Download magnetostatic (cont`d)

3.2 AMPERE’S CIRCUITAL LAW
In magnetostatic problems with sufficient
symmetry, we can employ Ampere’s Circuital
Law more easily that the law of Biot-Savart.
The law says that the integration of H
around any closed path is equal to the net
current enclosed by that path. i.e.
 H  dL  Ienc
1
AMPERE’S CIRCUITAL LAW (Cont’d)
• The line integral of H around the path is termed
the circulation of H.
• To solve for H in given symmetrical current
distribution, it is important to make a careful
selection of an Amperian Path (analogous to
gaussian
surface)
that
is
everywhere
either
tangential or normal to H.
• The direction of the circulation is chosen such
that the right hand rule is satisfied.
2
DERIVATION 4
Find
the
magnetic
field
intensity
everywhere resulting from an infinite
length line of current situated on the
z-axis using Ampere’s Circuital Law.
3
DERIVATION 4 (Cont’d)
Select the best Amperian
path,
Figure 3-15 (p. 113)
Two possible Amperian paths
around an infinite length line
of current.
where here are two possible
Amperian paths around an
infinite length line of current.
Choose path b which has a
constant
value
of
Hφ
around the circle specified
by the radius ρ
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
4
DERIVATION 4 (Cont’d)
Using Ampere’s circuital law:
 H  dL  Ienc
We could find:
H  H a
dL  da
So,
2
 H  dL  I enc   H a  da  I
 0
5
DERIVATION 4 (Cont’d)
Solving for Hφ:
H 
I
2
Where we find that the field resulting from an
infinite length line of current is the expected
result:
H
I
2
a
Same as applying
Biot-Savart’s Law!
6
DERIVATION 5
Use Ampere’s Circuital Law to find the
magnetic field intensity resulting from an
infinite extent sheet of current with current
sheet
K  K xa x
in the x-y plane.
7
DERIVATION 5 (Cont’d)
Rectangular amperian path of height Δh and width
Figure 3-16 (p. 113)
Calculating
H resulting from a current
K = K a hand
in the x–y plane.
Δw.
According
to sheet
right
rule, perform the
x x
circulation in order of a  b  c  d  a
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
8
DERIVATION 5 (Cont’d)
We have:
b
c
d
a
a
b
c
d
 H  dL  I enc   H  dL   H  dL   H  dL   H  dL
From symmetry argument, there’s only Hy
component exists. So, Hz will be zero and thus the
expression reduces to:
b
d
a
c
 H  dL  I enc   H  dL   H  dL
9
DERIVATION 5 (Cont’d)
So, we have:
b
d
a
0
c
 H  dL   H  dL   H  dL

w
 H y  a y   dya y   H ya y  dya y
w
0
 2 H y w
10
DERIVATION 5 (Cont’d)
The current enclosed by the path,
I   KdS 
w
This will give:
 H  dL  Ienc
2H y w  K x w
Kx
Hy 
2
 K x dy  K x w
0
Or generally,
1
H  K  aN
2
11
EXAMPLE 3
An infinite sheet of current with K  6a A
z
m
exists
on the x-z plane at y = 0. Find H at P (3,2,5).
12
SOLUTION TO EXAMPLE 3
Use previous expression, that is:
1
H  K  aN
2
is a normal vector from the sheet to the test
point P (3,4,5), where:
aN
aN  a y
So,
and
K  6a z
1
H  6a z  a y  3a x A
m
2
13
EXAMPLE 4
Consider the infinite length
cylindrical conductor
carrying a radially
dependent current J  J 0 a z
Find H everywhere.
14
SOLUTION TO EXAMPLE 4
What components of H will be present?
Finding the field at
some point P, the
field has both a 
and a components.
(a)
15
SOLUTION TO EXAMPLE 4 (Cont’d)
The field from the
second line current
results in a
cancellation of the a 
components
(b)
16
SOLUTION TO EXAMPLE 4 (Cont’d)
To calculate H everywhere, two amperian paths
are required:
Path #1 is for
a
Path #2 is for
 a
17
SOLUTION TO EXAMPLE 4 (Cont’d)
Evaluating the left side of Ampere’s law:
2
 H  dL   H a  da
 2H
0
This is true for both amperian path.
The current enclosed for the path #1:
I   J  dS   J 0 a z  dda z

2
3
2

J

0
 J 0    2 d d  
3
 0  0
18
SOLUTION TO EXAMPLE 4 (Cont’d)
Solving to get Hφ:
J0 2
H 
3
J0 2
H
a for   a
3
Or
The current enclosed for the path #2:
2
3
2

J
a
2
0
I   J  dS  J 0    dd 
3
 0  0
a
Solving to get Hφ:
J 0a3
H
a for
3
 a
19
EXAMPLE 5
Find H everywhere
for coaxial cable as
shown.
20
(a)
SOLUTION TO EXAMPLE 5
Even current
distributions are
assumed in the
inner and outer
conductor.
Consider four
amperian paths.
(a)
(b)
21
SOLUTION TO EXAMPLE 5 (Cont’d)
It will be four amperian paths:




a
a  b
b c
c
Therefore, the magnetic field intensity, H will
be determined for each amperian paths.
22
SOLUTION TO EXAMPLE 5 (Cont’d)
As previous example, only Hφ component is
present, and we have the left side of ampere’s
circuital law:
2
 H  dL   H a  da
 2H
0
 For the path #1:
Ienc   J  dS
23
SOLUTION TO EXAMPLE 5 (Cont’d)
We need to find current density, J for inner
conductor because the problem assumes an event
current distribution (ρ<a is a solid volume where
current distributed uniformly).
Where,
I
J
az
dS
dS  dd , S 
2

a
2

d

d



a

 0  0
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SOLUTION TO EXAMPLE 5 (Cont’d)
So, J  I a  I a
z
2 z
dS
a
We therefore have:
2

I
I enc   J  dS   
a  dda z
2 z
  0   0 a
I 2
 2
a
25
SOLUTION TO EXAMPLE 5 (Cont’d)
Equating both sides to get:
I 2
I
H  2

a 2 2a 2
for
a
 For the path #2:
The current enclosed is just I, I enc  I
Therefore: H  dL  2H  I
I

H 

I
2
for
enc
a  b
26
SOLUTION TO EXAMPLE 5 (Cont’d)
 For the path #3:
For total current enclosed by path 3, we need to
find the current density, J in the outer
conductor because the problem assumes an
event current distribution (a<ρ<b is a solid
volume where current distributed uniformly)
given by:
I
I
 a z   2 2  a z 
J
dS
 c b


27
SOLUTION TO EXAMPLE 5 (Cont’d)
We therefore have (for AP#3):
2

I
 J  dS     c 2  b2  a z   dda z
  0  b
 I


 2  b2
c2  b2
But, the total current enclosed is:
I enc  I   J  dS
2
2
2
2



b
c


I
 I    I 2
2 
2
2
c

b
c

b


28
SOLUTION TO EXAMPLE 5 (Cont’d)
So we can solve for path #3:
 H  dL  2H
 I enc
c2   2
I 2
c  b2
I  c 2   2  for b    c
H 
2  c 2  b 2 
 For the path #4, the total current is zero. So,
H  0 for   c This shows the shielding
ability by coaxial cable!!
29
SOLUTION TO EXAMPLE 5 (Cont’d)
Summarize the results to have:
I

a
2 

2a

I

a
2
H
 I  c2   2 

a

 c2  b2 
2





0
a
a  b
b c
 c
30