Chap I
Chap I
Chap I
Course Content
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Core Reading Material
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CHAPTER ONE: VECTOR ANALYSIS
1.1 INTRODUCTION
Electromagnetic theory is a discipline concerned with the study of charges
at rest and in motion. Electromagnetic principles are fundamental to the
study of electrical engineering and physics.
Electromagnetic theory is also indispensable to the understanding,
analysis and design of various electrical, electromechanical and electronic
systems.
Some of the branches of study where electromagnetic principles find
application are: RF communication, Microwave Engineering, Antennas,
Electrical Machines, Satellite Communication, Atomic and nuclear
research, Radar Technology, Remote sensing, EMI EMC, Quantum
Electronics.
What is a field?
Consider a magnet. It has its own effect a region surrounding the magnet,
there exists a particular value for that physical function, at every point
describing the effect of the magnet. So field can be defined as the region
in which, at each point there exists a corresponding value of some
physical function. If the field produced is due to a magnetic effect it’s
called magnetic field
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There are two types of electric charges, positive and negative. Such an
electric charge produces a field surrounding it which is called an electric
field
Moving charges produce current and current carrying conductor produces
a magnetic field. In such cases electric and magnetic fields are related to
each other. Such a field is called electromagnetic field. Such fields may
be time varying or time independent.
Electromagnetic theory deals directly with the electric and magnetic field
vectors whereas circuit theory deals with the voltages and currents.
Voltages and currents are integrated effects of electric and magnetic fields
respectively.
Electromagnetic field problems involve three space variables along with
the time variable and hence the solution tends to become correspondingly
complex. Vector analysis is a mathematical tool with which
electromagnetic concepts are more conveniently expressed and best
comprehended.
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Since use of vector analysis in the study of electromagnetic field theory
results in real economy of time and thought, we first introduce the concept
of vector analysis.
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REPRESENTATION OF A VECTOR
In two dimensional, a vector can be
represented by a straight line with an
arrow in a plane. The length of the
segment is the magnitude of a vector
while the arrow indicates the
direction of the vector The vector
shown in the figure is symbolically
⃗⃗⃗⃗⃗ .
denoted as 𝑂𝐴
Its length is called as magnitude,
which is R for the vector OA
⃗⃗⃗⃗⃗ |=R
It is represented as |𝑂𝐴
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UNIT VECTOR
A unit vector has a function to indicate the direction. Its magnitude is always
unity, irrespective of its direction. Thus for any vector, to indicate its direction
a unit vector can be used.
Consider a unit vector 𝑎𝑂𝐴 as shown in the figure.
This indicates the direction of 𝑂𝐴 ⃗⃗⃗⃗⃗ but its
magnitude is unity. So vector ⃗⃗⃗⃗⃗ can be
𝑂𝐴
represented completely as its magnitude R and the
direction as indicated by the unit vector along its
direction.
⃗⃗⃗⃗⃗ =|𝑂𝐴
𝑂𝐴 ⃗⃗⃗⃗⃗ |𝑎𝑂𝐴 = R 𝑎𝑂𝐴
𝑎𝑂𝐴 is a unit vector along the direction OA and| 𝑎𝑂𝐴 | = 1
letter a⃗ is used to indicate the unit vector
QUESTION
Mention the purpose of unit in vector algebra
Incase if a vector is known then the unit vector along that vector can be
obtained by dividing the vector by its magnitude. Thus unit vector can be
expressed as,
⃗⃗⃗⃗⃗⃗
𝑂𝐴
Unit vector 𝑎𝑂𝐴 =
|𝑂𝐴|
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VECTOR ALGEBRA
SCALING OF VECTOR
This is multiplication by a scalar to a vector
This changes the magnitude (length) of a vector but not its direction, when
scalar is positive.
When scalar =-1, the magnitude remains same but direction of the vector
reverses
ADDITION OF VECTORS
The vectors which lie on the same plane are coplanar vectors
PARALLELOGRAM RULE:
Complete the parallelogram as shown in the figure. Then the diagonal of the
parallelogram represents the addition of the two vectors.
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HEAD TO TRIAL RULE:
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SUBTRACTION OF VECTORS
⃗ is
The subtraction of vectors can be obtained from the rules of addition. If 𝐵
to be subtracted from 𝐴 then based on addition it can be represented as
⃗)
𝐶 = 𝐴+ (-𝐵
Identical vectors:
⃗ =0
𝐴-𝐵 ⃗ hence identical
𝐴=𝐵
VECTOR MULTIPLICATION:
Consider two vectors 𝐴 and 𝐵 ⃗ .There are two types of products existing
depending upon the result of the multiplication. These two types of products
are
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1.Scalar or dot product
2.Vector or cross product
The result of such a dot product is scalar hence it is also called a scalar
product.
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3 If the dot product of vector with itself is performed, the result is square of
the magnitude of that vector.
𝐴.𝐴 = | 𝐴| |𝐴| cos 𝜃 =| 𝐴| 2
4 Any unit vector dotted with itself is unity
𝑎𝑥 . 𝑎𝑥 = 1 = 𝑎𝑦 . 𝑎𝑦 = 𝑎𝑧 . 𝑎𝑧
Question
Given two vectors how to identify whether they are perpendicular or parallel to
each other
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VECTOR (OR) CROSS PRODUCT OF VECTORS
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⃗ ≠𝐵
𝐴× 𝐵 ⃗×𝐴
2 ⃗ , a unit vector
Reversing the order of the vectors 𝐴 and 𝐵
𝑎N reverses its direction hence we can write
⃗ = −(𝐵
𝐴× 𝐵 ⃗ × 𝐴)
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Then
𝑎𝑥 × 𝑎𝑦 = |𝑎𝑥 ||𝑎𝑦 | sin 900 𝑎N
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𝑎𝑥 𝑎𝑦 𝑎𝑧
⃗ =|𝐴𝑥
𝐴× 𝐵 𝐴𝑦 𝐴𝑧 |
𝐵𝑥 𝐵𝑦 𝐵𝑧
⃗ × 𝐶 )= 𝐵
𝐴.( 𝐵 ⃗ .( 𝐶 × 𝐴)= 𝐶 .( 𝐴 × 𝐵
⃗)
𝐴𝑥 𝐴𝑦 𝐴𝑧
⃗ × 𝐶 )=| 𝐵𝑥
𝐴.( 𝐵 𝐵𝑦 𝐵𝑧 |
𝐶𝑥 𝐶𝑧 𝐶𝑧
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2 If two or three vectors are equal, then the result of the scalar triple product
is zero
3 Cyclic order a, b, c is to be followed. If the order is changed, the sign is
reversed
⃗ × 𝐶 )= − 𝐵
𝐴.( 𝐵 ⃗ .( 𝐴 × 𝐶 )
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=14
Vector triple product
𝐴× ( 𝐵 ⃗ ( ⃗𝐴⃗⃗ . 𝐶 ) −𝐶( ⃗𝐴⃗⃗ . 𝐵
⃗ × 𝐶 )= 𝐵 ⃗)
⃗ × 𝐶 )= 3𝐵
𝐴× ( 𝐵 ⃗ − 2𝐶
=3(2𝑎𝑥 − 𝑎𝑦 + 2𝑎𝑧 ) − 2(2𝑎𝑥 − 3𝑎𝑦 + 𝑎𝑧 )
=2𝑎𝑥 + 3𝑎𝑦 + 4𝑎𝑧
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There are two types of such systems, they are
If x axis is rotated towards y axis through a small angle, thus this rotation
causes an upward movement of right handed screw in ‘z’ axis direction
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REPRESENTING A POINT IN RECTANGULAR COORDINATE SYSTEMS
Consider a point P (x1, y1, z1) in Cartesian co-ordinate system as shown in Fig
(c) below. Then the position vector of point ‘P’ is represented by the distance
of point P from the origin directed from the origin to point P. This is called
RADIUS VECTOR.
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The three components of the position vector 𝑟𝑂𝑃 are three vectors oriented along
the three co-ordinate axes with the magnitude x1, y1 and z1. Thus the position
vector of point P can be represented as
⃗ 𝑶𝑷 = 𝑥1 𝑎𝑥 + 𝑦1 𝑎𝑦 + 𝑧1 𝑎𝑧
𝒓
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Problem
Given three points in a Cartesian coordinate system as A (3, -2,1), B (-3, -3, 5)
and C (2, 6, -4)
Find
Solution
𝐴= 3 𝑎𝑥 − 2𝑎𝑦 + 𝑎𝑧
i. Vector from 𝐴 to 𝐶
⃗⃗⃗⃗⃗
𝐴𝐶 =𝐶 -𝐴 =(2-3) 𝑎𝑥 +(6- -2) 𝑎𝑦 +( -4-(+1)) 𝑎𝑧
= −𝑎𝑥 + 8𝑎𝑦 − 5𝑎𝑧
ii. ⃗ to 𝐴
Unit vector form from 𝐵
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⃗⃗⃗⃗⃗ =𝐴 − 𝐵
𝐵𝐴 ⃗ = 6𝑎𝑥 + 𝑎𝑦 − 4𝑎𝑧
⃗⃗⃗⃗⃗⃗⃗ = √62 + 12 + 42 =7.2801
|𝐵𝐴|
⃗⃗⃗⃗⃗
𝐵𝐴 6𝑎⃗𝑥 +𝑎⃗𝑦 −4𝑎⃗𝑧
𝑎𝐵𝐴 = ⃗⃗⃗⃗⃗⃗⃗ = =0.8241𝑎𝑥 +0.1373𝑎𝑦 − 0.5494𝑎𝑧
|𝐵𝐴| 7.2801
iii. ⃗ to 𝐶
The distance from 𝐵
⃗⃗⃗⃗⃗ = 𝐶 -𝐵
𝐵𝐶 ⃗ = 5𝑎𝑥 + 9𝑎𝑦 − 9𝑎𝑧
iv. Let B (x1, y1, z1) and C(x2, y2, z2) then coordinates of midpoint of BC are
𝑥1 +𝑥2 𝑦1 +𝑦2 𝑧1 +𝑧2 −3+2 −3+6 5−4
( , , )= ( , , ) =(-0.5, 1.5 , 0.5)
2 2 2 2 2 2
⃗⃗⃗
|𝒓𝑶𝑷 | = √(𝒙𝟏 )𝟐 + (𝒚𝟏 )𝟐 + (𝒛𝟏 )𝟐
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If the point P has co-ordinates (1,2,3) then its position vector is
⃗⃗⃗
|𝒓𝑶𝑷 |= √(𝟏)𝟐 + (𝟐)𝟐 + (𝟑)𝟐 = ⃗⃗⃗
|𝒓𝑶𝑷 | = √𝟏𝟒 = 𝟑. 𝟕𝟒𝟏𝟔
Now consider two points in a Cartesian coordinate system, P and Q with the
co-ordinate (x1, y1, z1) and (x2, y2, z2) respectively. The points are shown in Fig1.
The individual position vectors of the points are
⃗⃗ = 𝑥1 𝑎𝑥 + 𝑦1 𝑎𝑦 + 𝑧1 𝑎𝑧
𝑷
⃗𝑸
⃗ = 𝑥2 𝑎𝑥 + 𝑦2 𝑎𝑦 + 𝑧2 𝑎𝑧
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This is also called separation vectors.
The magnitude of this vector is given by
⃗⃗⃗⃗⃗⃗ | = √(𝑥2 − 𝑥1 )𝟐 + (𝑦2 − 𝑦1 )𝟐 + (𝑧2 − 𝑧1 )𝟐
|𝑷𝑸
⃗⃗⃗⃗⃗ | 𝑖𝑠 𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑃𝑄
|𝑃𝑄
Unit vector along direction of PQ is
⃗⃗⃗⃗⃗
𝑃𝑄
𝑎𝑂𝑃 =
⃗⃗⃗⃗⃗ |
|𝑃𝑄
Problem:
Obtain the unit vector in the direction from the origin towards the point P (3, -
3, 2)
Solution
⃗⃗⃗⃗⃗ is
The origin O(0,0,0) while P(3,-3, 2) hence the distance vector 𝑂𝑃
⃗⃗⃗⃗⃗ = (3 − 0)𝑎𝑥 + (−3 − 0)𝑎𝑦 + (−2 − 0)𝑎𝑧
𝑂𝑃
= 3𝑎𝑥 − 3𝑎𝑦 − 2𝑎𝑧
⃗⃗⃗⃗⃗ | = √32 + (−3)2 + (−2)2 = 4.6904
|𝑂𝑃
Hence the unit vector along the direction OP is
⃗⃗⃗⃗⃗
𝑂𝑃 3𝑎𝑥 − 3𝑎𝑦 − 2𝑎𝑧
𝑎𝑂𝑃 = =
⃗⃗⃗⃗⃗ |
|𝑂𝑃 4.6904
= 0.639𝑎𝑥 − 0.6396𝑎𝑦 − 0.4264𝑎𝑧
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DIFFERENTIAL ELEMENTS IN CARTESIAN CO-ORDINATOR SYSTEM
Hence differential vector length also called elementary vector length can be
represented as
⃗⃗⃗
𝑑𝑙 = 𝑑𝑥𝑎𝑥 + 𝑑𝑦𝑎𝑦 + 𝑑𝑧𝑎𝑧
⃗⃗⃗
𝑑𝑙 is the vector joining P to new point P’
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The distance of P` from P is given by magnitude of the differential vector length
⃗⃗⃗ | = √(𝑑𝑥)2 + (𝑑𝑦)2 + (𝑑𝑧)2
|𝑑𝑙
𝑑𝑣 = 𝑑𝑥𝑑𝑦𝑑𝑧
Note: ⃗⃗⃗
𝑑𝑙 is a vector but 𝑑𝑣 is a scalar
⃗⃗⃗⃗ is
Let us define differential surface areas, the differential surface element 𝑑𝑠
represented as
⃗⃗⃗⃗ = 𝑑𝑠 𝑎
𝑑𝑠 ⃗⃗⃗⃗𝑛
𝑎
⃗⃗⃗⃗𝑛 = unit vector normal to surace 𝑑𝑠
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1.3.2 CYLINDRICAL CO-ORDINATE SYSTEM
In this system of coordinates any point in a space is considered as the point of
intersection of the following surfaces
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The point P is a cylindrical
coordinate system has three
coordinates r, φ and z whose
values lie in the respective
ranges given in (1). The
points 𝑃(𝑟, 𝜑, 𝑧) can be seen
in Fig 4.
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BASE VECTORS
Similar to Cartesian coordinates system, there are three unit vectors in the r,
φ and z directions denoted as 𝑎𝑟 , 𝑎𝜑 and 𝑎𝑧 as shown in figure below. These are
mutually perpendicular to each other.
In the Cartesian coordinate system, the unit vectors are not dependent on the
coordinates. But in cylindrical coordinate system 𝑎𝑟 , and 𝑎𝜑 are functions of φ
coordinates as their direction changes as φ changes. Hence the differentiation
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or integration with respect to component in 𝑎𝑟 and 𝑎𝜑 should not be treated as
constants.
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Hence the differential vector length in cylindrical coordinate system is given
by
⃗⃗⃗
𝑑𝑙 = 𝑑𝑟𝑎𝑟 + 𝑟𝑑φ𝑎φ + 𝑑𝑧𝑎𝑧
𝑑𝑣 = 𝑑𝑟 × 𝑟𝑑φ × 𝑑𝑧
The differential surface areas in the three directions are as shown below
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RELATIONSHIP BETWEEN CARTESIAN AND CYLINDRICAL SYSTEMS
𝑥 = 𝑟 𝑐𝑜𝑠 𝜑
𝑦 = 𝑟 𝑠𝑖𝑛 𝜑
𝑧=𝑧
It can be seen that r can be
expressed in terms of x, y as:
𝑟 = √𝑥 2 + 𝑦 2
𝑦
tan 𝜑 =
𝑥
While
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1.3.3 SPHERICAL COORDINATE SYSTEM
The surfaces which are used to define the spherical coordinate system on the
three Cartesian axes are
Sphere of radius r, origin as the
center of the sphere
A right circular cone with its apex
at the origin and its axis as z axis.
Its half angle is θ. It rotates about z
axis and θ varies from 0 to 1800
A half plane perpendicular to the
𝑥𝑦 plane contains z axis, making
an angle φ with the x z plane
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BASE VECTORS
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DIFFERENTIAL ELEMENTS IN SPHERICAL COORDINATE SYSTEM
rsinθdφ
= differential length in φ direction
⃗⃗⃗
𝑑𝑙 = 𝑑𝑟𝑎𝑟 + 𝑟𝑑𝜃𝑎𝜃 + 𝑟 𝑠𝑖𝑛𝜃 𝑑𝜑𝑎𝜑
𝑑𝑣 = 𝑟 2 𝑠𝑖𝑛𝜃 𝑑𝑟 𝑑𝜃 𝑑𝜑
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The differential surface areas in the three directions are as shown below
⃗⃗⃗⃗ 𝑟 = 𝑟 2 𝑠𝑖𝑛𝜃 𝑑𝜃 𝑑𝜑
𝑑𝑠 ⃗⃗⃗⃗ 𝜃 = 𝑟𝜃𝑑𝑟𝑑𝜑
𝑑𝑠 ⃗⃗⃗⃗ 𝜑 = 𝑟𝑑𝑟𝑑𝜃
𝑑𝑠
𝑧
cos 𝜃 = 𝑧 = 𝑟 cos 𝜃
𝑟
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Hence the transformation from spherical to Cartesian can be obtained from
the equation
𝑿 = 𝒓 𝒔𝒊𝒏 𝜽 𝒄𝒐𝒔𝝋
𝒀 = 𝒓 𝒔𝒊𝒏 𝜽 𝒄𝒐𝒔𝝋
𝒛 = 𝒓 𝒄𝒐𝒔 𝜽
𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑟 2 [sin2 𝜃 + cos 2 𝜃]
𝑥2 + 𝑦2 + 𝑧2 = 𝑟2
𝑟 = √𝑥 2 + 𝑦 2 + 𝑧 2
𝑦 𝑧
while tan 𝜑= 𝑧 and 𝑐𝑜𝑠 𝜃 = 𝑟 and r is known , θ can be obtained.
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1.4 TRANSFORMATION OF VECTORS
𝐴=𝐴𝑥 𝑎𝑥 + 𝐴𝑦 𝑎𝑦 + 𝐴𝑧 𝑎𝑧
𝐴=𝐴𝑟 𝑎𝑟 + 𝐴𝜑 𝑎𝜑 + 𝐴𝑧 𝑎𝑧
From the dot product it is known that the component of vector in the
direction of unit vector is it dot product with that unit vector.
∴ 𝐴𝑟 = 𝐴 . 𝑎𝑟
= (𝐴𝑥 𝑎𝑥 + 𝐴𝑦 𝑎𝑦 + 𝐴𝑧 𝑎𝑧 ). 𝑎𝑟
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= 𝐴𝑥 𝑎𝑥 . 𝑎𝑟 + 𝐴𝑦 𝑎𝑦 . 𝑎𝑟 + 𝐴𝑧 𝑎𝑧 . 𝑎𝑟
𝑎𝑧 . 𝑎𝑟 = 𝑎𝑧 . 𝑎𝜑 = 0
𝑎𝑧 . 𝑎𝑧 = 1 𝐴𝑟 = 𝐴𝑥 cos 𝜑 + 𝐴𝑦 sin 𝜑
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𝐴𝜑 = 𝐴. 𝑎𝜑 𝐴𝜑 = −𝐴𝑥 sin 𝜑 + 𝐴𝑦 cos 𝜑
𝐴𝑧 = 𝐴. 𝑎𝑧 = 𝐴𝑧 𝐴𝑧 = 𝐴𝑧
𝑨𝒓 𝐜𝐨𝐬 𝝋 𝐬𝐢𝐧 𝝋 𝟎 𝑨𝒙
∴ [ 𝑨𝝋 ] = [− 𝐬𝐢𝐧 𝝋 𝐜𝐨𝐬 𝝋 𝟎] [ 𝑨𝒚 ]
𝑨𝒛 𝟎 𝟎 𝟏 𝑨𝒛
𝐴𝑥 = 𝐴 . 𝑎𝑥 = (𝐴𝑟 𝑎𝑟 + 𝐴𝜑 𝑎𝜑 + 𝐴𝑧 𝑎𝑧 ). 𝑎𝑥
= 𝐴𝑟 𝑎𝑟 . 𝑎𝑥 + 𝐴𝜑 𝑎𝜑 . 𝑎𝑥 + 𝐴𝑧 𝑎𝑧 . 𝑎𝑥
𝑎𝑟 . 𝑎𝑥 = 𝑎𝑥 . 𝑎𝑟
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𝐴=𝐴𝑥 𝑎𝑥 + 𝐴𝑦 𝑎𝑦 + 𝐴𝑧 𝑎𝑧
𝐴𝑟 = 𝐴 . 𝑎𝑟 = (𝐴𝑥 𝑎𝑥 + 𝐴𝑦 𝑎𝑦 + 𝐴𝑧 𝑎𝑧 ). 𝑎𝑟
𝐴𝑟 = 𝐴𝑥 𝑎𝑥 . 𝑎𝑟 + 𝐴𝑦 𝑎𝑦 . 𝑎𝑟 + 𝐴𝑧 𝑎𝑧 . 𝑎𝑟
𝐴𝜃 = 𝐴 . 𝑎𝜃 = (𝐴𝑥 𝑎𝑥 + 𝐴𝑦 𝑎𝑦 + 𝐴𝑧 𝑎𝑧 ). 𝑎𝜃
𝐴𝜃 = 𝐴𝑥 . 𝑎𝜃 + 𝐴𝑦 𝑎𝑦 . 𝑎𝜃 + 𝐴𝑧 𝑎𝑧 . 𝑎𝜃
𝐴𝜑 = 𝐴 . 𝑎𝜑 = (𝐴𝑥 𝑎𝑥 + 𝐴𝑦 𝑎𝑦 + 𝐴𝑧 𝑎𝑧 ). 𝑎𝜑
𝐴𝜑 = 𝐴𝑥 𝑎𝑥 . 𝑎𝜑 + 𝐴𝑦 𝑎𝑦 . 𝑎𝜑 + 𝐴𝑧 𝑎𝑧 . 𝑎𝜑
. 𝑎𝑟 𝑎𝜃 𝑎𝜑
𝑎𝑧 cos 𝜃 − sin 𝜃 0
𝐴=𝐴𝑟 𝑎𝑟 + 𝐴𝜃 𝑎𝜃 + 𝐴𝜑 𝑎𝜑
𝐴𝑥 = 𝐴 . 𝑎𝑥 = 𝐴𝑟 𝑎𝑟 . 𝑎𝑥 + 𝐴𝜃 𝑎𝜃 . 𝑎𝑥 + 𝐴𝜑 𝑎𝜑 . 𝑎𝑥
𝐴𝑦 = 𝐴 . 𝑎𝑦 = 𝐴𝑟 𝑎𝑟 . 𝑎𝑦 + 𝐴𝜃 𝑎𝜃 . 𝑎𝑦 + 𝐴𝜑 𝑎𝜑 . 𝑎𝑦
𝐴𝑧 = 𝐴 . 𝑎𝑧 = 𝐴𝑟 𝑎𝑟 . 𝑎𝑧 + 𝐴𝜃 𝑎𝜃 . 𝑎𝑧 + 𝐴𝜑 𝑎𝜑 . 𝑎𝑧
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𝑑 = √𝑟22 + 𝑟12 − 2𝑟1 𝑟2 cos(𝜑2 − 𝜑1 ) + (𝑧2 − 𝑧1 )2 → 𝐶𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙
𝐴=𝐴𝑟 𝑎𝑟 + 𝐴𝜃 𝑎𝜃 + 𝐴𝜑 𝑎𝜑
𝐴𝜌 = 𝐴𝑟 𝑎𝑟 . 𝑎𝜌 + 𝐴𝜃 𝑎𝜃 . 𝑎𝜌 + 𝐴𝜑 𝑎𝜑 . 𝑎𝜌
𝐴𝜑 = 𝐴𝑟 𝑎𝑟 . 𝑎𝜑 + 𝐴𝜃 𝑎𝜃 . 𝑎𝜑 + 𝐴𝜑 𝑎𝜑 . 𝑎𝜑
𝐴𝑧 = 𝐴𝑟 𝑎𝑟 . 𝑎𝑧 + 𝐴𝜃 𝑎𝜃 . 𝑎𝑧 + 𝐴𝜑 𝑎𝜑 . 𝑎𝑧
𝑎𝑟 . 𝑎𝜌 = sin 𝜃 𝑎𝜃 . 𝑎𝜌 = cos 𝜃 𝑎𝜑 . 𝑎𝜌 = 0
𝑎𝑟 . 𝑎𝜑 = 0 𝑎𝜃 . 𝑎𝜑 = 0 𝑎𝜑 . 𝑎𝜑 = 1
𝑎𝑟 . 𝑎𝑧 = cos 𝜃 𝑎𝜃 . 𝑎𝑧 = − sin 𝜃 𝑎𝜑 . 𝑎𝑧 = 0
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𝑨𝝆 𝐬𝐢𝐧 𝜽 𝐜𝐨𝐬 𝜽 𝟎 𝑨𝒓
∴ [𝑨𝝋 ] = [ 𝟎 𝟎 𝟏] [ 𝑨𝜽 ]
𝑨𝒛 𝐜𝐨𝐬 𝜽 − 𝐬𝐢𝐧 𝜽 𝟎 𝑨𝝋
𝐴=𝐴𝜌 𝑎𝜌 + 𝐴𝜑 𝑎𝜑 + 𝐴𝑧 𝑎𝑧
𝐴𝑟 = 𝐴𝜌 𝑎𝜌 . 𝑎𝑟 + 𝐴𝜑 𝑎𝜑 . 𝑎𝑟 + 𝐴𝑧 𝑎𝑧 . 𝑎𝑟
𝐴𝜃 = 𝐴𝜌 𝑎𝜌 . 𝑎𝜃 + 𝐴𝜑 𝑎𝜑 . 𝑎𝜃 + 𝐴𝑧 𝑎𝑧 . 𝑎𝜃
𝐴𝜑 = 𝐴𝜌 𝑎𝜌 . 𝑎𝜑 + 𝐴𝜑 𝑎𝜑 . 𝑎𝜑 + 𝐴𝑧 𝑎𝑧 . 𝑎𝜑
𝑨𝒓 𝐬𝐢𝐧 𝜽 𝟎 𝐜𝐨𝐬 𝜽 𝑨𝝆
∴ [ 𝑨𝜽 ] = [𝐜𝐨𝐬 𝜽 𝟎 − 𝐬𝐢𝐧 𝜽] [𝑨𝝋 ]
𝑨𝝋 𝟎 𝟏 𝟎 𝑨𝒛
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1.5 TYPES OF INTEGRAL RELATED TO ELECTROMAGNETIC THEORY
In electromagnetic theory a charge can exist in point form, line form, surface
form or volume form. Hence for charge distribution analysis, the following
types of integrals are required.
1. Line integral
2. Surface integral
3. Volume integral
∫ ⃗F. ⃗⃗⃗
𝑑𝑙 = ∫|𝐹 |𝑑𝑙 cos 𝜃
𝐿 𝑃
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This is called line integral of ⃗F around the curved path L. The closed path can
be of two types:
The closed path is also called contour. The corresponding integral is called
contour integral, closed integral (or) circular integral, and mathematical
defined as
∮𝐿 ⃗F. ⃗⃗⃗
𝑑𝑙 =circular integral
If there exists a charge along a line as shown in figure below, then the total
charge obtained by calculating a line integral.
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𝑄 = ∫ 𝜌𝐿 . 𝑑𝑙
𝐿
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Similarly a flux Φ may pass through a surface as shown in figure (b). While
doing analysis of such cases an integral is required called SURAFCE
INTEGRAL, to be carried out over a surface related to a vector field.
For a charge distribution shown in fig (a), we can write total charge existing on
the surface as
𝑐
𝑄 = ∫𝑠 𝜌𝑠 . 𝑑𝑠 𝜌𝑠 → 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑐ℎ𝑎𝑟𝑔𝑒 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑖𝑛 ,
𝑚2
From fig (b), the total flux density crossing the surface S can be expressed as
⃗⃗⃗⃗ = ∫|F
𝑄 = ∫ ⃗F. 𝑑𝑠 ⃗ |𝑑𝑠 cos 𝜃
𝑠 𝑠
⃗⃗⃗⃗
𝑄 = ∮ ⃗F. 𝑑𝑠
𝑠
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Thus if 𝜌𝑣 is the volume charge density over volume v then the volume integral
is defined as
𝑄 = ∫ 𝜌𝑣 . 𝑑𝑣
𝑉
𝑑𝑣 = 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑣𝑜𝑙𝑢𝑚𝑒
1.5.4 DIVERGENCE
It is seen that∮𝑠 F ⃗⃗⃗⃗ gives the flux flowing across the surface s. Then
⃗ . 𝑑𝑠
mathematically divergence is defined as the net outward flow of the flux
per unit volume over a closed incremental surface. It is denoted as div 𝐹
and given by
⃗⃗⃗⃗
∮𝒔 𝐅.𝒅𝒔
div ⃗𝑭 = 𝐥𝐢𝐦 ∆𝒗
= divergence of ⃗𝑭
𝚫𝒗→𝟎
symbolically is denoted as
⃗ = divergence of 𝑭
𝛁. 𝑭
∂ 𝜕 𝜕
Where ∇= vector operator = ∂x 𝑎𝑥 + 𝜕𝑦 𝑎𝑦 + 𝜕𝑧 𝑎𝑧
But 𝐹 = 𝐹𝑥 𝑎𝑥 + 𝐹𝑦 𝑎𝑦 + 𝐹𝑧 𝑎𝑧
Therefore, in Cartesian form
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∂𝐹𝑥 𝜕𝐹𝑦 𝜕𝐹𝑧
∇. 𝐹 = + +
∂x 𝜕𝑦 𝜕𝑧
Cylindrical:
1∂ 1 𝜕𝐹𝜑 𝜕𝐹𝑧
∇. 𝐹 = ( )
𝑟𝐹 + +
𝑟 ∂r 𝑟 𝑟 𝜕𝜑 𝜕𝑧
Spherical:
1 ∂ 2 1 𝜕 1 𝜕𝜑
∇. 𝐹 = 2 ( )
𝑟 𝐹𝑟 + ( )
sin 𝜃𝐹𝜃 +
𝑟 ∂r 𝑟 sin 𝜃 𝜕𝜃 𝑟 sin 𝜃 𝜕𝜑
The vector field having its divergence zero is called solenoidal field
∇. 𝐴 = 0 for 𝐴 to be solenoidal
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∂ 𝜕 𝜕
∇ (𝑑𝑒𝑙 ) =
𝑎 + 𝑎 + 𝑎
∂x 𝑥 𝜕𝑦 𝑦 𝜕𝑧 𝑧
The operation of the vector operator del (∇) on a scalar function is called
gradient of a scalar
∂ 𝜕 𝜕
Grad w = ∇𝑤 = ( 𝑎𝑥 + 𝑎𝑦 + 𝑎𝑧 ) 𝑤
∂x 𝜕𝑦 𝜕𝑧
In Cartesian coordinates
∂w 𝜕𝑤 𝜕𝑤
∇𝑤 = 𝑎𝑥 + 𝑎𝑦 + 𝑎
∂x 𝜕𝑦 𝜕𝑧 𝑧
In cylindrical coordinates
∂w 1 𝜕𝑤 𝜕𝑤
∇𝑤 = 𝑎𝑟 + 𝑎𝜑 + 𝑎
∂r 𝑟 𝜕𝜑 𝜕𝑧 𝑧
In spherical coordinates
∂w 1 𝜕𝑤 1 𝜕𝑤
∇𝑤 = 𝑎𝑟 + 𝑎𝜃 + 𝑎
∂r 𝑟 𝜕𝜃 𝑟 sin 𝜃 𝜕𝜑 𝜑
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𝜕𝐹𝑧 𝜕𝐹𝑦 𝜕𝐹𝑥 𝜕𝐹𝑧 𝜕𝐹𝑦 𝜕𝐹𝑥
∇ × ⃗F = [ − ]𝑎 + [ − ]𝑎 + [ − ]𝑎
𝜕𝑦 𝜕𝑧 𝑥 𝜕𝑧 𝜕𝑥 𝑦 𝜕𝑥 𝜕𝑦 𝑧
In Cartesian
𝑎𝑥 𝑎𝑦 𝑎𝑧
𝜕 𝜕 𝜕
∇ × ⃗F =
𝜕𝑥 𝜕𝑦 𝜕𝑧
[ 𝐹𝑥 𝐹𝑦 𝐹𝑧 ]
In cylindrical:
𝑎𝑟 𝑟𝑎𝜑 𝑎𝑧
⃗ = 𝜕
∇×F
𝜕 𝜕
𝜕𝑟 𝜕𝜑 𝜕𝑧
[ 𝐹𝑟 𝐹𝜑 𝐹𝑧 ]
In spherical:
𝑎𝑟 𝑟𝑎𝜃 𝑟 sin 𝜃 𝑎𝜑
𝜕 𝜕 𝜕
⃗ =
∇×F
𝜕𝑟 𝜕𝜃 𝜕𝜑
[ 𝐹𝑟 𝑟𝐹𝜃 𝑟 sin 𝜃 𝐹𝜑 ]
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Assignment I
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