ECE 2317
Applied Electricity and Magnetism
Spring 2014
Prof. David R. Jackson
ECE Dept.
Notes 27
1
Magnetic Field
NN r
N
S
Lorentz force Law:
v
q
S
F  qv B
N
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
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)










F
R

q

v




ˆ 0
B   zB
v  ˆ v
0
mv02
 qv0 B0
R
F  ˆ  qv0 B0 
3
Magnetic Field (cont.)
The most general path is a helix.
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
B
S
Magnetic pole (not possible) !
Note: Magnetic flux lines come out of a
north pole and go into a south pole.
S
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:
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
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
Hence
1
0
B
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.)
 H  dr 
C

y
ˆ H   ˆ  d  ˆ d   zˆ dz 




C
 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.)
y
0
C
I
x
A current (wire) is outside a closed path.
I
C H  dr  0 2 d
0
Note that the angle 
smoothly goes from
zero back to zero as
we go about 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
C
This is now an experimental law.
“Right-Hand Rule”
(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
encl
C
C
    H   nˆ dS  I
S
encl
(from Stokes’s theorem)
S
nˆ
    H   nˆ dS   J  nˆ dS
S
S is a small
planar surface.
S
   H   nˆ S  J  nˆ S
nˆ is constant
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 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
h
C
 =
encl
0
C
Hz    h  Hz   h  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.
(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
C
z

r
x
b
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
c
I
J z  J zA  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
The other components are zero,
as in the wire example.
 H  dr  I
y
encl
b
C


ˆ  d  I
H


encl

C
a

r
C
x
2
  H d  I encl
0
c
2 H  Iencl
I encl
H 
2
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
Iencl  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
Iencl  I   I   0
Note: There is no magnetic field outside of the coax
(a perfect “shielding property”).
26
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

27
Example (cont.)
>a
H z h  I encl
z
I encl  0
Hz
C
H d cancels
Hzh  0
h

so
Hz = 0 at 
Hz  0
Hence
H z   nI 
 A/m  ,
a
 0,   a
28
Example (cont.)
The other components of the magnetic field are zero:
1) H = 0 since
Iencl  0
2) H = 0 from
 B  nˆ dS  0
H 2  Iencl
C
S
B 2 h  0
S
29
Example (cont.)
Summary
Note: A right-hand rule can be used to
determine the direction of the magnetic
field (same as for a straight wire).
n = # turns/meter
z
  0  r
a
I
H  zˆ  nI 
 A/m ,
a
 0,   a
Note:
A larger relative permeability
will give a larger magnetic
flux density. This results is a
stronger magnet, or a larger
inductance.
B  0 r H ,   a
30
Example
z
J s  zˆ J sz
Calculate H
A/m
y
- side
x
x
Infinite sheet of current
+ side
Top view
y
31
Example (cont.)
Hxx
Hx+
H  xˆ H x  y 
y
Hz  0
(superposition with line currents)
Hy  0
(magnetic Gauss Law)
H y   y   H y   y 
By symmetry:
H x   y   H x   y 
32
Example (cont.)
w
- side
x
C
H  xˆ H x  y 
H

x
 y
+ side
y
 H  dr  I encl
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 

H x dx  H x w
 w /2
33
Example (cont.)
 H x w  H x w  J sz w
 H x  H x  J sz
1
H   J sz
2

x
 J sz 
ˆ
H  x
 [A/m], y  0
 2 
 J sz 
ˆ
H  x
 [A/m], y  0
 2 
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.
34
Example
Parallel-plate transmission line
Find H everywhere
y
I
I
h
x
w
z
Assume
w >> h
(We neglect fringing here.)
J
bot
s
J
top
s
I
 zˆ    A/m
 w
 I 
ˆ
 z    A/m
 w
35
Example (cont.)
y
h
x
Two parallel sheets of
opposite surface current
J
bot
sz
J sztop
I
    A/m
 w
 I 
    A/m
 w
36
Example (cont.)
y
“bot”
“top”
h
x
Magnetic field due to
top sheet
Magnetic field due to
bottom sheet
H H
bot
H
top
37
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
38
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   I
 w
H x  I / w
Hence
I

H   xˆ   [A/m], 0  y  h
 w
39