University of Missan
College of Engineering
Electrical Engineering
Department
1st Semester Year
2013-2014
2nd Lesson Stage
Engineering Electromagnetic Fields
Subject: Vector Analysis
Lecture No. 1
Dr. Ahmed Thamer Radhi
2013 - 2014
Electromagnetic Fields
Vector Analysis
Lesson Year 1 st Semester:2013-2014
Stage
2 nd Year
Subject
Introduction
Lecture No.
1
Lecturer
Dr. Ahmed Thamer
University of Missan
College of Engineering
Electrical Engineering Dept.
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Engineering
Electromagnetic Fields
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Lecture No.1
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Introduction :
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In electromagnetic fields study, we shall establish a few basic principles of electricity and
attempt to describe them in terms of it. If we had used vector calculus for several years and
already had a few correct ideas about electricity and magnetism, we might jump in now with
both feet and present a handful of equations, including Maxwell's equations and a few other
auxiliary equations, and proceed to describe them physically by virtue of our knowledge of
vector analysis. This is perhaps the ideal way, starting with the most general results and then
showing that Ohm's, Gauss's, Coulomb's, Faraday's, Ampere's, and a few less familiar laws are
all special cases of these equations. It is philosophically satisfying to have the most general
result and to feel that we are able to obtain the results for any special case at will.
The experimental laws mentioned above, expressing each in vector notation, and use
these laws to solve the problems. In this way with both vector analysis and electric and
magnetic fields and by the time we have finally reached our handful of general equations. The
entire field of electromagnetic theory is then open to us, and we may use Maxwell's equations
to describe wave propagation, radiation from antennas, transmission lines, … etc.
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Syllabuses :
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




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Vector Analysis
Coulomb's Law and Electric Field Intensity
Electric Flux Density, Gauss' Law, and Divergence
Energy and Potential
Conductors, Dielectrics, and Capacitance
Suggested References :
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• William H. Hayt, Jr.: "Engineering Electromagnetics," Fourth Ed., McGraw-Hill Book
Company, 1981.
• Della Torre, E., and Longo, C. L.: "The Electromagnetic Field," Allyn and Bacon, Inc.,
Boston, 1969.
• Schelkunoff, S. A.: "Electromagnetic Fields," Blaisdell Publishing Company, New York,
1963.
• Spiegel, M. R.: "Vector Analysis," Schaum Outline Series, McGraw-Hill Book
Company, New York, 1959.
• Boast, W. B.: "Vector Fields,"Harper and Row, Publishers, Incorporated, New York,
1964.
Dr. Ahmed Thamer
Vector Analysis
Page 1
Electromagnetic Fields
Vector Analysis
Lecture No.1
Lesson Year 1 st Semester:2013-2014
Stage
2 nd Year
Subject
Vector Analysis
Lecture No.
1
Lecturer
Dr. Ahmed Thamer
University of Missan
College of Engineering
Electrical Engineering Dept.
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Engineering
Electromagnetics Fields
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Vector Analysis
1- Scalars and vectors
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 The term Scalar refers to a quantity defined by magnitude only, whose value may be
represented by a single (positive or negative) real number, for examples [distance,
voltage, density, pressure, etc.].
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 The term Vector refers to a quantity defined by magnitude and direction in space, for
examples [velocity, current, force, acceleration, etc.].
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We shall be mostly concerned with scalar and vector fields. A field (scalar or vector) may be
defined mathematically as some function of that vector which connects an arbitrary origin to a
general point in space. We usually find it possible to associate some physical effect with a field,
such as the force on a compass needle in the earth's magnetic field, ….. etc.
2- Position and Distance Vector
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For a given point such as P in Cartesian coordinates may be represented by (x, y, z) as shown in
�⃗ P ) of point P is a directed distance from the origin O to
Fig.1.1 below. The Position Vector (R
point P.
�⃗ P = x + y + z
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R
R
Z
•
P(x, y, z)
�⃗P
R
O
(0, 0, 0)
Y
X
Figure 1.1
Dr. Ahmed Thamer
Vector Analysis
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Electromagnetic Fields
Vector Analysis
Lecture No.1
The Distance Vector is the displacement from one point to another. If two points P and N are
given by P(x P , y P , z P ) and N(x N , y N , z N ), the distance vector is the displacement from P to N
that is:
�R⃗ PN = (x N -x P ) + (y N -yP ) + (z N -z P )
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N (xN, yN, zN)
�R⃗PN
P(xP, yP, zP)
To describe a vector in the Cartesian coordinate system, let us first consider a vector r
extending outward from the origin. If the component vectors of the vector r are x, y, and z, then
r=x+y+z as shown in Fig. 1.2a. The using of unit vectors having unit magnitude to identify the
direction of vector components x, y, and z. Thus ā x , ā y , and ā z are the unit vectors in the
Cartesian coordinate system are directed along the x, y, and z axes, respectively, as shown in
Fig. 1.2b. Thus, 𝐫𝐫⃗ =x ā x + y ā y + z ā z .
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Figure 1.2 Vector in the Cartesian Coordinate System
For example, if 𝐫𝐫⃗ P is a position vector from the origin to point P(1, 2, 3) is written as 𝐫𝐫⃗ P =ā x +2
ā y +3ā z and 𝐫𝐫⃗ Q is a position vector from the origin to point Q(2, -2, 1) is written as 𝐫𝐫⃗ Q =2ā x -2
ā y + ā z , then the distance vector from P to Q is �R⃗ PQ equal to:
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�R⃗ PQ =𝐫𝐫⃗ Q - 𝐫𝐫⃗ P = (2-1) ā x + (-2-2) ā y + (1-3) ā z = ā x - 4 ā y - 2ā z
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�⃗ PQ are shown in Fig. 1.3. The difference between a point P and a
The Vectors 𝐫𝐫⃗ P , 𝐫𝐫⃗ Q and R
vector �A⃗ should be noted. Vector �A⃗ may be depending on point P, for example, if:
�A⃗=2xy ā x + y 2 ā y – xz 2 ā z and P is (2, -1, 4), then vector A
�⃗ at point P would be:
�A⃗(P)= (2*2*-1) ā x + (-1) 2 ā y – (2*4 2 ) ā z = -4 ā x + ā y – 32 ā z
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Electromagnetic Fields
Vector Analysis
Lecture No.1
��⃗PQ
Figure 1.3 Vectors 𝐫𝐫⃗P, 𝐫𝐫⃗Q and 𝐑𝐑
The vector field already defined as a vector function of a position vector. The vector field is
said to be constant or uniform if it does not depend on space variable x, y and z. For example,
�⃗=2xy ā x + y 2 ā y – xz 2 ā z is not
�⃗ =3ā x - 2ā y + 10ā z is a uniform vector while vector A
vector B
�⃗ varies from point to point.
�⃗ is the same everywhere where as A
uniform vector, because B
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Let �A⃗ = A x ā x + A y ā y + A z ā z is a vector with specific direction, then the magnitude or length
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�⃗�, is given by:
of �A⃗ written �A
�⃗�=√A2𝑥𝑥 + A2𝑦𝑦 + A2𝑧𝑧
�A
A unit vector is a vector with a specific direction has unit magnitude (i.e. |unit vector|=1).
Assume that, a��⃗A is a unit vector in the direction of �A⃗ with a magnitude is unity (i.e. |a��⃗A |=1),
�A�⃗
Ax
Ay
Az
�⃗�=√A2𝑥𝑥 + A2𝑦𝑦 + A2𝑧𝑧
then, �A⃗=A a��⃗A , Such that, a��⃗A = = ā x + ā y + ā z ; Where �A
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��⃗�
�A
��⃗�
�A
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��⃗�
�A
R
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R
��⃗�
�A
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Example 1.1 : Determine the unit vector extending from the origin toward point H(2, -2, -1).
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Solution : We first construct the position vector extending from the origin to point H,
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�⃗= 2ā x - 2ā y - ā z
�H
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We continue by finding the magnitude of H,
��⃗�= √ (2) 2 + (-2) 2 + (-1) 2 = 3
�H
P
P
P
P
P
P
and finally expressing the desired unit vector as:
a�⃗ H =
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�H
�⃗
��⃗�
�H
2
2
1
= ā x - ā y - ā z = 0.667ā x – 0.677ā y – 0.333ā z
3
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Q1.1 (H.W ): Given points M(-1, 2, 1), N(3, -3, 0), and P(-2, -3, -4), find:
�⃗ MN ; (b) R
�⃗ MN + R
�⃗ MP ; (c) �R
�⃗ M �;
�⃗ P − 3R
�⃗ N �
(d) a�⃗ RMP ;
(a) R
(e) �2R
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Electromagnetic Fields
Vector Analysis
Lecture No.1
3- Vector Addition and Subtraction
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�⃗. Two
�⃗ can be added together to give another vector �C⃗, that is �A⃗ + B
�⃗ =C
Two vectors �A⃗ and B
vectors may be added graphically either by drawing both vectors from a common origin and
completing the parallelogram or by beginning the second vector from the head of the first and
completing the triangle such as shown in Fig. 1.4 below:
Figure 1.4 Vector Addition
�⃗=A x ā x + A y ā y + A z ā z and B
�⃗ is:
�⃗ =B x ā x + B y ā y + B z ā z then, the vector addition C
If A
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�C⃗=(A x + B x ) ā x + (A y + B y) ā y + (A z + B z ) ā z
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�⃗. Vector addition also
�⃗ = B
�⃗ +A
The vector addition obeys the commutative law, such that �A⃗ + B
�⃗) = (A
�⃗ +B
�⃗ +C
�⃗) + �C⃗
obeys the associative law, such that �A⃗ + (B
The rule for the subtraction of vectors follows easily from that for addition, for we may always
�⃗ +(- B
�⃗ - B
�⃗ = A
�⃗); the sign, or direction, of the second vector is reversed, and this vector
express A
is then added to the first by the rule for vector addition as shown graphically in Fig. 1.5.
�⃗ - B
��⃗=A
�⃗
D
�⃗
−B
�A⃗
Figure 1.5 Vector Subtraction
�⃗ - B
�D
�⃗=A
�⃗
�⃗
B
�⃗
�⃗ - B
�⃗ + D
��⃗ = A
��⃗=A
�⃗
From Fig. 5 above, we see that: B
D
��⃗ =(A x - B x ) ā x + (A y - B y ) ā y + (A z - B z ) ā z
Such that, D
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Electromagnetic Fields
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Vector Analysis
Lecture No.1
3.1 - Some Properties of Vector Algebra
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 Vectors may be multiplied by scalars. The magnitude or length of the vector changes, but
its direction does not change when the scalar is positive, although it reverses direction
�⃗ is a vector and C is a positive
when multiplied by a negative scalar. For example, if A
�⃗ that is: B x ā x + B y ā y + B z ā z = C(A x ā x + A y ā y + A z ā z )
�⃗ = CA
scalar then, B
�⃗= CA x ā x + CA y ā y + CA z ā z
Hence, B
�⃗ is a vector and C is a scalar, then,
 If A
��⃗
A
1
�⃗) = 1 (A x ā x + A y ā y + A z ā z ) = 1 A x ā x + 1 A y ā y + 1 A z ā z
= (A
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 Multiplication of a vector by a scalar also obeys the associative and distributive laws of
�⃗ +B
�⃗ +B
�⃗ + rB
�⃗ + 𝑠𝑠B
�⃗ +B
�⃗)= r(A
�⃗) + s(A
�⃗) = rA
�⃗ + 𝑠𝑠A
�⃗
algebra, leading to (r+s)( (A
�⃗ = B
�⃗ if �A⃗ − B
�⃗ = 0.
 Two vectors are said to be equal if there difference is zero, or A
4- Vector Multiplication
�⃗ are multiplied, the result is either a scalar or a vector
When two vectors �A⃗ and B
depending on how they are multiplied. Thus there are two types of vector multiplication
are:
�⃗ . 𝐁𝐁
��⃗
4-1 Scalar or dot product: 𝐀𝐀
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�⃗ X 𝐁𝐁
��⃗
4-2 Vector or cross product: 𝐀𝐀
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4-1 Dot Product
�⃗ written as �A⃗.B
�⃗ , is defined geometrically as the product
The dot product of two vectors �A⃗ and B
�⃗ and B
�⃗ and the cosine of the angle between them; thus:
of the magnitudes of A
�A⃗.B
�⃗� �B
�⃗� cosθ
�⃗ = �A
�⃗ = √A2𝑥𝑥 + A2𝑦𝑦 + A2𝑧𝑧
�⃗�=magnitude of A
Where �A
�⃗ = √B𝑥𝑥2 + B𝑦𝑦2 + B𝑧𝑧2
�⃗� = magnitude of B
�B
θ
�⃗ and B
�⃗
θ= is the angle between A
�⃗.B
�⃗ is called scalar value.
The result of A
�A⃗
�⃗
B
�⃗ = B x ā x + B y ā y + B z ā z ; then,
If �A⃗ = A x ā x + A y ā y + A z ā z and B
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�⃗.B
�⃗ = A x B x + A y B y + A z B z
A
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�⃗.B
�⃗
�⃗ =B
�⃗.A
Commutative law: A
�⃗ . (B
�⃗) = �A⃗.B
�⃗
�⃗ +C
�⃗ + �A⃗.C
Distributive law: A
Dr. Ahmed Thamer
Vector Analysis
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Electromagnetic Fields
Vector Analysis
Lecture No.1
�⃗� 2 .
�⃗ = �A
A vector dotted with itself yields the magnitude squared, or �A⃗.A
P
P
Since the angle between two different unit vectors of Cartesian coordinate system is 90 o , we
P
P
then have:
ā x . ā y = ā y . ā x = ā x . ā z= ā z . ā x = ā y . ā z = ā z . ā y = 0
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The dot product of a unit vector with itself, which is unity, ā x . ā x = ā y . ā y = ā z . ā z =1 and any
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unit vector dotted with itself is unity, such as: a��⃗A . a��⃗A =1
One of the most important applications of the dot product is that of finding the component of a
�⃗ in
vector in a given direction. Referring to Fig. 1.6a, we can obtain the component (scalar) of B
the direction specified by the unit vector a��⃗ as:
�⃗. a��⃗= �B
�⃗� |a�⃗| cosθ Ba = �B
�⃗� cosθ Ba ; where |a�⃗|=1 (is a unit vector)
B
R
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R
R
�⃗ in the direction of a��⃗, we simply multiply the
In order to obtain the component vector of B
�⃗. a��⃗)a��⃗, as illustrated by Fig. 1.6b.
component (scalar) by a��⃗, i.e. (B
Figure 1.6
The geometrical term Projection is also used with dot product, such as:
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�⃗ along vector B
�⃗ , is equal to:
 Scalar component of projection of vector A
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AB=
R
R
���⃗
��⃗
A. B
��⃗
B
�⃗ . a��⃗B
=A
�⃗ , is equal to:
 Vector component of projection of vector �A⃗ along vector B
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�A⃗ B = A B a��⃗B = (A
�⃗ . a��⃗B ) a��⃗B
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4-2 Cross Product
�⃗ written as �A⃗ X B
�⃗ , is a vector quantity whose
The cross product of two vectors �A⃗ and B
�⃗ and B
�⃗. Thus:
magnitude is the area of the parallelepiped formed by A
Dr. Ahmed Thamer
�⃗XB
�⃗� �B
�⃗� sinθ a��⃗n
�⃗ = �A
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Vector Analysis
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Electromagnetic Fields
Vector Analysis
Lecture No.1
�⃗. The direction of a��⃗n is taken
Where a��⃗n is a unit vector normal to the plane containing �A⃗ and B
�⃗ as shown in
as the direction of right thumb when the fingers of right hand rotate from �A⃗ to B
Fig. 1.7 below.
�A⃗XB
�⃗
a��⃗n
�⃗
B
θ
�A⃗
Figure 1.7
�⃗ = B x ā x + B y ā y + B z ā z ; then,
If �A⃗ = A x ā x + A y ā y + A z ā z and B
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āx āy āz
�A⃗XB
�⃗ = �Ax Ay Az�
Bx By Bz
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R
�⃗ = (A y B z - A z B y ) ā x – (A x B z - A z B x ) ā y + (A x B y - A y B x ) ā z
Hence, �A⃗XB
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Basic properties of the cross product:
�⃗XB
�⃗ , it is anticommutative, A
�⃗XB
�⃗
�⃗ ≠ B
�⃗ XA
�⃗ = - B
�⃗ XA
 It is not commutative, A
�⃗X(B
�⃗) = (A
�⃗XB
�⃗XC
�⃗)
�⃗+C
�⃗) + (A
 It is distributive, A
 The cross product of any vector with itself is zero, since the included angle is zero.
�⃗ = 0
 �A⃗XA
āx
 ā x X ā x = ā y X ā y = ā z X ā z =0
+  āx X ā y= ā z ; ā y X āz= ā x ; ā z X ā x= ā y
 ā y X ā x = -ā z ; ā z X ā y= -ā x ; ā x X ā z = -ā y
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āz
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āy
�⃗XB
�⃗ = -4 ā x - 2 ā y + 5 ā z , find A
�⃗.
Example 1.2 : If �A⃗ = 2 ā x - 3 ā y + ā z and B
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Solution :
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āx āy āz āx
�A⃗XB
�⃗ = �Ax Ay Az�=� 2
Bx By Bz −4
āy āz
−3 1 �
−2 5
�A⃗XB
�⃗ = [(-3)(5)- (1)(-2)] ā x – [(2)(5)- (1) (-4)] ā y + [(2)(-2)- (-3)(-4)] ā z
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�⃗XB
�⃗ = -13 ā x – 14 ā y - 16 ā z
A
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