Download R - Steady Server Pages

Physics 441
Electro-Magneto-Statics
M. Berrondo
Physics BYU
1
1. Introduction
• Electricity and Magnetism as a single field
(even in static case, where they decouple)
Maxwell:
* vector fields
* sources (and sinks)
• Linear coupled PDE’s
* first order (grad, div, curl)
* inhomogeneous (charge & current distrib.)
~
F  J

~
F  J
Physics BYU
1
2
Tools
Math
• trigonometry
• vectors (linear combination)
dot, cross, Clifford
• vector derivative operators
 / xi ,
,
,
,

• Dirac delta function
• DISCRETE TO CONTINUUM
• INTEGRAL THEOREMS:
* Gauss, Stokes, FundThCalc
• cylindrical, spherical coord.
• LINEARITY
Physics
• trajectories: r(t)
• FIELDS: * scalar, vector
* static, t-dependent
• SOURCES: charge, current
• superposition of sources =>
• superposition of fields
• unit point sources
• Maxwell equation
• field lines
• potentials
• charge conservation
interpretation of equations and their solutions
Physics BYU
3
2. Math. Review
•
•
•
•
sum of vectors: A , B => A + B
dilation: c , A => c A
linear combinations: c1 A + c2 B
scalar (dot) product:
A , B => A  B = A B cos ( ), a scalar
where A2 = A  A (magnitude square)
• cross product:
A , B => A  B = n A B |sin( ) |, a new vector,
with n  A and B, and nn = n2 = 1
• orthonormal basis: {e1 , e2 , e3 } = {i , j , k}
Physics BYU
4
Geometric Interpretation:
Dot product
Physics BYU
5
Cross Product and triple dot product
Physics BYU
6
Rotation of a vector (plane)
• Assume s in the x-y plane
• Vector
s‘ = k x s
• Operation k x rotates s
by 90 degrees
s  xi  yj
s'  k  ( xi  yj)   yi  xj
• k x followed by k x again
equivalent to multiplying
by -1 in this case!!
Physics BYU
y
s’
s
x
7
Rotation of a vector (3-d)
n unit vector: n2 = 1 defines rotation axis
 = rotation angle
vector r  r’
r'  e n r
r'  e n (r//  r )  r//  e nr
r'  r//  cos  r  sin  n  r
where
r  n (n  r )
r  r  r//  n  (n  r )
Physics BYU
8
Triple dot product
A  (B  C)  C  (A  B)  B  (C  A) 
e1
 A  Bx
Cx
e2
By
Cy
e3
Bz
Cz

Ax
Bx
Cx
Ay
By
Cy
Az
Bz
Cz
is a scalar and corresponds to the (oriented)
volume of the parallelepiped {A, B, C}
Physics BYU
9
Triple cross product
• The cross product is not associative!
• Jacobi identity:
A  (B  C)  C  ( A  B)  B  (C  A)  0
• BAC-CAB rule:
A  (B  C)  B( A  C)  C( A  B)
is a vector linear combination of B and C
Physics BYU
10
Inverse of a Vector
0
|
vector
1
|
n-1 = n as a unit
n A
A  ( An)   2
A A
1
1
along any direction in R2 or R3
Physics BYU
11
Clifford Algebra Cl3 (product)
• starting from R3 basis {e1 , e2 , e3 }, generate
all possible l.i. products 
• 8 = 23 basis elements of the algebra Cl3
• Define the product as:
AB  A  B  iA  B
• with
i  e1e 2e3
Physics BYU
12
1
0
scalar
1
vector
iê1= ê2ê3 iê2= ê3ê1 iê3=ê1ê2
2
bivector
i = ê 1ê 2ê 3
3
pseudoscalar
ê1
ê2
ê3
e1  e 2  e3  1,
2
2
2
e1e 2  e 2e1 (e1e 2 ) 2  1,
eˆ i , eˆ k   eˆ ieˆ k  eˆ k eˆ i  2 ik
C    i  A  i B
Physics BYU
R3
R
iR
i R3
13
Clifford Algebra Cl3
•
•
•
•
Non-commutative product w/ A A = A2
Associative:
(A B) C = A (B C)
Distributive w.r. to sum of vectors
*symmetric part  dot product
*antisym. part  proportional cross product
• Closure: extend the vector space until every
product is a linear combination of
elements of the algebra
Physics BYU
14
Subalgebras
• R
Real numbers
• C = R + iR
Complex numbers
• Q = R + iR3
Quaternions
Product of two vectors is a quaternion:
AB  A  B  iA  B
AB scalar  A  B  (AB  BA) / 2
AB
bivector
 iA  B  (AB  BA) / 2  A  B
represents the oriented surface (plane)
orthogonal to A x B.
Physics BYU
15
Bivector: oriented surface
b
a
i ab  a  b
sweep
b
a
i b  a  b  a  a  b
• antisymmetric, associative
• absolute value  area
Physics BYU
16
Differential Calculus
• Chain rule:
In 2-d:
f
df  
dxi
i xi
f
f
df  dx  dy
x
y
f  f ( x, y )
 xdx  2 ydy

• Is 2 ydx  xdy an “exact differential”?
 ydx  xdy


A

B
• given A( x, y)dx  B( x, y)dy ,

y x
Physics BYU
17
• In 3-d:
1. 
 f
df  f  dr  
 x
2. 

f
y
 dx 
f   
  dy 
z   
 dz 
df  0
b
3. 
 df
 F (b)  F ( a )
independen t of path
a
• Geometric interpretation:
 0 when T  dr
dT  T dr cos   
max when T // dr
T
points in direction of steepest ascent
Physics BYU
18



  eˆ1  eˆ 2  eˆ 3
x
y
z
Examples:
z  kˆ ,
del operator
dg ˆ
g ( x ) 
i
dx
r  rˆ
df
f (r )  rˆ
dr
1
1
r
   2 rˆ   3
r
r
r
Physics BYU
19
Gradient of r
• Contour surfaces: spheres
•  gradient is radial
f (r )  r
r // rˆ
and
df  dr  (r )  d r  r dr
 r  1
r  r̂
• Algebraically:
r  2rr  ( x  y  z )  2r  r  rˆ
2
and, in general,
2
2
2
df
f (r )  f ' (r )rˆ  rˆ
dr
Physics BYU
20
Divergence of a Vector Field
• E (r)  scalar field (w/ dot product)
E x E y E z
E 


x
y
z
 Ex 


  
 E y 
 
 x y z  E 
 z
• It is a measure of how much the filed lines
diverge (or converge) from a point (a line, a
plane,…)
Physics BYU
21
• Divergence as FLUX:

E
x
( )
E x ( x  dx)  E x ( x) 
dx

x
dx
Examples:
  kˆ  0
  ( z kˆ )  1
 r  3
1 d 2
  (rˆ F (r ))  2
(r F (r ))
r dr
 rˆ 
  2   0
for r  0
r 
Physics BYU
22
Curl of a Vector Field
ˆi

E 
x
Ex
ˆj

y
Ey
kˆ

1
 2
z r sin 
Ez
rˆ

r
Er
r θˆ


rE
r sin  φˆ


r sin  E
The curl measures circulation about an axis.
Examples:
E(r )  xˆj
  E  kˆ
E(r )   yˆi  xˆj
  E  2kˆ
Physics BYU
23
Clifford product del w/ a cliffor
• For a scalar field
T = T( r ),
T  grad T
• For a vector field
E = E( r ),
E    E  i  E
• For a bivector field
i B = i B( r ),
(i B)  i  B    B
Physics BYU
24
Second order derivatives
• For a scalar:
 2T  (T )    (T )  i  (T )
  2T    (T ),
   (T )  0
• For a vector:
 2E  (E)  (  E)  i(  E)
  E  (  E)    (  E),
   (  E)  0
2
Physics BYU
25
What do we mean by “integration”?
cd
c
dw = c(cd)/ 2
dr
dg = (2pr) dr
||
dl
rd
dr
da= (rd)dr
2
 dy 
dl  (dx)  (dy )  1    dx
 dx 
2
Physics BYU
2
26
Cliffor differentials
• dk is a cliffor representing the “volume”
element in k dimensions
• k = 1  dl is a vector  e1 dx
(path integral)
• k = 2  i nda bivector  e1 dx e2 dy
(surface integral)
• k = 3  i dt ps-scalar  e1 dx e2 dy e3 dz
(volume integral)
Physics BYU
27
Fundamental Theorem of Calculus
 d  F ( )  ()  d
k 1
k
 F ( )
k 1
V
V
b
 (T )  d l  T (b)  T (a)
Particular cases:
a
   E dt   E  da
V
V
   B d a   B  d l
S
Gauss’s theorem
Stokes’ theorem
S
Physics BYU
28
Delta “function” (distribution)
• 1-d:
 ( x)  0 for x  0,


  ( x' )dx'   ( x' )dx' 1,






  ( x' ) f ( x' )dx'  f (0),
[  f ]( x)  f ( x)
d ( x)
 ( x) 
dx
1
 (ax) 
 ( x)
|a|
unit area
  ( x  x' ) f ( x' )dx'  f ( x)
unit convolutio n

as a distributi on
scaling
step “function”
1
x
Physics BYU
29
• Divergence theorem and unit point source
apply
   E dt   E  da
V
V
to
rˆ
E (r )  2
r
for a sphere of radius 
rˆ
rˆ
2
ˆ
E

d
a


d
a


r
r
V
V r 2
V r 2 dΩ  V dΩ  4p

rˆ 

V   E dt  V    r 2  dt  4p
Physics BYU
30
 rˆ 
   2   4p  ( 3) (r )
r 

and

1 
   (3) (r).
  
 4p r 
2
Displacing the vector r by r‘:


1
   (3) (r  r' ).
  
 4p | r  r' | 
2
Physics BYU
31
Inverse of Laplacian
• To solve
 2 A(r)  B(r)
 1
1 
B(r ' )dt '  B(r )
   
 4p | r  r ' | 
2
1
1
1
1
A(r )  2 B(r )  
B(r ' )dt '  


4p | r  r ' |
4p
so
1

  B(r ) 
r

• In short-hand notation:
1
A(r )  
4p
1
 r B(r' )dt ' ,
Physics BYU
where r | r  r ' |
32
Orthogonal systems of coordinates
•
•
•
•
•
coordinates: (u1, u2, u3 )
orthogonal basis:
(e1, e2, e3 )
scale factors:
(h1, h2, h3 )
volume:
d (vol)  dt  h1h2 h3 du1du2 du3
area  u3 :
da3  h1h2 du1du2 e3
• displacement vector:
d l  h1du1e1  h2 du2e 2  h3du3e3
Physics BYU
33
Scale Factors
Cartesian
Cylindrical
Spherical
du1
du2
du3
h1
h2
h3
dx
ds
dr
dy
d
d
dz
dz
d
1
1
1
1
s
r
1
1
r sin
• polar (s,  ):
s  s sˆ
• cylindrical (s, , z ):
r  s sˆ  z eˆ 3
• spherical (r, ,  ):
r  r rˆ
Physics BYU
34
• Grad:
• Div:   E 
• Curl:
• Laplacian:
uˆ 1 T uˆ 2 T uˆ 3 T
T 


h1 u1 h2 u2 h3 u3
1   (E1 h2 h3 )  h1 E2 h3   (h1 h2 E3 ) 




h1 h2 h3 
u1
u2
u3

h1 uˆ 1 h2 uˆ 2
1


E 
h1 h2 h3 u1
u2
h1 E1 h2 E 2
h3 uˆ 3

u3
h3 E3
1    h2 h3  T    h1 h3  T    h1 h2  T  



 

 
T
h1 h2 h3  u1  h1 u1  u2  h2 u2  u3  h3 u3  
2
Physics BYU
35
Maxwell’s Equations
1
• Electro-statics:
E(r ) 
• Magneto-statics:
(iB(r))  0 J(r)
• Maxwell:
0
 (r )
1
F  ( E  icB ) 
(c  J )
 0c
1
 0 c  377
 0c
Physics BYU
36
Formal solution
~
F (r )  J (r )
~
F  J
1

separates into:
1 
1 1 
E(r)   (r)  E        2   
0
  0 
 0 
1
1
and
iB(r )    0 J (r )

Physics BYU
1
iB   2  0 J 

37
Electro-statics
1

   (r)   E0 (r)   (r)
Convolution: E(r)  
4p0  r

1
rˆ
E 0 (r ) 
4p0 r 2
1
where
for point charge.
r  r'
1
rˆ
E(r ) 
 r ' dt ' 
 r ' dt '
3
2


4p0 | r  r' |
4p0 r
1
1
1
 (r' ) dt '
E(r )  V (r ) where V (r ) 

4p0 | r  r' |
Physics BYU
38
Superposition of charges
• For n charges
V (ri ) 
1
4p0
{dq1, dq2, …, dqn }
n
dq j
j 1
ri j


1
4p0
n
dq j
j 1
i
|r r
j
|
• continuum limit:
1
dq'
V (r ) 
4p0  | r  r ' |
• where
E(r )  V (r )
  (r' ) d l '

dq '   (r' ) da'
  (r' ) dt '

Physics BYU
39
• linear uniform charge density  (x’) from –L to L
• field point @ x = 0, z variable
z
dq '   dx'

r  z k  x' i
r  x'  z  z sec 
x'  z tan 
2
|
-L
|L x
dq’
E(r ) 
1
4p0
L

L
r
r
3
2
dx'
Physics BYU
40
with
dx'  z sec 2  d ,
0 2
ˆ
2 k z sec 2 
1 2 kˆ
E(r) 
d 
sin  0
3
3

4p0 0 z sec 
4p0 z
so
• Limits:
2
E(r ) 
4p0 z
1
L
L2  z 2
kˆ .
1 

ˆ
s
L

 2p0 s
E(r )  
q 1

rˆ z  , L fixed
2

 4p0 r
Physics BYU
41
Gauss’s law
ΦE   E  da
• Flux of E through a surface S:
S
volume V enclosed by surface S.
• Flux through the closed surface:
1
 E  da  
S
Qenc 
0
1
  dt
0 V
• choose a “Gaussian surface” (symmetric case)
E Area(G.S .) 
1
0
Qenc
ˆ determined by symmetry
E
Physics BYU
42
Examples:
• Charged sphere (uniform density) radius R
A  4p r 2
Gaussian surface:
3
r
a) r < R:
Q   (r' )dt '  Q
enc

R
Vol ( r )
Q
r
E(r ) 
rˆ
3
4p0 R
3
for r  R
rˆ
 E(r ) 
4p0 r 2
Q
b) r > R: Qenc  Q
as if all Q is concentrated @ origin
Physics BYU
for r  R
43
• Thin wire: linear uniform density (C/m)
Gaussian surface: cylinder
A  2p sl
Qenc   l
l
 sˆ
E (2p sl ) 

E(s) 
0
2p  0 s
• Plane: surface uniform density (C/m2)
Gaussian surface: “pill box” straddling plane
A
 ˆ
E (2 A) 

E
k
0
2 0
CONSTANT, pointing AWAY from surface (both
sides)
Physics BYU
44
Boundary conditions for E
• Gaussian box w/ small area D A // surface w/
charge density 
1
with n pointing away from 1
 nˆ E  
2
• Equivalently:

E 1  E 2  nˆ
0
• Component parallel to surface is continuous
• Discontinuity for perp. component = /0
Physics BYU
45
Electric Potential (V = J/C)
• Voltage
1
1
1
dq'
V (r )  V0 (r ) *  (r ) 
 (r ' )dt ' 

4p0 r
4p0  r
solves Poisson’s equation:
 V (r )  
2
1
 (r )
0
• point charge Q at the origin  (r )  Q (r )
1 Q
V (r ) 
4p0 r
Physics BYU
46
Potential Difference (voltage)
r
• in terms of E:
Edl  0
V (r)  V (a)   E  d l
and
a
• Spherical symmetry: V = V (r)
dV (r )
E(r )  
rˆ
dr
where r  r
• Potential energy: U = q V (joules)
• Equi-potential surfaces: perpendicular to field
lines
Physics BYU
47
• Example: spherical shell
radius R, uniform surface charge density 
Gauss’s law  E(r) = 0 inside (r < R)
For r > R:
1
q
E(r ) 
rˆ
2
4p0 r
r
wher e q  4p R 
2
1
and
r
q
1 q
V (r )    E  d l  
dr 
2

4p0  r
4p0 r

Physics BYU
48
• Example: infinite straight wire,
uniform line charge density 
Gauss’s law:
E(r ) 
r
2p0 s
V (r )    E  d l  
a

1
1
2p0
Physics BYU
sˆ
s

a
and


a
ds 
ln
s
2p0 s
49
Electric-Magnetic materials
• conductors
– surface charge
– boundary conditions
– 2nd order PDE for V (Laplace)
• dielectrics
– auxiliary field D (electric displacement field)
• non-linear electric media
• magnets
– auxiliary field H
• ferromagnets
– non-linear magnetic media
Physics BYU
50
Perfect conductors
•
•
•
•
•
•
•
•
Charge free to move with no resistance
E = 0 INSIDE the conductor
 = 0 inside the conductor
NET charge resides entirely on the surface
The conductor surface is an EQUIPOTENTIAL
E perpendicular to surface just OUTSIDE
E = /0 locally
Irregularly shaped   is GREATEST where
the curvature radius is SMALLEST
Physics BYU
51
Induced charge
• Charge held close to a conductor will induce
charge displacement due to the attraction
(repulsion) of the inducing charge and the
mobile charges in the conductor
Example: find charge distribution
1. Inner: conducting sphere,
radius a, net charge 2 Q
2. Outer: conducting thick
shell b < r < c, net charge -Q
Physics BYU
52
Work and Energy
W   Q E  d l  Q [V (b)  V (a)]  Q DV
b
• Work:
a
• for n interacting charges:
1 1 n qi q j 1 n
W
  qiV (ri )

4p0 2 i  j ri j
2 i
• continuous distribution 
0
0
1
W    Vdt   ( E) Vdt 
2
2
2
• Energy density:
also electric PRESSURE
Physics BYU
 E dt   V E  da
2
u (r ) 
0
2
E 2 (r )
53
Capacitors (vacuum)
• Capacitor: charge +Q and –Q on each plate
- - - - - d
E
++++++

Q
d
E 
, so DV  E d  Q
 0 A 0
A 0
• define capacitance C :
with units 1F = 1C /1V
Physics BYU
Q = C DV
C
0 A
d
54
• charging a capacitor C: move dq at a given
time from one plate to the other adding to the
q already accumulated
2
2
q
1Q
CV
dW  dq 
W

C
2 C
2
• in terms of the field E
1 A
1
2
2
W   0 ( Ed )   0 E (Vol)
2 d
2

u (r ) 
0
2
2
E (r )
Physics BYU
energy density
55
Laplace’s equation
• V at a boundary (instead of )
 V 0
2
one dimension:
capacitor
d 2V
0
2
dx
V0  V1
V1  V0
V ( x) 
x  V0
d

V ( x)  mx  b
(0 to d)
from
0d
V(x) is the AVERAGE VALUE of V(x-a) and V(x+a)
Physics BYU
56
• Harmonic function has NO maxima or minima
inside. All extrema at the BOUNDARY
• Average value (for a sphere):
1
V0  V (r  0) 
V (r )da

Area
Proof:
2

 Vdt    (V )dt  (V )  da 
V
2 d
R 
d  R
Vd   0

r
dR
 Vd   V0 4p for R  0
2

2
Vda

V
4
p
R
0

for finite R
Physics BYU
57
Method of images
• Problem: find V for a point charge q facing an
infinite conducting plane
q
equivalent to
potential:
r
q
-q
charge density:
1 q q
  
V (r) 
4p0  r1 r2 
V
   0
n
Physics BYU
plane
58
Laplace’s equation in 2-d
Complex variable z = x + iy  w’(z) well defined
w
w
dx 
dy  w
 x
dw( z ) x
y

  w
dz
dx  idy
 i
 y
for dy  0
for dx  0
2
 2
 
 2  2 w( x  iy )  0

 x y 
both real and imaginary parts of w fulfill
Laplace’s equation
Physics BYU
59
Examples and equipotentials in 2-d
•
w(z) = z
equipotential lines:
V(x, y) = y
y = a (const)
•
w(z) = z2
e. l.:
V(x, y) = 2xy, or x2 - y 2
xy = a (hyperbolae)
•
w(z) = ln z
e. l.:
V(s, ) = ln s, arctan(y/x)
s = exp(a)
•
w(z) = 1/z = exp(-i  /s
V(s, ) = cos /s
e. l.:
s = cos /a
(circles O)
Physics BYU
60
 z 1/ 2 
• w( z )  ln 

 z 1/ 2 
• w(z) = z + 1/z
e. l.:
V(x, y) = finite dipole
V(s, ) = (s – 1/s) sin 
s = 1 for a = 0
• w(x) = exp(z)
V(x, y) = exp(x) cos y
Vk(x, y) = exp(k x) cos (k y) & exp(k x) sin (k y)
Physics BYU
61
Separation of variables (polar)
z s e
n
• Powers of z:
n
Vn ( s,  )  S n ( s) n ( )
S n ( s)  s
n
and
in
whe re
cos( n )
 n ( )  
 sin( n )
1   V  1  2V
0
s
 2
2
s s  s  s 
• Solution of
V  V0  a0 ln s   (an s n  bn s  n )[cn cos( n )  d n sin( n )]
n 1
Physics BYU
62
Separation of variables (x, y)
• k 2 = separation constant V ( x, y )  X ( x)Y ( y )
1 2
1 d 2 X 1 d 2Y
2
2
V


k

k
0
2
2
V
X dx
Y dy
• Eigenvalue equation for X and Y
2
2
d X ( x)
d Y ( y)
2
2
 k X ( x)
&
 k Y ( y )
2
2
dx
dy
• Construct linear combination, determine
coefficients w/ B.C. using orthogonality
Physics BYU
63
• Example: i) V ( x  0)  V0 ii) V ( x  )  0
iii) V ( y  0)  0
iv) V ( y  p )  0
solution:
X k ( x)  A e kx  B e  kx
Yk ( y)  C sin( ky)  D cos(ky)
B.C.: ii) A = 0;
iii) D = 0;
iv) quantization k = 1, 2, 3, …

V ( x, y)   Cn e
 nx
sin( nx)
n 1
Physics BYU
64
• orthonormality:
p
p
 sin( ny) sin( my)dy  2 
nm
0
i)

 40V m even
Cm   V0 sin( my)dy   0
m odd
p 0

 mp
solution:
2
p
4V0
odd
1  nx
V ( x, y ) 
e sin( ny )

p n 1 n
Physics BYU
65
Separation of variables - spherical
• Assume axial symmetry, i.e. no  dependence
1   2  
1
 
 
  2
r
 2
 sin 

r r  r  r sin   
 
2
separation constant l (l +1)
V (r , )  R(r ) ( )
d  2 dR(r ) 
r
  l (l  1) R(r )
dr 
dr 
1 d 
d ( ) 
 sin 
  l (l  1)( )
sin  d 
d 
Physics BYU
66
Legendre polynomials
• change of variable
( )  Pl ( )  Pl (cos  )
d
d

2 d Pl ( ) 
(1   ) d   l (l  1) Pl ( )  0


• orthogonality
1
2
1 Pl ( ) Pm ( ) d   2l  1  l m
• “normalization”
Pl (1) = 1
Physics BYU
67
• Radial function
R(r) = r l or r – l – 1
Bl 

l
V (r ,  )    Al r  l 1 Pl (cos  )
r 
l 0 

 A0  A1r cos   ...  A0  A1 z  ...



B0 B1
 2 cos   ...

r r

Physics BYU
68
• Given V0( ) on the surface of a hollow sphere
(radius R), find V(r)
• Soln: Bl = 0 for r < R, and Al = 0 for r > R
B.C. @ r = R:
 
l
A
R
Pl (cos  )

l

 l 0
V0 ( )   
 Bl R l 1 Pl (cos  )

 l 0
using orthogonality:
p
2l  1
Al 
V ( ) Pl (cos  ) sin  d
rR
l  0
2R 0
p
2l  1 l 1
Bl 
R  V0 ( ) Pl (cos  ) sin  d
rR
2
0
Physics BYU
69
• Uncharged conducting sphere in uniform
electric field E = E0 k
B0
B1
V  ( A0  )  ( A1r  2 ) cos   ...
r
r
Bl
l
2 l 1
A
R


0

B


A
R
V(r = R) = 0:
l
l
l
l 1
R
V  -E0 r cos  r >> R:
A0  B0  0
B1   A1 R 3
A1   E0
3

R 
 V (r )   E0  r  2  cos 
r 

Physics BYU
70
V
 ( )   0
r
 3 0 E0 cos 
r R
Physics BYU
71
Electric Dipole
• finite dipole
q 1 1
  
V (r ) 
4p0  r r 
1
1
s

 1  cos  
as r  
r
r  2r

1 1
s
and
  2 cos 
r r
r
qs 1
+q
V (r ) 
cos 
2
4p0 r
r+
r
s

r-q
p = qs is the dipole moment for a finite dipole
Physics BYU
72
Multipole expansion
• Legendre polynomials: coefficients of 1/ r
expansion in powers of   r ' / r  1
r  r  r'  r  r ' 2 r  r'  r 1  
2
2
where    2  2 cos 
and
1
3 2
1 / 2
(1   )
 1      ...
2
8
2

3
cos
 1 
2
  ...
 1   cos    
2


Physics BYU
73
• Multipole expansion of V:
using
1
1  r' 
1 rˆ  r '
    Pl (cos  ' )   2  ...
r l 0 r  r 
r r
1
1
l
V (r ) 
r ' Pl (cos  ' ) dq'

l 1 
4p0 l 0 r
l
1  Q p  rˆ

V (r) 
  2  ...
4p0  r
r

Q   dq'
and
Physics BYU
with
p   r ' dq'
74
Electric field of a dipole
• Choosing p along the z axis
1 p  rˆ
1 pz
Vdip (r ) 

2
4p0 r
4p0 r 3
1
z

and Edip (r)  Vdip (r)  
p  3 
4p0
r 
1 1
E dip (r ) 
[3(p  rˆ )rˆ  p]
3
4p0 r
Clifford form:
1
1
E dip (r ) 
(3 rˆprˆ  p)
3
4p0 2r
Physics BYU
75
Polarization
• P(r) polarization, vector field =
= dipole moment/volume.
1
 (r ' )
Compare:
Vch dist (r ) 
dt '

4p0
r
w/ potential due to dipole distribution:
rˆ  P(r ' )
V pol (r ) 
dt '
2

4p0
r
1
Physics BYU
76
Equivalent “bound charges”
• Integration by parts:
 P(r ' ) 
1 1
'
  P(r ' )  '    'P(r ' ) 
 r 
r r
rˆ  P(r ' ) 1

 P
2
r
r
so that:
1
1
1
1
V pol (r ) 
 b (r ' )dt '
 b (r ' )da'


4p 0 r
4p 0 r
Physics BYU
77
where
and
b    P and  b  nˆ  P
  P  0 only if P is not uniform
• surface charge:
 -• In a dielectric charge does not migrate.
It gets polarized.
Physics BYU
+
78
Electric Displacement field D(r)
• Total charge density: free plus bound
 (r )   f (r )  b (r )   f (r )    P
• Defining
we get:
D(r)   0E(r)  P(r)
D  f
• Gauss’s law for D:
and
E  0
 D  da  Q
Physics BYU
f enc
79
• Example: Find the electric field E produced
by a uniformly polarized sphere of radius R
1
rˆ
V (r) 
P   2 dt '
4p0 sph r
where
and
rˆ
4p
sph r 2 dt '  3
 r
 3
2
ˆ
R r / r
 1
P

E  V   3 0

dipole field
Physics BYU
rR
rR
80
• Example of Gauss’s law for D:
Long straight wire, uniform line charge ,
surrounded by insulation radius a. Find D.
D(2p sh )   h


D
sˆ
2p s
Outer region, we can calculate the electric
field:
1

P0  E D
sˆ for s  a
0
2p  0 s
Physics BYU
81
Linear Dielectrics
• D, E, and P are proportional to each other for
linear materials
0

k
ce


 / 0
k1
permittivity space
permittivity material
dielectric constant
susceptibility
D   E  k  0E
C  k C0
P  D   0E  (   0 )E   0 c e E
Physics BYU
82
• Theorem:
In a linear dielectric material with constant
susceptibility the bound charge distribution is
proportional to the free charge distribution.
ce
D

b    P      0 c e   
f

1  ce

Physics BYU
83
Boundary conditions
• Boundary between two linear dielectrics
 1 n   D   f and  E  0
2
Boundary
conditions:
E  D / 1 in 1, and E  D /  2 in 2
D  D   f n̂

1

2
V
V
1
2
n 1
n
Physics BYU
  f
2
84
• Example: Homogeneous dielectric sphere,
radius R in uniform external electric field E0
• Solution:
1
 b   f  0  E sph  
P
3 0
in terms of the unknown P.
Total field: E  E sph  E 0 and polarization:
eliminating P:
P  (k  1) 0E
Ε
3
E
E0
k 2
k 1
and P  3 0
E0
k 2
Physics BYU
85
Energy in dielectrics
• vacuum energy density: u (r )  1 ( E)  E
0
2
• dielectric: u (r )  1 D  E
2
• Energy stored in dielectric system:
1
W   D  E dt
2
• energy stored in capacitor:
1
k
2
2
W  CV  C0V
2
2
Physics BYU
86
Fringing field in capacitors
l
F
for fixed charge Q:
x
dW
1 2 d  1  1 2 dC

  V
F 
 Q
dx
2
dx  C ( x)  2
dx
where
so
 0w
w
C ( x)  [ 0 x   (l  x)] 
(k l  c e x)
d
d
F 
 0 wc e
2d
V2 0
Physics BYU
87
Magnetostatics
duality in Clifford space
•
•
•
•
•
•
•
Electric
Field lines: from + to E(r): vector field
pos. &neg. charge
 (r): scalar source
V (r): scalar potential
units [E] = V/m
E 
1
0
•
•
•
•
•
•
•
Magnetic
Field lines: closed
i B: bivector field
no magnetic monopoles
J(r): vector source
A (r): vector potential
units [B] = Vs/m2
iB  0 J

Physics BYU
88
Lorentz force
• Point charge q in an external magnetic field B
F  qv  B
• velocity dependent (but no-friction)
• perpendicular to motion: no work done
v
F  d l  q ( v  B)  v d t  0
Example: uniform magnetic field B
define cyclotron (angular)
qB

frequency:
m
Physics BYU
89
equation of motion:
v   nˆ  v where
solution:
B
nˆ 
B
v(t )  exp(  t nˆ ) v(0)
 v // (0)  cos( t ) v  (0)  sin(  t ) nˆ  v  (0)
with
v  (t ) 2  v  (0) 2  const .
2
and radius:
v
qv B  m
R
Physics BYU
so R 
v

90
Add uniform electric field E perpendicular to B:
qE
F  q (E  v  B )
v   nˆ  v 
m
a particular solution is given by:
En
v(t ) 
given that
En  0
B
and
E B
v(t )  exp( t nˆ ) v(0)  2
B
Applications:
• cyclotron motion
• velocity selector
Physics BYU
91
Physics BYU
92
Charge current densities
dq
I
dt
• current
[ I ] = Amp = C/s 
flow of charge per unit time
2

J


v
volume
(A/m
)
• current density

qv  K   v surface (A/m)
   v
wire
(A)

 J  B dt


F   v  B dq    K  B da

  Ι  B dl
• magnetic force
Physics BYU
93
• Relation between I and J
• a) consider cylinder, radius R, with uniform
steady current I
I
J
p R2
• b) for I  k s
2p 3
I   J  da   (ks)sdsd  2p  ks ds 
kR
3

0
R
2
Physics BYU
94
Charge conservation
• current across any closed surface
d
I   J  da    J dt     dt
dt
• local (differential) form:

  J  0
u
• steady currents:
 J  0
Physics BYU
95
Solving Magnetostatic Equation
• Two Maxwell equations rewritten as
 B  i0 J
formally solved as:
1
1
 1
B  i  0 2 J  i  0 2 i   J    0    2



B   A
Explicitly:
0
A(r ) 
4p
where
1
A   0 2 J

1
 0
  J (r )  
r
 4p
Physics BYU

J


1
J (r ') dt '
r
96
Using
J (r ') 
1  J (r ')  rˆ


 
  J (r ')     
2
r
r
r


 
the magnetic field is given in terms of the
vector source as:
0 J (r ')  rˆ
0 J (r ')  (r  r ')
B(r ) 
dt ' 
dt '
2
3


4p
r
4p
|r r'|
For current on a wire (Biot-Savart expression):
0 I(r ')  rˆ
0 I d l ' rˆ
B(r ) 
dl ' 
2

4p
r
4p  r 2
Physics BYU
97

Calculate the magnetic field due to a uniform
steady current I
Trig:
1 sin 2 
sd
z '   s cot 
I
in
s

r
2

s
2
dz ' 
sin 2 
Biot-Savart:
d l ' rˆ  dz 'sin  ˆ
0 I d l ' rˆ 0 I  sin 2    sd

ˆ
dB 

dl’
 2   2 sin   
2
4p r
4p  s   sin 

0 I sin 
dB 
d ˆ
4p s
p
0 I ˆ
0 I ˆ
Integration:
B(r) 
  sin  d 

4p s 0
2p s
Physics BYU
98
Force between two // currents
I2
I1
f
a
in
at wire 2:
I 2  B1
0 I1
B1 
2p a
F12  I 2 B1l
attractive force per unit length:
F12 0 I1 I 2
f

( sˆ2 )
l
2p a
Physics BYU
99
Ampere’s Law
• using Stokes’ theorem



(

B
)
d
a

B

 dl
S
S
 Bdl   I
we get
over “Amperian loop”.
Symmetric case:
0 enc

B L  0 I enc  0
J da
surf
Physics BYU
100
Applications of Ampere’s law
B tangent to Amperian loop
BL  0 I
where L  2p s
0 I ˆ
B

2p s
Current in x direction
K  Kiˆ so B(2l )  0  0 Kl
B  (sgn z )
Physics BYU
0
2
K ˆj
101
• Solenoid:
•
•
•
•
n turns/length
current I, K  nIˆ
magnetic field in z direction
Outer loop: B(a)  B(b)  0
Straddling loop: BL  0 nIL
B   nI kˆ inside; B  0 outside
0
• Torus:
Amperian loop inside the torus
N = TOTAL number of turns
B(2p s)  0 I B  0 N I ˆ inside; B  0 outside
2p s
Physics BYU
102
Vector Potential A
• Poisson equation:
• Solution:
 2 A   0 J
0 J (r ')
A(r )  
dt '

4p
r
• Disadvantages:
integral diverges
it is a vector field (not scalar)
• Advantage: dA  dJ in the same direction
Physics BYU
103
Examples:
• Long thin wire with current I
0 I
A
4p
L

L
0 I
dz '

2 ln( z  z 2  s 2 )
z '2  s 2 4p
0
0 I
 2L 

2 ln 

4p
 s 
L
so
0 I  s  ˆ
0 I ˆ
A(r)  
ln   k and B(r) 

2p  s0 
2p s
Physics BYU
104
• Example: Find the vector field corresponding
to a uniform magnetic field.
• Using Stokes’ theorem:
 A  d l   ( A)  da   B  da  magnetic2 flux
A(2p s)  B(p s )
• Amperian loop FOR A
1
1 ˆ
1
ˆ
A(r )  Bs   B(k  s)  B  r
2
2
2
• For a solenoid
B  0 nI inside, and
2

1 ˆ s 0 nI ˆ  s
A(r)  B   2 
 2
2s
2
a / s
a
Physics BYU
105
Boundary conditions for B
• Integrating Maxwell’s equation for B
B  i  0 J
1
ˆ 2  i 0 K
nB
1
2
or, equivalently B1  B2  i0nK
ˆ  0K  nˆ
• B  is CONTINUOUS
//
//
B

B
while
1
2  0 K
Physics BYU
106
Multipole expansion: magnetic moment
• Applying the expansion
1
1  r' 
1 rˆ  r '
    Pl (cos  ' )   2  ...
r l 0 r  r 
r r
to the vector potential for a CLOSED loop:
l
0 I 1
A(r ) 
dl ' 

4p
r
0 I  1
1


  d l '  2  rˆ  r ' d l '  ... 
4p  r
r

Physics BYU
107
• The dipole term is:
0 I 1
A dipole (r ) 
4p r 2

(rˆ  r ')d l '
• Using the fundamental theorem in 2-d
i  da      d l, (r)  C  r a scalar field
S
S
  (kˆ  r )  kˆ and  ' (r ')   '(rˆ  r ')  rˆ
Physics BYU
108
• Finally:
m
Area
 (rˆ  r ')d l '  i  d a ' rˆ    d a '   rˆ
0 m  rˆ
A dipole (r ) 
2
4p r
where the magnetic dipole moment is:
m  I  da  I Area
Curl:
mr  1
3

  3   3  (m  r )  4 rˆ  (m  r )
r
 r  r
Physics BYU
109
Magnetic field for a dipole
0 1
B   A 
[3(m  rˆ ) rˆ  m]
3
4p r
compared to the electric dipole field:
1
1
E
[3(p  rˆ ) rˆ  p]
3
4p 0 r
Physics BYU
110
Magnetism in Matter
• Paramagnets
• Diamagnets
• Ferromagnets
M // B
M // -B
nonlinear - permanent M
Dipoles
p  qs
mIa
F  0 uniform Ε
F  0 uniform B
N  pΕ
torque
N  mB
torque
W  p  Ε
W  m  B
F  (p  Ε) point dipole F  (m  B) point dipole
Physics BYU
111
Atomic diamagnetism
B
add B 
-e
ev
I 
2p a
ev
2ˆ
m
pa k
2p a
m0 v 2
Fcent 
a
Energy:
-e
m
m0v
2m0
DF  D

v Dv  evB
a
a
eBa
Dv 
2m0
2 2
e ˆ
ea
Dm   a k Dv  
B
2
4m0
2
eva
e
ˆ B  
W  m  B 
m
LB
2m0
Physics BYU 2
112
Magnetization
• Magnetization = (dipole moment)/volume
• Vector potential: d A '  0 d m(r ')  rˆ
2
4p
r
0
0
rˆ
 1
A(r ) 
M (r ')  2 dt ' 
M (r ')    '  dt '


4p
r
4p
 r
0  ' M (r ') 0 M (r ')  d a '
A(r ) 




4p
r
4p
r
0 J b (r ') 0 K b (r ')d a '



4p
r
4p 
r
Physics BYU
113
Sphere - constant magnetization
• sphere radius R - uniform M
 r inside
0
0
rˆ
 3
A(r ) 
M   2 dt ' 
MR
4p
r
3
 2 rˆ outside
r
2

0 M inside

B   A  
3
(dipole @ center) outside
with magnetic moment
Physics BYU
4
3
m  pR M
3
114
A useful integral over the sphere
I 0 (r ) 

sph
rˆ
dt '
2
r
corresponds to electric
field w/ constant density:
  4p 0
using Gauss’s law:
 r inside
4p  3
I 0 (r ) 
R
3  2 rˆ outside
r
Applications: constant charge density,
polarization, and magnetization
Physics BYU
115
• Bound current densities:
J b (r)   M(r)
and
K b (r)  M(r)  nˆ
• Auxiliary field H
1
0
 B  J  J f  J b  J f    M
• Defining
H
1
0
BM
we get
Physics BYU
 H  J f
116
• An equivalent problem: sphere with uniform
surface charge  spinning with constant
angular velocity .
Tangential velocity & surface current

v  ωs  ωR
R
K   v   ωR
equivalent to uniform M w/bound Kb
K b  M  rˆ , so M    R
2
B  0 R ω inside spinning sphere
3
Physics BYU
117
• Ampere’s law for magnetic materials

H  d l  I f enc
• Experimentally: I & V  H & E
Example: copper rod of radius R carrying a
uniform free current I
p s 2 / p R 2 inside
Amperian loop H (2p s)  I 
1 outside

radius s:
0 I ˆ
outside M = 0 and B  0 H 

(s  R)
2p s
Physics BYU
118
Boundary conditions for B & H
• Integrating Maxwell’s equations:
 H  J f
 B  0
nˆ  H 2  K f
nˆ  B 2  0
1
1
or, equivalently
H1//  H2//  K f  nˆ
Physics BYU
and B1  B2
119
Linear Magnetic Materials
• Susceptibility and permeability:
M  cm H
and B  0 (H  M)   H
  0 1  c m 
magnetic permeability
BH
• diamagnetic
paramagnetic
Gadolinium
~ -10-5
~ +10-5
~ +0.5
Physics BYU
120
• Theorem:
In a linear homogeneous magnetic material
(constant susceptibility) the volume bound
current density is proportional to the free
current density :
J b   M   ( c m H )  c m J f
Bound surface current:
K b  M  nˆ  c m H  nˆ
example: solenoid
K b  cm n I ˆ
Physics BYU
121
Ferromagnetism
• Permanent magnetization (no external field
necessary)
• non-linear relation between M and I
• hysteresis loop
• dipole orientation in DOMAINS
• magnetic domain walls disappear with
increasing magnetization
• phase transition (Curie point): iron T ~ 770 C
Physics BYU
122