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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
The Steady Magnetic Field
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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Chapter 8
The Steady Magnetic Field
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
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Chapter 8
The Steady Magnetic Field
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
The Steady Magnetic Field
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
conductor b, leads immediately to
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
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Chapter 8
The Steady Magnetic Field
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
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Chapter 8
The Steady Magnetic Field
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
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Chapter 8
The Steady Magnetic Field
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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Chapter 8
The Steady Magnetic Field
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
The Steady Magnetic Field
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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EEM 11/10
Chapter 8
The Steady Magnetic Field
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
H0
( z 0, z h)
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Chapter 8
The Steady Magnetic Field
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
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EEM 11/12
Chapter 8
The Steady Magnetic Field
Ampere’s Circuital Law
For an infinitely long
solenoid of radius a and
uniform current density
Ka aφ, the result is
H Ka a z
( a)
H0
( 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
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EEM 11/13
Chapter 8
The Steady Magnetic Field
Ampere’s Circuital Law
For a toroid with ideal case
0 a
inside
H Ka
a (toroid
)
H0
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For the N-turn toroid, we have
the good approximations
(outside
toroid)
Erwin Sitompul
NI
H
a
2
H0
(inside
toroid)
(outside
toroid)
EEM 11/14
Chapter 8
The Steady Magnetic Field
End of the Lecture
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