Transcript ECE 1100 Introduction to Electrical and Computer

ECE 3318
Applied Electricity and Magnetism
Spring 2016
Prof. David R. Jackson
ECE Dept.
Notes 28
1
Magnetic Field
S
NN r
N
v
q
S
N
Note: Flux lines come out of north poles!
Lorentz force Law:
F  qv B
This experimental
law defines B.
B is the magnetic flux density vector.
In general, (with both E and B present):
F  q  E  v  B
The units of B
are Webers/m2
or Tesla [T].
2
Magnetic Field (cont.)
F  q v  B
y
q>0







Beam of electrons moving in a circle, due to the
presence of a magnetic field. Purple light is emitted
along the electron path, due to the electrons colliding
with gas molecules in the bulb.
(From Wikipedia)
A stable orbit is a circular path.
mv02
 qv0 B0
R

F


R

q

v




x
ˆ 0
B   zB
v  ˆ v
0
F  ˆ  qv0 B0 
3
Magnetic Field (cont.)
The most general stable path is a helix,
with the helix axis aligned with the magnetic field.
B
F  qv B
There is no force in the axis direction (hence a constant velocity in this direction).
Magnetic field lines thus “guide” charged particles.
4
Magnetic Field (cont.)
The earth's magnetic field protects us from charged particles from the sun.
The particles spiral along the directions of the magnetic field,
and are thus directed towards the poles.
5
Magnetic Field (cont.)
This also explains the auroras seen near the north pole (aurora borealis)
and the south pole (aurora australis).
The particles from the sun that reach the earth are directed towards the poles.
6
Magnetic Gauss Law
z
S (closed surface)
 B  nˆ ds  0
S
N
y
N
x
S
B
S
Magnetic pole (not possible) !
Note: Magnetic flux lines come out of a
north pole and go into a south pole.
No net magnetic flux out!
7
Magnetic Gauss Law: Differential Form
 B  nˆ dS  0
S
From the definition of divergence we then have
1
  B  lim
V 0 V
 B  nˆ dS
S
Hence
 B  0
8
Ampere’s Law
z
I
Iron filings
y
x
Experimental law:
Note: A right-hand rule predicts the
direction of the magnetic field.
 I 
ˆ
B 
 0
 2 
0  4 107  H/m 
(This is an exact value:
please see next slide.)
9
Ampere’s Law (cont.)
Note: The definition of the Amp* is as follows:
7
1 [A] current produces: B  2  10
at   1
T 
m
*This is one form of the
definition. Another form
involves the force between
two infinite wires.
Hence
 I 
B  
0

 2 
so
0  4 107
1 A
2  10 T  
0
2 1 m
7
H/m
10
Ampere’s Law (cont.)
Define:
H
1
0
B
(This is the definition in free space.)
Hence
B  0 H
H is called the “magnetic field”
The units of H are [A/m].
 I 
ˆ
H 

 2 
 A/m
(for single infinite wire)
11
Ampere’s Law (cont.)
Wire inside path
 H  dr 
ˆ H   ˆ  d  ˆ d   zˆ dz 



C
C

y

 H   d
C
C
2
I
x
 I 
 
  d
2 
0 
I

2
A current (wire) is inside a closed path.
2
I
0 d  2  2 
I
We integrate counterclockwise.
12
Ampere’s Law (cont.)
Wire outside path
0
I
C H  dr  0 2 d
y
C
0
I
x
A current (wire) is outside a closed path.
Note: The angle 
smoothly goes from
zero back to zero as
we go around the path,
starting on the x axis.
13
Ampere’s Law (cont.)
Hence
 H  dr  I
encl
Ampere’s Law (DC currents)
C
Although the law was derived for an infinite
wire of current, the assumption is made
that it holds for any shape current.
Iencl
This is now an experimental law.
C
“Right-Hand Rule”
Right-hand rule for Ampere’s law: The fingers of the right hand are in the direction of the path
C, and the thumb gives the reference direction for the current that is enclosed by the path.
(The contour C goes counterclockwise if the reference direction for current is pointing up.)
Note: The same DC current Iencl goes through any surface that is attached to C.
14
Amperes’ Law: Differential Form
 H  dr  I
C
encl
C
    H   nˆ dS  I
S
encl
(from Stokes’s theorem)
S
n̂
    H   nˆ dS   J  nˆ dS
S
S is a small
planar surface.
n̂ is constant
S
   H   nˆ S  J  nˆ S
Let S  0
 H   nˆ  J  nˆ
Since the unit normal is arbitrary (it could be any of the three unit vectors), we have
 H  J
15
Maxwell’s Equations (Statics)
  D  v
Electric Gauss law
 B  0
Magnetic Gauss law
 E  0
Faraday’s law
 H  J
Ampere’s law
16
Maxwell’s Equations (Dynamics)
  D  v
Electric Gauss law
 B  0
Magnetic Gauss law
B
 E  
t
D
 H  J 
t
Faraday’s law
Ampere’s law
17
Displacement Current
D
 H  J 
t
Ampere’s law:
Capacitor being charged with DC current
I
A
h
--------------
dQ
dt
“Displacement current”
(This term was added by Maxwell.)
z
Insulator
+++++++++++
J
I
Q
A current density vector J exists
inside the lower plate.
The charge Q on the lower plate
increases with time.
Motivation:
 s
I
 1  dQ
J  zˆ    zˆ  
 zˆ
t
 A
 A  dt
  D  zˆ 
  Dz    zˆ Dz   D
 zˆ
 zˆ


t
t
t
t
18
Ampere’s Law: Finding H
 H  dr  I
encl
C
Rules:
1) The “Amperian path” C must be a closed path.
2) The sign of Iencl is from the right-hand rule.
3) Pick C in the direction of H (to the extent possible).
19
Example
Calculate H
z
Note : H  H   
I
First solve for H .
y

x
r
C
“Amperian path”
An infinite line current along the z axis
20
Example (cont.)
 H  dr  I encl
z
C
ˆ  d   I  I
H





encl

I
C
2
 H   d  I
0
H   2   I
y

r
C
“Amperian path”
H 
I
2
21
Example (cont.)
H d cancels
2) Hz = 0
Hz = 0 at 
I

 H  dr  I
encl
0
C
h
C
 =
H z    h   H z   h   0
 Hz     0
3) H = 0
I
Magnetic Gauss law:
 B  nˆ dS  0
h
S
B  2 h   0
S
22
Example (cont.)
z
I

 I 
ˆ
H 

2



 A/m
y
r
x
Note:
There is a “right-hand rule” for predicting the
direction of the magnetic field from a wire.
(The thumb is in the direction of the current, and
the fingers are in the direction of the field.)
23
Example
y
Coaxial cable
DC current
b
I
c
a
a

I
r
C
z
x
b
c
This inner wire is solid.
The outer shield (jacket) of the coax
has a thickness of t = c – b.
Note: The permittivity of
the material inside the
coax does not matter here.
 <a
b<<c
J z  J zA 
I
2


A/m
2 
a
J z  J zB 
I
2


A/m
2
2 
c b
Note: At DC, the current density is uniform, since the electric field is uniform
(due to the fact that the voltage drop is path independent).
24
Example (cont.)
H  ˆ H
 H  dr  I
The other components are zero, as in the wire example.
y
encl
C


ˆ  d  I
H


encl

C
2
  H  d  I encl
b
a

C
r
x
0
2 H  I encl
I encl
H 
2
c
This formula holds for any radius,
as long as we get Iencl correct.
25
Example (cont.)
y
<a
I encl
a < < b
b < < c
 I  2
 J     2  
a 
A
z
2
b
I encl  I
I encl  I  J
a
B
z

2
b
2

I
2
2
I 2



b

2 
c b
>c

r
x
C
c
I encl  I    I   0
Note: There is no magnetic field outside of the coax
(a perfect “shielding property”).
26
Example (cont.)
y
Summary
<a
a < < b
2


I

ˆ
H 
 2  [A/m]
2  a 
b
a

x
I
ˆ
H 
[A/m]
2
c
b < < c
>c
2
2


I



b
ˆ
H 
[A/m]
1  2
2 
2   c   b 
H  0 [A/m]
27
Example
Solenoid
Calculate H
n = # turns/meter
  0  r
a
z
Ideal solenoid: n 
I
Find Hz
<a
 H  dr  I
C
C
H d cancels
h
encl
H z h  I encl  I  nh 
H z  nI

Hz is zero at 
28
Example (cont.)
>a
H z h  I encl
z
Hz
C
H d cancels
I encl  0
h
Hzh  0

Hz = 0 at 
so
Hz  0
Hence
H z   nI 
 A/m ,
 a
 0,   a
29
Example (cont.)
The other components of the magnetic field are zero:
1) H = 0 since
I encl  0
2) H = 0 from
 B  nˆ dS  0
H 2  I encl
C
S
B 2 h  0
S
30
Example (cont.)
Summary
n = # turns/meter
z
Note: A right-hand rule can be used to
determine the direction of the magnetic
field (the same rule as for a straight wire).
  0  r
a
I
H  zˆ  nI 
 A/m ,
 a
 0,   a
B  0 r H ,   a
Note:
A larger relative permeability will
give a larger magnetic flux density.
This results is a stronger magnet,
or a larger inductance.
31
Example
z
J s  zˆ J sz
Calculate H
A/m
y
- side
x
x
Infinite sheet of current
+ side
Top view
y
32
Example (cont.)
Hxx
Hx+
H  xˆ H x  y 
y
Hz  0
(superposition with line currents)
Hy  0
(magnetic Gauss Law)
Also, by symmetry:
H y   y   H y   y 
H x   y   H x   y 
2H y   y  xz  0
33
Example (cont.)
w
- side
x
C
H  xˆ H x  y 
H

x
 y
+ side
y
 H  dr  Iencl
Note: There is no contribution from the left and right edges
(the edges are perpendicular to the field).
C

 w/2
H  dr 
front

back

H x dx   H x w
w /2
w/2
H  dr 

 w/2
H x dx  H x w
34
Example (cont.)
 H x w  H x w  J sz w
 H x  H x  J sz
1
H   J sz
2

x
 J
H  xˆ   sz
 2
 J
H  xˆ   sz
 2

 [A/m], y  0


 [A/m], y  0

Note:
We can use a right hand-rule to quickly
determine the direction of the magnetic field:
Put your thumb is in the direction of the
current, and your fingers will give the overall
direction of the magnetic field.
Note: The magnetic field does not depend on y.
35
Example
Parallel-plate transmission line
Find H everywhere
y
I
I
h
x
w
z
Assume
w >> h
J
bot
s
top
(We neglect fringing here.)
Js
I
ˆ
 z    A/m 
 w
 I 
 zˆ    A/m 
 w
36
Example (cont.)
y
h
x
Two parallel sheets (plates)
of opposite surface current
J
bot
sz
J sztop
I
    A/m 
 w
 I 
    A/m 
 w
37
Example (cont.)
y
“bot”
“top”
h
x
Magnetic field due to
top plate
Magnetic field due to
bottom plate
H H
bot
H
top
38
Example (cont.)
We then have
   J szbot  
2   xˆ 
, 0  y  h

H     2 

0, otherwise

Recall that
J
Hence
bot
sz
I
 
 w
 A/m
I
H   xˆ   [A/m], 0  y  h
 w
H  0, otherwise
39
Example (cont.)
We could also apply Ampere’s law directly:
y
x
H 0
C
Note:
It is convenient to put one
side (top) of the Amperian
path in the region where
the magnetic field is zero.
H  xˆ H x
h
x
 H  dr  I
C
H 0
encl
 I
H x x  J sztop x     x
 w
H x  I / w
Hence
I
ˆ
H   x   [A/m], 0  y  h
 w
40
Low Frequency Calculations
At low frequency, the DC formulas should be accurate,
as long as we account for the time variation in the results.
Example (line current in time domain)
z
i  t   I 0 cos t   
  0  c / f
i(t)

x
r
y
 i t  
ˆ
H t    

2



 A/m
41
Low Frequency Calculations (cont.)
We can also used the DC formulas in the phasor domain.
Example (line current in the phasor domain)
z
i  t   I 0 cos t   
  0  c / f
i(t)

x
r
I  I 0e j (phasor current )
y
 I 
ˆ
H 

2



 A/m
(phasor H field)
42
Low Frequency Calculations (cont.)
Example (cont.)
Converting from phasor domain to the time domain:
 I 
ˆ
H 

2



H  t   Re  He
jt
(phasor H field)
 ˆ  I  jt  ˆ  1 
  Re    2  e     2  Re  Ie jt 



 

 1 
j
jt
 ˆ 
Re
I
e
e


0

 2 


 1 
ˆ
H t    
I 0 cos t    [A/m]

 2 
43