### Chapter 8 - Erwin Sitompul

```Engineering Electromagnetics
Lecture 10
Dr.-Ing. Erwin Sitompul
President University
http://zitompul.wordpress.com
President University
Erwin Sitompul
EEM 10/1
Chapter 6
Dielectrics and Capacitance
Capacitance of a Two-Wire Line
 The configuration of the two-wire line consists of two parallel
conducting cylinders, each of circular cross section.
 We shall be able to find complete information about the electric
field intensity, the potential field, the surface charge density
distribution, and the capacitance.
 This arrangement is an important type of transmission line.
President University
Erwin Sitompul
EEM 10/2
Chapter 6
Dielectrics and Capacitance
Capacitance of a Two-Wire Line
 The potential field of two
infinite line charges, with
a positive line charge in
the xz plane at x = a and
a negative line at x = –a,
is shown below.
 The potential of a single
line charge with zero
R0 is:
R0
L
V
ln
2
R
 The combined potential field can be written as:
R10 R2
L
R20 
 L  R10
ln
V
 ln
 ln

2  R1
R2  2 R20 R1
President University
Erwin Sitompul
EEM 10/3
Chapter 6
Dielectrics and Capacitance
Capacitance of a Two-Wire Line
 We choose R10 = R20, thus placing the zero reference at equal
distances from each line.
 Expressing R1 and R2 in terms of x and y,
 L ( x  a) 2  y 2
L
( x  a)2  y 2

ln
V
ln
2
2
4 ( x  a)2  y 2
2
( x  a)  y
 To recognize the equipotential surfaces, some algebraic
manipulations are necessary.
 Choosing an equipotential surface V = V1, we define a
dimensionless parameter K1 as:
K1  e4V1
L
( x  a) 2  y 2
K1 
( x  a) 2  y 2
President University
Erwin Sitompul
EEM 10/4
Chapter 6
Dielectrics and Capacitance
Capacitance of a Two-Wire Line
 After some multiplications and algebra, we obtain:
K1  1 2
2
x  2ax
 y  a2  0
K1  1
2
2
 2a K1 

K1  1 
2

xa
  y  

K1  1 
K

1

1


 The last equation shows that the
V = V1 equipotential surface is
independent of z and intersects
the xz plane in a circle of radius b,
2a K1
b
K1  1
 The center of the circle is x = h, y = 0, where:
K1  1
ha
K1  1
President University
Erwin Sitompul
EEM 10/5
Chapter 6
Dielectrics and Capacitance
Capacitance of a Two-Wire Line
 Let us now consider a zero-potential conducting plane located
at x = 0, and a conducting cylinder of radius b and potential V0
with its axis located a distance h from the plane.
 Solving the last two equations for a and K1 in terms of b and h,
a  h2  b2
h  h2  b2
K1 
b
 The potential of the cylinder is V0, so that:
K1  e2V0
L
 Therefore,
2V0
L 
ln K1
President University
Erwin Sitompul
EEM 10/6
Chapter 6
Dielectrics and Capacitance
Capacitance of a Two-Wire Line
 Given h, b, and V0, we may determine a, K1, and ρL.
 The capacitance between the cylinder and the plane is now
available. For a length L in the z direction,
L L
4 L 2 L
C


V0
ln K1 ln K1
C

2 L
ln  h  h2  b2


2 L

1
cosh
( h b)

b

• Prove the equity by solving
where cosh(α)=h/b.
President University
Erwin Sitompul
EEM 10/7
Chapter 6
Dielectrics and Capacitance
Capacitance of a Two-Wire Line
 Example
The black circle shows the
cross section of a cylinder of
5 m radius at a potential of
100 V in free space. Its axis is
13 m away from a plane at
zero potential.
b  5, h  13, V0  100
a  h2  b2  132  52  12
h  h2  b2 13  12
 5  K1  25
K1 

b
5
4V0 4 (8.854 1012 )(100)
L 

 3.46 nC m
ln K1
ln 25
2 (8.854 1012 )
2

 34.6 pF m
C
1
1
cosh (13 5)
cosh (h b)
President University
Erwin Sitompul
EEM 10/8
Chapter 6
Dielectrics and Capacitance
Capacitance of a Two-Wire Line
 We may also identify the
cylinder representing the 50 V
equipotential surface by
finding new values for K1, b,
and h.
K1  e4V1 L
4 8.8541012 50 3.46109
e
5
2a K1 2 12 5
 13.42 m
b

5 1
K1  1
K1  1
5 1
 18 m
ha
 12
K1  1
5 1
President University
Erwin Sitompul
EEM 10/9
Chapter 6
Dielectrics and Capacitance
Capacitance of a Two-Wire Line
 L
( x  a)2  y 2 
E   
ln
2
2
2

(
x

a
)

y


 L  2( x  a)a x  2 ya y 2( x  a)a x  2 ya y 




2
2
2  ( x  a)  y
( x  a)2  y 2 

D  E =  L
2
 2( x  a)a x  2 ya y 2( x  a)a x  2 ya y 



2
2
2
2
(
x

a
)

y
(
x

a
)

y


S ,max  Dx, xhb, y 0 =
S ,max
3.46 109

2
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L
2
 hb a
h b a 

 ( h  b  a) 2 (h  b  a) 2 


 13  5  12
13  5  12 
2

0.165
nC
m

 (13  5  12)2 (13  5  12)2 


Erwin Sitompul
EEM 10/10
Chapter 6
Dielectrics and Capacitance
Capacitance of a Two-Wire Line
S ,min  Dx, xhb, y 0 =
 hba
hba 

 ( h  b  a) 2 ( h  b  a) 2 


 13  5  12
13  5  12 
2

0.073
nC
m

 (13  5  12)2 (13  5  12)2 


+
+
+
+
+
+
+
+
-
-
S ,min
3.46 109

2
L
2
-
-
-
S ,max  Dx, xhb, y 0
S ,min  Dx, xhb, y0
S ,max  2.25S ,min
President University
Erwin Sitompul
EEM 10/11
Chapter 6
Dielectrics and Capacitance
Capacitance of a Two-Wire Line
 For the case of a conductor with b << h, then:

ln  h  h 2  b 2

2 L
C
ln(2h b)
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
b

ln  h  h  b 
ln  2h b 
(b  h)
Erwin Sitompul
EEM 10/12
Chapter 6
Dielectrics and Capacitance
Capacitance of a Two-Wire Line
 For the case of a conductor with b << h, then:

ln  h  h 2  b 2

2 L
C
ln(2h b)
President University

b

ln  h  h  b 
ln  2h b 
(b  h)
Erwin Sitompul
EEM 10/13
Engineering Electromagnetics
Chapter 8
President University
Erwin Sitompul
EEM 10/14
Chapter 8
 At this point, we shall begin our study of the magnetic field with
a definition of the magnetic field itself and show how it arises
from a current distribution.
 The relation of the steady magnetic field to its source is more
complicated than is the relation of the electrostatic field to its
source.
 The source of the steady magnetic field may be a permanent
magnet, an electric field changing linearly with time, or a direct
current.
 Our present concern will be the magnetic field produced by a
differential dc element in the free space.
President University
Erwin Sitompul
EEM 10/15
Chapter 8
Biot-Savart Law
 Consider a differential current element as a vanishingly small
section of a current-carrying filamentary conductor.
 We assume a current I flowing in a differential vector length of
the filament dL.
 The law of Biot-Savart then states that
“At any point P the magnitude of the magnetic field intensity
produced by the differential element is proportional to the
product of the current, the magnitude of the differential length,
and the sine of the angle lying between the filament and a
line connecting the filament to the point P at which the field is
desired; also, the magnitude of the magnetic field intensity is
inversely proportional to the square of the distance from the
differential element to the point P.”
President University
Erwin Sitompul
EEM 10/16
Chapter 8
Biot-Savart Law
 The Biot-Savart law may be written concisely using vector
notation as
IdL  a R IdL  R
dH 

2
4 R
4 R3
 The units of the magnetic field intensity H are evidently
amperes per meter (A/m).
indicate the point to which each of
the quantities refers,
I1dL1  a R12
dH 2 
4 R122
President University
Erwin Sitompul
EEM 10/17
Chapter 8
Biot-Savart Law
 It is impossible to check experimentally the law of Biot-Savart
as expressed previously, because the differential current
element cannot be isolated.
 It follows that only the integral form of the Biot-Savart law can
be verified experimentally,
H

IdL  a R
4 R 2
President University
Erwin Sitompul
EEM 10/18
Chapter 8
Biot-Savart Law
 The Biot-Savart law may also be expressed in terms of
distributed sources, such as current density J (A/m2) and
surface current density K (A/m).
 Surface current flows in a sheet of vanishingly small thickness,
and the sheet’s current density J is therefore infinite.
 Surface current density K, however, is measured in amperes
per meter width. Thus, if the surface current density is uniform,
the total current I in any width b is
I  Kb
where the width b is measured
perpendicularly to the direction in
which the current is flowing.
President University
Erwin Sitompul
EEM 10/19
Chapter 8
Biot-Savart Law
 For a nonuniform surface current density, integration is
necessary:
I   KdN
where dN is a differential element of the path across which the
current is flowing.
 Thus, the differential current element I dL may be expressed in
terms of surface current density K or current density J,
IdL  KdS  Jdv
and alternate forms of the Biot-Savart law can be obtained as
K  a R dS
and
H
2
4 R
s
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J  a R dv
H 
2
4

R
vol
Erwin Sitompul
EEM 10/20
Chapter 8
Biot-Savart Law
 We may illustrate the application of the Biot-Savart law by
considering an infinitely long straight filament.
 Referring to the next figure, we should recognize the symmetry
of this field. As we moves along the filament, no variation of z or
 occur.
 The field point r is given by r = ρaρ,
and the source point r’ is given by
r’ = z’az. Therefore,
R12  r  r  a  zaz
a R12 
 a   z a z
 2  z 2
President University
Erwin Sitompul
EEM 10/21
Chapter 8
Biot-Savart Law
 We take dL = dz’az and the current is directed toward the
increasing values of z’. Thus we have
Idza z  ( a   za z )
H2  
2
2 32

4

(


z
)


I

4

 (

 dza
2
 z2 )3 2
• The resulting magnetic field intensity
is directed to aφ direction.
President University
Erwin Sitompul
EEM 10/22
Chapter 8
Biot-Savart Law
 Continuing the integration with respect to z’ only,
I a

dz
H2 
 (  2  z 2 ) 3 2
4 

H2 
I a

z
4  2 (  2  z2 )
I
2

a
• The magnitude of the field is not a
function of φ or z.
• It varies inversely with the distance
from the filament.
• The direction of the magnetic-fieldintensity vector is circumferential.
President University
Erwin Sitompul
EEM 10/23
Chapter 8
Biot-Savart Law
 The formula to calculate the magnetic field intensity caused by
a finite-length current element can be readily used:
H
I
4
(sin  2  sin 1 )a
• Try to derive this formula
President University
Erwin Sitompul
EEM 10/24
Chapter 8
Biot-Savart Law
 Example
Determine H at P2(0.4,0.3,0) in
the field of an 8 A filamentary
current directed inward from
infinity to the origin on the
positive x axis, and then outward
to infinity along the y axis.
1x  90, 2 x  53.1
1y  36.9, 2 y  90
12
12
8
H2 x 
(sin 53.1  sin(90))a  a  H 2 x   a z A m


4 (0.3)
8
8
8
H2 y 
(sin 90  sin(36.9))a  a  H 2 x   a z A m


4 (0.4)
• What if the line goes
20
H2  H2 x  H2 y   a z  6.37a z A m
onward to infinity

along the z axis?
President University
Erwin Sitompul
EEM 10/25
Chapter 8