Transcript Lecture 3: Electrostatic Fields

1
Lecture 3: Static Fields
Instructor:
Dr. Gleb V. Tcheslavski
Contact:
[email protected]
Office Hours:
Room 2030
Class web site:
www.ee.lamar.edu/gleb/
em/Index.htm
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1. Electrostatic Fields
1.1. Coulomb’s Law
Something known from the ancient time (here comes amber): two charged
particles exert a force on each other…
Electrostatic
(Coulomb’s)
force:
F
Q1Q2
u [ N ]
2 R
4 0 R
where Q1 and Q2 are charges,
R –distance between particles,
uR – the unit-vector
 0  8.854 1012 
1
109 F / mthe permittivityof  freespace
36
In this notation, negative force means attraction, positive – repelling.
ELEN 3371 Electromagnetics
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(3.2.1)
3
1. Electrostatic Fields
1.1. Coulomb’s Law (Example)
Find the magnitude of the Coulomb force that exists between an electron
and a proton in a hydrogen atom. Compare the Coulomb force and the
gravitational force between the two particles. The two particles are
separated approximately by 1 Ångström 1Å 10-10 m.
FC 
Q1Q2

4 0 R 2
FG  G
me mp
R
Ratio :
2
1.602 1019 
2
 1

4 
109  10
 36


10 2
 2.3 108  N
9.1110 1836  9.1110   1.02 10
10 
31
 6.67 1011
(3.3.1)
31
10 2
47
N
(3.3.2)
FC
 2.27 1039 times
FG
This is why chemical bounds are so strong!
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1. Electrostatic Fields
1.2. Electric (electrostatic) Field
Electrostatic field due to the
charge Q:
F
Q
N
V




E 
u


R
 C   m 
q 4 0 R 2
An “alternative definition”:
Fq  q  Fq
dF
E
 lim
q 0
dq
q
(3.4.2)
What’s wrong with it?
For a system of two charges:
ELEN 3371 Electromagnetics
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(3.4.1)
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1. Electrostatic Fields
1.3. Superposition
For several charges placed at different locations in space, the total electric field
at the particular location would be a superposition (vector summation) of
individual electric fields:
N
Etot   En
(3.5.1)
a vector sum!
n 1
(Example): find the EF at P
Q1 = +1C, Q1 = +2C, Q3 = -3C
Etot , P  EQ1 , P  EQ1 , P  EQ1 , P
Q3
Q1
Q2

u 
u 
u
2 R1
2 R2
2 R3
4 0 R1
4 0 R2
4 0 R3

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1(3u x  4u y )
4 0 53
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2u y
3u x


2
4 0 4 4 0 32
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1. Electrostatic Fields
1.3. Superposition (cont)
Volume charge density:
v 
Q
[C / m3 ]
v
(3.6.1)
Surface charge density:
s 
Q
[C / m2 ]
s
(3.6.2)
Linear charge density:
l 
Q
[C / m]
l
(3.6.3)
if  vi  0,numberof volumes  
1
4 0
v
R
2
uR dv '
(3.6.4)
v
There is a differential electric field directed radially from each differential charges
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1. Electrostatic Fields
1.3. Superposition (Example)
Calculate the electric field from a
finite charge uniformly distributed
along a finite line.
Linear charge density: (z’)l
Theunitvector :u R 
 z 'u z   u 
z '2   2
We assume a symmetry along z with respect to the observation point. Therefore, it
will be a charge element at –z’ for every charge element at +z’. As a result, fields
in z direction will cancel each other:

Ezi  0duetosymmetry
i
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(3.7.1)
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1. Electrostatic Fields
1.3. Superposition (Example, cont)
The radial component:
dE  dE cos   dE

R
 dE

 2  z '2
(3.8.1)
Combining (3.4.1) and (3.6.3), we arrive to:
l dz '
dE 
4 0 (  2  z '2 )
(3.8.2)
which, combined with (3.8.1) and integrated leads to:
E 
1
4 0
l 
a

a

2

2 3
 z'
dz ' 
l
a
2 0   2  a 2
l
2a  E 
2 0 
ELEN 3371 Electromagnetics
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(3.8.3)
(3.8.4)
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1. Electrostatic Fields
1.3. Superposition (Example 2)
Calculate the electric field
from an infinite plane charged
with s and consisting of an
infinite number of parallel
charged lines.
Symmetry leads to cancellation
of tangent components.
R  x 2  y 2 . The linear charge density:
Utilize (3.8.4) and that
l  s dx
(3.9.1)
Ey 




 dEcos  dx   2
s
0
x '2  y 2

s 
s
s
y
1  x 
dx 
dx

tan

 
 x2  y 2
2 0 
2 0
x2  y 2
 y   2 0
y
(3.9.2)
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1. Electrostatic Fields
1.4. Gauss’s Law
A charge Q is uniformly distributed within a
sphere of radius a.
We can assume first that the charge is
located at the center. Than, by (3.4.1):
E
Q
4 0 a
2
ur
(3.10.1)
By evaluating surface integrals of both sides
 E ds 
Q
 4 a
2
ur ds
(3.10.2)
0
At the surface of the sphere, the unit-vector associated with the differential surface
area ds points in the radial direction. Therefore, ur ur  1 and the closed
surface integral is 4 a 2
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1. Electrostatic Fields
1.4. Gauss’s Law (cont)
Therefore, the
integral form:
 E ds 
Qencl
(3.11.1)
0
Here Qencl is the charge enclosed within the closed surface.
By using divergence theorem and volume charge density concept:
 E ds    Edv 
s
Differential form:
ELEN 3371 Electromagnetics
  dv
v
v
v
v
 E
0
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0
(3.11.2)
(3.11.3)
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1. Electrostatic Fields
1.4. Gauss’s Law (cont 2)
For the charge Q uniformly distributed within the spherical
volume
v  4 a3 / 3
v 
The volume charge density:
The total charge
enclosed:
  dv
v
v
0

Q
Q

v  4 a3 


 3 
Qencl
(3.12.2)
0
a) Outside the sphere: r > a, Qencl = Q
Er 
ELEN 3371 Electromagnetics
(3.12.1)
Gauss Law
  dv
v

Q
4 0 r
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2
v
0

Q
0

2
E
ds

4

r
Er

(3.12.3)
s
(3.12.3)
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1. Electrostatic Fields
1.4. Gauss’s Law (cont 3)
a) Inside the sphere: r < a
v
Qenc

0

  dv
v
0
 v dv
v
0
Q 4 r 3 Q  r 

  
3
4

a
0
3
0  a 
3
1
3
Gauss Law
3
Qr
   
0  a 
Qr
Er 
4 0 a 3
ELEN 3371 Electromagnetics
2
E
ds

4

r
Er (3.13.2)

s
(3.13.3)
Fall 2008
(3.13.1)
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1. Electrostatic Fields
1.5. Gaussian Surface
A Gaussian surface is a closed two-dimensional surface through which a flux or
electric field is calculated. The surface is used in conjunction with Gauss's law (a
consequence of the divergence theorem), allowing to calculate the total enclosed
electric charge by exploiting a symmetry while performing a surface integral.
Commonly used are:
a)
•
•
•
b)
•
•
Spherical surface for
A point charge;
A uniformly distributed spherical shell of charge;
Other charge distribution with a spherical symmetry
Cylindrical surface for
A long, straight wire with a uniformly distributed charge;
Any long, straight cylinder or cylindrical shell with uniform charge distribution.
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1. Electrostatic Fields
1.6. Potential Energy and Electric Potential
A charged particle will gain a certain amount of potential energy as the
particle is moved against an electric field.
b
b
We   F dl  Q  E dl[ J ]
a
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a
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(3.15.1)
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1. Electrostatic Fields
1.6. Potential Energy and Electric Potential (cont)
Imaginary experiment: compute a total work required to bring three charged
particles from - to the shaded region. No electric field exists at - and there
are no friction, no gravity, and no other forces.
I
There are no forces here, therefore, no work is required! W1 = 0;
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1. Electrostatic Fields
1.6. Potential Energy and Electric Potential (cont 2)
II
xb
xa
We need to overcome the Coulomb’s force, therefore, some work is required.
xb
Q1Q2
Q1Q2
dx

 Q2V1
2
4 0 ( x  xa )
4 0 x  xa

W2   
(3.17.1)
since both charges are positive
V1 is an absolute electric potential caused by the charge Q1
xb
Q1
Q1
dx

2
4

(
x

x
)
4 0 x  xa
0
a

V1   
ELEN 3371 Electromagnetics
Fall 2008
(3.17.2)
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1. Electrostatic Fields
1.6. Potential Energy and Electric Potential (cont 3)
III
We need again to overcome the Coulomb’s force, therefore, some work is required.
xc
x
c
Q1Q3
Q2Q3
Q1Q3
Q2Q3
W3   
dx

dx


 Q3V1  Q3V2
2
2

4 0 ( x  xa )
4 0 ( x  xb )
4 0 xc  xa 4 0 xc  xb


(3.18.1)
Totally, for the three particles: Wtot  W1  W2  W3  0  Q2V12  Q3 (V13  V23 )
(3.18.2)
Qi Q j
1 N N
1 N
  QV
Or, for N particles: Wtot   
i i
2 i 1 j 1, j i 4 0 xij 2 i 1
(3.18.3)
Here xij is the distance between charges i and j;
ELEN 3371 Electromagnetics
Fall 2008
Vi 
N

j 1, j i
Qi
4 0 xij
(3.18.4)
19
1. Electrostatic Fields
1.6. Potential Energy and Electric Potential (cont 4)
Note: the total work in our case is equal to the total electrostatic energy
stored in the shaded region.
Note: the total work (and the total energy) do not depend on the order, in
which particles are brought.
The electrostatic energy can also be evaluated as
1
We   vVdv[ J ]
2 v
ELEN 3371 Electromagnetics
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(3.19.1)
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1. Electrostatic Fields
1.6. Potential Energy and Electric Potential (cont 5)
Electric potential difference between points a and b is the work required to
move the charge from point a to point b divided by that charge.
We can express the electric potential difference or voltage as:
a
b
b
 1 a

1
Va  Vb  Vab    F dl   F dl    Q  E dl  Q  E dl 
Q  


 Q  

(3.20.1)
b
or
J
Vab   E dl   [V ]
 C 
a
ELEN 3371 Electromagnetics
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(3.20.2)
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1. Electrostatic Fields
1.6. Potential Energy and Electric Potential (Example)
Evaluate the work (charge times potential
difference) required to move a charge q from
a radius b to a radius a.
The electric field is
E
Q
4 0 r
2
ur
(3.21.1)
The potential difference between the two
spherical surfaces is
b
Vab  
a
Q
4 0 r
u ur dr  
2 r
Q
4 0 r
b

a
Q 1 1
   (3.21.2)
4 0  a b 
4’
3’
6’
2’
5’
1’
The potential at r =  is assumed to be 0 and is called a ground potential.
The electric potential defined with respect to the ground potential is called
an absolute potential.
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1. Electrostatic Fields
1.6. Potential Energy and Electric Potential (Example, cont)
Considering the path 1-2-3-4, we notice that there are only potential differences
while going 1  2 and 3  4. Therefore, these are the only paths where some
work is required. When moving 2  3, the potential is constant, therefore no
work is required.
A surface that has the same potential is called an equipotential surface.
If the separation between two equipotential surfaces and the voltage between
them are small:
dV   E dl   Ex dx  E y dy  Ez dz 
Theelectric field :E  
ELEN 3371 Electromagnetics
V
V
V
dx 
dy 
dz
x
y
z
V
V
V
V
ux 
uy 
uz  
x
y
z
m
Fall 2008
(3.22.1)
(3.22.2)
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1. Electrostatic Fields
1.6. Potential Energy and Electric Potential (cont 6)
We can modify (3.22.2) as following:
E  V
Since E 
(3.23.1)
v
 V
   V 2V   v  
0
 0  m 
(3.23.2)
2
Poisson’s equation
Laplace’s eqn. when v = 0
An absolute potential caused by a volume distribution that is not at the origin:
V ( x, y , z ) 
ELEN 3371 Electromagnetics
1
4 0

v
v ( x ', y ', z ')
 x  x '   y  y '   z  z '
2
2
Fall 2008
2
dx ' dy ' dz '
(3.23.3)
24
1. Electrostatic Fields
1.6. Potential Energy and Electric Potential (cont 7)
The potential energy would be
We 
0
1
1

Vdv

(


E
)
Vdv

v
0
2 v
2 v
2
   VE  E V  dv 
v
0
0
VE
ds

2 v
2
 E Vdv
v
(3.24.1)
We 
0
2
 E Edv 
v
0
2
for R  
2
E
 dv
(3.24.2)
v
Note that when a charged particle is moved along a closed contour, no work is required
We
   E ds  0   E ds E  0
Q
s
Electrostatic field is conservative and irrotational.
ELEN 3371 Electromagnetics
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(3.24.3)
25
1. Electrostatic Fields
1.6. Potential Energy and Electric Potential (Example 2)
Find the potential V due to
two equal charges that have
opposite signs and are in the
vacuum. The distance from
the point of interest is much
greater than the separation.
The configuration is known
as an electric dipole.
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1. Electrostatic Fields
1.6. Potential Energy and Electric Potential (Example 2, cont)
Due to superposition:
V
Q
4 0 r1

Q
4 0 r2
Since r >> d; r, r1, and r2 are almost parallel.
d
d
r1  r  cos  and r2  r  cos 
2
2
(3.26.1)
(3.26.2)
Q
Q
V

d
d




4 0  r  cos   4 0  r  cos  
2
2





Q 
d
d
Qd

1

cos


1

cos


cos 


2
4 0 r  2r
2r
 4 0 r
A vector p = Qd is a dipole moment.
ELEN 3371 Electromagnetics
Fall 2008
(3.26.3)
27
1. Electrostatic Fields
1.6. Potential Energy and Electric Potential (Example 2, cont)
Electric potential distribution plot of (3.26.3)
0
0.01
30
330
0.008
0.006
60
300
0.004
V
0.002
0.01 0.008
0.006 0.004
0.002
0
90
270
-0.002
-0.004
120
-0.006
240
-0.008
-0.01
-4
150
-3
-2
-1
0
Angle, rad
1
2
3
4
plot
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210
180
polar
Fall 2008
28
1. Electrostatic Fields
1.7. On numerical integration
When no symmetry can be used to simplify the problem, numerical integrations
are quite helpful. Numerical integration = APPROXIMATION.
trapz
quad
dblquad
Self-study
triplequad
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1. Electrostatic Fields
1.8. Dielectric materials
A material can be considered as a collection of randomly (in general) oriented
small electric dipoles.
If an external electric
field is applied, the
dipoles may orient
themselves.
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1. Electrostatic Fields
1.8. Dielectric materials (cont)
We may suggest that an external electric
field causes a “thin layer of charge” of the
opposite sign at either edge of the material.
This charge is called a polarization charge.
The density of the polarization charge:
 p   P
where P is the polarization field:
(3.30.1)
1 N

P  lim   p j 
v0
 v j 1 
(3.30.2)
Here pj = Qdud is the dipole moment of individual dipole, N – number of atoms
(dipoles)
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1. Electrostatic Fields
1.8. Dielectric materials (cont 2)
Let us add the polarization charge density to the real charge density. The
Gauss’s Law will take a form:
v   p
 E
0
which leads to
where
(3.31.1)
 D  v
C
D 0 E  P
 m
2
(3.31.2)
electric(displacement ) fluxdensity (3.31.3)

The total flux that passes through the surface
e 
 D ds[C ]
s
ELEN 3371 Electromagnetics
Fall 2008
(3.31.4)
32
1. Electrostatic Fields
1.8. Dielectric materials (cont 3)
Integrating (3.31.2) over a volume, leads to
 D ds  Q
(3.32.1)
enc
s
Dielectric materials are susceptible to polarization. Usually, polarization is linearly
(3.32.2)
proportional to the applied (small) electric field. Then P   0 e E
where e is the electric susceptibility
D   0 (1  e ) E   0 r E E
for linear and isotropic materials
r is the relative dielectric constant
We consider only linear materials here.
ELEN 3371 Electromagnetics
Fall 2008
(3.32.3)
33
1. Electrostatic Fields
1.9. Capacitance
Q
C
[ F ]
V
A parallel-plate capacitor
Area A  w z
(3.33.2)
Assume A >> d
ELEN 3371 Electromagnetics
Fall 2008
(3.33.1)
34
1. Electrostatic Fields
1.9. Capacitance (cont)
Qenc
s A
s
s E  ds   0 2EA   0 E  2 0 between plates
(3.34.1)
b
Vab   E dl  Ed
(3.34.2)
a
C
Q s A  0 A
A


,orincaseof dielectricC 
V d s
d
d
(3.34.3)
0
Stored energy:
0
0  V 
2
2
CV
2
We 
E
dv    Ad 

2 v
2 d
2
(3.34.4)
Assumed uniform field in the capacitor and uniform distribution of charge on plates
ELEN 3371 Electromagnetics
Fall 2008
35
1. Electrostatic Fields
1.9. Capacitance (Example)
Calculate the mutual capacitance of a coax cable
with dielectric r inside…
From the Gauss’s Law:
 D ds  Q
enc
D  2 L  l L
(3.35.1)
Potential difference:
b
l

b 
d   l ln  
2
2  a 
a
b
Vab   E dl  
a
Thetotalcharge:Q  l L
C 
HW 2 is ready
ELEN 3371 Electromagnetics
l L
Q
2


L
l
Vab
b 
b 
ln   ln  
2  a 
a
Fall 2008
(3.35.2)
(3.35.3)
(3.35.4)
36
2. Magnetostatic Fields
2.1. Electric currents
Let’s consider a wire…
The generalized Ohm’s Law specifies a current density:
I
J 
A
Alternative:
V
R  V L 1   E A 
 m 2 
A
L AR
(3.36.1)
J  vvdrift  v m E   E
(3.36.2)
where v is the electron volume charge density, vdrift is an average
electron drift velocity, m is the mobility of the material.
  v m  1/ Rtheconductivity
ELEN 3371 Electromagnetics
Fall 2008
(3.36.3)
37
2. Magnetostatic Fields
2.1. Electric currents (cont)
The total current that passes through the wire
I
 J ds
(3.37.1)
A
In our case f the current is distributed uniformly in a cylindrical wire:
I  J a2
(3.37.2)
The power density (density of power dissipating within a conductor):
p  J EW m2 
(3.37.3)
The total power absorbed within the volume:
P
 pdv[W ]
v
ELEN 3371 Electromagnetics
Fall 2008
(3.37.4)
38
2. Magnetostatic Fields
2.1. Electric currents (Examples)
a) Calculate the current flowing through the wire of radius a; current density:
J  I0
2

a

a
uz
(3.38.1)
“skin effect”
a2
I    I 0 uz d duz  2 I 0
a
3
 0  0
(3.38.2)
b) Calculate the power dissipated within a resistor with a uniform conductivity .
The voltage across the resistor is V, a current passing through is I.
P
 J Edv 
v
ELEN 3371 Electromagnetics
L 2
I V
z 0 0 0  2 L d ddz  VI
a
Fall 2008
(3.38.3)
39
2. Magnetostatic Fields
2.2. Fundamentals of Magnetic Fields
Magnetic loops at the
surface of the Sun, as seen
with the TRACE solar
spacecraft. (©TRACE
operation team, Lockheed
Martin)
?Magnetic monopole?
ELEN 3371 Electromagnetics
Fall 2008
40
2. Magnetostatic Fields
2.2. Fundamentals of Magnetic Fields (cont)
Magnetic field lines are continuous, don’t originate nor terminate at a point.
There is no “magnetic monopole”…
 B ds  0
(3.40.1)
B is a magnetic flux density, [T] = [Wb/m2]
By applying the divergence theorem to (3.40.1):
 B ds    Bdv
(3.40.2)
v
 B0
ELEN 3371 Electromagnetics
Fall 2008
(3.40.3)
41
2. Magnetostatic Fields
2.2. Fundamentals of Magnetic Fields (cont 2)
As a result, we can
split a bar magnet into
tiny pieces and all of
them will have both
north and south poles.
ELEN 3371 Electromagnetics
Fall 2008
42
2. Magnetostatic Fields
2.2. Fundamentals of Magnetic Fields (cont 3)
Magnetic field can be created by the electric current:
Ampere’s Law:
 B dl   I
0 enc
(3.42.1)
electric current enclosed within a closed loop
Permiabilityof  freespace :0  4 107 12.566 107  H m
A cylindrical wire caring a current
creates a magnetic field.
“Right Hand Rule” (RHR).
ELEN 3371 Electromagnetics
Fall 2008
43
2. Magnetostatic Fields
2.2. Fundamentals of Magnetic Fields (cont 4)
By applying Stokes’s theorem to (3.42.1), we arrive to
 B dl     B ds    J
0
s
Therefore, the differential
form of Ampere’s Law is:
(3.43.1)
s
 B  0 J
It appears that even very small currents generate
magnetic fields…
An axial view of the cortically-generated magnetic
field of a human listener, measured using wholehead magnetoencephalography (MEG) – from the
journal “Cerebral Cortex”
ELEN 3371 Electromagnetics
ds
Fall 2008
(3.43.2)
44
2. Magnetostatic Fields
2.2. Fundamentals of Magnetic Fields (Example)
A symmetry in the system greatly simplifies
evaluation of integrals in (3.42.1).
2
 B dl   B u
du  2 B
(3.44.1)
0
The right-hand side of (3.42.1) for the radius
greater than a is just 0I.
B 
0 I
,  a
2
(3.44.2)
Assuming a uniformly distribution of current within the wire, the current density
inside the wire is
I
J  2 uz
a
ELEN 3371 Electromagnetics
Fall 2008
(3.44.3)
45
2. Magnetostatic Fields
2.2. Fundamentals of Magnetic Fields (Example 2)
The total current enclosed within the circle of radius   a
2
I enc
Therefore:

I

  J ds   
r

dr

d


I
 
2

a
a
s
 0 r 0
B 
0 I 
,  a
2
2 a
We notice that at the edge of the
wire, two solutions given by
(3.44.1) and (3.45.2) are equal.
Can you further explain the
dependence of magnetic flux
on the radius?
ELEN 3371 Electromagnetics
Fall 2008
2
(3.45.1)
(3.45.2)
46
2. Magnetostatic Fields
2.3. Magnetic Vector Potential
A magnetic vector potential A such that:
Ampere’s Law:
 A  B
(3.46.1)
  A  0
(3.46.2)
   A  0 J
(3.46.3)
2 A  0 J
In the Cartesian coordinates:
A(r ) 
0
4
J (r ')
v R dv
whereR  ( x  x ') 2  ( y  y ') 2  ( z  z ') 2
ELEN 3371 Electromagnetics
Fall 2008
(3.46.4)
(3.46.5)
(3.46.6)
47
2. Magnetostatic Fields
2.3. Magnetic Vector Potential (cont)
Magnetic vector potential, magnetic flux, and current element
ELEN 3371 Electromagnetics
Fall 2008
48
2. Magnetostatic Fields
2.3. Magnetic Vector Potential (cont 2)
The Magnetic flux density:
 0
B(r )   A(r )   
 4
J (r ')  0
v R dv   4
 J (r ') 

v  R  dv
Since a  a  B  a Band J (r ')  0
0
0
1
B( r ) 
 J (r ')dv  
4 v R
4
0
uR
v R2 J (r ')dv  4
J (r ')  uR
v R2 dv
(3.48.1)
(3.48.2)
(3.48.3)
If the current is passing through a wire
0
B(r ) 
4

I dl ' uR
R2
the Biot-Savart Law
ELEN 3371 Electromagnetics
Fall 2008
(3.48.4)
49
2. Magnetostatic Fields
2.3. Magnetic Vector Potential (Example)
Find the magnetic field on the axis perpendicular to
the loop of current. Use the Biot-Savart Law.
We identify the terms appearing in (3.48.4):
dl '  ad ' u ,uR   au  zuz  R ,R  a 2  z 2
0 I
B( z ) 
4

(ad 'u )  (au  zuz )
(a  z )
2
2 32
0 I

4

a 2 d 'uz  azd 'u
(a  z )
2
2 32
(3.49.1)
Due to symmetry, the terms with the unit vector u are zero.
0
0 m
I a2
B( z ) 
u

z
2 (a 2  z 2 )3 2
2 R3
m  I a2uz themagneticdipolemoment
ELEN 3371 Electromagnetics
Fall 2008
(3.49.2)
(3.49.3)
50
2. Magnetostatic Fields
2.3. Magnetic Vector Potential (Cont 3)
We have learned the following analytical methods to find the magnetic flux
density at a point in space from a current element:
1. Application of Ampere’s Law, which requires considerable
symmetry.
2. Determination of the vector magnetic potential and the calculation
of a magnetic flux density. No symmetry is required.
3. Application of the Biot-Savart law. No symmetry is required.
ELEN 3371 Electromagnetics
Fall 2008
51
2. Magnetostatic Fields
2.4. Magnetic forces
If a charged particle is moving with a constant velocity v in a region that ONLY
contains a magnetic field with the density B, the force that acts upon the particle is
Fm  q  v  B 
Direction of the force
- RHR!
F+ stands for a
positively charged
particle;
F- represents a
negatively charged
one.
ELEN 3371 Electromagnetics
Fall 2008
(3.51.1)
52
2. Magnetostatic Fields
2.4. Magnetic forces (cont)
When a charged particle is going through an area with both: uniform electric field
and uniform magnetic field, the force exerted on it would be the Lorentz Force:
F  q( E  v  B)[ N ]
(3.52.1)
Recall that the work done by a charged particle moving in a field is
b
W   F dl
(3.52.2)
a
A differential charge dQ = vdv moving at a constant velocity creates a current. If
this current flows in a closed loop:
dFm  dQ(v  B)  v (v  B)dv  J  Bdsdl  Idl  B
The total magnetic force:
ELEN 3371 Electromagnetics
Fm    B  I dl
Fall 2008
(3.52.3)
(3.52.4)
53
2. Magnetostatic Fields
2.4. Magnetic forces (cont 2)
Example: a charged particle entered a constant magnetic
field will move along a circular orbit. Find the radius…
Magnetic force:
Centripetal force:
Fm  qvB
(3.53.1)
Fc  mac  m
v2
j
, m – particle’s mass
mv
j 
qB
This radius is called the Larmor radius or gyro radius. This effect is used in
mass spectroscopy.
ELEN 3371 Electromagnetics
Fall 2008
(3.53.2)
(3.53.3)
54
2. Magnetostatic Fields
2.4. Magnetic forces (Example)
Evaluate the force existing
between two parallel wires caring
currents.
B1 will go up at the location of wire 2.
From (3.52.4) force on the wire 2:
a)
F2  B1u y  I 2dluz
(3.54.1)
to the left
b)
F2  B1uy  (I 2 )dluz
(3.54.2)
to the right
ELEN 3371 Electromagnetics
Fall 2008
55
2. Magnetostatic Fields
2.4. Magnetic forces (Example, cont)… ”Alternative approach”
Let’s re-state the force on wire 1 caused by the magnetic field generated by
the current in wire 2 (from 3.52.4)
F12   I1  B12  dl
(3.55.1)
L1
From the Biot-Savart law:
I
B12   0 2
4
Finally:
II
F12  0 1 2
4


2
21
R
L2
u
L1 L2
This is what’s called as Ampere’s force.
ELEN 3371 Electromagnetics
u R21  dl2
Fall 2008
R21

(3.55.2)
 dl2  dl1
2
21
R
(3.55.3)
56
2. Magnetostatic Fields
2.4. Magnetic forces (Example 2)
Consider a current-caring loop in a constant
magnetic field B = B0 uz.
We assume the separation between the In/Out
wires to be infinitely small.
Parallel wires carry the same current in the
opposite direction. Therefore, the net force will
be a vector sum of all forces, which is zero!
However, there will be a torque on the loop
that will make it to rotate (say, about x for
simplicity).
ELEN 3371 Electromagnetics
Fall 2008
57
2. Magnetostatic Fields
2.4. Magnetic forces (Example 2, cont)
The torque on the loop is given by
y
y
T  F1
sin   F3
sin 
2
2
(3.57.1)
Assumptions?
whereF1  IB0 x,F3  IB0 x
T  IB0  x ysin 
Finally:
(3.57.2)
(3.57.3)
T  m B
(3.57.4)
magnetic moment
ELEN 3371 Electromagnetics
Fall 2008
58
2. Magnetostatic Fields
2.5. Magnetic materials
Two sources of magnetism inside an atom:
1) an electron rotating around a nucleus;
2) an electron spinning about its own axis.
Types of material:
1. Diamagnetic: 1) and 2) cancel each other almost
completely, magnetic susceptibility m  -10-5.
magnetic dipoles
2. Paramagnetic: 1) and 2) do not cancel each other
oriented randomly
completely, magnetic susceptibility m  10-5.
3. Ferromagnetic: domain structure, very high m (hundreds
and higher)
magnetic
dipoles in
each domain
are oriented
ELEN 3371 Electromagnetics
Fall 2008
59
2. Magnetostatic Fields
2.5. Magnetic materials: Ferromagnetics
External magnetic field may change dipole orientation “permanently” - HDD.
1 N
M  lim  m j  A m
v 0 v
j 1
Total magnetization (magnetic dipole
moment per unit volume):
(3.59.1)
There is a current created inside domains (magnetization current):
Im 
M
dl 
J
m
ds     M ds
s
(3.59.2)
s
Therefore,J m   M
(3.59.3)
We may modify the Ampere’s Law by adding the magnetization current:
B

 B  J  J m  J   M J     M 
0
 0

1
ELEN 3371 Electromagnetics
Fall 2008
(3.59.4)
60
2. Magnetostatic Fields
2.5. Magnetic materials: Ferromagnetics (cont)
We introduce a new quantity, the Magnetic Field Intensity:
H
B
0
 M  A m
(3.60.1)
Therefore, the Ampere’s circular law is
H
dl  I enc
Magnetization :M  m H
Therefore:
B  0 (1  m ) H  0 r H   H
where r is the relative permeability.
ELEN 3371 Electromagnetics
Fall 2008
(3.60.2)
(3.60.3)
(3.60.4)
61
2. Magnetostatic Fields
2.5. Magnetic materials: Ferromagnetics (cont 2)
Example: a magnetic flux density B = 0.05 T appears in a material with r = 50.
Find the magnetic susceptibility and the magnetic field intensity.
m  r 1  50 1  49
H
B
 r 0

0.05
 796[ A m]
7
50  4 10
saturation
Hysteresis
Magnetic flux density B exhibits nonlinear
dependence on the magnetic field intensity H.
Because of hysteresis, magnetic materials
“remember” the magnitude and direction of
magnetic flux density. They can be used as
memory elements.
saturation
ELEN 3371 Electromagnetics
Fall 2008
62
2. Magnetostatic Fields
2.6. Magnetic circuits
Just like electrical circuits, we can build magnetic circuits where magnetic flux “flows”.
L – mean length
of the iron region
g – length of the
gap.
m – the reluctance.
Assumptions:
1) the gap is very small;
2) the cross-sectional area of the gap is identical to the cross-sectional area of the
magnetic material.
ELEN 3371 Electromagnetics
Fall 2008
63
2. Magnetostatic Fields
2.6. Magnetic circuits (cont)
The Ampere’s circular law leads to
H ( L  g )  I enc  NI
(3.63.1)
where I is the current flowing through the N turns of a wire.
 L
g 

L  g  m 

  NI
0  r
0
 A0 r A0 
B
B
(3.63.2)
Here A is the cross-sectional area of the iron
Fm  NI is the magnetomotive force [A-turns]
L
g
 m,iron   m, gap 

is the reluctance
A0 r A0
The Hopkinson’s Law (aka Ohm’s):
ELEN 3371 Electromagnetics
m  Fm m
Fall 2008
(3.63.3)
(3.63.4)
(3.63.5)
64
2. Magnetostatic Fields
2.6. Magnetic circuits (Examples)
ELEN 3371 Electromagnetics
Fall 2008
65
2. Magnetostatic Fields
2.7. Inductance (an ability to create magnetic flux)
L jk 
j
Ik
 H   Wb A
where  is a magnetic flux linkage.
When j = k – self –inductance; otherwise – mutual inductance.
Ex. 1: Let us consider a
solenoid of the length d,
cross-section area A,
and having N turns.
It may also have a core
made from a magnetic
material.
z is the solenoid’s axis.
ELEN 3371 Electromagnetics
Fall 2008
(3.65.1)
66
2. Magnetostatic Fields
2.7. Inductance (cont)
The magnetic flux density at the center of the solenoid is:
Bz 
The total magnetic flux:
m 
 NI
(3.66.1)
d
 NI
d
The magnetic flux linkage:   N  m 
A
 N 2I
d
(3.66.2)
A
(3.66.3)
Therefore, the self-inductance of a solenoid is
L
ELEN 3371 Electromagnetics
N2
d
Fall 2008
A
(3.66.4)
67
2. Magnetostatic Fields
2.7. Inductance (cont 2)
Example 2: a self-inductance of a coaxial cable
What is the main difference as compared to a
solenoid?
Here, the magnetic flux linkage equals to the
total magnetic flux.
I
(3.67.1)
2 r
z b
I
I  b 
  m   
drdz 
ln   z
2 r
2  a 
z 0 r  a
mfd:
Therefore:
ELEN 3371 Electromagnetics
B 
 b
L
ln   z
2  a 
Fall 2008
(3.67.2)
(3.67.3)
68
2. Magnetostatic Fields
2.7. Inductance (cont 3)
Example 3: a mutual inductance between two
circular solenoids, whose individual lengths are
d and areas S1 and S2, separated by x; x << d
B1 
First coil:
0 N1 I1
x
(3.68.1)
d
m,1  B1S1
(3.68.2)
Assuming that the magnetic flux has the
same value in the second solenoid:
2  N2 B1S1
(3.68.3)
 2 0 N1 N 2 S1
M

I1
d
(3.68.4)
Therefore, the mutual inductance:
ELEN 3371 Electromagnetics
Fall 2008
69
2. Magnetostatic Fields
2.7. Inductance (cont 4)
Example 4: consider a transformer with N1 and N2 turns.
correction coefficient
permeance of the space
occupied by the flux
Alternative formulas:
coefficient of coupling
ELEN 3371 Electromagnetics
M12  N1 N2 P12
(3.69.1)
M  k L1L2
(3.69.2)
Fall 2008
70
2. Magnetostatic Fields
2.7. Inductance (cont 5)
Solenoids, transformers, etc. can store magnetic energy:
LI 2
 dI 
Wm   pdt '   IV dt '   I  L dt '  L I dI 
dt ' 
2
0
0
0 
0
t
for a solenoid:
t
t
Wm 
The total magnetic energy
stored within a volume
t
Bd
1 N
A z 
2 d
 N 
2
2
B2
Wm  
dv J 
2
v
Your homework 3 is available through “My Lamar” 
ELEN 3371 Electromagnetics
Fall 2008
(3.70.1)
(3.70.2)
(3.70.3)