Engineering Electromagnetics
Lecture 1
Dr.-Ing. Erwin Sitompul
President University
http://zitompul.wordpress.com
President University
Erwin Sitompul
EEM 1/1
Engineering Electromagnetics
Textbook and Syllabus
Textbook:
“Engineering Electromagnetics”,
William H. Hayt, Jr. and John A. Buck,
McGraw-Hill, 2006.
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
President University
Erwin Sitompul
EEM 1/2
Engineering Electromagnetics
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.
President University
Erwin Sitompul
EEM 1/3
Engineering Electromagnetics
Grade Policy
• Heading of Homework Papers (Required)
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).
President University
Erwin Sitompul
EEM 1/4
Engineering Electromagnetics
Grade Policy
Grade Policy:
 Extra 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.
http://zitompul.wordpress.com
President University
Erwin Sitompul
EEM 1/5
Engineering Electromagnetics
What is Electromagnetics?
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.
President University
Erwin Sitompul
EEM 1/6
Engineering Electromagnetics
Electromagnetic Wave Spectrum
President University
Erwin Sitompul
EEM 1/7
Engineering Electromagnetics
Why do we learn Engineering Electromagnetics
 Electric and magnetic field exist nearly everywhere.
President University
Erwin Sitompul
EEM 1/8
Engineering Electromagnetics
Applications
 Electromagnetic principles find application in various disciplines
such as microwaves, x-rays, antennas, electric machines,
plasmas, etc.
President University
Erwin Sitompul
EEM 1/9
Engineering Electromagnetics
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.
President University
Erwin Sitompul
EEM 1/10
Engineering Electromagnetics
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.
President University
Erwin Sitompul
EEM 1/11
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 threedimensional 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)
President University
Erwin Sitompul
EEM 1/12
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
President University
Erwin Sitompul
EEM 1/13
Chapter 1
Vector Analysis
Rectangular Coordinate System
• Differential surface units:
dx  dy
dy  dz
dx  dz
• Differential volume unit :
dx  dy  dz
President University
Erwin Sitompul
EEM 1/14
Chapter 1
Vector Analysis
Vector Components and Unit Vectors
RPQ ?
r  xyz
r  xax  ya y  zaz
ax , a y , az : unit vectors
RPQ  rQ  rP
 (2ax  2a y  az )  (1ax  2a y  3az )
 ax  4a y  2a z
President University
Erwin Sitompul
EEM 1/15
Chapter 1
Vector Analysis
Vector Components and Unit Vectors
 For any vector B, B  Bxax  Bya y + Bzaz :
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.
RMN  (3ax  3a y  0az )  (1ax  2a y 1az )  4a x  5a y  a z
a MN
4a x  5a y  1a z
R MN
 0.617a x  0.772a y  0.154a z


2
2
2
R MN
4  (5)  (1)
President University
Erwin Sitompul
EEM 1/16
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  Axax  Aya y + Azaz and B  Bxax  Bya y + Bzaz ,
A  B  Ax Bx  Ay By + Az Bz
President University
Erwin Sitompul
EEM 1/17
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
B  a  B a cosBa  B cosBa
President University
Erwin Sitompul
EEM 1/18
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  (2ax  3a y  4az )  (6ax  a y  2az )  8a x  4a y  6a z
R AC  (3ax  1a y  5az )  (6ax  a y  2az )  9a x  2a y  3a z
BAC
R AB  R AC  R AB R AC cosBAC
 cos  BAC
R R
 AB 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
President University
Erwin Sitompul
EEM 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.
R AB on R AC   R AB  a AC  a AC

(9a x  2a y  3a z )
  (8a x  4a y  6a z )
(9)2  (2)2  (3)2



 (9a x  2a y  3a z )

2
2
2
 (9)  (2)  (3)

62 (9a x  2a y  3a z )

94
94
 5.963a x  1.319a y  1.979a z
President University
Erwin Sitompul
EEM 1/20
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  aN 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)
President University
Erwin Sitompul
EEM 1/21
Chapter 1
Vector Analysis
The Cross Product
 Example
Given A = 2ax–3ay+az and B = –4ax–2ay+5az, find AB.
A  B  ( Ay Bz  Az By )ax  ( Az Bx  Ax Bz )a y  ( Ax By  Ay Bx )az
  (3)(5)  (1)(2)  ax   (1)(4)  (2)(5)  a y  (2)(2)  (3)(4)  az
 13a x  14a y  16a z
President University
Erwin Sitompul
EEM 1/22
Chapter 1
Vector Analysis
The Cylindrical Coordinate System
President University
Erwin Sitompul
EEM 1/23
Chapter 1
Vector Analysis
The Cylindrical Coordinate System
• Differential surface units:
d   dz
 d  dz
d    d
• Relation between the
rectangular and the cylindrical
coordinate systems
x    cos 
y    sin 
• Differential volume unit :
d    d  dz
President University
zz
Erwin Sitompul
  x2  y 2
1 y
  tan
x
zz
EEM 1/24
Chapter 1
Vector Analysis
The Cylindrical Coordinate System
?
az
az
A  Ax a x  Ay a y + Az a z  A  A a   A a + Az a z
ay
A  A  a
 ( Axax  Aya y + Azaz )  a
 Axax  a  Aya y  a + Azaz  a
 Ax cos  Ay sin 
• Dot products of unit vectors in
cylindrical and rectangular
coordinate systems
A  A  a
 ( Axax  Aya y + Azaz )  a
 Axax  a  Aya y  a + Azaz  a
  Ax sin   Ay cos 
a
a
ax
Az  A  a z
 ( Axax  Aya y + Azaz )  az
 Axax  az  Aya y  az + Azaz  az
 Az
President University
Erwin Sitompul
EEM 1/25
Chapter 1
Vector Analysis
The Spherical Coordinate System
President University
Erwin Sitompul
EEM 1/26
Chapter 1
Vector Analysis
The Spherical Coordinate System
• Differential surface units:
dr  rd
dr  r sin  d
rd  r sin  d
• Differential volume unit :
dr  rd  r sin  d
President University
Erwin Sitompul
EEM 1/27
Chapter 1
Vector Analysis
The Spherical Coordinate System
• Relation between the rectangular and
the spherical coordinate systems
x  r sin  cos 
r  x2  y2  z 2 , r  0
y  r sin  sin 
  cos
1
z  r cos 
  tan
1
z
x y z
2
2
2
, 0    180
y
x
• Dot products of unit vectors in spherical and
rectangular coordinate systems
President University
Erwin Sitompul
EEM 1/28
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
z
x2  y 2  z 2
 cos 1
1
 74.50
3.742
y
1 2
 tan
 33.69  180  146.31
x
3
 C (r  3.742,   74.50,   146.31)
 D( x  0.585, y  1.607, z  4.698)
President University
Erwin Sitompul
EEM 1/29
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.
President University
Erwin Sitompul
EEM 1/30