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CHAPTER 3
MAGNETOSTATICS
MAGNETOSTATICS
3.1
BIOT-SAVART’S LAW
3.2
AMPERE’S CIRCUITAL LAW
3.3
MAGNETIC FLUX DENSITY
3.4
MAGNETIC FORCES
3.5
BOUNDARY CONDITIONS
2
INTRODUCTION
Magnetism
and
electricity
were
considered distinct phenomena until
1820 when Hans Christian Oersted
introduced an experiment that showed a
compass
needle
deflecting
when
in
proximity to current carrying wire.
3
INTRODUCTION (Cont’d)
He used compass to show that current produces
magnetic fields that loop around the conductor.
The field grows weaker as it moves away from
the source of current.
A
 represents current
coming out of paper.
A  represents current
heading into the paper.
Figure 3-7 (p. 102)
Oersted’s experiment with a compass placed in several positions in close proximity
to a current-carrying wire. The inset shows  used to represent the cross section
for current coming out of the paper: this represents the head of an arrow. A 
4
INTRODUCTION (Cont’d)
The principle of magnetism is widely used in
many applications:

Magnetic memory

Motors and generators

Microphones and speakers

Magnetically levitated high-speed
vehicle.
5
INTRODUCTION (Cont’d)
Magnetic fields can be easily visualized by
sprinkling iron filings on a piece of paper
suspended over a bar magnet.
(a)
(b)
6
INTRODUCTION (Cont’d)
The field lines are in terms of the magnetic field
intensity, H in units of amps per meter.
This is analogous to the volts per meter units
for electric field intensity, E.
Magnetic field will be introduced in a manner
paralleling our treatment to electric fields.
7
3.1 BIOT-SAVART’S LAW
Jean Baptiste Biot and Felix Savart arrived a
mathematical relation between the field and
current.
dH 
I1dL1  a12
4 R12
2
Figure 3-8 (p. 103)
Illustration of the law of Biot–Savart showing magnetic field arising from a
8
BIOT-SAVART’S LAW (Cont’d)
To get the total field resulting from a
current, sum the contributions from each
segment by integrating:
IdL  a R
H
2
4R
9
BIOT-SAVART’S LAW (Cont’d)
Due to continuous current distributions:
Line current
Surface current
Volume current
10
BIOT-SAVART’S LAW (Cont’d)
In terms of distributed current sources, the
Biot-Savart’s Law becomes:
IdL  a R
H
4R 2
KdS  a R
H
4R 2
JdV  a R
H
4R 2
Line current
Surface current
Volume current
11
DERIVATION
Let’s apply
IdL  a R
H
2
4R
to determine the magnetic field, H everywhere
due to straight current carrying filamentary
conductor of a finite length AB .
12
DERIVATION (Cont’d)
DERIVATION (Cont’d)
We assume that the conductor is along the zaxis with its upper and lower ends respectively
subtending angles
1
and
2
at point P where
H is to be determined.
The field will be independent of z and φ and
only dependant on ρ.
14
DERIVATION (Cont’d)
The term dL is simply
dza z
and the vector
from the source to the test point P is:

R  Ra R   za z  a 
Where the magnitude is:
R  z2   2
And the unit vector:
aR 
 za z  a 
z2   2
15
DERIVATION (Cont’d)
Combining these terms to have:
IdL  a R
IdL  R
H

2
3
4R
4R
B Idza   za  a 
z
z


2
2 32
A
4 z  


16
DERIVATION (Cont’d)


Cross product of dza z   za z  a  :
a

dL  R  0

a
0
0
az
dz  dza
z
This yields to:
B
H
A 4
z
Idz
2


2 32
a
17
DERIVATION (Cont’d)
Trigonometry from figure,
tan  

So, z   cot 
z
Differentiate to get:
I
H  
4
2

1

dz    cos ec 2d
 2 cos ec 2d
2
  cot 
2
2

32
a
18
DERIVATION (Cont’d)
Remember!



u
2
cot(u )    cos ec (u )
x
x
2
2
1  cot (u )  cos ec (u )
19
DERIVATION (Cont’d)
Simplify the equation to become:
I
H
4

I
2

 2 cos ec 2d
1
 3 cos ec 3
a
2
sin d a

4
1

I
4
cos  2  cos 1 a
20
DERIVATION 1
Therefore,
H
I
4
cos 2  cos1 a
This expression generally applicable for any
straight filamentary conductor of finite length.
21
DERIVATION 2
As a special case, when the conductor is semifinite
with respect to P,
A at 0,0,0
B at 0,0,   or 0,0, 
The angle become:
So that,
H
I
4
1  900 , 2  00
a
22
DERIVATION 3
Another special case, when the conductor is
infinite with respect to P,
A at 0,0, 
B at 0,0,  
The angle become:
So that,
H
I
2
1  180 0 , 2  00
a
23
HOW TO FIND UNIT VECTOR aaφ ?
From previous example, the vector H is in
direction of aφ, where it needs to be determine
by simple approach:
a  al  a 
Where,
al
unit vector along the line current
a
unit vector perpendicular from the
line current to the field point
24
EXAMPLE 1
The conducting triangular loop carries of 10A.
Find H at (0,0,5) due to side 1 of the loop.
25
SOLUTION TO EXAMPLE 1
• Side 1 lies on the x-y
plane
and
treated as a
straight conductor.
• Join the point of interest
(0,0,5) to the beginning and
end of the line current.
26
SOLUTION TO EXAMPLE 1 (Cont’d)
This will show how H 
I
4
cos 2  cos1 a
is
applied for any straight, thin, current carrying
conductor.
1  900  cos 1  0
2
and   5
cos 2 
29
From figure, we know that
and from trigonometry
27
SOLUTION TO EXAMPLE 1 (Cont’d)
To determine a by simple approach:
al  a x
and
a  az
so that,
a  al  a   a x  a z  a y
H 
I
4
cos 2  cos1 a


10  2


 0   a y  59.1 a y m A m

4 5  29

28
EXAMPLE 2
A ring of current with radius a lying in the x-y
plane with a current I in the  a direction. Find
an expression for the field at arbitrary point a
height h on z axis.
29
SOLUTION TO EXAMPLE 2
Can we use H 
I
4
cos 2  cos1 a
?
Solve for each term in
the Biot-Savart’s Law
30
SOLUTION TO EXAMPLE 2 (Cont’d)
We could find:
dL  ada

R  Ra R  ha z  aa 
R  h a
2
a R 
2
ha z  aa 
h a
2
2
31
SOLUTION TO EXAMPLE 2 (Cont’d)
It leads to:
IdL  a R
IdL  R
H

2
3
4R
4R
2 Iada  ha  aa 

z

 
2
2 32
 0
4 h  a


The differential current element will give a field
with:
a  from a  a z
az
from
a   a  
32
a)
SOLUTION TO EXAMPLE 2 (Cont’d)
(b) the problem:
However, consider the symmetry of
The radial components
cancel but the a z
components adds, so:
3-10a (p. 105)
nt to find H a height h 2above a ring
2
a z (b) The
entered in the x – Ia
y plane.
H
d
3
2
values are shown for
use
in
the
2
2
 0
4

h

a
t equation. (c) The radial
s of H cancel by symmetry.



Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
33
SOLUTION TO EXAMPLE 2 (Cont’d)
This can be easily solved to get:
H

Ia 2
2 h2  a 2

a
32 z
At h=0 where at the center of the loop, this
equation reduces to:
I
H  az
2a
34
BIOT-SAVART’S LAW (Cont’d)
• For many problems involving surface current
densities and volume current densities, solving for
the magnetic field using Biot-Savart’s Law can be
quite cumbersome and require numerical
integration.
• There will be sufficient symmetry to be able to
solve for the fields using Ampere’s Circuital Law.
35
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
36
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.
37
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.
38
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.
39
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
40
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!
41
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.
42
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.
43
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
44
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
45
DERIVATION 5 (Cont’d)
The current enclosed by the path,
I   KdS 
This will give:
w
 K x dy  K x w
0
 H  dL  Ienc
2 H y w  K x w
Kx
Hy 
2
Or generally,
1
H  K  aN
2
46
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).
47
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
48
EXAMPLE 4
Consider the infinite length
cylindrical conductor
carrying a radially
dependent current J  J 0 a z
Find H everywhere.
49
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)
50
SOLUTION TO EXAMPLE 4 (Cont’d)
The field from the
second line current
results in a
cancellation of the a 
components
(b)
51
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
52
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
2J 0 
 J 0    dd 
3
 0  0
2
3
53
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
54
EXAMPLE 5
Find H everywhere
for coaxial cable as
shown.
55
(a)
SOLUTION TO EXAMPLE 5
Even current
distributions are
assumed in the
inner and outer
conductor.
Consider four
amperian paths.
(a)
(b)
56
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.
57
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
58
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
59
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
60
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
61
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


62
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
c 2  b2
But, the total current enclosed is:
I enc  I   J  dS
2
2
2
2



b
c


 I    I 2 2   I 2 2
c b 
c b

63
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  b2 
 For the path #4, the total current is zero. So,
H  0 for   c This shows the shielding
ability by coaxial cable!!
64
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
65
AMPERE’S CIRCUITAL LAW (Cont’d)
Expression for curl by applying Ampere’s
Circuital Law might be too lengthy to derive, but
it can be described as:
H  J
The expression is also called the point form of
Ampere’s Circuital Law, since it occurs at
some particular point.
66
AMPERE’S CIRCUITAL LAW (Cont’d)
The Ampere’s Circuital Law can be rewritten in
terms of a current density, as:
 H  dL   J  dS
Use the point form of Ampere’s Circuital Law to
replace J, yielding:
 H  dL     H  dS
This is known as Stoke’s Theorem.
67
3.3 MAGNETIC FLUX DENSITY
In electrostatics, it is convenient to think in terms
of electric flux intensity and electric flux density.
So too in magnetostatics, where magnetic flux
density, B is related to magnetic field intensity by:
B  H
  0  r
Where μ is the permeability with:
0  4  10 7 H m
68
MAGNETIC FLUX DENSITY (Cont’d)
The amount of magnetic flux, φ in webers
from
magnetic
field
passing
through
a
surface is found in a manner analogous to
finding electric flux:
   B  dS
69
MAGNETIC FLUX DENSITY (Cont’d)
Fundamental features of magnetic fields:
• The field lines form a
closed loops. It’s different
from electric field lines,
where it starts on positive
charge and terminates on
negative charge
Figure 3-26 (p. 125)
Magnetic field lines form closed loops, sot he netflux through a Gaussian surface is
zero.
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
70
MAGNETIC FLUX DENSITY (Cont’d)
• The magnet cannot be
divided in two parts, but it
results in two magnets.
The magnetic pole cannot
be isolated.
Figure 3-27 (p. 125)
Dividing a magnet in two parts results in two magnets. You cannot isolate a
magnetic pole.
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
71
MAGNETIC FLUX DENSITY (Cont’d)
The net magnetic flux passing through a
gaussian surface must be zero, to get Gauss’s
Law for magnetic fields:
 B  dS  0
By applying divergence theorem, the point form
of Gauss’s Law for static magnetic fields:
B  0
72
EXAMPLE 6
Find the flux crossing the portion of the
plane φ=π/4 defined by 0.01m < r < 0.05m
and 0 < z < 2m in free space. A current
filament of 2.5A is along the z axis in the az
direction.
Try to sketch this!
73
SOLUTION TO EXAMPLE 6
The relation between B and H is:
B  0 H  0
I
2
a
To find flux crossing the portion, we need to use:
   B  dS
where dS is in the aφ direction.
74
SOLUTION TO EXAMPLE 6 (Cont’d)
So,
dS  ddza
Therefore,
   B  dS
0 I
  
a  ddza
z  0   0.01 2
2
0.05
20 I 0.05
6

ln
 1.61  10 Wb
2
0.01
75
3.4 MAGNETIC FORCES
Upon application of a magnetic field, the wire is
deflected in a direction normal to both the field and the
direction of current.
(a)
(b)
76
MAGNETIC FORCES (Cont’d)
The force is actually acting on the individual
charges moving in the conductor, given by:
Fm  qu  B
By the definition of electric field intensity, the
electric force Fe acting on a charge q within an
electric field is:
Fe  qE
77
MAGNETIC FORCES (Cont’d)
A total force on a charge is given by Lorentz force
equation:
F  qE  u  B 
The force is related to acceleration by the
equation from introductory physics,
F  ma
78
MAGNETIC FORCES (Cont’d)
To find a force on a current element, consider a
line conducting current in the presence of
magnetic field with differential segment dQ of
charge moving with velocity u:
dF  dQu  B
But,
dL
u
dt
79
MAGNETIC FORCES (Cont’d)
So,
dQ
dF 
dL  B
dt
Since dQ
the line,
dt corresponds to the current I in
 dF  IdL  B
We can find the force from a collection of
current elements
F12   I 2dL2  B1
80
MAGNETIC FORCES (Cont’d)
Consider a line of current in +az direction on the z
axis. For current element a,
IdL a  Idz aa z
But, the field cannot exert magnetic force
on the element producing it. From field of
second element b, the cross product will
be zero since IdL and aR in same
direction.
Figure 3-28 (p. 128)
81
(a) Differential current elements on a line. (b) A pair
EXAMPLE 7
If there is a field from a
second line of current
parallel to the first, what
will be the total force?
Figure 3-28 (p. 128)
(a) Differential current elements on a line. (b) A pair of current-carrying lines will
exert magnetic force on each other.
82
SOLUTION TO EXAMPLE 7
The force from the magnetic field of line 1 acting
on a differential section of line 2 is:
dF12  I 2 dL 2  B1
Where,
0 I1
B1 
a
2
By inspection from figure,
  y, a  a x
Why?!?!
83
SOLUTION TO EXAMPLE 7
Consider
dL 2  dza z
, then:
0 I1
0 I1I 2
 a x  
dF12  I 2dza z 
dz a y 
2y
2y
0
0 I1I 2

F12 
 a y  dz
2y
L
0 I1I 2 L
 F12 
ay
2y
84
MAGNETIC FORCES (Cont’d)
Generally,
0
dL 2  dL1  a12 
F12 
I 2 I1  
4
R12 2
• Ampere’s law of force between a pair of current-
carrying circuits.
• General case is applicable for two lines that are not
parallel, or not straight.
• It is easier to find magnetic field B1 by Biot-Savart’s
law, then use
F12   I 2dL2  B1 to find F12
.
85
EXAMPLE 8
The magnetic flux density in a region of free space
is given by B = −3x ax + 5y ay − 2z az T. Find the
total force on the rectangular loop shown which
lies in the plane z = 0 and is bounded by x = 1, x =
3, y = 2, and y = 5, all dimensions in cm.
Try to sketch this!
86
SOLUTION TO EXAMPLE 8
The figure is as shown.
87
SOLUTION TO EXAMPLE 8 (Cont’d)
First, note that in the plane z = 0, the z component
of the given field is zero, so will not contribute to the
force. We use:
F  loop IdL x B
Which in our case becomes with,
I  30 A and B  3xa x  5 ya y  2 za z
88
SOLUTION TO EXAMPLE 8 (Cont’d)
So,
F
0.03



30 dxa x   3 xa x  5 y y  0.02 a y 
0.01
0.05
 30dya y   3x x 0.03 a x  5 ya y  
0.02
0.01



30 dxa x x  3 xa x  5 y y  0.05 a y 
0.03
0.02
 30dya y x  3x x 0.01 a x  5 ya y 
0.05
89
SOLUTION TO EXAMPLE 8 (Cont’d)
Simplifying these becomes:
F
0.03
0.05
0.01
0.01
0.02
0.02
0.03
0.05
 30(5)(0.02)a z dx    30(3)(0.03) a z dy

 30(5)(0.05)a z dx    30(3)(0.01) a z dy
 0.06  0.081  0.150  0.027  a z N
 F  36a z mN
90
3.5 BOUNDARY CONDITIONS
We could see how the fields behave at the
boundary between a pair of magnetic materials
which derived using Ampere’s Circuital Law and
Gauss’s Law for magnetostatic fields:
 H  dL  Ienc
 B  dS  0
91
BOUNDARY CONDITIONS (Cont’d)
Consider,
Figure 3-38 (p. 142)
Boundary between a pair of magnetic media, and the placement of a rectangular
path for performing the circulation of H.
92
BOUNDARY CONDITIONS (Cont’d)
A pair of magnetic media separated by a sheet
current density K. Choose a rectangular Amperian
path of width Δw and height Δh centered at the
interface. The current enclosed by the path is:
I enc   KdW  Kw
93
BOUNDARY CONDITIONS (Cont’d)
The sheet current is heading into the page and
use the right hand rule to determine the direction
of integration around the loop. So,
b
c
d
a
a
b
c
d
 H  dL         (H  dL)  Kw
94
BOUNDARY CONDITIONS (Cont’d)
For first and second integral,
a b
b
w
 H  dL   HT aT  dLaT
1
a
 HT w
1
0
bc
c
h 2
0
 H  dL   H N a N  dLa N   H N a N  dLa N
b
1
h / 2

2
0
  HN  HN
1
2

h
2
95
BOUNDARY CONDITIONS (Cont’d)
For third and fourth integral,
cd
d
0
 H  dL   HT aT  dLaT
c
2
w
  HT w
2
d a
a
h 2
0
 H  dL   H N a N  dLa N   H N a N  dLa N
d
h 2

2
 HN  HN
1
1
0
2

h
2
96
BOUNDARY CONDITIONS (Cont’d)
Combining the result, we get the first boundary
condition for magnetostatic field,
HT  HT  K
1
2
In more general case,
a 21  H1  H 2   K
Where a21 is unit vector normal from medium 2
to medium 1
97
BOUNDARY CONDITIONS (Cont’d)
Second boundary condition can be determined by
applying Gauss’s Law over a small pillbox shaped
Gaussian surface,
Figure 3-39 (p. 143)
98
BOUNDARY CONDITIONS (Cont’d)
The Gauss’s Law,
 B  dS  0
Where,
 B  dS   B  dS   B  dS   B  dS
top
bottom
side
The pillbox is short enough, so the flux out of
the side is negligible.
99
BOUNDARY CONDITIONS (Cont’d)
We have
 B  dS   BN a N  dSa N   BN a N  dS  a N 

1

2
 BN  BN S  0
1
2
Since ΔS can be chosen unequal to zero, it
follows that:
BN  BN
1
2
100
EXAMPLE 9
The magnetic field intensity is given as:
H1  6a x  2a y  3a x A m
In a medium with µr1=6000 that exist for z
< 0. Find H2 in a medium with µr2=3000 for
z>0.
101
SOLUTION TO EXAMPLE 9
102
BOUNDARY CONDITIONS (Cont’d)
Recall that, for a conductor-dielectric interface:
ET  0
DN   S
Generally, it is not exist for magnetostatic fields.
If one of the media is superconductor, where the
magnetic field rapidly attenuates away from the
surface, such that:
B0
103
BOUNDARY CONDITIONS (Cont’d)
If medium 2 is superconductor, the equations for
magnetostatic fields become:
a N  H1  K
BN  0
1
2
The second expression is logical since the magnetic field
lines must form closed loops and cannot suddenly
terminate even on a superconductor.
104
CHAPTER 3
END
PRACTICAL APPLICATION
 Loudspeakers
 Maglev (Magnetically Levitated
Trains)
106
LOUDSPEAKERS
• Paper or plastic cone
affixed to a voice coil
(electromagnet) suspended
in a magnetic field.
136)
•AC Signals to the voice
coil  moves back and
forth, resulting vibration of
the cone and producing
sound waves of the same
frequency as the AC signal
ving-coil loudspeaker.
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
107
MAGLEV
igure 3-53 (p. 159)
108
MAGLEV (Cont’d)
amentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
• Interaction between
electromagnets in the train and
the current carrying coils in the
guide rail provide levitation.
• By sending waves along the
guide rail coils, the train magnet
pushed/pulled in the direction of
travel. The train is guided by
magnet on the side of guide rail.
• Computer algorithms maintain
the separation distance.
109
SUMMARY (1)
•For a differential current element I1dL1 at point 1,
the magnetic field intensity H at point 2 is given by
the law of Biot-Savart,
dH 
I1dL1  a12
4 R12
2
Where R12  R12 a12 is a vector from the source
element at point 1 to the location where the field is
desired at point 2. By summing all the current
elements, it can rewritten as:
IdL  a
H
R
4R 2
110
SUMMARY (2)
•The Biot-Savart law can be written in terms of
surface and volume current densities:
KdS  a R
H
4R 2
Surface current
Jdv  a R
H
4R 2
Volume current
•The magnetic field intensity resulting from an
infinite length line of current is:
H
I
2
a
111
SUMMARY (3)
and from a current sheet of extent it is:
Where aN is a unit vector normal from
1
H  K  a N the current sheet to the test point.
2
•An easy way to solve the magnetic field intensity
in problems with sufficient current distribution
symmetry is to use Ampere’s Circuital Law,
which says that the circulation of H is equal to the
net current enclosed by the circulation path
 H  dL  Ienc
112
SUMMARY (4)
• The point or differential form of Ampere’s circuital
Law is:
H  J
• A closed line integral is related to surface integral by
Stoke’s Theorem:
 H  dL     H  dS
• Magnetic flux density, B in Wb/m2 or T, is related
to the magnetic field intensity by
B  H
113
SUMMARY (5)
Material permeability µ can be written as:
and the free space permeability is:
  0  r
0  4  10 7 H m
• The amount of magnetic flux Φ in webers through
a surface is:
   B  dS
Since magnetic flux forms closed loops, we have
Gauss’s Law for static magnetic fields:
 B  dS  0
114
SUMMARY (6)
• The total force vector F acting on a charge q moving
through magnetic and electric fields with velocity u is
given by Lorentz Force equation:
F  qE  u  B 
The force F12 from a magnetic field B1 on a current
carrying line I2 is:
F12   I 2dL2  B1
115
SUMMARY (7)
• The magnetic fields at the boundary between
different materials are given by:
a 21  H1  H 2   K
Where a21 is unit vector normal from medium 2
to medium 1, and:
BN  BN
1
2
116
VERY IMPORTANT!
From electrostatics and magnetostatics, we can
now present all four of Maxwell’s equation for
static fields:
 D  dS  Qenc
 B  dS  0
 E  dL  0
 H  dL  I enc
  D  v
Integral Form
Differential Form
B  0
E  0
H  J
117