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Syllabus
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Syllabus for Electromagnetic Theory
Elements of Vector Calculus, Divergence and Curl, Gauss and Stoke’s Theorems, Maxwell’s
Equations, Differential and Integral Forms. Wave Equation, Poynting Vector. Plane Waves,
Propagation Through Various Media, Reflection and Refraction, Phase and group Velocity, Skin
Depth. Transmission Lines, Characteristic Impedance, Impedance Transformation, Smith Chart,
Impedance Matching, S Parameters, Pulse Excitation. Waveguides, Modes in Rectangular
Waveguides, Boundary Conditions, Cut-Off Frequencies, Dispersion Relations. Basics of Propagation
in Dielectric Waveguide and Optical Fibers. Basics of Antennas, Dipole Antennas, Radiation Pattern,
Antenna Gain.
Analysis of GATE Papers
Year Percentage of Marks Overall Percentage
2015 8.70
8.862%
2014 8.25
2013 5.00
2012 12.00
2011 9.00
2010 7.00
2009 8.00
2008 8.00
2007 10.67
2006 12.00
Contents
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CCoonntteennttss
Chapters Page No.
#1. Electromagnetic Field 1 – 43 Introduction 1
Operators 2 – 7
Material and Physical Constants 7 – 8
Electromagnetic (EM Field) 8 – 9
Electric Field Intensity 9 – 12
Electric Dipole 12 – 17
Divergence of Current Density and Relaxation 18
Boundary Conditions 19 – 21
The Magnetic Vector Potential 21 – 25
Faraday’s Law 25 – 27
Maxwell’s Equation’s 27 – 28
Solved Examples 28 – 34
Assignment 1 35 – 37
Assignment 2 37 – 38
Answer Keys & Explanations 39 – 43
#2. EM Wave Propagation 44 – 71 Introduction 44
General Wave Equations 44 – 45
Plane Wave in a Dielectric Medium 45 – 49
Loss Tangent 49 – 54
Wave Polarization 54 – 61
Assignment 1 62 – 64
Assignment 2 64 – 66
Answer Keys & Explanations 67 – 71
#3. Transmission Lines 72 – 103 Introduction 72 – 73
Classification of Transmission Lines 73 – 77
Transients on Transmission Lines 77 – 78
Bounce Diagram 78 – 81
Transmission & Reflection of Waves on a Transmission Line 81 – 82
Impedance of a Transmission Line 83 – 88
The Smith Chart 88 – 90
Scattering Parameters 90 – 91
Contents
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Strip Line 91 – 93
Balun 93 – 94
Assignment 1 95 – 96
Assignment 2 97 – 98
Answer Keys & Explanations 99 – 103
#4. Guided E.M Waves 104 – 133 Introduction 104
Wave Guide 104
Rectangular Wave Guide 105 – 107
Transverse Magnetic (TM) Mode: (Hz = 0) 107 – 110
Transverse Electric (TE) Mode: (Ez = 0) 110 – 115
Non-Existence of TEM Waves in Wave Guides 116
Circuter Wave Guide 116 – 117
Cavity Resonators 117 – 119
Optical Fiber 120 – 121
Stub Matching Technique 121 – 122
Solved Examples 122 – 127
Assignment 1 128 – 129
Assignment 2 129 – 130
Answer Keys & Explanations 131 – 133
#5. Antennas 134 – 166 Inroduction 134
Hertzian Dipole 135 – 137
Field Regions 137
Antenna Characteristics 138 – 141
Radiation Pattern 141 – 142
Antenna Arrays 142 – 147
Solved Examples 148 – 159
Assignment 1 160 – 161
Assignment 2 162 – 163
Answer Keys & Explanations 163 – 166
Module Test 167 – 177
Test Questions 167 – 172
Answer Keys & Explanations 172 – 177
Reference Books 178
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“Picture yourself vividly as winning and that
alone will contribute immeasurably to success."
…Harry Fosdick
Electromagnetic
Field
Learning Objectives After reading this chapter, you will know:
1. Elements of Vector Calculus
2. Operators, Curl, Divergence
3. Electromagnetic Coulombs’ law, Electric Field Intensity, Electric Dipole, Electric Flux Density
4. Gauss's Law, Electric Potential
5. Divergence of Current Density and Relaxation
6. Boundary Conditions
7. Biot-Savart’s Law, Ampere Circuit Law, Continuity Equation
8. Magnetic Vector Potential, Energy Density of Electric & Magnetic Fields, Stored Energy in
Inductance
9. Faraday’s Law, Motional EMF, Induced EMF Approach
10. Maxwell’s Equations
Introduction Cartesian coordinates (x, y, z), −∞ < x < ∞, −∞ < y < ∞, −∞ < z < ∞ Cylindrical coordinates (ρ , ϕ, z), 0 ≤ ρ < ∞, 0 ≤ ϕ < 2π, −∞ < z < ∞ Spherical coordinates (r, θ, ϕ ) , 0 ≤ r < ∞, 0 ≤ θ ≤ π, 0 ≤ ϕ < 2π Other valid alternative range of θ and ϕ are-----
(i) 0 ≤ θ < 2π, 0 ≤ ϕ ≤ π (ii) −π ≤ θ ≤ π, 0 ≤ ϕ ≤ π
(iii) −π
2≤ θ ≤ π
2⁄ , 0 ≤ ϕ < 2π
(iv) 0 < θ ≤ π,−π ≤ ϕ < π
Vector Calculus Formula
SL. No Cartesian Coordinates Cylindrical Coordinates Spherical Coordinates
(a) Differential Displacement dl = dx ax + dy ay + dz az
dl = dρaρ + ρdϕaϕ +dzaz dl = drar + rdθaθ + r sin θdϕaϕ
(b) Differential Area dS = dy dz ax = dx dz ay
= dx dy az
dS = ρ dϕ dz aρ
= d ρ dz aϕ
= ρdρd ϕ az
ds = r2sin θ dθ dϕ ar = r sin θ dr dϕ aθ = r dr dθ aϕ
(c) Differential Volume dv = dx dy dz
dv = ρ dρ dϕ dz dv = r2sin θ dθ dϕ dr
CH
AP
TE
R
1
Electromagnetic Field
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Operators 1) ∇ V – Gradient, of a Scalar V
2) ∇ V – Divergence, of a Vector V
3) ∇ × V – Curl, of a Vector V
4) ∇2 V – Laplacian, of a Scalar V
DEL Operator:
∇ =∂
∂xax +
∂
∂yay +
∂
∂z az(Cartesian)
= ∂
∂ρaρ +
1
ρ
∂
∂ϕaϕ +
∂
∂z az(Cylindrical)
=∂
∂rar +
1
r
∂
∂θaθ +
1
rsi n θ ∂
∂ϕ aϕ(Spherical)
Gradient of a Scalar field
V is a vector that represents both the magnitude and the direction of maximum space rate of
increase of V.
∇V =∂V
∂xax +
∂V
∂yay +
∂V
∂z az For Cartisian Coordinates
=∂V
∂ρaρ +
1
ρ
∂V
∂ϕaϕ +
∂V
∂z az For Spherical Coordinates
=∂V
∂rar +
1
r
∂V
∂θaθ +
1
rsi n θ ∂V
∂ϕ aϕ For Cylindrical Coordinates
The following are the fundamental properties of the gradient of a scalar field V
1. The magnitude of ∇V equals the maximum rate of change in V per unit distance.
2. ∇V points in the direction of the maximum rate of change in V.
3. ∇V at any point is perpendicular to the constant V surface that passes through that point.
4. If A = ∇V, V is said to be the scalar potential of A.
5. The projection of ∇V in the direction of a unit vector a is ∇V. a and is called the directional
derivative of V along a. This is the rate of change of V in direction of a.
Example: Find the Gradient of the following scalar fields:
(a) V = e−z sin 2x cosh y
(b) U = ρ2z cos 2ϕ
(c) W = 10r sin2θ cos ϕ
Solution:
(a) ∇V =∂V
∂xax +
∂V
∂yay +
∂V
∂zaz
= 2e−z cos 2x cosh y ax + e−z sin 2x sinh y ay − e−z sin 2x cosh y az
(b) ∇U =∂U
∂ρaρ +
1
ρ
∂U
∂ϕaϕ +
∂U
∂zaz
= 2ρz cos 2ϕ aρ − 2ρz sin 2ϕ aϕ + ρ2 cos 2 ϕ az
(c) ∇W =∂W
∂rar +
1
r
∂W
∂θaθ +
1
r sinθ
∂W
∂ϕaϕ
= 10 sin2θ cosϕ ar + 10 sin 2θ cosϕ aθ − 10 sin θ sinϕ aϕ
Electromagnetic Field
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Divergence of a Vector
Statement: Divergence of A at a given point P is the outward flux per unit volume as the volume
shrinks about P.
Hence,
DivA = ∇. A = lim∆v→0
∮ A . ds
S
∆v ………………………………………(1)
Where, ∆v is the volume enclosed by the closed surface S in which P is located. Physically, we may
regard the divergence of the vector field A at a given point as a measure of how much the field
diverges or emanates from that point.
∇. A =∂Ax
∂x+
∂Ay
∂y
∂Az
∂z Cartisian System
=1
ρ
∂
∂ρ(ρAρ) +
1
ρ
∂Aϕ
∂ϕ+
∂Az
∂z Cylindrical System
=1
r2
∂
∂r(r2Ar) +
1
r sin θ ∂
∂θ(Aθ sin θ) +
1
r sin θ ∂Aϕ
∂ϕ Sphearical System
From equation (1),
∮A . dS
S
= ∫∇ . A dv
V
This is called divergence theorem which states that the total outward flux of the vector field A
through a closed surface S is same as the volume integral of the divergence of A.
Example: Determine the divergence of these vector field
(a) P = x2yzax + xzaz
(b) Q = ρ sinϕ aρ + ρ2zaϕ + z cos ϕ az
(c) T =1
r2 cos θ ar + r sin θ cosϕ aθ + cos θ aϕ
Solution:
(a) ∇. P =∂
∂xPx +
∂
∂yPy +
∂
∂zPz
=∂
∂x(x2yz) +
∂
∂y(0) +
∂
∂z(xz)
= 2xyz + x
(b) ∇ . Q =1
ρ
∂
∂ρ(ρQρ) +
1
ρ
∂
∂ϕQϕ +
∂
∂zQz
=1
ρ
∂
∂ρ(ρ2 sinϕ) +
1
ρ
∂
∂ϕ(ρ2z) +
∂
∂z (z cosϕ)
= 2 sinϕ + cosϕ
(c) ∇. T =1
r2
∂
∂r(r2Tr) +
1
r sinθ
∂
∂θ(Tθ sin θ) +
1
r sinθ
∂
∂ϕ (Tϕ)
=1
r2
∂
∂r(cos θ) +
1
r sin θ
∂
∂θ(r sin 2θ cosϕ) +
1
r sin θ
∂
∂ϕ(cos θ)
= 0 +1
r sin θ2r sin θ cos θ cosϕ + 0
= 2 cos θ cosϕ
P
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Curl of a Vector
Curl of a Vector field provides the maximum value of the circulation of the field per unit area and
indicates the direction along which this maximum value occurs.
That is,
Curl A = ∇ × A = limΔS→0
(∮ A . dl
L
∆S)
max
an ………………… . . (2)
∇ × A = ||
ax ay az
∂
∂x
∂
∂y
∂
∂zAx Ay Az
||
= 1
ρ ||
aρ ρaϕ az
∂
∂ρ
∂
∂ϕ
∂
∂zAρ ρAϕ Az
||
= 1
r2 sin θ ||
aρ raθ r sin θ aϕ
∂
∂r
∂
∂θ
∂
∂ϕAr rAθ r sin θ Aϕ
||
From equation (2) we may expect that
∮ A dl = ∫(∇ × A
S
) . ds
L
This is called stoke’s theorem, which states that the circulation of a vector field A around a (closed)
path L is equal to the surface integral of the curl of A over the open surface S bounded by L, Provided
A and Δ × A are continuous no s.
Example: Determine the curl of each of the vector fields.
(a) P = x2yz ax + xzaz
(b) Q = ρ sinϕaρ + ρ2 zaϕ + z cosϕaz
(c) T =1
r2 cos θ ar + r sinθ cosϕaθ + cosϕ aϕ
Solution:
(a) ∇ × P = (∂Pz
∂y−
∂Py
∂z) ax + (
∂Px
∂z−
∂Pz
∂x) ay + (
∂Py
∂x−
∂Px
∂y) az
= (0 − 0)ax + (x2y − z)ay + (0 − x2z)az
= (x2y − z)ay − x2zaz
(b) ∇ × Q = [1
ρ
∂Qz
∂ϕ−
∂Qϕ
∂z] aρ + [
∂Qρ
∂z−
∂Qz
∂ρ] aϕ +
1
ρ [
∂
∂ρ(ρQϕ) −
∂Qρ
∂ϕ] az
= (−z
ρsinϕ − ρ2) aρ + (0 − 0)aϕ +
1
ρ(3ρ2z − ρ cosϕ)az
= −1
ρ(z sinϕ + ρ3)aρ + (3ρz − cosϕ)az
Electromagnetic Field
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(c) ∇ × T =1
r sin θ[∂
∂θ(Tϕ sin θ) −
∂
∂ϕTθ] ar
+1
r[
1
sin θ
∂
∂ϕTr −
∂
∂r(rTϕ)] aθ +
1
r [
∂
∂r(rTθ) −
∂
∂θTr] aϕ
=1
r sin θ[∂
∂θ(cos θ sin θ) −
∂
∂ϕ(r sin θ cosϕ)] ar
+1
r [
1
sin θ
∂
∂ϕ
(cos θ)
r2−
∂
∂r(r cos θ)] aθ
+1
r[∂
∂r(r2 sin θ cosϕ) −
∂
∂θ
(cos θ)
r2] aϕ
=1
r sin θ(cos 2θ + r sin θ sinϕ)ar +
1
r(0 − cos θ)aθ
+1
r(2r sin θ cosϕ +
sin θ
r2) aϕ
= (cos 2θ
r sin θ+ sinϕ) ar −
cos θ
raθ + (2 cosϕ +
1
r3) sin θ aϕ
Laplacian
(a) Laplacian of a scalar field V, is the divergence of the gradient of V and is written as ∇2V.
∇2V =∂2V
∂x2+
∂2V
∂y2+
∂2V
∂z2→ For Cartisian Coordinates
∇2V =1
ρ
∂
∂ρ(ρ
∂V
∂ρ) +
1
ρ2
∂2V
∂ϕ2+
∂2V
∂z2→ For Cylindrical Coordinates
=1
r2
∂
∂r(r2
∂V
∂r) +
1
r2 sin θ
∂
∂θ(sin θ
∂V
∂θ) +
1
r2 sin θ
∂2V
∂ϕ2→ For Spherical Coordinates
If ∇2V = 0, V is said to be harmonic in the region.
A vector field is solenoid if ∇.A = 0; it is irrotational or conservative if ∇ × A = 0
∇. (∇ × A) = 0
∇ × (∇V) = 0
(b) Laplacian of Vector A
∇2A = ⋯ is always a vector quantity
∇2A = (∇2Ax)ax + (∇2Ay)ay + (∇2Az)az
∇2Ax → Scalar quantity
∇2Ay → Scalar quantity
∇2Az → Scalar quantity
∇2V =−p
ϵ........Poission’s E.q.
∇2V = 0 ........Laplace E.q.
∇2E = μσ ∂E
∂t+ μE
∂2E
∂t2. . . . . . . wave E. q.
Example: The potential (scalar) distribution in free space is given as V = 10y4 + 20x3.
If ε0: permittivity of free space what is the charge density ρ at the point (2,0)?
Electromagnetic Field
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𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧: Poission’s Equation ∇2 V = −ρ
ε
(∂2
∂x2+
∂2
∂y2+
∂2
∂z2) (10 y4 + 20x3) =
−ρ
ε0
∵ ε = εr ε0[ε = ε0 as εr = 1]
20 × 3 × 2x + 10 × 4 × 3y2 =−ρ
ε0
At pt(2, 10) ⇒ 20 × 3 × 2 × 2 =−ρ
ε0ρ = −240ε0
Example: Find the Laplacian of the following scalar fields
(a) V = e−z sin 2x cosh y
(b) U = ρ2z cos 2ϕ
(c) W = 10r sin2 θ cos ϕ
Solution: The Laplacian in the Cartesian system can be found by taking the first derivative and later
the second derivative.
(a) ∇2V =∂2V
∂x2+
∂2V
∂y2+
∂2V
∂z2
=∂
∂x(2e−z cos 2x cosh y) +
∂
∂y(e−z sin 2x sinh y) +
∂
∂z(−e−z sin 2x cosh y)
= −4e−z sin 2x cosh y + e−z sin 2x cosh y + e−z sin 2x cosh y
= −2e−z sin 2x cosh y
(b) ∇2U =1
ρ
∂
∂ρ(ρ
∂U
∂ρ) +
1
ρ2
∂2U
∂ϕ2+
∂2U
∂z2
=1
ρ
∂
∂ρ(2ρ2z cos 2ϕ) −
1
ρ24ρ2z cos 2ϕ + 0
= 4z cos 2ϕ − 4z cos 2ϕ
= 0
(c) ∇2W =1
r2
∂
∂r (r2
∂W
∂r) +
1
r2 sin θ
∂
∂θ(sin θ
∂W
∂θ) +
1
r2 sin2θ
∂2W
∂ϕ2
=1
r2
∂
∂r(10 r2 sin2θ cosϕ) +
1
r2 sinθ
∂
∂θ(10r sin 2θ sin θ cosϕ) −
10r sin2θ cosϕ
r2 sin2θ
=20 sin2 θ cosϕ
r+
20r cos 2θ sin θ cosϕ
r2 sin θ+
10r sin 2θ cos θ cosϕ
r2 sin θ−
10 cosϕ
r
=10 cosϕ
r (2 sin2 θ + 2 cos 2θ + 2 cos2 θ − 1)
=10 cosϕ
r(1 + 2 cos 2θ)
Stoke’s Theorem
Statement: Closed line integral of any vector A integrated over any closed curve C is always equal to
the surface integral of curl of vector A integrated over the surface area ‘s’ which is enclosed by the
closed curve ‘c’.