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President University

Erwin Sitompul

EEM

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1

Dr.
-
Ing. Erwin Sitompul

President University

Lecture
1

Engineering Electromagnetics

http://zitompul.wordpress.com

President University

Erwin Sitompul

EEM

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Textbook:

“Engineering Electromagnetics”,

William H. Hayt, Jr. and John A. Buck,
McGraw
-
Hill, 2006.

Textbook and Syllabus

Syllabus:

Chapter 1:

Vector Analysis

Chapter 2:

Coulomb’s Law and Electric Field Intensity

Chapter 3:

Electric Flux Density, Gauss’ Law, and




Divergence

Chapter 4:

Energy and Potential

Chapter 5:

Current and Conductors

Chapter 6:

Dielectrics and Capacitance

Chapter 8:

The Steady Magnetic Field

Chapter 9:

Magnetic Forces, Materials, and Inductance

Engineering Electromagnetics

President University

Erwin Sitompul

EEM

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Grade Policy

Grade Policy:

Final Grade =

10% Homework + 20% Quizzes +





30% Midterm Exam + 40% Final Exam +





Extra Points





Homeworks will be given in fairly regular basis. The average

of homework grades contributes 10% of final grade.




Homeworks are to be written on
A4 papers
, otherwise they

will not be graded.




Homeworks must be submitted
on time
. If you submit late,



< 10 min.



No penalty



10
–

60 min.



–
20 points



> 60 min.




–
40 points




There will be 3 quizzes. Only the best 2 will be counted.


The average of quiz grades contributes 20% of final grade.


Engineering Electromagnetics

President University

Erwin Sitompul

EEM

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Grade Policy:



Midterm and final exam schedule will be announced in time.




Make up of quizzes and exams will be held
one week

after

the schedule of the respective quizzes and exams.




The score of a make up quiz or exam
can be
multiplied by 0.9

(
i
.
e
., the
maximum score for a make up
will be
90).


Engineering Electromagnetics

Grade Policy

•
Heading of Homework Papers (Required)

President University

Erwin Sitompul

EEM

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Grade Policy

Grade Policy:


Ex
tra points will be given every time you solve a problem in
front of the class. You will earn
1

or
2

points.


Lecture slides can be copied during class session. It also will
be available on internet around 3 days after class. Please
check the course homepage regularly.




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Engineering Electromagnetics

President University

Erwin Sitompul

EEM

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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.

Engineering Electromagnetics

What is Electromagnetics?

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Engineering Electromagnetics

Electromagnetic Wave Spectrum

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Engineering Electromagnetics


Electric and magnetic field exist nearly everywhere.

Why do we learn Engineering Electromagnetics

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Engineering Electromagnetics


Electromagnetic principles find application in various disciplines
such as microwaves, x
-
rays, antennas, electric machines,
plasmas, etc.

Applications

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Engineering Electromagnetics


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.

Applications

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Erwin Sitompul

EEM

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Engineering Electromagnetics

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.

Applications

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EEM

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Chapter 1

Vector Analysis


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)

Scalars and Vectors

→

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Chapter 1

Vector Analysis

  
A B B A
( ) ( )
  
A B+C A B +C
( )
   
A B A B
1
n n

A
A
0
   
A B A B
Vector Algebra

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Chapter 1

Vector Analysis

Rectangular Coordinate System

•
Differential surface units:

dx dy

dy dz

dx dz

•
Differential volume unit :

dx dy dz
 
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Chapter 1

Vector Analysis

Vector Components and Unit Vectors

  
r x y z
x y z
x y z
  
r a a a
, , :
x y z
a a a
unit vectors
?
PQ
R
PQ Q P
 
R r r
(2 2 ) (1 2 3 )
x y z x y z
     
a a a a a a
4 2
x y z
  
a a a
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
For any vector
B
, :

Chapter 1

Vector Analysis

Vector Components and Unit Vectors

x x y y z z
B B B
 
B a a + a
2 2 2
x y z
B B B
  
B
Magnitude of
B

B

2 2 2
B
x y z
B B B

 
B
a

B
B
Unit vector in the direction of
B


Example


Given points
M
(
–
1,2,1) and
N
(3,
–
3,0), find
R
MN

and
a
MN
.

(3 3 0 ) ( 1 2 1 )
MN x y z x y z
      
R a a a a a a
4 5
x y z
  
a a a
MN
MN
MN

R
a
R
2 2 2
4 5 1
4 ( 5) ( 1)
x y z
 

   
a a a
0.617 0.772 0.154
x y z
  
a a a
President University

Erwin Sitompul

EEM

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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:

cos
AB

 
A B A B

The dot product is a scalar, and it obeys the commutative law:

  
A B B A

For any vector and ,

x x y y z z
A A A
 
A a a + a
x x y y z z
B B B
 
B a a + a
x x y y z z
A B A B A B
  
A B +
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
One of the most important applications of the dot product is that of
finding the component of a vector in a given direction.

Chapter 1

Vector Analysis

The Dot Product

cos
Ba

 
B a B a
cos
Ba


B
•
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

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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
)
R
AB
;

(
b
)
R
AC
; (
c
) the angle

θ
BAC

at vertex
A
; (
d
) the vector projection of
R
AB

on
R
AC
.

( 2 3 4 ) (6 2 )
AB x y z x y z
      
R a a a a a a
8 4 6
x y z
   
a a a
( 3 1 5 ) (6 2 )
AC x y z x y z
      
R a a a a a a
9 2 3
x y z
   
a a a
A
B
C
BAC

cos
AB AC AB AC BAC

 
R R R R
cos
AB AC
BAC
AB AC


 
R R
R R
2 2 2 2 2 2
( 8 4 6 ) ( 9 2 3 )
( 8) (4) ( 6) ( 9) (2) (3)
x y z x y z
      

      
a a a a a a
62
116 94

1
cos (0.594)
BAC


 
53.56
 
0.594

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Chapter 1

Vector Analysis

The Dot Product



on
AB AC AB AC AC
 
R R R a a
2 2 2 2 2 2
( 9 2 3 ) ( 9 2 3 )
( 8 4 6 )
( 9) (2) (3) ( 9) (2) (3)
x y z x y z
x y z
 
     
 
   
 
     
 
 
a a a a a a
a a a
( 9 2 3 )
62
94 94
x y z
  

a a a
5.963 1.319 1.979
x y z
   
a a a

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
)
R
AB
;

(
b
)
R
AC
; (
c
) the angle

θ
BAC

at vertex
A
; (
d
) the vector projection of
R
AB

on
R
AC
.

President University

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sin
N AB

 
A B a A B
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

䈠
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
.


The cross product is a vector, and it is
not commutative:

( ) ( )
   
B A A B
x y z
y z x
z x y
 
 
 
a a a
a a a
a a a
President University

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Chapter 1

Vector Analysis

The Cross Product


Example


Given
A
= 2
a
x
–
3
a
y
+
a
z

and
B

=
–
4
a
x
–
2
a
y
+5
a
z
, find
A

B
.

( ) ( ) ( )
y z z y x z x x z y x y y x z
A B A B A B A B A B A B
      
A B a a a






( 3)(5) (1)( 2) (1)( 4) (2)(5) (2)( 2) ( 3)( 4)
x y z
           
a a a
13 14 16
x y z
   
a a a
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Chapter 1

Vector Analysis

The Cylindrical Coordinate System

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Chapter 1

Vector Analysis

The Cylindrical Coordinate System

•
Differential surface units:

d dz


d dz


d d


•
Differential volume unit :

d d dz

 
cos
x
 
 
sin
y
 
 
z z

2 2
x y

 
1
tan
y
x



z z

•
Relation between the
rectangular and the cylindrical
coordinate systems

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
a
z
a

a
Chapter 1

Vector Analysis

The Cylindrical Coordinate System

•
Dot products of unit vectors in
cylindrical and rectangular
coordinate systems

y
a
z
a
x
a
A
 
 
A a
( )
x x y y z z
A A A

  
a a + a a
x x y y z z
A A A
  
    
a a a a + a a
cos sin
x y
A A
 
 
A
 
 
A a
( )
x x y y z z
A A A

  
a a + a a
x x y y z z
A A A
  
    
a a a a + a a
sin cos
x y
A A
 
  
z z
A
 
A a
( )
x x y y z z z
A A A
  
a a + a a
x x z y y z z z z
A A A
    
a a a a + a a
z
A

?
x x y y z z z z
A A A A A A
   
    
A a a + a A a a + a
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Chapter 1

Vector Analysis

The Spherical Coordinate System

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Chapter 1

Vector Analysis

The Spherical Coordinate System

•
Differential surface units:

dr rd


sin
dr r d


sin
rd r d
 

•
Differential volume unit :

sin
dr rd r d
 
 
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sin cos
x r
 

sin sin
y r
 

cos
z r


2 2 2
, 0
r x y z r
   
1
2 2 2
cos, 0 180
z
x y z
 

    
 
1
tan
y
x



•
Relation between the rectangular and
the spherical coordinate systems

Chapter 1

Vector Analysis

The Spherical Coordinate System

•
Dot products of unit vectors in spherical and
rectangular coordinate systems

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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.

2 2 2
r x y z
  
1
2 2 2
cos
z
x y z



 
1
tan
y
x



2 2 2
( 3) (2) (1)
   
3.742

1
1
cos
3.742


74.50
 
1
2
tan
3



33.69 180
  
146.31
 
( 3.742,74.50,146.31 )
C r
 
     
( 0.585,1.607,4.698)
D x y z
    
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Chapter 1

Vector Analysis

Homework 1


D1.4.


D1.6.


D1.8.

All homework problems from Hayt and Buck, 7th Edition.



Due: Next week
17 April 2012,
at
08:00.