Electromagnetics Wave and Fields
Lecture 1
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /1
Chapter 1
Vector Analysis
Course Description
• Review of Vector analysis, coordinate systems and
solutions to static field problems.
• Electrostatics: Coulomb's law, force, electric field
intensity, electrical flux density. Gauss's theorem,
Electrostatic potential, boundary conditions, method of
images, Laplace's and Poisson's equations, energy of
an electrostatic system, conductor and dielectrics.
• Magnetostatics: Concepts of magnetic field, Ampere's
law, Bio-Savart law, vector magnetic potential, energy of
magnetostatic system, Mechanical forces and torques in
Electric and Magnetic fields.
• Graphical field mapping with applications, solution to
Laplace equations, rectangular, cylindrical and spherical
harmonics with applications.
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /2
Chapter 1
Vector Analysis
Course Description (contd.)
• Relation between circuit theory and field theory.
• Time Varying Fields: Maxwell's equations.
• Polarization: Propagation and reflection of
electromagnetic waves in unbounded media: plane
wave propagation, polarization, power flow and
Poynting's theorem.
• Transmission line analogy, reflection from
conducting and dielectric boundary display lines ion
in dielectrics, plane wave propagation through the
ionosphere. Introduction to radiation.
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /3
Chapter 1
Vector Analysis
Textbook
• William H. Hayt, Jr., John A. Buck,
“Engineering Electromagnetics”, 6th Edition
Onward
AIUB
Dr. M. Tanseer Ali
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Chapter 1
Vector Analysis
Objectives
• Apply vector calculus to solve simple Electrostatic
and Magnetostatics problem.
• Use Gauss’s law, Coulomb’s law and Poisson’s
equation to calculate electric fields and potential.
• Describe the interaction between time varying
electric and magnetic fields and how this interaction
leads to Maxwell’s Equation.
• Apply Maxwell’s Equations to solve simple
electromagnetic problems.
• Analyze plane wave propagation and the effects of
material parameters on the wave propagation.
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /5
Chapter 1
Vector Analysis
Evaluation
• At least 80% class attendance is necessary to sit for the
exam. If there is any assignment given to the students, they
have to submit it before the deadline decided by the course
teacher.
• Marking system (Midterm and Final term):
Quiz:
20%
Assignments
20%
Attendance & class performance:
20%
Midterm/Final term exam:
40%
Total:
100%
• Final Grade/ Grand Total:
Midterm:
40%
Final Term:
60%
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /6
Chapter 1
Vector Analysis
What is Electro-magnetics?
Electric field
Produced by the presence of
electrically charged particles,
and gives rise to the electric
force.
Magnetic field
Produced by the motion of
electric charges, or electric
current, and gives rise to the
magnetic force associated
with magnets.
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /7
Chapter 1
Vector Analysis
Electromagnetic Wave Spectrum
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /8
Chapter 1
Vector Analysis
Importance of Electromagnetic Engineering
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /9
Chapter 1
Vector Analysis
Applications
• Electromagnetic principles find application in various disciplines such
as microwaves, x-rays, antennas, electric machines, plasmas, etc.
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /10
Chapter 1
Vector Analysis
Applications
◼ Electromagnetic fields are used in induction heaters for melting, forging, annealing,
surface hardening, and soldering operation.
◼ Electromagnetic devices include transformers, radio, television, mobile phones, radars,
lasers, etc.
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /11
Chapter 1
Vector Analysis
Applications: Transrapid Train
• A magnetic traveling field moves the vehicle without contact.
• The speed can be continuously regulated by varying the frequency of the
alternating current.
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /12
Chapter 1
Vector Analysis
Scalars and Vectors
◼ Scalar refers to a quantity whose value may be represented by a
single (positive or negative) real number.
◼ Some examples include distance, temperature, mass, density,
pressure, volume, and time.
◼ A vector quantity has both a magnitude and a direction in space. We
especially concerned with two- and three-dimensional spaces only.
◼ Displacement, velocity, acceleration, and force are examples of
vectors.
• Scalar notation: A or A (italic or plain)
• Vector notation: A or A (bold or plain with arrow)
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /13
Chapter 1
Vector Analysis
Vector Algebra
A+B =B+A
A + (B + C) = ( A + B) + C
A − B = A + ( −B )
A 1
= A
n n
A−B = 0 → A = B
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /14
Chapter 1
Vector Analysis
Rectangular Coordinate System
• Differential surface units:
dx  dy
dy  dz
dx  dz
• Differential volume unit :
dx  dy  dz
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /15
Chapter 1
Vector Analysis
Vector Components and Unit Vectors
r = x+y+z
r = xa x + ya y + za z
a x , a y , a z : unit vectors
R PQ = rQ − rP
= (2a x − 2a y + a z ) − (1a x + 2a y + 3a z )
= a x − 4a y − 2a z
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /16
Chapter 1
Vector Analysis
Vector Components and Unit Vectors
◼ For any vector B,
B = Bxa x + By a: y + Bz a z
B = Bx2 + By2 + Bz2 = B
aB =
B
Bx2 + By2 + Bz2
=
B
B
Magnitude of B
Unit vector in the direction of B
◼ Example
Given points M(–1,2,1) and N(3,–3,0), find RMN and aMN.
R MN = (3a x − 3a y + 0a z ) − (−1a x + 2a y + 1a z ) = 4a x − 5a y − a z
4a x − 5a y − 1a z
R MN
= 0.617a x − 0.772a y − 0.154a z
a MN =
=
2
2
2
R MN
4 + (−5) + (−1)
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /17
Chapter 1
Vector Analysis
The Dot Product
◼ Given two vectors A and B, the dot product, or scalar product, is defines as the
product of the magnitude of A, the magnitude of B, and the cosine of the smaller
angle between them:
A  B = A B cos  AB
◼ The dot product is a scalar, and it obeys the commutative law:
AB = BA
◼ For any vector
A = Ax a x + Ay a y + Az a z and B = Bxa x + By a y +,Bz a z
A  B = Ax Bx + Ay By + Az Bz
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /18
Chapter 1
Vector Analysis
The Dot Product
◼ One of the most important applications of the dot product is that
of finding the component of a vector in a given direction.
• The scalar component of B in the
direction of the unit vector a is Ba
• The vector component of B in the
direction of the unit vector a is (Ba)a
AIUB
B  a = B a cos  Ba = B cos  Ba
Dr. M. Tanseer Ali
EMWF Lec 1 /19
Chapter 1
Vector Analysis
The Dot Product
◼ Example
The three vertices of a triangle are located at A(6,–1,2),
B(–2,3,–4), and C(–3,1,5). Find: (a) RAB; (b) RAC; (c) the angle
θBAC at vertex A; (d) the vector projection of RAB on RAC.
B
R AB = (−2a x + 3a y − 4a z ) − (6a x − a y + 2a z ) = −8a x + 4a y − 6a z
R AC = (−3a x + 1a y + 5a z ) − (6a x − a y + 2a z ) = −9a x + 2a y + 3a z
 BAC
R AB  R AC = R AB R AC cos  BAC
 cos  BAC =
R AB  R AC
=
R AB R AC
C
A
(−8a x + 4a y − 6a z )  (−9a x + 2a y + 3a z )
(−8) + (4) + (−6)
2
2
2
(−9) + (2) + (3)
2
2
2
=
62
116
= 0.594
94
  BAC = cos −1 (0.594) = 53.56
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /20
Chapter 1
Vector Analysis
The Dot Product
◼ Example
If
 
A• B = 0
, What is the θ between them
 
 A• B 


 AB  = 0


 

A• B 
 = 90
and  = cos −1 

 AB 


Therefore A is perpendicular to B and vice versa.
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /21
Chapter 1
Vector Analysis
The Cross Product
◼ Given two vectors A and B, the magnitude of the cross product, or vector
product, written as AB, is defines as the product of the magnitude of A, the
magnitude of B, and the sine of the smaller angle between them.
◼ The direction of AB is perpendicular to the plane containing A and B and is in
the direction of advance of a right-handed screw as A is turned into B.
A  B = a N A B sin  AB
◼ The cross product is a vector, and it is not
commutative:
ax  a y = az
a y  az = ax
az  ax = a y
(B  A ) = −( A  B )
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /22
Chapter 1
Vector Analysis
The Cross Product
◼ Example
Given A = 2ax – 3ay + az and B = –4ax – 2ay + 5az, find AB.
ax
 
A  B = Ax
Bx
ay
Ay
By
az
Az
Bz
A  B = ( Ay Bz − Az By )a x + ( Az Bx − Ax Bz )a y + ( Ax By − Ay Bx )a z
= ( (−3)(5) − (1)(−2) ) a x + ( (1)(−4) − (2)(5) ) a y + ( (2)(−2) − (−3)(−4) ) a z
= −13a x − 14a y − 16a z
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /23
Chapter 1
Vector Analysis
The Cross Product
◼ Example
 
If A  B = 0
, What is the θ between them
 
 A B 


 AB  = 0


 

A• B 
 = 0
and  = sin −1 

 AB 


Therefore A is parallel to B
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /24
Chapter 1
Vector Analysis
Rectangular Coordinate System
• Differential surface units:
dx  dy
dy  dz
dx  dz
• Differential volume unit :
dx  dy  dz
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /25
Chapter 1
Vector Analysis
Rectangular Coordinate System
Example:
Given, 3 ≤ x ≤ 6 and 2 ≤ y ≤ 4; Find the surface.
Solution:
6
4
S z =  dS z =  dx  dy = x 3  y 2 = (6 − 3)(4 − 2) = 6
6
3
4
2
Therefore the surface is Sz = 6 az
Example:
Given, 3 ≤ x ≤ 4, 3 ≤ y ≤ 6 and 4 ≤ z ≤ 7; Find the volume.
4
Solution:
7
V =  dx  dy  dz = x 3  y 3 z 4 = 9
4
3
AIUB
6
3
6
7
4
Dr. M. Tanseer Ali
EMWF Lec 1 /26
Chapter 1
Vector Analysis
The Cylindrical Coordinate System
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Dr. M. Tanseer Ali
EMWF Lec 1 /27
Chapter 1
Vector Analysis
The Cylindrical Coordinate System
• Differential surface units:
d   dz
 d  dz
d    d
• Differential volume unit :
d    d  dz
AIUB
• Relation between the rectangular
and the cylindrical coordinate
systems
x =   cos 
y =   sin 
z=z
Dr. M. Tanseer Ali
 = x2 + y 2
y
 = tan −1
x
z=z
EMWF Lec 1 /28
Chapter 1
Vector Analysis
The Cylindrical Coordinate System
Example:
Given, 3 ≤ ρ ≤ 4 and 300 ≤ ϕ ≤ 600; Find the surface.
 3
4
  2   3  4 2 32     7
Solution:
S z =  dS z =  d  d =     6 =  −  −  = 
 2 3
 2 2  3 6  12
3
 6
4
Example:
Given, 3 ≤ ρ ≤ 4, 300 ≤ ϕ ≤ 600 and 3 ≤ z ≤ 7; Find the volume.
Solution:
 3
4
  2   3 7  4 2 32    
7
S z =  dS z =  d  d  dz =     6 z 3 =  −  − (7 − 3) = 
3
 2 3
 2 2  3 6 
3
 6
3
4
AIUB
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Dr. M. Tanseer Ali
EMWF Lec 1 /29
Chapter 1
Vector Analysis
The Cylindrical Coordinate System
Example:
Given, ρ = 3 cm and 0 ≤ z ≤ 5; Find the volume for the full cylinder.
Solution:
3
2
5
0
0
0
S z =  dS z =  d  d  dz
3
 
2
5
=    0 z 0
 2 0
2
 32 
=  (2 )(5)
2
= 45
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /30
Chapter 1
Vector Analysis
The Cylindrical Coordinate System
?
A = Axa x + Ay a y + Az a z  A = A a  + A a + Az a z
az
az
A = A  a 
= ( Ax a x + Ay a y + Az a z )  a 
a
= Ax a x  a  + Ay a y  a  + Az a z  a 
ay
a
= Ax cos  + Ay sin 
ax
A = A  a
• Dot products of unit vectors in
cylindrical and rectangular coordinate
systems
= ( Ax a x + Ay a y + Az a z )  a
= Ax a x  a + Ay a y  a + Az a z  a
= − Ax sin  + Ay cos 
Az = A  a z
= ( Ax a x + Ay a y + Az a z )  a z
= Ax a x  a z + Ay a y  a z + Az a z  a z
= Az
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /31
Chapter 1
Vector Analysis
The Spherical Coordinate System
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Dr. M. Tanseer Ali
EMWF Lec 1 /32
Chapter 1
Vector Analysis
The Spherical Coordinate System
• Differential surface units:
dr  rd
dr  r sin  d
rd  r sin  d
AIUB
• Differential volume unit :
dr  rd  r sin  d
Dr. M. Tanseer Ali
EMWF Lec 1 /33
Chapter 1
Vector Analysis
The Spherical Coordinate System
Example:
Given, r =3, 300 ≤ ϕ ≤ 600 and 600 ≤ θ ≤ 900 Find the surface of the share.
Solution:
S z =  dS r = r
 3
2
 2
 sin d  d = 3 − cos   
 3
 6
2
6
3
 2
 3
 1
3   
  = 0.55
= 9 − +

 2 2  6 
Example:
Given, 3 ≤ ρ ≤ 4, 300 ≤ ϕ ≤ 600 and 7 ≤ z ≤ 7; Find the volume.
Solution:
 3
4
  2   3 7  4 2 32    
7
S z =  dS z =  d  d  dz =     6 z 3 =  −  − (7 − 3) = 
3
 2 3
 2 2  3 6 
3
 6
3
4
AIUB
7
Dr. M. Tanseer Ali
EMWF Lec 1 /34
Chapter 1
Vector Analysis
The Spherical Coordinate System
• Relation between the rectangular and
the spherical coordinate systems
x = r sin  cos 
y = r sin  sin 
z = r cos 
r = x2 + y 2 + z 2 , r  0
z
 = cos −1
, 0    180
2
2
2
x +y +z
y
 = tan −1
x
• Dot products of unit vectors in spherical and
rectangular coordinate systems
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /35
Chapter 1
Vector Analysis
The Spherical Coordinate System
◼ Example
Given the two points, C(–3,2,1) and D(r = 5, θ = 20°, Φ = –70°), find: (a) the
spherical coordinates of C; (b) the rectangular coordinates of D.
r = x 2 + y 2 + z 2 = (−3) 2 + (2) 2 + (1) 2 = 3.742
 = cos
−1
 = tan −1
1
= cos
= 74.50
2
2
2
3.742
x +y +z
z
−1
2
y
= tan −1
= −33.69 + 180 = 146.31
−3
x
 C (r = 3.742,  = 74.50,  = 146.31)
 D( x = 0.585, y = −1.607, z = 4.698)
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /36
Chapter 1
Vector Analysis
Field Due to a Continuous Volume Charge Distribution
◼ We denote the volume charge density by ρv, having the units of coulombs per cubic
meter (C/m3).
◼ The small amount of charge ΔQ in a small volume Δv is
Q = v v
◼ We may define ρv mathematically by using a limit on the above equation:
Q
v = lim
v →0 v
◼ The total charge within some finite volume is obtained by integrating throughout that
volume:
Q =  v dv
vol
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /37
Chapter 1
Vector Analysis
Field Due to a Continuous Volume Charge Distribution
◼ Example
Find the total charge inside the volume indicated by ρv = 4xyz2, 0 ≤ ρ ≤ 2, 0 ≤ Φ ≤
π/2, 0 ≤ z ≤ 3. All values are in SI units.
x =  cos 
y =  sin 
v = 4   sin    cos   z 2
3  2 2
Q =  v dv =  
vol
2
(4


sin



cos


z
)(d    d  dz )

z =0  =0  =0
3 22
=    4  3 z 2 sin  cos  d  d dz
sin 2 = 2sin  cos 
0 0 0
3 2
=   16 z 2 sin  cos  d dz
0 0
3
=  8z 2 dz = 72 C
0
AIUB
Dr. M. Tanseer Ali
EMWF Lec 1 /38
Chapter 1
Vector Analysis
Practice Problems
◼ D1.4. The three vertices of a triangle located at A(6,-1,2), B(-2,3,-4) and C(-3,1,5).
Find: (a) RAB ˣ RAC; (b) the area of the triangle; (c) unit vector perpendicular to the
plane in which the triangle is located.
◼ D1.5: (a) Give the Cartesian coordinates of the point C(ρ = 4.4, Φ = -11.5º, z = 2).
(b) Give the cylindrical co-ordinates of the point D(x = - 3.1, y = 2.6, z = -3). (c)
Specify the distance from C to D.
◼ D1.6. Transform to cylindrical coordinates: (a) F = 10ax - 8ay + 6az at point P(10,8,6); (b) G = (2x + y)ax - (y - 4x)ay at point Q(ρ,ϕ,z); (c) Give the rectangular
components of the vector H = 20aρ - 10aϕ + 3az at point P(x=5,y=2,z=-1).
◼ D1.8. Transform the following vectors to spherical co-ordinates at the point given:
(a) 10ax at P(x=-3,y=2,z=4); (b) 10ay at Q(ρ=5,ϕ=30o,z=4); (c) 10az at
M(r=4,ϕ=110o,θ=120o).
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Dr. M. Tanseer Ali
EMWF Lec 1 /39