Download The Dot Product of Two Vectors

Chapter 10 Electric Field, Magnetic Field,
Maxwell’s Equations and Plane Waves
In a wireless environment, most of the signals, are transmitted in the form of
electromagnetic waves. It is therefore important to have a basic understanding of
electromagnetism. Since Maxwell’s equations are based upon electric and magnetic
fields, we shall first introduce these two fields. Before doing that, we first have to
introduce some vector operations, which are needed for the discussion of them..
10.1 The Dot and Cross Products of Vectors
In this chapter, we shall use two terms which are actually simple. They are scalar
field and vector field. We shall not give formal mathematical definitions of these
two terms. Consider a mountainous terrain. At every point characterized by x y
and z, there is an altitude associated with it. This illustrates a scalar field. Similarly,
in a region, every location has a temperature associated with it. This is another case
of scalar field. Except this time, it may change with time. Thus the temperature
can be expressed as T ( x, y, z, t ) . Now, consider the case where water flows through,
say a tube. In this case, at each point, we not only have to note the speed of the flow,
also the direction of the flow. This is a typical vector field. In this chapter, many
parameters are vectors, such as electric and magnetic field intensities and so on. On
the other hand, the voltage, for instance, is a scalar.
Since we are living in a three-dimensional space, we shall use the
three-dimensional coordinate system in this chapter. In this system, there are three
vectors representing the x, y and z directions. These three vectors are denoted as
xˆ , yˆ and ẑ representing vectors in the x, y, and z directions respectively and all of
these vectors have magnitudes equal to unity. Every vector A will have three
components, which are the projection of A onto x, y, and z coordinates.
Thus, a vector A may be denoted as
A  Ax xˆ  Ay yˆ  Az z .
10-1
The Addition and Subtraction of Vectors
Let A and B be two vectors.
A  Ax xˆ  Ay yˆ  Az zˆ .
B  Bx xˆ  B y yˆ  Bz zˆ .
Then
A  B  ( Ax  Bx )xˆ  ( Ay  B y )yˆ  ( Az  Bz )zˆ .
A  B  ( Ax  Bx )xˆ  ( Ay  B y )yˆ  ( Az  Bz )zˆ .
(10.1-1)
A schematic diagram of the addition and subtraction operations is now shown in
Fig. 10.1-1.
Fig. 10.1-1
The Addition and Subtraction of Vectors.
Example 10.1-1.
Let A  (2,6,1) and B  (1,3,2).
Then
A  B  (1,9,1)
and A  B  (3,3,3).
The Dot Product of Two Vectors
10-2
(a)A+B. (b)A-B.
The dot product of two vectors is defined as follows:
A  B  Ax B x  Ay B y  Az B z .
(10.1-2)
Let θ denote the angle between vectors A and B.
also equivalent to the following:
The dot product of A and B is
A  B  A B cos  .
(10.1-3)
Note that the result of the dot product of two vectors is not a vector, rather a
scalar.
Example 10.1-2.
Let A  (1,1) and B  (5,0).
Then A  B  (1  5)  (1  0)  5  0  5.
Another way of determining A  B is to use Formula (10.1-3). Note that
A  2 , B  5 and  

4
.
Thus

2
A  B  2  5  cos( )  2  5 
 5.
4
2
The Cross Product of Two Vectors
The cross product between two vectors A and B is also a vector.
as follows:
It is defined
A×B  ( Ay B z  Az B y )xˆ  ( Az B x  Ax B z )yˆ  ( Ax B y  Ay B x )zˆ .
(10.1-4)
It is sometimes not easy to remember the above formula. A convenient way is
to use the following equivalent formula using the determinant notation:
10-3
xˆ
A×B  Ax
yˆ
Ay
Bx
By
zˆ
Ay
Az 
By
Bz
Az
Bz
xˆ 
Ax
Az
Bx
Bz
yˆ 
Ax
Ay
Bx
By
zˆ .
(10.1-5)
The meaning of the cross product of two vectors can be understood by the
so-called right-hand rule. Note that two vectors form a plane. Let n̂ be the vector,
which is normal to this plane as shown in Fig. 10.1-2. Then the direction of A×B is
in the direction of n̂ and A×B can be found by the following formula:
A×B= A B sin(  )nˆ
(10.1-6)
where  is the angle between A and B.
Fig. 10.1-2 The Cross Product of Two Vectors.
Example 10.1-3.
Let A  (1,0,0) and B  (0,1,0). Then it can be easily found that, by using either
Equation (10.1-5) or Equation (10.1-6),
A×B  (0,0,1).
10-4
Note that the direction of A×B is normal to the plane consisting of A and B.
10.2 The ▽ Operator, the Gradient of a Scalar Field and
the Line Integral of a Vector Field
The ▽ operator is defined as follows:
▽



xˆ  yˆ  zˆ .
x
y
z
(10.2-1)
The Gradient of a Scalar.
Let V be a scalar field function. Then, the gradient of V is defined as follows:
▽V 
V
V
V
x̂ 
ŷ 
ẑ.
x
y
z
(10.2-2)
Note that while V is a scalar, ▽V is a vector.
Example 10.2-1.
Let V  x 2  y 2  2 .
1
Then ▽V=
x
x y
2
xˆ 
y
x y
2
yˆ .
The Line Integral of a Vector
Let us consider a contour as shown in Fig. 10.2-1. There is a force along this
contour. This force is a vector and therefore it is denoted as F. To move an object
from an initial point p to a terminal point q requires energy E as follows:
E   F dl .
q
p
10-5
The integral in the above equation is called a line integral.
Line integrals play
important roles in deriving many electromagnetic formulas.
Example 10.2-1.
Let F  ( x  y)xˆ  xyˆ  yzzˆ .
Since dl  dxxˆ  dyyˆ  dzzˆ , we have
F  dl  ( x  y )dx  xdy  yzdz.
Consider the following two cases as shown in Fig. 10.2-1(a) and (b) respectively.
y
y
c
c
(1,1,0)
(1,1,0)
a
b
(0,0,0)
x
(1,0,0)
z
x
a
(0,0,0)
z
(a) Case 1
(b) Case 2
Fig. 10.2-1 Two Cases for Line Integral.
(a)Case1. (b)Case2.
Case 1:

c
a
A  dl  
1
x0
( x  y)dx  
1
y 0
xdy.
(10.2-3)
For the first integral in Equation (10.2-3), y  0 and for the second integral,
x  1. Thus

b
a
F  dl   xdx   dy 
1
1
0
0
1
1 2
1
3
1
x  y 0  1 
2 0
2
2.
10-6
(10.2-4)
Case 2:

c
a
F  dl  
1
x0
( x  y)dx  
1
y0
( x  y)dy,
For the first integral, y=x and in the second integral, x=0. Therefore, we have
c
1
1
2 1
a
0
0
0
 F  dl   2xdx  (1  y)dy  x
1
 (y  y2
2
1
 11
0
1 3

2 2
(10.2-5)
Let A be a scalar function. Let A be the gradient of A. Let c denote a closed
tour. Then we can show that the following holds:
 A  dl   A  dl  0.
c
(10.2-6)
c
Equation (10.2-6) also indicates that if the line integral of a vector A is zero,
the vector A can always be expressed as the gradient of a scalar function A.
Example 10.2-3.
y
c
a
(1,1,0)
b
(0,0,0)
(1,0,0)
x
z
Fig. 10.2-2 A Tour for Example 10.2-3.
Consider A  xy . Let the closed tour be the one shown in Fig. 10.2-2. This
tour c consists of three tours, namely c1 , c2 and c3 . First, we compute the gradient
10-7
of A.
A=▽ A  yxˆ  xyˆ .
A  dl  ydx  xdy.
For tour c1 ,

q
p
1
0
0
0
A  dl   ydx   xdy  0 .
The first integral of the above equation disappears because y  0 for this tour.
For tour c2 , we can show that

r
q
1
1
1
1
0
0
A  dl   ydx   xdy   dy  1
For tour c3 , we can use the same kind of reasoning to show that

p
r
0
0
0
A  dl   ydx   xdy   ydy  
1
1
1
0
1
0
0
1
1
xdx  y 2  x 2  1 .
2 1 2 1
Note that in the above integrals, since we integrate along the line y  x , x is
substituted by y and y is substituted by x.
the three tours, we conclude that
By adding up the values corresponding to
 A  dl  0.
c
10.3 The Divergence of a Vector Field
In the above section, the ▽ operator was applied to a scalar field and the result
is a vector. In this section, we shall introduce the result of applying the ▽ operator
to a vector field. Let A be a vector. The operator, ▽  A , called the divergence of
A, is defined as follows:
10-8
▽ A 
Ax Ay Az


x
y
z .
(10.3-1)
Note that the divergence of a vector is a scalar.
The physical meaning of divergence is explained as follows. Consider Fig.
10.3-1 in which there is a vector field.
surface s is defined as follows:
The flux of this vector field A through a
   A  ds .
(10.3-2)
s
A
ds
S
Fig. 10.3-1 Flux through a Surface S
Since flux goes in and out of a closed surface, there is a net outward flux through a
closed surface s defined as follows
   A  ds.
(10.3-3)
s
Equation (10.3-3) tells us whether there is a sink or a source. If the amount of
flux going into the surface is equal to the amount of flux going out of the surface, the
net outward flux is equal to 0. Let us pay attention to the case where the enclosed
surface reduces to zero. In doing so, the associated volume is also reduced to zero.
We are therefore interested in a point. The net outward flux is:
10-9
lim
v  0
 A  ds
s
v
.
Now, it can be shown that the divergence of A, as defined in Equation (10.3-1), is
related to the above term as follows:
▽  A  lim
v  0
 A  ds
s
v
(10.3-4)
.
The proof of Equation (10.3-4) will not be given in this book.
It suffices for the
reader to remember, from Equation (10.3-4), that the divergence of a vector field A is
the net outward flux A per unit volume as the volume reduces to zero.
Example 10.3-1.
Let A  cos(2x)xˆ  sin( 3 y)yˆ  sin( 2x  5z)zˆ.
Then
▽  A  2 sin( 2 x)  3 cos(3 y )  5 cos( 2 x  5 z ).
Consider the following two cases: A  yxˆ and A  xxˆ .
In the first case, the
divergence is 0 and in the second case, it is 1. The reader is encouraged to see why
they are so by using Equation (10.3-4). In both cases, flux is in the x-direction. For
the first case, although the flux changes with respect to y, in the x-direction, when a
flux goes into the volume, the same flux goes out of it without any change. But, for
the second case, it seems that the amount of flux is increased as it goes through the
volume. This is why the divergence is not 0 in this case.
Given a vector field A and a closed surface s, the net outward flux Ψ of A
through the surface s is defined in Equation (10.3-3). We can use Equation (10.3-3)
to find Ψ directly. But, based upon Equation (10.3-4), we can prove the following
Divergence Theorem:
 A  ds     Adv .
s
(10.3-5)
v
10-10
We shall not give a formal proof of the above theorem.
The reader can gain some
feeling about the above equation through the two examples given below:
Example 10.3-2.
Let A  yxˆ  zyˆ  xzˆ and s be the surface as illustrated in Fig. 10.3-2.
y
(0,1,0)
(0,1,1)
(1,1,0)
(1,1,1)
(1,0,0)
(0,0,1)
x
(1,0,1)
z
Fig. 10.3-2
 A  ds
s
The Closed Surface for Example 10.3-2.
can be computed as follows:
 A  ds   
Ax


Ay


Az


ydydz  
1 1

ydydz


zdxdz  

zdxdz


xdxdy  

xdxdy
1 1
s
0 0
1 1
0 0
1 1
0 0
1 1
0 0
1 1
0 0
1 1
0 0
x 1
y 1
z 1
dydz  
 A
x
x0
 A
y
y0
z
z 0
1 1
0 0
dxdz  
1 1
0 0
dxdy  
 A
1 1
0 0
0 0
1 1
0 0
1 1
0 0
 0.
10-11
dydz
dxdz
dxdy
If we use Equation (10.3-5), we first compute ▽  A as follows:
▽  A =0.
   Adv  0.
v
From Equation (10.3-5), we conclude that
 A  ds =0.
s
Example 10.3-3.
Let A  xxˆ  yyˆ  zzˆ and s be the same surface as used in Example 10.3-2.
 A  ds   
x


y


z
1 1
s
0 0
1 1
0 0
1 1
0 0
dydz  
1 1

x
dxdz  
1 1

y
dxdy  

z
x 1
y 1
z 1
0 0
0 0
1 1
0 0
x0
y 0
z 0
dydz
dxdz
dxdy
 1 0 1 0 1 0
 3.
It can be easily seen that ▽  A =3 and
   Adv  3 .
v
The physical meaning of ▽  A can be best understood by assuming the case
where ▽  A =0. Under such a condition, according to the Divergence Theorem
expressed in Equation (10.3-5),
   Adv  0 .
v
As we shall see later, this result is
quite significant.
10.4 The Curl of a Vector Field
Let A be a vector.
The ▽  A operator is called the curl of A, and is defined as
follows:
10-12
 A Ay 
 Ay Ax 
 Ax Az 
▽ A   z 

ŷ  

 x̂  
 ẑ.

z 
x 
y 
 z
 y
 x
(10.4-1)
In some sense, which we shall not elaborate here, ▽  A , is related to
"circulation".
If ▽  A =0, this indicates that A has no circulation; otherwise, A
has some circulation. This concept will become useful as we study the electrostatic
field intensity which has no circulation and magnetic field intensity which has
circulation.
Example 10.4-1.
Let A  cos( kz)xˆ  sin( kz)yˆ
Then, we have
▽ A  
Ay
z
xˆ 
Ay Ax
Ax
yˆ  (

)zˆ  k cos(kz)xˆ  k sin( kz)yˆ .
z
x
y
We like to emphasize here that ▽  A is a vector.
What is the physical meaning of ▽  A ?
In Section 10.2, we introduced the
concept of line integral of a vector field. Let c denote a closed tour and let A be a
vector field defined all along this tour. This tour c, since it is closed, is associated
with a surface s, as shown in Fig. 10.4-1.
A
c
Fig. 10.4-1 An Illustration of ▽  A
.
We can prove that ▽  A is related to the integral as follows:
10-13
▽  A  lim
s 0
 A  dl nˆ .
c
(10.4-2)
s
Note that ▽  A is a vector and its direction, denoted as n̂ , is perpendicular to
the surface Δs determined by the right hand rule. A formal proof of Equation
(10.4-2) will not be given.
Based upon Equation (10.4-2), it is easy to prove the
Stoke's Theorem, as expressed below:
 (  A)  ds   A  dl.
s
(10.4-3)
c
The physical meaning of ▽  A can be best understood by considering the
special case where ▽  A =0. In this special case, we conclude immediately that
 A  dl  0 or A has no circulation.
Later, after we have introduced the electrostatic
c
field intensity E, we will prove that
 E  dl  0 .
c
This is quite significant.
The term ds in Equation (10.4-3) is a vector and the direction of it is that of the
normal perpendicular to it and follows the right hand rule. In the following, we shall
give two examples to give the reader some feeling about Equation (10.4-3).
Example 10.4-2.
Let A be the same as that in Example 10.3-2. That is, A  yxˆ  zyˆ  xzˆ .
the tour c be the one illustrated in Fig. 10.4-2.
10-14
Let
y
c3
(0,1,0)
(1,1,0)
c4
c2
(0,0,0)
c1
x
(1,0,0)
z
Fig. 10.4-2 The Tour for Example 9.4-2.
▽  A   xˆ - yˆ - zˆ .
ds  dxdyzˆ .
(  A)  ds  -dxdy .
1 1
 (  A)  ds  
s
 dxdy  1.
0 0
If we use Equation (10.4-3), we can conclude now that
can be verified by computing
 A  dl
c
 A  dl  1 .
c
In fact, this
directly.
From Fig. 10.4-2, we can see that the tour consists of four tours, denoted as
c1 , c2 , c3 and c4 . The line integrals along these four tours are as follows:
1
 A  dl  
c
0
y
1
y 0
dx   z
0
0
z 0
dy   y
1
Example 10.4-3.
10-15
0
y 1
dx   z
1
z 0
dy  0  0  (1)  0  1 .
In this example, let A be the one used in Example 10.3-3. That is,
A  xxˆ  yyˆ  zzˆ . We will again use the same closed tour shown in Fig. 10.4-2. In
this case, we can easily see that
▽  A =0.
Thus, we can immediately conclude that
proved by computing
 A  dl
c
1
1
0
0
 A  dl   xdx  
c
 A  dl  0 .
c
Again, this can be
directly as follows:
0
0
ydy   xdx   ydy 
1
1
1 1 1 1
   0
2 2 2 2
From the above two examples, we can see that the curl of a vector does give us
valuable information about a line integral. It is often easier to obtain the line integral
of a vector along a closed tour by using the surface integral curl of the vector on the
surface associated with the closed tour, as demonstrated in the above two examples.
Up to now, we have introduced three terms, namely gradient, denoted as ▽A,
divergence, denoted as ▽  A , and curl, denoted as ▽×A. For each of these three
terms, there is an equation related to it. These equations are Equation (10.2-6),
related to gradient, Equation (10.3-5), related to divergence and Equation (10.4-3),
related to curl. Let us summarize and relabel these equations as below:
 A  dl   A  dl  0
(10.4-4)
 A  ds     Adv
(10.4-5)
 (  A)  ds   A  dl .
(10.4-6)
c
s
s
c
v
c
10-16
10.5 The Electrostatic Fields
In this section, we shall study various aspects of the electric field. We call this
electrostatic field because it will not change with time. The electric field which
changes with time will be discussed later.
Electric Field
First of all, let us consider a positive charge Q1 . Suppose that there is another
positive charge Q2 , which is a test charge, in certain sense. Then there is a repulsive
force between these charges as follows according to Coulomb’s Law:
F
Q1Q2
r̂.
4  r 2
(Coulomb’s Law)
where F = the repulsion force between Q1 and Q2 , in Newtons
Q1 = charge 1, in coulombs
Q2  charge 2, in coulombs
r = the distance between charge 1 and charge 2, in meters
r̂ = a unit vector in the direction linking Q1 and Q2 , dimensionless
ε = the permittivity (or dielectric constant), in Farads per meter.
Fig. 10.5-1 depicts the Coulomb’s law.
10-17
(10.5-1)
Fig. 10.5-1 Coulomb’s Law.
Coulomb’s Law was invented by Charles Coulomb, a French army colonel
(1736-1706). He retired from the army because of the French revolution. His
elaborate experiments showed a force, either repulsion or attraction, exists between
two charges. Coulomb’s Law is quite similar to Newton’s gravitational law.
The permittivity ε for vacuum, denoted as  0 , is equal to 8.85  10 12.
For air
at atmospheric pressure, ε is equal to 1.0006  0 . For dielectric materials, ε >  0 .
Having presented Coulomb’s law, we can now introduce the electric field E.
Given a charge Q1 , for any location in space and a second test charge Q2 located
there, there will be a force forced by Q2 which is
F
Q1Q2
rˆ .
4  r 2
The electric field E, or also called electric field intensity, defined as force per unit
charge, is thus
10-18
E
Q1
F

rˆ
Q2 4  r 2
(10.5-2)
where E = electric field in Newton per Coulomb.
If there is more than one charge, for any location in the space, each charge causes
an electric field intensity vector and the summation of the vectors is the resulting
electric field at that location.
Example 10.5-1.
Y
E2
E1
(1,1)
Q
(0,0)
Q
X
(2,0)
Fig. 10.5-2 The Electric Field Resulting from Two Charges.
Consider Fig. 10.5-2. Two charges are located at (0,0) and (0,2) respectively.
For the point (1,1), there will be two vectors as shown and it is clear that the addition
of these two vectors creates a vector in the positive y-axis direction.


E  E1 cos( )yˆ  E 2 cos( )yˆ
4
4
2Q
2

yˆ
2
4 ( 2 ) 2

Q
4 2
yˆ
10-19
Example 10.5-2.
Y
dE
(0,y)
d
(-1,0)
Fig. 10.5-3
x
X
(1,0)
The Electric Field Induced by a Line Charge.
In Fig. 10.5-3, there is a line charge starting from (-1,0) to (1,0). The charge
density of this line charge is denoted as ρ. Consider an infinitesimal section dx.
This section contains  dx charge and causes an electric field at (0,y) as follows:
dE 
 dx
4  ( x 2  y 2 )
.
As shown in Fig. 10.5-3, there is also an electric field symmetrical to the above
electric field, which cancels out the electric field in the x-direction. Thus, only the
y-direction component electric field exists. The combined y-direction electric field is
dE y  2dE cos 

2  dx
cos 
4  ( x 2  y 2 )

2  dx
4 ( x 2  y 2 )

y
x2  y2
 ydx
2 ( x 2  y 2 )
3
2
.
y1
dx
.
Ey 

2 0 ( x 2  y 2 ) 3 2
(10.5-3)
10-20
To find the solution of Equation (9.5-3), let x  y tan  .
Ey 


y
2
y
2
y
2
1
y
tan 1

1
y
tan 1


1
y
0
2
3
2
1
cos  d
y2
tan


sin

0
2  y


3
y sec 2  d
( y 2 sec 2  )
0
tan 1
y sec 2  d
( y 2 tan 2   y 2 )
0
Then
2  y y 2  1
(10.5-4)
1 1
y
.
Suppose the line charge extends from (-a,0) to (a,0), it can be proved that in this
case,
Ey 

y2
2  y 2  1
a
.
(10.5-5)
If a  ∞, Equation (8.2-5) becomes
Ey 

.
2  y
(10.5-6)
Compare Equations (10.5-6) and (10.5-2). We now can see that the electric
field intensity for a line charge is inversely to the distance while that for a point
charge is inversely to the square of the distance.
This is expected intuitively.
Electric Flux Density
Having introduced the concept of electric field, or electric field intensity, we can
now introduce another related concept, called, the electric flux density. Note that the
electric field intensity is inversely proportional to the permittivity. In other words, it
is dependent of the medium. The electric flux density is something independent of
10-21
the medium.
The famous British scientist, Michael Faraday (1791-1867), performed a very
interesting experiment. We shall not give details of this experiment. But it can be
briefly illustrated as follows: Consider Fig. 10.5-4. In Fig. 10.5-4, there are two
concentric spheres. Between these two spheres, there is an insulation material. The
surface of the inner sphere is charged positively. The outer sphere was temporarily
grounded so that it was free of charge. Then Faraday found that the outer sphere
was charged negatively. It was also observed that the amount of negative charge
induced on the outer sphere was exactly the same as that in the inner sphere.
Fig. 10.5-4
An Experiment with Two Concentric Spheres
Somehow, we can visualize that there are lines emanating from the inner sphere
to the outer sphere. This leads to the concept of electric flux. Denote the flux by
 . We have
  Q.
We may imagine that the inner sphere shrinks to a point charge. Then the
electric flux density D, at any point r meters from the center, as illustrated in Fig.
8.2-5, is defined as the flux per square meter as follows:
D
Q
rˆ ,
4 r 2
(10.5-7)
where D = electric flux density in Coulomb per m2.
10-22
Fig. 10.5-5 The Electric Flux.
Comparing Equation (10.5-7) and (10.5-2), we now have
D  E,
where
(10.5-8)
D = electric flux density in Coulomb per m2
E = electric field in Newton per Coulomb
ε = permittivity in Faraday per m.
Gauss Law
Let s denote a closed surface enclosing some charges. Let Q denote the total
charges enclosed by s. We can show that the following is true:
 D  ds  Q .
(10.5-9)
s
This is referred to as the Gauss' law. By using the divergence theorem expressed in
Equation (10.3-5), we have
 D  ds   (  D)dv  Q .
s
(10.5-10)
v
Let  v denote the volume charge density. That is,
10-23
Q    v dv .
(10.5-11)
v
Then, we have
▽  D  v .
(10.5-12)
Equation (10.5-12) is also called the differential form of Gauss' law.
Electrostatic Potential
Consider Fig. 10.5-6. Assume that there is an electric field E and a charge q
is moved along a tour from a to b. At any point, there is a force qE inserted on the
charge q. The total work required by performing this action is
Wab   q  E  dl.
b
(10.5-13)
a
b
E
a
Fig. 10.5-6. The Moving of a Charge along a Tour.
If the tour is a closed one, as shown in Fig. 10.5-7, the total work done is zero.
That is,
 E  dl  0.
(10.5-14)
c
Equation (10.5-14) shows that the electrostatic field is a conservative field.
10-24
b
a
Fig. 10.5-7 The Moving of a Charge along a Closed Tour.
As discussed at the end of Section 10.2, since the line integral of E along a
closed tour is zero, E can be expressed as the gradient of a scalar function.
customary to define the gradient of E as the potential function V as follows:
E  V .
It is
(10.5-15)
Given two points a and b along a tour c, the potential difference between a and b is
Vba  Vb  Va    E  dl.
b
(10.5-16)
a
Note that we are usually concerned about the potential difference between two
points. Of course, an absolute potential can be defined by setting point a to be
infinity and forces V a to be 0. Note that if there is no electrostatic field E involved,
there will be no potential difference. Thus, to detect the existence of E, often make
use of the potential difference.
In the above discussion, we defined the potential difference with the
specification of a tour. Now, given two points, there can be many tours between
them. Will different tours produce different potential differences? The answer is
“No”. It can be easily proved that the potential difference is independent of tours.
We omit the proof.
10-25
10.6 Magnetostatic Field
The magnetic field is an important topic. In this section, we shall study the case
where the magnetic field does not change with time. Thus, the field is called
magnetostatic field. The magnetic field which changes with time is another
important topic and will be discussed later.
The electric field is, as we discussed before in Section 10.5, caused by the
presence of charges. The magnetic field is caused by the presence of currents. Let
us now first introduce currents.
Currents
When charges move in a conductor, we say that there is a current. We denote a
current by I which is a scalar function. In this section, we assume that I does not
change with time. While current I is a scalar function, we can define a vector J,
called the current density vector, out of it as illustrated in Fig. 10.6-1.
J
ds
Fig.10.6-1 The Current Density Vector.
The relationship between I and J is as follows:
I   J  ds .
(10.6-1)
s
10-26
It is important to note that in the above equation, the surface, as illustrated in Fig.
10.6-1, is not a closed one. For a closed surface, the net current through the surface
is zero as we are all familiar with and this is known as the Kirchhoff’s current law.
Ampere’s Law
In Section 10.5, we introduced the Coulomb’s law which describes the force
between two charges. Interesting enough, there is an analogy in the current case.
That is, if there are two currents in the space, there is also a force exerting on them.
Ampere found this result by executing many exceedingly elaborate experiments.
One of these experiments was so elegantly designed that it was highly praised by
Maxwell later. In space, current segments are oriented. Let us consider the
simplest case where two current segments are parallel to each other as illustrated in
Fig. 10.6-2.
I 1 dl 1
r
I 2 dl 2
Fig. 10.6-2 An Illustration of the Ampere’s Law for a Simple Case
In the case as illustrated in Fig. 10.6-2, the force between the two current
segments is
dF 
 I1dl1 I 2 dl 2
rˆ
4
r2
(10.6-2)
where F is the force between I 1 and I 2 , in Newtons
I 1 = current segment 1
I 2 = current segment 2
r = the distance between I 1 and I 2 , in meters
r̂ = a unit vector in the direction linking I 1 and I 2 , dimensionless
  the permeability, in Henry per meter.
In free space, the permeability  , denoted as  0 , is equal to 4  10 7
10-27
Henry/m.
For a general case, the two current segments are necessarily parallel to each
other as shown in Fig. 10.6-3.
I 2 dI 2
I1dI1
r

dF
Fig. 10.6-3 An Illustration of Ampere’s Law for the General Case.
The force between these two current segments is now expressed as follows:
dF 
 0 I 2 dl 2  ( I1dl1  rˆ)
.
4
r2
(10.6-3)
Magnetic Flux Density
As we did in the case of Coulomb’s law, we now consider the case where I1
is fixed and I 2 is moved around. At an any point in space and with any orientation,
I 2 experiences a force. The normalized force can be obtained as:
dF2 
 0 I1dl1  rˆ
.
4
r2
(10.6-4)
For reasons which can be understood later, we shall define this dF2 as the
magnetic flux density produced by current segment I 1 at a location specified by the
direction vector r̂ and distance r as illustrated in Fig. 10.6-4. Let us use IdL ,
instead of I 1 dl 1 . We denote dB to represent this magnetic flux density and we have
the Biot-Savart law as below:
dB 
 0 IdL  rˆ
.
4 r 2
(10.6-5)
10-28
Fig. 10.6-4. The Biot-Savart Law.
The Biot-Savart law, expressed in Equation (10.6-5), states that a magnetic
field will be induced by a current segment. The reader may simply think the
magnetic flux density B as the magnetic field intensity which signifies the strength of
the magnetic field. It is important to note that Equation (10.6-5) is valid only in free
space. If there is magnetic material present, this formula will not be valid.
In many equations introduced in this section, cross products are used. The
reader is encouraged to go back to Section 10.1 to review the definition of the cross
product of two vectors. The two vectors dL and r̂ certainly define a plane.
According to the definition of the cross product of two vectors, the direction of the
resulting vector is normal to the plane and according to the right-hand rule. Using
Equation (10.1-6), we have
dB 
 0 Idl sin 
.
4 r 2
(10.6-6)
10-29
Example 10.6-1.
(0,y)
d
dx sin θ
r
y


dx
Fig.10.6-5
The Magnetic Flux Density Induced by a Line Current.
Now, let us consider Fig. 10.6-5.
dB 

I
In this case,
 0 Idx sin 
.
4 r 2
But, dx sin   rd and r 
y
.
sin 
Therefore,
o I 
B
sin d
4 y 0
 I
 0 .
2y
(10.6-7)
We can see that the magnetic flux density induced by a line current has the same
form as the electric field intensity induced by a line charge, expressed in Equation
(10.5-6). But the direction of the magnetic flux density is entirely different from that
of the electric field. As shown in Fig. 10.6-6, it is circular with respect to the line
current.
10-30
Fig. 10.6-6 Magnetic Flux Density Induced by a Current.
In Section 10.5, we indicated that the line integral of the electric field intensity E
along a closed tour is zero. Will this happen in the magnetic field case? That is, if
we calculate the line integral of B along the circles in Fig. 10.6-6, will the result be
also zero? Of course not, because the value of B is a constant along this circle. As
discussed in Section 10.4, we may say that the circulation of the magnetic field is not
zero, or there is circulation in the magnetic field. This leads to significant results as
presented below.
Magnetic Field Intensity
We note that the magnetic flux density is caused by current.
show the following formula which holds only in free space:
B  0J .
In fact, we can
(10.6-8)
The proof of Equation (10.6-8) is quite complicated and we shall omit it in this book.
Thus it is convenient to define a new term, namely magnetic field intensity H, as
follows:
10-31
H
B
(10.6-9)
0
Using Equations (10.6-8) and (10.6-9), we have the following formula:
H  J
.
(10.6-10)
The above equation is called the Ampere’s law in differential form as the curl of
a vector field involves differentiation. Equation (10.6-10) not only holds in free
space. If there is magnetic material present, as often is the case in transformers,
usually the magnetic material will cause the value of the magnetic flux density B to
increase while the value of H remains the same. This is due to the fact that there is
also so called secondary current in the magnetic material. Thus we have the
following formula relating B and H if there is some magnetic material present:
B   0 (1  xm )H  H
(10.6-11)
where  = permeability of medium in Henry per m.
We now present Ampere’s law in integral form.
Equation (10.6-10) and Stoke’s Theorem.
closed tour c. From (10.6-10), we have
Let s be a surface whose boundary is a
 ( H)  ds   J  ds .
s
We do this essentially by using
(10.6-12)
s
Applying Stoke’s Theorem, which is expressed in Equation (10.4-3), we have
 H  dl   J  ds .
c
(10.6-13)
s
The above equation is called the Ampere’s law in integral form. Note that J is the
current density function. We therefore have another form of Equation (10.6-13) as
follows:
 H  dl  I
c
(10.6-14)
s
10-32
where I s is the net current flowing through the surface s.
Another important property concerning with the divergence of B.
easily prove that
  B  0.
We can
(10.6-15)
This is the Gauss’ law for magnetostatic field. We shall omit the proof here. The
reader should understand the physical meaning of this law. It essentially states that
there is no magnetic unit in the magnetic field, as a charge in the electric field, unless
permanent magnets are present. Note that we have never even mentioned permanent
magnets. Thus, if we examine a closed volume, we will note that the total flux going
through this volume is always zero.
10.7 A Review of Electrostatic and Magnetostatic Fields
In the above two sections, we introduced electric and magnetic fields. They were
discussed under the condition that they are time-invariant. That is why they are
called electrostatic and magnetostatic fields. It is interesting to compare these fields.
We shall do a review of them in this section and hopefully the reader can see the
similarity and difference between them.
1. The electrostatic field is always induced by charges. If there are no charges,
there will be no electrostatic field. Since charges do not accumulate in a current,
no electrostatic field is induced by it. Essentially, the electrostatic field is
governed by Coulomb’s law.
2. The magnetostatic field is always induced by a current segment. Without
current, there can be no magnetostatic field. This field is governed by
Biot-Savart law or Ampere’s law.
3. Coulomb’s law states that there is a force between two charges. Thus if there is
a charge in the space and if we move a testing charge around, the testing charge
will always experience a force. We assume that the charge is of one unit. Then
the electric field intensity E at any point in the space is the force experienced by
this testing charge.
4. Similarly, the Biot-Savart law states that if there is a current segment in the space
10-33
and if we move a testing current segment around, the testing current segment will
experience a force. The magnetic field intensity H at any location in the space
is the force experienced by the testing current segment, up to a constant.
5. The line integral of the electric field intensity E along a closed tour is zero,
indicating that the electric field is a conservative field. This is why the curl of E
is zero. That is, ▽  E  0 ; there is no circulation in the electric field.
6. The line integral of the magnetic field intensity along a closed tour is not zero.
It is equal to the total current flowing through the surface defined by the tour.
That is,   H  J ,
 H  dl   J  ds
c
s
 H  dl  I
and
c
s
. There is circulation in
the magnetic field.
7. As for divergence, ▽  D   v and   B  0 .
This is because there are
charges in the electric fields and no similar magnetic units in the magnetic fields.
The following equations summarize the results:
Differential form
Integral form
 E  dl  0
▽ E  0
(10.7-1)
c
▽ H  J
 H  dl   J  ds
(10.7-2)
▽  D  v
 D  ds    dv
(10.7-3)
▽B  0
 B  ds  0
(10.7-4)
c
s
s
s
v
v
The above four equations are valid for the time-invariant case. If the fields are
time-varying, these four equations will be modified and the modified ones are the
Maxwell’s equations which will be introduced in the next section.
10.8 Maxwell’s Equations
10-34
In the above sections, we introduced the static electric and magnetic fields. The
critical equations governing these fields were grouped together as Equations (10.7-1)
to (10.7-4). In this section, we shall get into the time-varying case. Let us first
note that for Equations (10.7-3) and (10.7-4), they remain valid under the
time-varying case. But Equations (10.7-1) and (10.7-2) will be changed. Both
Equations (10.7-1) and (10.7-2) will be modified to relate the electric field and
magnetic field. That is, we shall now answer the following question: Will the
change of magnetic field affect the electric field and vice versa?
Let us first discuss the case where the changing of magnetic field will affect the
electric field. This is stated in Faraday’s law. We shall not give a formal derivation
of this law. Instead, we shall only present an informal presentation of it with
emphasis of its physical meaning.
Fraday’s Law
Consider Fig. 10.8-1.
E
B
E
C
(a)
(b)
Fig. 10.8-1 Illustration of the Relationship between Electric and Magnetic Fields: (a)
A Loop without Time-Varying Magnetic Flux Going Through. (b) A Loop with
Time-Varying Magnetic Flux Going Through.
In Fig. 10.8-1(a), there is no time-varying magnetic flux going through the
closed surface s defined by the tour c. In this case, we have
 E  dl  0 .
c
10-35
In Fig. 10.8-1(b), there is magnetic flux going through the surface s and it is
time-varying. Thus there is a time-varying magnetic field inside this surface. In
this case, Faraday observed that
B
 E  dl   t
c
s
 ds .
(10.8-1)
The above equation is called Faraday’s law in integral form. Using the Stoke’s law,
we can also have Faraday’s law in differential form as follows:
E
B
.
t
(10.8-2)
Essentially, Faraday’s law states that a time-varying magnetic field induces an
electric field. Note that this electric field is no longer conservative as in the
electrostatic case.
Faraday’s law is actually easy to understand. The major point is that the
induced electric field may be hard to measure. But the electric filed in turn produces
a voltage and that can be measured. Let us consider Fig. 10.8-2. If the permanent
magnet is fixed in space, there will be no electric field induced and there is no
indication in the voltmeter. If we move the magnet inside the coil, this will cause a
time-varying magnetic field and induce an electric field. Finally, this will cause the
voltmeter indicator to reflect.
Voltmeter
N
S
Fig. 10.8-2 The Voltage Caused by the Movement of a Magnet Inside a Coil
10-36
The Modification of Ampere’s Law by Maxwell
Ampere’s law, as expressed in Equation (10.7-2), states that a magnetic field will
be induced by a current.
It was Maxwell who conjectured that in time-varying case,
Equation (10.7-2) should be modified to the following equation:
D
 H  dl   J  ds   t
c
s
s
 ds .
(10.8-3)
Using the divergence theorem, we have
▽ H  J 
D
.
t
(10.8-4)
It should be noted here that Maxwell proposed the addition of
D
without any
t
experimental results. He simply thought that this should be the case although he
could not prove it by any experiment. Still, as we shall see later, this addition was a
milestone in the development of electromagnetic theory. Through this equation,
Maxwell said that a time-varying electric field will induce a magnetic field.
But, what is the physical meaning of
D
?
t
An examination of Equation
(10.8-4) will inform us that its dimension must be the same as J, which is volume
current density. Or, to put it another way, we may say that
D
 t
s
 ds is a kind of
current. In electromagnetic theory, this is called the displacement current.
consider Fig. 10.8.3.
10-37
Let us
capacitor
Voltage
Source
Voltmeter
Fig. 10.8-3 The Displacement Current
In Fig. 10.8-3, there is a capacitor connected to a voltage source. We all know
that charges do not move through the capacitor. Why is there still current in this
circuit. We note that there must be an electric field inside the capacitor because of
the voltage source. If the voltage source is a static one, the electric field will not
change with respect to time. Then there is no current induced. If the voltage
source is a sinusoidal one, the electric field changes with time. According to
Equation (10.8-4), there is a
D
which will induce a time-varying magnetic field B.
t
Then, according to Faraday’s law expressed in equation (10.8-2), this time-varying
magnetic field B induces an electric field E. How can we verify the existence of E?
Suppose that we insert a solenoid coil inside the capacitor.. Inside the solenoid coil,
B is circular. The direction of E induced by B is along side of the coil. Note that
the integration of E along the coil produces a potential difference. Thus this field E
can be measured by a voltmeter.
From the above discussion, we know that
D
can be considered a current,
t
which is called displacement current because in some sense, charges are displaced.
In fact, this is why the term electric flux density is called D.
10-38
Maxwell’s Equations
The equations presented above are summarized as follows and they are called
Maxwell’s equations for historical reasons.
Differential form
▽ E  
Integral form
B
t
▽ H  J 
B
 E  dl   t
c
D
t
s
 ds
(10.8-5)
D
 H  dl   J  ds   t
c
s
s
 ds
(10.8-6)
▽  D  v
 D  ds    dv
(10.8-7)
▽B  0
 B  ds  0
(10.8-8)
s
v
v
s
A brief history about Maxwell’s equations is in order.
(1) As early as 1813, Equations (10.8-7) and (10.8-8) were discovered by Gauss.
(2) In 1825, Ampere discovered a law similar to Equation (10.8-6).
But Ampere
only mentioned the first quantity on the right side of Equation (10.8-6), namely J.
That is, Ampere in 1825, did not understand the significance of the second term,
namely D
t
.
(3) Farady, in 1831, discovered the relationship between E and B.
Equation (10.8-5) was actually discovered by Farady.
Therefore
(4) Maxwell, in 1865, modified Ampere’s law to its present form by introducing the
term D
t
and presented the second Maxwell’s equation in the form of Equation
(10.8-6).
What do the above equations tell us? By examining Equation (10.8-5), we can
understand that the changing of the magnetic field with time influences the electric
field. But more interestingly, Equation (10.8-6) tells us that the changing of the
10-39
electric field in turn influences the magnetic field.
This is the basic message of
Maxwell’s Equations: The electric and magnetic fields influence each other. This is
how electromagnetic waves are generated. Although Maxwell predicted the existence
of electromagnetic waves, he did not live long enough to physically see the waves.
(5) In 1888, Hertz successfully invented an oscillator and verified the correctness of
Maxwell’s equations.
10.9 The Plane Wave Equation Derived from the
Maxwell’s Equations
By a plane wave, we mean a wave, which is uni-directional. Without losing
generality, we may assume that there is only one electric field component, namely
E x and the wave propagates in the z-direction. Let us now derive the equation
governing such an electromagnetic wave from the Maxwell’s equations. We start
from Equation (10.8-5) and we shall first find the electric field of the electromagnetic
wave.
The Electric Field of the Plane Electromagnetic Wave
From Equation (10.8-5), we have
E  

(  B) .
t
(10.9-1)
To find the left side of Equation (10.9-1), we first note that   E is defined by
Equation (10.4-1) as follows:
E y 
 E
 E y E x 
 E x E z 
E   z 

yˆ  

 xˆ  
 zˆ

z 
x 
y 
 z
 y
 x
10-40
   E y E x    E x E z  
E   


 xˆ

y  z  z
x  
 y  x
   E
E y    E y E x  
   z 

 
  yˆ
z  x  x
y  
 z  y
   E
E    E z E z  
   x  z

 zˆ .
x  y  y
z  
 x  z
The right side of Equation (10.8-5) is



(  B)    (  H )
t
t

D
   (J 
)
t
t
by (10.8-6)
by Equation (10.8-6).
In the free space, we have J  0.
Thus the right side of Equation (10.8-5) is
 2D
 2E
  2    2 .
t
t
Thus we have the following wave equation:
(
 2Ey
 (
yx

 2 Ex  2 Ex  2 Ez


)xˆ
zx
y 2
z 2
2
2
 2 Ez  E y  E y  2 Ex



)yˆ
 z y
yx
z 2
x 2
 2 Ex  2 Ez  2 Ez  2 Ez
 (



)zˆ
xz
yx
x 2
y 2
(10.9-2)
 2Ey
 2 Ex
 2 Ez
ˆ
   ( 2 xˆ 
y

zˆ ).
t
t 2
t 2
For our plane waves, E y  E z  0 . We also assume that E x is a function of z
and t. Thus, from Equation (10.8-5), for plane waves, we have the following
equation:
10-41
 2 Ex
 2 Ex
 
z 2
t 2
Let v 
1

(10.9-3)
. We now show that
E x  E0 cos k ( z  vt)
(10.9-4)
is a solution of Equation (10.9-3).
The left side of Equation (10.9-3) is
 2 Ex
 k 2 Eo cos k ( z  vt)  k 2 E x .
2
z
The right side of Equation (10.9-3) is

 2 Ex
   E0 k 2 v 2 cos k ( z  vt)  k 2 E x .
2
t
Thus E x  E0 cos k ( z  vt) is a solution of Equation (10.9-3) provided v 
1
 .
In free space,
  4 (10 7 )
  10 9 / 36 .
That is,
1

 9 x(1016 )  3x10 8 m/s
which is the speed of light. This means that the parameter v in Equation (10.9-4)
is equal to the speed of light, which implies that the electromagnetic wave travels in
the free space with the speed of light.
What is the meaning of Equation (10.9-4)? Let us assume that the transmitter is
located at z  0 and the receiver is located at z  a. At t  t1 , the signal
10-42
transmitted at z = 0 is
E0 cos( vt1 ) .
Let t 2 denote the time when the receiver, located at z = a, receives this signal.
Then
E0 cos k (a  vt2 )  E0 cos k (vt1 ) .
Thus, this indicates that it takes a
v
time units for the wave to travel from z = 0 to
z = a. This is exactly the time needed for light to travel such a distance. Since the
speed of light is exceedingly fast, so far as we human beings are concerned, the
receiver receives the signal almost instantly.
In previous chapters, a signal is only a function of time. Now, a wave is a
function of both time and space. Besides, the wave travels in space. The physical
meaning of this kind of waves can be better understood by considering the waveform
when t  0. The waveform is E0 cos kz.
Fig. 10.9-1 illustrates the traveling of this wave. Consider the particular case
where z = 0. The signal strength at this point is E0 . At any time t, the first
location z where E0 cos k ( z  vt)  E0 can be found by setting z  vt  0. Thus
z  vt . This means that the peak of the wave travels from z  0 to z  vt.
we may say that the wave travels with speed v.
10-43
Or,
E
t=0
t=t
z
z=vt
Fig. 10.9-1 The Traveling of a Wave.
The Value of k.
Finally, let us ask a very puzzling question: Our transmitter transmits a signal
with a certain frequency, say, f. That is, the transmitter sends out a signal with the
following form:
cos 2ft.
But our electromagnetic wave has the following form:
cos k ( z  vt)  cos( kvt  kz).
We may rewrite the above equation as cos( kvt   ).
be considered a phase shift. The problem is:
the following equation gives the answer:
That is, the term kz can
What is the meaning of kv ?
kv  2f .
Indeed,
(10.9-5)
That is, the electromagnetic wave is still of the form:
10-44
cos( 2ft   ).
From (10.9-5), we have
k
2f
v
(10.9-6)
where f is the frequency of the transmitted signal and v  3  10 8 is the speed of
light. That is, after f is determined, the parameter k will be determined. A high
frequency f implies a high value of k and a low frequency f implies a small value of k.
The Wavelength
In the following, we shall introduce a very important term of waves, namely,
the wavelength of a wave. Note that at any particular time t, so far as the z-axis is
concerned, the wave is a sinusoidal function of z because
E  E0 cos(2ft  kz)  E0 cos(kz   ) .
Let us suppose at z  z1 , E  E1 and at z  z 2 , E  E1 again.
kz1  kz2  2 .
Then
(10.9-7)
The wavelength of the wave is defined as follows:
  z 2  z1 .
(10.9-8)
By using Equations (10.9-8), (10.9-7) and (10.9-6), we have
  z 2  z1 
2
2
v

 .
2f
k
f
v
(10.9-9)
Note that v is the speed of light and f is the frequency of the signal. A higher
frequency implies a shorter wavelength and a lower frequency implies a longer
wavelength. The frequency of human voice is 3k. Thus the wavelength is
10-45
3  10 8
 10 5 m  100km .
3
3  10
We shall show in Chapter 11 that the antenna length
should be  . This is why we cannot broadcast the human voice directly.
2
In
Chapter 5, we showed the techniques of modulation which lift the baseband frequency
to a higher one. The concept of wavelength can be visualized by Fig. 10.9-2.
E
z

Fig. 10.9-2. The Wavelength.
The Magnetic Field of the Plane Electromagnetic Wave
We have discussed the electric field. The magnetic field can be found from
Equations (10.8-5) and (10.4-1). Let us rewrite Equation (10.8-5) as follows:
E  
B
.
t
Note that
E  E x xˆ  E0 cos k ( z  vt)xˆ .
10-46
According to (10.4-1),
E 
E x
yˆ  kE0 sin k ( z  vt)yˆ .
z
We thus have
 kE0 sin k ( z  vt)ŷ  
B
.
t
(10.9-10)
The existence of the directional basic vector ŷ means that the B only has the
B y component.
Noting k 
1

, we can easily see that
B   E0 cos k ( z  vt)ŷ
(10.9-11)
is a solution of Equation (10.9-10). Another equation, which we often use, is
H

 cos k ( z  vt)y.
ˆ
(10.9-11)
Fig. 10.9-3 illustrates how the electric and magnetic waves propagate.
10-47
Fig. 10.9-3. The Electric and Magnetic Fields in a Plane Electromagnetic Wave
10-48