) TM 10
Answer: b
Explanation: TE 10 is the dominant mode in the rectangular waveguide. This is because it gives the minimum cut off frequency required for transmission.
3. The number of modes in a waveguide having a V number of 10 is
a) 10
b) 25
c) 100
d) 50
Answer: d
Explanation: The number of modes is given by m = V 2 /2, where V is the v number. On substituting for V = 10, we get m = 100/2 = 50.
4. Does the waveguide with dimensions 3 cm x 5.5 cm exist?
a) Yes
b) No
Answer: b
Explanation: For a waveguide, the dimension a should be greater than b. Here a = 3 and b = 5.5, thus such waveguide does not exist.
5. The mode which has the highest wavelength is called
a) Dominant mode
b) Evanescent mode
c) Generate mode
d) Degenerate mode
Answer: a
Explanation: Dominant modes are the modes having least cut off frequency. This implies they have highest cut off wavelength.
6. The intrinsic impedance of a TE wave having a cut off frequency of 6 GHz at a frequency of 7.5 GHz in air is
a) 628.33
b) 338.62
c) 498.76
d) 342.24
Answer: a
Explanation: The intrinsic impedance of a TE wave is given by η TE = η/cos θ, where cos θ is given by √ 2 ). On substituting for fc = 6 GHz, f = 7.5 GHz and η = 377, we get the intrinsic impedance as 628.33 units.
7. The cut off frequency of a rectangular waveguide of dimensions 3 cm x 1.5 cm is
a) 12 GHz
b) 6 GHz
c) 4 GHz
d) 5 GHz
Answer: d
Explanation: The cut off frequency in dominant mode will be fc = mc/2a. On substituting for c = 3 x 10 8 and a = 0.03, we get the cut off frequency as 5 GHz.
8. The propagation constant for a lossless transmission line will be
a) Real
b) Complex
c) Real and equal to phase constant
d) Complex and equal to phase constant
Answer: d
Explanation: The propagation constant is given by γ = α + jβ, where α and β are the attenuation and phase constants respectively. For a lossless line, the attenuation constant is zero. Thus γ = jβ. It is clear that γ is complex and equal to β.
9. The attenuation of a 50 ohm transmission line having a resistance of 100 ohm is
a) 0.01
b) 0.1
c) 1
d) 10
Answer: c
Explanation: The attenuation of a wave is given by α = R/2Z 0 . On substituting for R = 100 and Z 0 = 50, we get α = 100/ = 1 unit.
10. The cut off frequency of a TE wave with waveguide dimension of a= 3.5 cm in a medium of permittivity 2.2 is
a) 2.88 GHz
b) 3.32 GHz
c) 4.5 GHz
d) 2.12 GHz
Answer: a
Explanation: The cut off frequency of a TE wave in any other medium is mc/2a√εr. On substituting for a = 0.035 and εr = 2.2, we get the cut off frequency as 2.88 GHz.
11. The phase constant and frequency are related by
a) Phase constant α ω
b) Phase constant α 1/ω
c) Phase constant α 1/2ω
d) Phase constant α ω/2
Answer: a
Explanation: The phase constant is given by β = ω√LC. Thus the relation is β is directly proportional to ω.
Answer: d
Explanation: The attenuation constant, resistance and conductance are zero for a lossless line. Only the phase constant is non zero.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Transverse Magnetic Waves”.
1. In transverse magnetic waves, which of the following is true?
a) E is parallel to H
b) H is parallel to wave direction
c) H is transverse to wave direction
d) E is transverse to H
Answer: c
Explanation: In transverse magnetic waves, the magnetic field strength is transverse to the wave direction. They are also called E waves.
2. The dominant mode in the TM waves is
a) TM 01
b) TM 10
c) TM 20
d) TM 11
Answer: d
Explanation: The modes TM 10 , TM 01 and TM 20 does not exist in any waveguide. The TM 11 mode is the dominant mode in the waveguide.
3. The modes in a waveguide having a V number of 20 is
a) 400
b) 200
c) 100
d) 40
Answer: b
Explanation: The number of modes in a waveguide is given by m = V 2 /2. On substituting for V = 20, we get m = 400/2 = 200 modes.
4. The v number of a waveguide having 120 modes is
a) 15.5
b) 18
c) 24
d) 12
Answer: a
Explanation: The number of modes in a waveguide is given by m = V 2 /2. On substituting for m = 120, we get V = √ = 15.5.
5. The intrinsic impedance of a TM wave will be
a) Greater than 377 ohm
b) Equal to 377 ohm
c) Lesser than 377 ohm
d) Does not exist
Answer: c
Explanation: The intrinsic impedance of the transverse magnetic wave is given by η TM = η √ 2 ). Here the term √ 2 ) is always lesser than unity. Thus the intrinsic impedance of the TM wave is lesser than 377 ohms.
6. The modes TM mo and TM on are called
a) Generate modes
b) Degenerate modes
c) Dominant modes
d) Evanescent modes
Answer: d
Explanation: The modes TM mo and TM on does not exist. These modes are said to be evanescent mode.
7. Two modes with same cut off frequency are said to be
a) Generate modes
b) Dominant modes
c) Degenerate modes
d) Regenerate modes
Answer: c
Explanation: Two modes with same cut off frequency are called as degenerate modes. These modes have same field distribution.
8. Does the mode TM 30 exists?
a) Yes
b) No
Answer: b
Explanation: Modes in the format of TM mo and TM on does not exist. The given mode is in the form of TM mo , which is does not exist. It is an evanescent mode.
9. The boundary between the Fresnel and Fraunhofer zones having a length of 12 units and a wavelength of 3 units is
a) 96
b) 48
c) 192
d) 36
Answer: a
Explanation: The Fresnel- Fraunhofer boundary is related by the wavelength as R = 2L 2 /λ. On substituting for L = 12 and λ = 3, we get R = 2 x 12 2 /3 = 96 units.
10. The reflection coefficient, when a resonant cavity is placed between the waveguide is
a) 0
b) 1
c) -1
d) Infinite
Answer: b
Explanation: When the waveguide is shorted by conducting plates, the reflection coefficient will be unity. This will lead to the occurrence of standing waves.
11. The distance between two terminated plates is given by the
a) Guided wavelength
b) 2
c) Guided wavelength/2
d) /4
Answer: c
Explanation: The distance between the terminating plates is given by Vmin = λg/2, where λg is the guided wavelength.
Answer: a
Explanation: The distance between the terminating plates is given by Vmin = λg/2, where λg is the guided wavelength. On substituting for Vmin = 2cm, we get λg = 2Vmin = 2 x 0.02 = 4cm.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Transverse Electric Magnetic Waves”.
1. In a transverse electric magnetic wave, which of the following will be true?
a) E is transverse to H
b) E is transverse to wave direction
c) H is transverse to wave direction
d) E and H are transverse to wave direction
Answer: d
Explanation: In the transverse electric magnetic wave , both the electric and magnetic field strengths are transverse to the wave propagation.
2. The cut off frequency of the TEM wave is
a) 0
b) 1 GHz
c) 6 GHz
d) infinity
Answer: a
Explanation: The TEM waves have both E and H perpendicular to the guide axis. Thus its cut off frequency is zero.
3. Which component is non zero in a TEM wave?
a) Ex
b) Hz
c) Ez
d) Attenuation constant
Answer: a
Explanation: In a TEM wave, the wave propagates along the guided axis. Thus the components Ez and Hz are zero. The attenuation is also zero. The non-zero component will be Ex.
4. TEM wave can propagate in rectangular waveguides. State true/false.
a) True
b) False
Answer: b
Explanation: The rectangular waveguide does not allow the TEM wave. TEM mode can exist only in two conductor system and not in hollow waveguide in which the centre conductor does not exist.
5. The cut off wavelength in the TEM wave will be
a) 0
b) Negative
c) Infinity
d) 1/6 GHz
Answer: c
Explanation: The cut off frequency in a TEM wave is zero. Thus the cut off wavelength will be infinity.
6. The guided wavelength of a TEM wave in a waveguide having a wavelength of 5 units is
a) 0
b) Infinity
c) 5
d) 1/5
Answer: c
Explanation: The guided wavelength is same as the wavelength of the waveguide with a TEM wave. Thus the guided wavelength is 5 units.
7. The guided phase constant of a TEM wave in a waveguide with a phase constant of 2.8 units is
a) 2.8
b) 1.4
c) 0
d) Infinity
Answer: a
Explanation: The guided phase constant is same as the phase constant of the waveguide. For the given data, the guided phase constant is 2.8 units.
8. Which type of transmission line accepts the TEM wave?
a) Copper cables
b) Coaxial cable
c) Rectangular waveguides
d) Circular waveguides
Answer: b
Explanation: Hollow transmission lines support TE and TM waves only. The TEM wave is possible only in the coaxial cable transmission line, which is not hollow.
9. For a TEM wave to propagate in a medium, the medium has to be
a) Air
b) Insulator
c) Dispersive
d) Non dispersive
Answer: d
Explanation: The medium in which the TEM waves propagate has to be non- dispersive. This implies the phase velocity and the characteristic impedance has to be constant over a wide band.
Answer: a
Explanation: The stripline and parallel plate waveguides are not hollow and the dielectric is lossless. The medium is non dispersive. Thus the statement is true.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Phase and Group Velocity”.
1. In a waveguide, which of the following condition is true always?
a) phase velocity = c
b) group velocity = c
c) phase velocity > c
d) phase velocity < c
Answer: c
Explanation: The phase velocity is always greater than the speed of light in waveguides. This implies the group velocity is small.
2. The term cos θ is given by 2.5. Find the phase velocity.
a) 3
b) 5
c) 7.5
d) 2.5
Answer: c
Explanation: The phase velocity is given by Vp = c cos θ. On substituting for cos θ = 2.5 and the speed of light, we get the phase velocity as 7.5 x 10 8 m/s.
3. The cut off wavelength and the guided wavelength are given by 0.5 and 2 units respectively. Find the wavelength of the wave.
a) 0.48
b) 0.32
c) 0.45
d) 0.54
Answer: a
Explanation: The cut off wavelength and the guided wavelength are related as 2 = 2 + 2 . On substituting for λc = 0.5 and λg = 2, we get λ = 0.48 units.
4. The cut off wavelength of the rectangular waveguide in dominant mode with dimensions 6 cm x 4 cm is
a) 12cm
b) 6cm
c) 4cm
d) 2cm
Answer: a
Explanation: The cut off wavelength in the dominant mode is given by λc = 2a/m, where a is the broad wall dimension. On substituting for m = 1 and a = 6cm, we get the cut off wavelength as 12cm.
5. The product of the phase and the group velocities is given by the
a) Speed of light
b) Speed of light/2
c) 2 x Speed of light
d) /4
Answer: d
Explanation: The product of the phase and the group velocities is given by the square of the speed of the light. Thus Vp x Vg = c 2 is the relation.
6. The phase velocity of a wave having a group velocity of 6 x 10 6 is (in order of 10 8 m/s)
a) 2.4
b) 3
c) 15
d) 150
Answer: d
Explanation: We know that the phase and the group velocities are given by Vp x Vg = c 2 . On substituting for Vg = 6 x 10 6 and the speed of light, we get Vp = 150 x 10 8 m/s.
7. The group velocity of a wave with a phase velocity of 60 x 10 9 is (in 10 6 order)
a) 1.5
b) 2
c) 2.5
d) 3
Answer: a
Explanation: We know that the phase and the group velocities are given by Vp x Vg = c 2 . On substituting for Vp = 60 x 10 9 and the speed of light, we get Vg = 1.5 x 10 6 m/s.
8. The phase velocity of a wave having a phase constant of 4 units and a frequency of 2.5 x 10 9 radian/sec is (in 10 8 order)
a) 3.25
b) 3.75
c) 6.25
d) 6.75
Answer: c
Explanation: The phase velocity and the phase constant are related by Vp = ω/βg. On substituting for ω = 2.5 x 10 9 and β = 4, we get the phase velocity as 6.25 units.
9. The guided wavelength and the phase constant are related by
a) 2π/βg = λg
b) 1/βg = λg
c) 1/2πβg = λg
d) βg = λg
Answer: a
Explanation: The guided wavelength and the phase constant are related by 2π/βg = λg, where βg is the guided phase constant and λg is the guided wavelength.
10. The phase velocity refers to a group of waves and the group velocity refers to a single wave. State true/false.
a) True
b) False
Answer: b
Explanation: The phase velocity refers to a single wave and the group velocity refers to a group of waves.
11. The phase and group velocities does not depend on which of the following?
a) Frequency
b) Wavelength
c) Phase constant
d) Attenuation constant
Answer: d
Explanation: The phase and the group velocities are directly related by the frequency, wavelength and the phase constant. It is independent of the attenuation constant.
Answer: c
Explanation: The distance between two successive points in a waveguide is equal to half of the guided wavelength.
This set of Basic Electromagnetic Theory Questions and Answers focuses on “Waveguide Current and Excitation”.
1. The source voltage of a 75ohm transmission line is given by 150V. Find the load current.
a) 0.5
b) 2
c) 4
d) 1
Answer: b
Explanation: The load current is given by I L = V S /Z 0 . On substituting for V S = 150 and Z 0 = 75, we get I L = 150/75= 2A.
2. The guided terminations are used to
a) Increase reflection
b) Increase transmission
c) Eliminate reflection loss
d) Eliminate attenuation
Answer: c
Explanation: The guided termination refers to the waveguide shorted by conducting plates. This is done in order to eliminate the reflection losses.
3. Which type of wave does the resonant cavity produce?
a) Standing waves
b) Guided waves
c) Transmitted waves
d) Attenuated waves
Answer: a
Explanation: Resonant cavity is the waveguide shorted by a conducting plate. This is to reduce the reflection losses. Such arrangement leads to standing waves.
4. Which of the following parameter cannot be calculated from the standing waves?
a) Peak voltage and peak current
b) SWR
c) Reflection and transmission coefficients
d) Attenuation constant
Answer: d
Explanation: The peak voltage and current can be directly measured from the standing waves. The standing wave ratio, reflection coefficient and the transmission coefficient can also be calculated from it. Only the attenuation constant cannot be calculated directly.
5. For efficient transmission, the characteristic impedance of the transmission line has to be
a) 50 ohm
b) 75 ohm
c) Either 50 or 75 ohm
d) Neither 50 nor 75 ohm
Answer: c
Explanation: Generally, for ideal transmission lines, the characteristic impedance should be either 50 ohm or 75 ohm.
6. The cavity resonators are used in the klystron amplifiers for
a) Amplifying RF signals
b) Amplifying microwave signals
c) Attenuating RF signals
d) Attenuating microwave signals
Answer: b
Explanation: The cavity resonators are employed in the klystron amplifiers for amplifying the microwave signals.
7. The cavity resonators used in the reflex klystron oscillators are for
a) Generating RF signals
b) Generating microwave signals
c) Amplifying RF signals
d) Amplifying microwave signals
Answer: b
Explanation: Oscillators are devices that generate signal waveforms. The reflex klystron oscillator is used to generate microwave signals.
8. One of the applications of the cavity resonators is duplexer in RADAR systems. State true/false.
a) True
b) False
Answer: a
Explanation: Cavity resonator is used in duplexers of RADAR systems, as resonant cavity in transmit receive tubes and antitransmit receive tubes.
9. Cavity wave meters are used to measure which parameter of the wave?
a) Wavelength
b) Reflection factor
c) Phase
d) Frequency
Answer: d
Explanation: Cavity resonators are used in cavity wave meters for the measurement of frequency of the microwave signals.
Answer: c
Explanation: A waveguide is terminated by concept of tapered or exponential line and uses a dielectric having considerable conductivity to provide power absorbing properties. This will eliminate the reflection losses.
This set of Electromagnetic Theory Multiple Choice Questions & Answers focuses on “Transients”.
1. The resonant circuit in a waveguide refers to the
a) Tank circuit
b) RC circuit
c) Bridge circuit
d) Attenuator circuit
Answer: a
Explanation: The resonant circuit refers to the tank circuit. It is parallel combination of an inductor and a capacitor.
2. When a waveguide is terminated, the mode of the guided termination will be
a) zero
b) non-zero
c) infinite
d) does not exist
Answer: b
Explanation: A waveguide mode with a guided termination is represented by TEmnp and TMmnp. Here m and n are the orders of the waveguide and p is the order of the resonant cavity. It is always a non-zero value.
3. The cut off frequency of a waveguide with resonant cavity is given by
a) V√ 2 + 2 + 2 )/2
b) V√ 2 + 2 + 2 )
c) 2V√ 2 + 2 + 2 )
d) V√ + + )/2
Answer: a
Explanation: The cut off frequency of the waveguide of dimensions a x b with the resonant cavity of dimension d is given by fc = √ 2 + 2 + 2 )/2. Here m and n are the orders of the waveguide, p is the order of the cavity and v is the velocity.
4. The power of a wave in a transmission line, when the current and the resistance are 5A and 120 ohm respectively is
a) 3000
b) 4000
c) 2000
d) 1500
Answer: a
Explanation: The power of a line is given by P = I 2 R. On substituting for I = 5 and R = 120, we get P = 5 2 x 120 = 3000 units.
5. The power of a wave having a magnetic field intensity of 2.5 units is
a) 1.17 kilo watt
b) 1.17 mega watt
c) 1.17 watt
d) 11.7 watt
Answer: a
Explanation: The power of a wave is given by P = 0.5 η H 2 , where η is the intrinsic impedance and H is the magnetic field intensity. On substituting for η = 377 and H = 2.5, we get P = 0.5 x 377 x 2.5 2 = 1.17 kilo watts.
6. The power of a wave having an electric field strength of 12.8 units is
a) 0.217
b) 0.721
c) 0.127
d) 0.172
Answer: a
Explanation: The power of a wave is given by P = 0.5 E 2 /η, where E is the electric field intensity and η is the intrinsic impedance. On substituting for E = 12.8 and η = 377, we get P = 0.5 x 12.8 2 /377 = 0.217 units.
7. The form or mode of propagation is determined by which factors?
a) Type of excitation device
b) Location of excitation device
c) Type and location of the excitation device
d) Waveguide characteristics
Answer: c
Explanation: The form and the mode of propagation of the wave in a waveguide is determined by the type and location of the excitation device.
8. The phase of the wave after the installation of the guided terminations will be
a) 0
b) 45
c) 90
d) 180
Answer: a
Explanation: Using a guided termination, the guides serves as a reflector. If the distance between the exciter and the wall is properly adjusted, the transmitted and the reflected wave will be in phase.
9. The exciter of the waveguide in a transmission line is the
a) Transmitter
b) Receiver
c) Transponder
d) Antenna
Answer: d
Explanation: The waveguide is usually excited by the antenna rod. The reflection depends on the phase of excitation and the antenna current.
Answer: b
Explanation: A waveguide passes only high frequency waves and attenuates low frequencies. This is the characteristic of a high pass filter.
