Transcript magnetostatic - UniMAP Portal

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
MAGNETIC MATERIALS
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
a
az
dL  R  0
0
dz  dza

0
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
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
36
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 2dL 2  B1
37
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
38
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
39