### ρ - Erwin Sitompul

```Engineering Electromagnetics
Lecture 11
Dr.-Ing. Erwin Sitompul
President University
http://zitompul.wordpress.com
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Erwin Sitompul
EEM 11/1
Chapter 8
Ampere’s Circuital Law
 In solving electrostatic problems, whenever a high degree of
symmetry is present, we found that they could be solved much
more easily by using Gauss’s law compared to Coulomb’s law.
 Again, an analogous procedure exists in magnetic field.
 Here, the law that helps solving problems more easily is known
as Ampere’s circuital law.
 The derivation of this law will waits until several subsection
ahead. For the present we accept Ampere’s circuital law as
another law capable of experimental proof.
 Ampere’s circuital law states that the line integral of magnetic
field intensity H about any closed path is exactly equal to the
direct current enclosed by that path,
 H  dL  I
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Erwin Sitompul
EEM 11/2
Chapter 8
Ampere’s Circuital Law
• The line integral of H about the closed
path a and b is equal to I
• The integral around path c is less than I.
 The application of Ampere’s circuital law involves finding the
total current enclosed by a closed path.
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Erwin Sitompul
EEM 11/3
Chapter 8
Ampere’s Circuital Law
 Let us again find the magnetic field
intensity produced by an infinite long
filament carrying a current I. The
filament lies on the z axis in free
space, flowing to az direction.
 We choose a convenient path to any
section of which H is either
perpendicular or tangential and
along which the magnitude H is
constant.
 The path must be a circle of radius ρ, and Ampere’s circuital
law can be written as
2
 H  dL   H  d
0
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2
 H   d  H 2  I
 H 
0
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EEM 11/4
I
2
Chapter 8
Ampere’s Circuital Law
 As a second example, consider an infinitely long coaxial
transmission line, carrying a uniformly distributed total current I
in the center conductor and –I in the outer conductor.
 A circular path of radius ρ, where ρ is larger than the radius of
the inner conductor a but less than the inner radius of the outer
I
H 
( a    b)
2
 If ρ < a, the current enclosed is
I encl  2 H  I
2
a2
 Resulting

H  I
(   a)
2
2 a
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Erwin Sitompul
EEM 11/5
Chapter 8
Ampere’s Circuital Law
 If the radius ρ is larger than the outer radius of the outer
conductor, no current is enclosed and
H  0
(  c)
 Finally, if the path lies within the outer conductor, we have
  2  b2 
2 H  I  I  2 2 
 c b 
I c2   2
H 
2 c 2  b2
(b    c)
• ρ components cancel,
z component is zero.
• Only φ component of H
does exist.
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Erwin Sitompul
EEM 11/6
Chapter 8
Ampere’s Circuital Law
 The magnetic-field-strength variation with radius is shown
below for a coaxial cable in which b = 3a, c = 4a.
 It should be noted that the magnetic field intensity H is
continuous at all the conductor boundaries  The value of Hφ
does not show sudden jumps.
 Outside the coaxial cable, a complete cancellation of magnetic
field occurs. Such coaxial cable would not produce any
noticeable effect to the surroundings (“shielding”)
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Erwin Sitompul
EEM 11/7
Chapter 8
Ampere’s Circuital Law
 As final example, consider a sheet of current flowing in the
positive y direction and located in the z = 0 plane, with uniform
surface current density K = Kyay.
 Due to symmetry, H cannot vary with x and y.
 If the sheet is subdivided into a number of filaments, it is
evident that no filament can produce an Hy component.
 Moreover, the Biot-Savart law shows that the contributions to
Hz produced by a symmetrically located pair of filaments cancel
each other.  Hz is zero also.
 Thus, only Hx component is present.
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Erwin Sitompul
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Chapter 8
Ampere’s Circuital Law
 We therefore choose the path 1-1’-2’-2-1 composed of straightline segments which are either parallel or perpendicular to Hx
and enclose the current sheet.
 Ampere's circuital law gives
H x1L  H x 2 (L)  K y L
 H x1  H x 2  K y
 If we choose a new path 3-3’-2’-2’3, the same current is
enclosed, giving
H x3  H x 2  K y
and therefore
H x3  H x1
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Erwin Sitompul
EEM 11/9
Chapter 8
Ampere’s Circuital Law
 Because of the symmetry, then, the magnetic field intensity on
one side of the current sheet is the negative of that on the other
side.
 Above the sheet
H x  12 K y ( z  0)
while below it
H x   12 K y ( z  0)
 Letting aN be a unit vector normal (outward) to the current
sheet, this result may be written in a form correct for all z as
H  12 K  aN
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Erwin Sitompul
EEM 11/10
Chapter 8
Ampere’s Circuital Law
 If a second sheet of current flowing in the opposite direction,
K = –Kyay, is placed at z = h, then the field in the region
between the current sheets is
H  K  aN
(0  z  h)
and is zero elsewhere
H0
( z  0, z  h)
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Erwin Sitompul
EEM 11/11
Chapter 8
Ampere’s Circuital Law
 The difficult part of the application of Ampere’s circuital law is
the determination of the components of the field which are
present.
 The surest method is the logical application of the Biot-Savart
law and a knowledge of the magnetic fields of simple form (line,
sheet of current, “volume of current”).
• Solenoid
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• Toroid
Erwin Sitompul
EEM 11/12
Chapter 8
Ampere’s Circuital Law
 For an infinitely long
uniform current density
Ka aφ, the result is
H  Ka a z
(   a)
H0
(   a)
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 If the solenoid has a finite
length d and consists of N
closely wound turns of a
filament that carries a
current I, then
NI
well within
H
a z (the
solenoid)
d
Erwin Sitompul
EEM 11/13
Chapter 8
Ampere’s Circuital Law
 For a toroid with ideal case
0  a
inside
H  Ka
a (toroid
)

H0
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 For the N-turn toroid, we have
the good approximations
(outside
toroid)
Erwin Sitompul
NI
H
a
2
H0
(inside
toroid)
(outside
toroid)
EEM 11/14
Chapter 8