Post on 05-Jan-2016
CHAPTER 3
MAGNETOSTATICS
2
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
3
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.
4
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.
INTRODUCTION (Cont’d)
5
The principle of magnetism is widely used
in many applications:
Magnetic memory
Motors and generators
Microphones and speakers
Magnetically levitated high-speed
vehicle.
INTRODUCTION (Cont’d)
6
INTRODUCTION (Cont’d)
Magnetic fields can be easily visualized by sprinkling iron filings on a piece of paper suspended over a bar magnet.
7
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.
INTRODUCTION (Cont’d)
8
3.1 BIOT-SAVART’S LAW
Jean Baptiste Biot and Felix Savart arrived a
mathematical relation between the field and
current.
212
1211
4 R
aLH
dI
d
9
BIOT-SAVART’S LAW (Cont’d)
To get the total field resulting from a
current, sum the contributions from each
segment by integrating:
24 R
Id R
aL
H
10
BIOT-SAVART’S LAW (Cont’d)
Due to continuous current distributions:
Line current Surface current Volume current
11
In terms of distributed current sources, the
Biot-Savart’s Law becomes:
24 R
Id R
aL
H
24 R
dS R
aK
H
24 R
dV R
aJ
H
Line current
Surface current
Volume current
BIOT-SAVART’S LAW (Cont’d)
12
DERIVATION
Let’s apply
to determine the magnetic field, H
everywhere due to straight current carrying
filamentary conductor of a finite length AB .
24 R
Id R
aL
H
DERIVATION (Cont’d)
14
We assume that the conductor is along the z-
axis with its upper and lower ends
respectively subtending angles and at
point P where H is to be determined.
1 2
The field will be independent of z and φ and
only dependant on ρ.
DERIVATION (Cont’d)
15
The term dL is simply and the vector
from the source to the test point P is:
zdza
aaaR
zR zR
Where the magnitude is:
22 zR
And the unit vector:22
z
z zR
aaa
DERIVATION (Cont’d)
16
Combining these terms to have:
B
A
zz
R
z
zIdzR
Id
R
Id
2322
32
4
44
aaa
RLaLH
DERIVATION (Cont’d)
17
aaa zz zdzCross product of :
a
aaa
RL dz
z
dzdz
0
00
This yields to:
a H
B
A z
dzI23224
DERIVATION (Cont’d)
18
Trigonometry from figure,
z
tan So, cotz
Differentiate to get: decdz 2cos
a H
2
1
23222
22
cot
cos
4
decI
DERIVATION (Cont’d)
19
DERIVATION (Cont’d)
Remember!
)(cos)(cot1
)(cos)cot(
22
2
uecu
x
uuecu
x
20
Simplify the equation to become:
a
a
a H
12
33
22
coscos4
sin4
cos
cos
4
2
1
2
1
I
dI
ec
decI
DERIVATION (Cont’d)
21
Therefore,
aH 12 coscos4
I
This expression generally applicable for any
straight filamentary conductor of finite
length.
DERIVATION 1
22
As a special case, when the conductor is semifinite with respect to P,
,0,0,0,0
0,0,0
or at
at
B
A
The angle become: 02
01 0,90
So that,
aH4
I
DERIVATION 2
23
Another special case, when the conductor is infinite with respect to P,
,0,0
,0,0
at
at
B
A
The angle become: 02
01 0,180
So that,
aH2
I
DERIVATION 3
24
HOW TO FIND UNIT VECTOR aφ ?
From previous example, the vector H is in
direction of aφ, where it needs to be determine
by simple approach:
a
aaa l
Where,la unit vector along the line current
a unit vector perpendicular from the line current to the field point
25
EXAMPLE 1
The conducting triangular loop carries of 10A.
Find H at (0,0,5) due to side 1 of the loop.
26
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.
27
SOLUTION TO EXAMPLE 1 (Cont’d)
This will show how is
applied for any straight, thin, current carrying
conductor.
aH 12 coscos4
I
From figure, we know that 0cos90 10
1
and from trigonometry and 29
2cos 2 5
28
SOLUTION TO EXAMPLE 1 (Cont’d)
To determine by simple approach:a
xl aa zaa and so that,
yzxl aaaaaa
mAm
I
yy a a
aH
1.59029
2
54
10
coscos4 12
29
EXAMPLE 2
A ring of current with radius a lying in the x-y
plane with a current I in the direction. Find
an expression for the field at arbitrary point a
height h on z axis.
a
30
Can we use ?
aH 12 coscos4
I
SOLUTION TO EXAMPLE 2
Solve for each term in the Biot-Savart’s Law
31
SOLUTION TO EXAMPLE 2 (Cont’d)
We could find:
aL add
aaaR ahR zR
22 ah R
22 ah
ah zR
aa
a
32
It leads to:
2
02322
32
4
44
ah
ahIadR
Id
R
Id
z
R
aaa
RLaLH
SOLUTION TO EXAMPLE 2 (Cont’d)
The differential current element will give a field with:
a from
fromzazaa
aa
33
However, consider the symmetry of the problem:
SOLUTION TO EXAMPLE 2 (Cont’d)
The radial components cancel but the components adds, so:
za
2
02322
2
4d
ah
Ia zaH
34
This can be easily solved to get:
zah
IaaH
2322
2
2
At h=0 where at the center of the loop, this equation reduces to:
za
IaH
2
SOLUTION TO EXAMPLE 2 (Cont’d)
35
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.
36
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.
encId LH
37
• 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.
AMPERE’S CIRCUITAL LAW (Cont’d)
38
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.
39
DERIVATION 4 (Cont’d)
Select the best Amperian
path, 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
40
DERIVATION 4 (Cont’d)
Using Ampere’s circuital law:
encId LH
We could find:
aL
aH
dd
H
So,
IdHId enc
2
0
aaLH
41
Solving for Hφ:
2
IH
Where we find that the field resulting from an infinite length line of current is the expected result:
aH
2
I Same as applying
Biot-Savart’s Law!
DERIVATION 4 (Cont’d)
42
Use Ampere’s Circuital Law to find the
magnetic field intensity resulting from an
infinite extent sheet of current with current
sheet in the x-y plane.
DERIVATION 5
xxK aK
43
DERIVATION 5 (Cont’d)
Rectangular amperian path of height Δh and
width Δw. According to right hand rule, perform
the circulation in order of a b c d a
44
We have:
a
d
d
c
c
b
b
aenc ddddId LHLHLHLHLH
DERIVATION 5 (Cont’d)
From symmetry argument, there’s only Hy
component exists. So, Hz will be zero and thus
the expression reduces to:
d
c
b
aenc ddId LHLHLH
45
So, we have:
wH
dyHdyH
ddd
y
w
yyyw
yyy
d
c
b
a
20
0
aaaa
LHLHLH
DERIVATION 5 (Cont’d)
46
The current enclosed by the path,
wKdyKKdSI x
w
x
0
DERIVATION 5 (Cont’d)
This will give:
encId LH
wKwH xy 2
2x
yK
H
Or generally,
NaKH 2
1
47
EXAMPLE 3
An infinite sheet of current with exists
on the x-z plane at y = 0. Find H at P (3,2,5). m
Az aK 6
48
SOLUTION TO EXAMPLE 3
Use previous expression, that is:
NaKH 2
1
is a normal vector from the sheet to the test point P (3,4,5), where:Na
yN aa and zaK 6
So,
mA
xyz aaaH 362
1
49
Consider the infinite
length cylindrical
conductor carrying a
radially dependent current
Find H
everywhere.
zJ aJ 0
EXAMPLE 4
50
What components of H will be present?
Finding the field at
some point P, the
field has both
and
components.
a
a
SOLUTION TO EXAMPLE 4
51
The field from the
second line current
results in a
cancellation of the
components
a
SOLUTION TO EXAMPLE 4 (Cont’d)
52
To calculate H everywhere, two amperian paths are required:
Path #1 is for
Path #2 is for
a
a
SOLUTION TO EXAMPLE 4 (Cont’d)
53
Evaluating the left side of Ampere’s law:
HdHd 22
0
aaLH
This is true for both amperian path.
The current enclosed for the path #1:
3
2 30
0
2
0
20
0
JddJ
ddJdI zz
aaSJ
SOLUTION TO EXAMPLE 4 (Cont’d)
54
Solving to get Hφ:
3
20
J
H Or
aH3
20J for a
The current enclosed for the path #2:
3
2 30
0
2
0
20
aJddJdI
a
SJ
Solving to get Hφ:
aH3
30aJ
for a
SOLUTION TO EXAMPLE 4 (Cont’d)
55
EXAMPLE 5
Find H
everywhere for
coaxial cable as
shown.
56
Even current
distributions are
assumed in the
inner and outer
conductor.
Consider four
amperian paths.
SOLUTION TO EXAMPLE 5
57
It will be four amperian paths:
Therefore, the magnetic field intensity, H will be determined for each amperian paths.
aba cb
c
SOLUTION TO EXAMPLE 5 (Cont’d)
58
As previous example, only Hφ component is
present, and we have the left side of ampere’s circuital law:
HdHd 22
0
aaLH
For the path #1:
SJ dIenc
SOLUTION TO EXAMPLE 5 (Cont’d)
59
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).
zdS
IaJ
Where,
22
0 0
, addSdddSa
SOLUTION TO EXAMPLE 5 (Cont’d)
60
So,zz
a
I
dS
IaaJ
2
We therefore have:
2
2
2
0 02
a
I
dda
IdI zzenc
aaSJ
SOLUTION TO EXAMPLE 5 (Cont’d)
61
Equating both sides to get:
2
2
2 22 a
I
a
IH
for a
For the path #2:
The current enclosed is just I,
Therefore:
IIenc
IIHd enc 2LH
2
IH for ba
SOLUTION TO EXAMPLE 5 (Cont’d)
62
For the path #3:
SOLUTION TO EXAMPLE 5 (Cont’d)
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:
zzbc
I
dS
IaaJ
22
63
We therefore have (for AP#3):
22
22
2
022
bc
bI
ddbc
Id
bzz
aaSJ
But, the total current enclosed is:
22
22
22
22
bc
cI
bc
bII
dIIenc
SJ
SOLUTION TO EXAMPLE 5 (Cont’d)
64
SOLUTION TO EXAMPLE 5 (Cont’d)
So we can solve for path #3:
22
22
2bc
cIIHd enc
LH
22
22
2 bc
cIH
for cb
For the path #4, the total current is zero. So,
0H for c This shows the shielding ability by coaxial cable!!
65
SOLUTION TO EXAMPLE 5 (Cont’d)
Summarize the results to have:
c
cbbc
cI
baI
aa
I
02
2
2
22
22
2
a
a
a
H
66
Expression for curl by applying Ampere’s Circuital Law might be too lengthy to derive, but it can be described as:
JH The expression is also called the point form of Ampere’s Circuital Law, since it occurs at some particular point.
AMPERE’S CIRCUITAL LAW (Cont’d)
67
The Ampere’s Circuital Law can be rewritten in terms of a current density, as:
SJLH dd
Use the point form of Ampere’s Circuital Law to replace J, yielding:
SHLH dd
This is known as Stoke’s Theorem.
AMPERE’S CIRCUITAL LAW (Cont’d)
68
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:r 0 HB
Where μ is the permeability with:
mH 70 104
69
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:
SB d
70
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
MAGNETIC FLUX DENSITY (Cont’d)
71
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.
72
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:
0 SB d
By applying divergence theorem, the point
form of Gauss’s Law for static magnetic fields:
0 B
73
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!
74
SOLUTION TO EXAMPLE 6
The relation between B and H is:
aHB200I
To find flux crossing the portion, we need to use:
SB d
where dS is in the aφ direction.
75
So, aS dzdd Therefore,
WbI
dzdI
d
z
aa
SB
60
2
0
05.0
01.0
0
1061.101.0
05.0ln
2
2
2
SOLUTION TO EXAMPLE 6 (Cont’d)
76
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.
77
MAGNETIC FORCES (Cont’d)
The force is actually acting on the individual charges moving in the conductor, given by:
BuF qm
By the definition of electric field intensity, the
electric force Fe acting on a charge q within an
electric field is:
EF qe
78
A total force on a charge is given by Lorentz force equation:
BuEF q
MAGNETIC FORCES (Cont’d)
The force is related to acceleration by the equation from introductory physics,
aF m
79
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:
BuF dQd
dt
dLu
But,
80
BLF Idd
So,BLF d
dt
dQd
Since corresponds to the current I in the line,
dtdQ
MAGNETIC FORCES (Cont’d)
We can find the force from a collection of current elements
12212 BLF dI
81
Consider a line of current in +az direction on the z
axis. For current element a,
zaa IdzId aL
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.
MAGNETIC FORCES (Cont’d)
82
EXAMPLE 7
If there is a field from a
second line of current
parallel to the first, what
will be the total force?
83
The force from the magnetic field of line 1
acting on a differential section of line 2 is:
12212 BLF dId
Where,
aB
210
1I
By inspection from figure,
xy aa , Why?!?!
SOLUTION TO EXAMPLE 7
84
0210
12
21010212
2
22
Ly
yxz
dzy
II
dzy
II
y
IdzId
aF
aaaF
zdzd aL 2Consider , then:
yy
LIIaF
2
21012
SOLUTION TO EXAMPLE 7
85
Generally,
212
121212
012 4 R
ddII
aLLF
• 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 to find F12 . 12212 BLF dI
MAGNETIC FORCES (Cont’d)
86
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!
87
The figure is as shown.
SOLUTION TO EXAMPLE 8
88
SOLUTION TO EXAMPLE 8 (Cont’d)
B L F xIdloop
AI 30
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:
Which in our case becomes with,
zyx zyx aaaB 253 and
89
02.0
05.001.0
01.0
03.005.0
05.0
02.003.0
03.0
01.002.0
5330
5330
5330
5330
yxxy
yyxx
yxxy
yyxx
yxxdy
yxxdx
yxdy
yxdx
aa a
aa a
aa a
aa aFSo,
SOLUTION TO EXAMPLE 8 (Cont’d)
90
Simplifying these becomes:
N
dydx
dydx
z
zz
zz
a
aa
aaF
027.0150.0081.006.0
)01.0)(3(30)05.0)(5(30
)03.0)(3(30)02.0)(5(30
02.0
05.0
01.0
03.0
05.0
02.0
03.0
01.0
mNz aF 36
SOLUTION TO EXAMPLE 8 (Cont’d)
91
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:
0 SB d
encId LH
92
BOUNDARY CONDITIONS (Cont’d)
Consider,
93
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:
wKKdWIenc
BOUNDARY CONDITIONS (Cont’d)
94
b
a
c
b
d
c
a
d
wKdd )( LHLH
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,
95
221
21
11
2
0
0
2/
0
hHH
dLHdLHd
cb
wHdLHd
ba
NN
h
NNNNNh
N
c
b
b
a
w
TTTT
aaaaLH
aaLH
For first and second integral,
BOUNDARY CONDITIONS (Cont’d)
96
221
12
22
2
0
0
2
0
hHH
dLHdLHd
ad
wHdLHd
dc
NN
h
NNNh
NNN
a
d
wTTTT
d
c
aaaaLH
aaLH
BOUNDARY CONDITIONS (Cont’d)
For third and fourth integral,
97
KHH TT 21
Combining the result, we get the first boundary condition for magnetostatic field,
KHH 2121a
In more general case,
BOUNDARY CONDITIONS (Cont’d)
Where a21 is unit vector normal from medium 2 to medium 1
98
BOUNDARY CONDITIONS (Cont’d)
Second boundary condition can be determined by applying Gauss’s Law over a small pillbox shaped Gaussian surface,
99
The Gauss’s Law,
Where,
sidebottomtop
dddd SBSBSBSB
The pillbox is short enough, so the flux out of
the side is negligible.
0 SB d
BOUNDARY CONDITIONS (Cont’d)
100
We have
0
21
21
SBB
dSBdSBd
NN
NNNNNN
aaaaSB
Since ΔS can be chosen unequal to zero, it follows that:
21 NN BB
BOUNDARY CONDITIONS (Cont’d)
101
EXAMPLE 9
The magnetic field intensity is given as:
In a medium with µr1=6000 that exist for z
< 0. Find H2 in a medium with µr2=3000
for z>0.
mAxyx aaaH 3261
102
SOLUTION TO EXAMPLE 9
103
Recall that, for a conductor-dielectric interface:
0TE SND 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:
0B
BOUNDARY CONDITIONS (Cont’d)
104
If medium 2 is superconductor, the equations
for magnetostatic fields become:
0NB
KHa 1N 1 2
The second expression is logical since the magnetic field lines must form closed loops and cannot suddenly terminate even on a superconductor.
BOUNDARY CONDITIONS (Cont’d)
CHAPTER 3
END
106
PRACTICAL APPLICATION
Loudspeakers
Maglev (Magnetically
Levitated
Trains)
107
• Paper or plastic cone affixed to a voice coil (electromagnet) suspended in a magnetic field.
•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
LOUDSPEAKERS
108
MAGLEV
109
MAGLEV (Cont’d)
• 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.
110
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,
Where 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:
121212 aR R
212
1211
4 R
aLH
dI
d
24 R
Id R
aL
H
111
•The Biot-Savart law can be written in terms of surface and volume current densities:
SUMMARY (2)
24 R
dS R
aK
H
24 R
dv R
aJ
H
Surface current
Volume current
•The magnetic field intensity resulting from an infinite length line of current is:
aH
2
I
112
SUMMARY (3)
and from a current sheet of extent it is:
NaKH 2
1 Where aN is a unit vector normal from
the current sheet to the test point.
•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
encId LH
113
SUMMARY (4)
• The point or differential form of Ampere’s circuital Law is:
JH
SHLH dd
• A closed line integral is related to surface integral by Stoke’s Theorem:
• Magnetic flux density, B in Wb/m2 or T, is related to the magnetic field intensity by
HB
114
r 0
SUMMARY (5)
Material permeability µ can be written as:and the free space permeability is:
mH 70 104
• The amount of magnetic flux Φ in webers through a surface is:
SB d
Since magnetic flux forms closed loops, we have Gauss’s Law for static magnetic fields:
0 SB d
115
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:
BuEF q
The force F12 from a magnetic field B1 on a current
carrying line I2 is:
12212 BLF dI
116
SUMMARY (7)
• The magnetic fields at the boundary between different materials are given by:
KHH 2121a
Where a21 is unit vector normal from medium 2
to medium 1, and:
21 NN BB
117
VERY IMPORTANT!
From electrostatics and magnetostatics, we can now present all four of Maxwell’s equation for static fields:
enc
enc
Id
d
d
Qd
LH
LE
SB
SD
0
0
Integral Form
JH
E
B
D
0
0v
Differential Form