Physics BYU 1

Physics 441

Electro-Magneto-Statics

M. Berrondo

Physics BYU 2

JFJF~

~ 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.)

Physics BYU 3

ToolsMath

• trigonometry• vectors (linear combination)

dot, cross, Clifford• vector derivative operators

• Dirac delta function• DISCRETE TO CONTINUUM• INTEGRAL THEOREMS:

* Gauss, Stokes, FundThCalc• cylindrical, spherical coord.• LINEARITY

Physics

,,,,/ ix

• 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 4

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 nn = n2 = 1 • orthonormal basis: {e1 , e2 , e3 } = {i , j , k}

Physics BYU 5

Geometric Interpretation:Dot product

Physics BYU 6

Cross Product and triple dot product

Physics BYU 7

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 • k x followed by k x again

equivalent to multiplying by -1 in this case!!

y

s’

s x

jijiks

jis

xyyx

yx

)('

Physics BYU 8

n unit vector: n2 = 1 defines rotation axis = rotation angle vector r r’

wherernrrr sincos' //

Rotation of a vector (3-d)

rrrrr

rrnn

n

ee

e

//// )('

'

)(

)(

// rnnrrr

rnnr

Physics BYU 9

Triple dot product

is a scalar and corresponds to the (oriented) volume of the parallelepiped {A, B, C}

zyx

zyx

zyx

zyx

zyx

CCC

BBB

AAA

CCC

BBB

eee

A

A)(CBB)(ACC)(BA

321

Physics BYU 10

Triple cross product

• The cross product is not associative!• Jacobi identity:

• BAC-CAB rule:

is a vector linear combination of B and C

0)()()( ACBBACCBA

)()()( BACCABCBA

Physics BYU 11

0 1 | | n-1 = n as a unit vector

along any direction in R2 or R3

211 )(

AAA

AnnA

Inverse of a Vector

Physics BYU 12

• 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:

• with

BABAAB i

Clifford Algebra Cl3 (product)

321 eeei

Physics BYU 13

1 0 scalar

ê1 ê2 ê3 1 vector

iê1= ê2ê3 iê2= ê3ê1 iê3=ê1ê2 2 bivector

i = ê1ê2ê3 3 pseudoscalar

Ri R

R3

i R3 ikikkiki 2ˆˆˆˆˆ,ˆ

,1)(

,12

211221

23

22

21

eeeeee

eeeeee

eee

BA ii C

Physics BYU 14

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 15

Subalgebras • R Real numbers• C = R + iR Complex numbers• Q = R + iR3 QuaternionsProduct of two vectors is a quaternion:

represents the oriented surface (plane)orthogonal to A x B.

BABAAB i

2/)( BAABBAAB scalar

BABAABBAAB 2/)(ibivector

Physics BYU 16

Bivector: oriented surface

sweepbaba ia

b

a

bsw

eep

baabab i

• antisymmetric, associative• absolute value area

Physics BYU 17

Differential Calculus• Chain rule:

In 2-d:

• Is an “exact differential”?

• given

i

ii

dxx

fdf

dyy

fdx

x

fdfyxff

),(

xdyydx

xdyydx

ydyxdx

2

2

x

B

y

AdyyxBdxyxA

, ),(),(

Physics BYU 18

• In 3-d:

• Geometric interpretation:

hpat oft independen )()(.3

0.2

.1

b

a

aFbFdf

df

dz

dy

dx

z

f

y

f

x

fdfdf r

ascentsteepest ofdirection in points

// when max

when 0cos

T

dT

dTdTdT

r

rr

Physics BYU 19

r

r

ik

ˆ)(

ˆ

ˆ)( ,ˆ

dr

dfrf

rdx

dgxgz

32ˆ

11

rrr

rr

zyx

321 ˆˆˆ eee del operator

Examples:

Physics BYU 20

Gradient of r• Contour surfaces: spheres • gradient is radial

• Algebraically:

and, in general,

1

)(

ˆ// and )(

r

drrdrdrdf

rrf

r

rr

rr ˆ 2)(2 2222 rzyxrrr

rr ˆˆ)(')(dr

dfrfrf

r̂r

Physics BYU 21

Divergence of a Vector Field• E (r) scalar field (w/ dot product)

• It is a measure of how much the filed lines diverge (or converge) from a point (a line, a plane,…)

z

y

x

zyx

E

E

E

zyx

z

E

y

E

x

EE

Physics BYU 22

• Divergence as FLUX: ( ) dxExamples:

dxx

ExEdxxE x

xx

)()(

0for 0ˆ

))((1

))(ˆ(

3

1)ˆ( 0ˆ

2

22

rr

rFrdr

d

rrF

z

r

r

r

kk

Physics BYU 23

Curl of a Vector Field

The curl measures circulation about an axis.Examples:

ErrEE

r

rr

rEEEzyx

rzyx sin

ˆsinˆˆ

sin

1

ˆˆˆ

2

φθrkji

E

kEjirE

kEjrE

ˆ2 ˆˆ)(

ˆ ˆ)(

xy

x

Physics BYU 24

BBB ii )(

EEE i

TT grad• For a scalar field T = T( r ),

• For a vector field E = E( r ),

• For a bivector field i B = i B( r ),

Clifford product del w/ a cliffor

Physics BYU 25

Second order derivatives• For a scalar:

• For a vector:0)(

),(

)()()(2

2

T

TT

TiTTT

0)(

),()(

)()()(2

2

E

EEE

EEEE i

Physics BYU 26

What do we mean by “integration”? cdq

c dr

dw = c(cdq)/ 2 dg = (2pr) dr

| | dl

rdq dr

da= (rdq)dr

dxdx

dydydxdl

222 1)()(

Physics BYU 27

Cliffor differentials• dka 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 28

V

kk

V

k FdFd )()()( 11

Fundamental Theorem of Calculus

Particular cases: )()()( ablb

a

TTdT

VV

dd aEE

SS

dd lBaB

Gauss’s theorem

Stokes’ theorem

Physics BYU 29

Delta “function” (distribution)• 1-d:

q step “function”

1

x

areaunit ,1')'(')'( ,0for 0)(

dxxdxxxx

nconvolutiounit )()]([

)(')'()'( ,)0(')'()'(

xfxf

xfdxxfxxfdxxfx

scaling )(||

1)(

ondistributi a as )(

)(

xa

ax

dx

xdx

Physics BYU 30

• Divergence theorem and unit point source apply to

for a sphere of radius e

VV

dd aEE 2

ˆ)(

r

rrE

VVVV

dΩdΩrr

dr

d 4ˆˆˆ 222r

ra

raE

4ˆ2

d

rd

VV

rE

Physics BYU 31

and

Displacing the vector r by r‘:

)(4ˆ )3(2

rr

r

).(4

1 )3(2 r

r

).'(|'|4

1 )3(2 rrrr

Physics BYU 32

Inverse of Laplacian• To solve

so

• In short-hand notation:

)()(2 rr BA

)(

1

4

1')'(

|'|

1

4

1)(

1)(

2rr

rrrr B

rdBBA

)(')'(|'|

1

4

12 rrrr

BdB

|'| where,')'(1

4

1)( rrrr

rr

dBA

Physics BYU 33

Orthogonal systems of coordinates

• coordinates: (u1, u2, u3 )

• orthogonal basis: (e1, e2, e3 )

• scale factors: (h1, h2, h3 ) • volume: • area

• displacement vector:

333222111 eeel duhduhduhd

321321(vol) dududuhhhdd 3212133 : ea duduhhdu

Physics BYU 34

Scale Factorsdu1 du2 du3 h1 h2 h3

Cartesian dx dy dz 1 1 1Cylindrical ds df dz 1 s 1Spherical dr dq df 1 r r sinq

• polar (s, f ):

• cylindrical (s, f, z ):

• spherical (r, q, f ):

ss ˆs

3ˆˆ esr zs

rr ˆr

Physics BYU 35

• Grad:

• Div:

• Curl:

• Laplacian:

33

3

22

2

11

1 ˆˆˆ

u

T

hu

T

hu

T

hT

uuu

3

321

2

321

1

321

321

)()(1

u

hh

u

hh

u

hh

hhh

EEEE

332211

321

332211

321

ˆˆˆ1

EEE

uuu

E

hhhuuu

hhh

hhh

33

21

322

31

211

32

1321

2 1

u

T

h

hh

uu

T

h

hh

uu

T

h

hh

uhhhT

Physics BYU 36

Maxwell’s Equations

• Electro-statics:

• Magneto-statics:

• Maxwell:

)(1

)(0

rrE

)())(( 0 rJrB i

)(1

)(0

J

cc

icBEF

3771

00

cc

Physics BYU 37

Formal solution

separates into:

and

JFJF~

)(~

)( 1 rr

02

0

1

0

111 )(

1)( ErrE

JBrJrB 020

1 )()(

ii

Physics BYU 38

Electro-statics

Convolution:

where for point charge.

)()()(1

4

1)( 0

0

rrErrE

r

''ˆ

4

1''

||

'

4

1)(

20

30

dr

d rr

rr'r

rrrE

)()( rrE V

20

0

ˆ

4

1)(

r

rrE

')(||

1

4

1)( where

0

dV r'r'r

r

Physics BYU 39

Superposition of charges• For n charges {dq1, dq2, …, dqn }

• continuum limit:

• where

n

j ji

jn

j ji

ji

dq

r

dqV

1010 ||4

1

4

1)(

rrr

)()( |'|

'

4

1)(

0

rrErr

r Vdq

V

')(

')(

')(

'

d

da

ld

dq

r'

r'

r'

Physics BYU 40

• linear uniform charge density l (x’) from –L to L• field point @ x = 0, z variable z q | | x

-L Ldq’

tan'

sec'

'

''

22

zx

zzxr

xz

dxdq

ikr

'4

1)(

30

dxr

L

L

r

rE

Physics BYU 41

with

so

• Limits:

000

33

22

0

sinˆ2

4

1

sec

sec

4

ˆ2)(

0

zd

z

z kkrE

,sec' 2 dzdx

.ˆ2

4

1)(

220

krEzL

L

z

fixed , ˆ1

4

ˆ2

1

)(

20

0

Lzr

q

Ls

r

srE

Physics BYU 42

Gauss’s law• Flux of E through a surface S: volume V enclosed by surface S.• Flux through the closed surface:

• choose a “Gaussian surface” (symmetric case)

S

E dΦ aE

VS

enc dQd 00

11aE

symmetryby determined ˆ 1

.).Area(0

EencQSGE

Physics BYU 43

Examples:• Charged sphere (uniform density) radius R Gaussian surface:a) r < R:

b) r > R: as if all Q is concentrated @ origin

24 rA

RrR

rQ

R

rQdQ

rVol

enc

for ˆ4

)(

')(

30

3

3

)(

rrE

r'

Rrr

QQQenc for

ˆ

4)(

20

rrE

Physics BYU 44

• Thin wire: linear uniform density (C/m) Gaussian surface: cylinder

• Plane: surface uniform density (C/m2) Gaussian surface: “pill box” straddling plane

CONSTANT, pointing AWAY from surface (both sides)

lQslA enc 2

s

lslE

ssE

ˆ

2)( )2(

00

kE ˆ2

)2(00

A

AE

Physics BYU 45

Boundary conditions for E• Gaussian box w/ small area D A // surface w/

charge density s with n

pointing away from 1

• Equivalently:

• Component parallel to surface is continuous• Discontinuity for perp. component = /s e0

1

2ˆEn

nEE ˆ0

21

Physics BYU 46

Electric Potential (V = J/C)

• Voltage

solves Poisson’s equation:

• point charge Q at the origin

r

dqd

rVV

'

4

1')'(

1

4

1)(*)()(

000

rrrr

)(1

)(0

2 rr

V

)()( rr Q

r

QV

04

1)(

r

47

Potential Difference (voltage)• in terms of E: and

• Spherical symmetry: V = V (r)

• Potential energy: U = q V (joules)• Equi-potential surfaces: perpendicular to field

lines Physics BYU

r

a

lEar dVV )()(

rrrE rdr

rdV whereˆ

)()(

0lE d

Physics BYU 48

• Example: spherical shell radius R, uniform surface charge density s Gauss’s law E(r) = 0 inside (r < R) For r > R:

and

22

0

4 e wherˆ4

1)( Rq

r

q rrE

r

qdr

r

qdrV

r

02

0 4

1

4

1)(

r

lE

Physics BYU 49

• Example: infinite straight wire, uniform line charge density l Gauss’s law: and

ˆ2

1)(

0

srEs

s

ads

sdrV

s

a

ln22

1)(

00

r

a

lE

Physics BYU 50

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 51

Perfect conductors

• Charge free to move with no resistance• E = 0 INSIDE the conductor• r = 0 inside the conductor• NET charge resides entirely on the surface• The conductor surface is an EQUIPOTENTIAL• E perpendicular to surface just OUTSIDE• E = s/e0 locally• Irregularly shaped s is GREATEST where

the curvature radius is SMALLEST

Physics BYU 52

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 53

)(2

)( 20 rr Eu

Work and Energy• Work:• for n interacting charges:

• continuous distribution r

• Energy density: also electric PRESSURE

VQVVQdQW )]()([ ablEb

a

n

iii

n

ji ji

ji Vqr

qqW )(

2

1

2

1

4

1

0

r

aEE dVdEVdVdW 200

2)(

22

1

54

d

AC 0

Capacitors (vacuum)• Capacitor: charge +Q and –Q on each plate - - - - - - + + + + + +

• define capacitance C : Q = C DV with units 1F = 1C /1V

Physics BYU

000

so ,

A

dQdEV

A

QE

Ed

Physics BYU 55

• charging a capacitor C: move dq at a given time from one plate to the other adding to the q already accumulated

• in terms of the field E

energy density

22

1

22 CV

C

QWdq

C

qdW

(Vol)2

1)(

2

1 20

20 EEdd

AW

)(2

)( 20 rr Eu

Physics BYU 56

Laplace’s equation• V at a boundary (instead of s)

one dimension:

capacitor V0 V1 (0 to d)

V(x) is the AVERAGE VALUE of V(x-a) and V(x+a)

02 VbmxxV

dx

Vd )( 0

2

2

dVxd

VVxV

0 from )( 0

01

Physics BYU 57

• Harmonic function has NO maxima or minima inside. All extrema at the BOUNDARY

• Average value (for a sphere):

Proof:

daVrVV )(Area

1)0(0 r

0

)()(

22

2

Vd

dR

dRd

r

VR

dVdVVd a

RRVVda

RVVd

finitefor 4

0for 4

20

0

Physics BYU 58

Method of images• Problem: find V for a point charge q facing an

infinite conducting plane

equivalent to

potential: charge density:

q -q q

r

2104

1)(

r

q

r

qV

r

plane0 n

V

Physics BYU 59

Laplace’s equation in 2-dComplex variable z = x + iy w’(z) well defined

both real and imaginary parts of w fulfillLaplace’s equation

0)( 2

2

2

2

iyxwyx

0for

0for )(

dxy

wi

dyx

w

idydx

dyyw

dxxw

dz

zdw

Physics BYU 60

Examples and equipotentials in 2-d

• w(z) = z V(x, y) = yequipotential lines: y = a (const)

• w(z) = z2 V(x, y) = 2xy, or x2 - y 2

e. l.: xy = a (hyperbolae)

• w(z) = ln z V(s, f) = ln s, arctan(y/x) e. l.: s = exp(a)

• w(z) = 1/z = exp(-i f) /s V(s, f) = cosf /s e. l.: s = cosf /a (circles O)

Physics BYU 61

• V(x, y) = finite dipole

• w(z) = z + 1/z V(s, f) = (s – 1/s) sin f e. l.: s = 1 for a = 0

• w(x) = exp(z) V(x, y) = exp(x) cos yVk(x, y) = exp(k x) cos (k y) & exp(k x) sin (k y)

2/1

2/1ln)(

z

zzw

Physics BYU 62

Separation of variables (polar)• Powers of z:

• Solution of

innn sz e

)sin(

)cos()( and )(

re whe)()(),(

n

nssS

sSsV

nn

n

nnn

)]sin()cos()[(ln1

00 ndncsbsasaVV nnn

nn

nn

011

2

2

2

V

ss

Vs

ss

Physics BYU 63

Separation of variables (x, y)• k 2 = separation constant

• Eigenvalue equation for X and Y

• Construct linear combination, determine coefficients w/ B.C. using orthogonality

0111 22

2

2

2

22 kk

dy

Yd

Ydx

Xd

XV

V

)()(),( yYxXyxV

)()(

& )()( 2

2

22

2

2

yYkdy

yYdxXk

dx

xXd

Physics BYU 64

• Example:

solution:

B.C.: ii) A = 0; iii) D = 0; iv) quantization k = 1, 2, 3, …

0)( iv) 0)0( iii)

0)( ii) )0( i) 0

yVyV

xVVxV

)cos()sin()(

ee)(

kyDkyCyY

BAxX

k

kxkxk

)sin(e),(1

nxCyxVnx

nn

Physics BYU 65

• orthonormality:

i)

solution:

0 2)sin()sin( mndymyny

0

00 odd 4

even 0)sin(

2m

m

Vm

dymyVCm

odd

n

nx nyn

VyxV

1

0 )sin(e14

),(

Physics BYU 66

Separation of variables - spherical• Assume axial symmetry, i.e. no f dependence

separation constant l (l +1)

sinsin

112

22

2

rrr

rr

)()1()(

sinsin

1

)()1()(

)()(),(

2

lld

d

d

d

rRlldr

rdRr

dr

d

rRrV

Physics BYU 67

Legendre polynomials• change of variable

• orthogonality

• “normalization” Pl (1) = 1

)(cos)()( ll PP

0)()1()(

)1( 2

ll Plld

Pd

d

d

mlml ldPP

12

2)()(

1

1

Physics BYU 68

• Radial function R(r) = r l or r – l – 1

)(cos),(0

1 l

llll

l Pr

BrArV

...cos

......cos

210

1010

r

B

r

B

zAArAA

Physics BYU 69

• Given V0( q ) 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:

using orthogonality:

0

1

00

)(cos

)(cos)(

ll

ll

ll

ll

PRB

PRAV

RrdPVR

lA lll

sin)(cos)(

2

12

0

0

RrdPVRl

B ll

l

sin)(cos)(2

12

0

01

Physics BYU 70

• Uncharged conducting sphere in uniform electric field E = E0 k

V(r = R) = 0:

V -E0 r cos q r >> R:

121

0 l

lllll

l RABR

BRA

cos)(

0

2

3

0

013

1100

r

RrEV

EARABBA

r

...cos)()(21

10

0 r

BrA

r

BAV

Physics BYU 71

cos3)( 000 Er

V

Rr

Physics BYU 72

Electric Dipole• finite dipole

as

p = qs is the dipole moment for a finite dipole

+q -q

r+

rr-

qs

rr

qrV

11

4)(

0

cos11

and

cos2

111

2r

s

rr

r

s

rrr

cos1

4)(

20 r

qsV r

Physics BYU 73

Multipole expansion• Legendre polynomials: coefficients of 1

expansion in powers of

where and

12'22 r'rr'r rrrr

cos22

1/' rr

...2

1cos3cos1

...8

3

2

11)1(

22

22/1

r/

Physics BYU 74

• Multipole expansion of V: using

with

...'ˆ1

)'(cos'11

20

rrP

r

r

rr l

l

l

rr

0

10

')'(cos'1

4

1)(

ll

ll

dqPrr

V

r

...

ˆ

4

1)(

20 rr

QV

rpr

'' and ' dqdqQ rp

Physics BYU 75

]ˆ)ˆ(3[1

4

1)(

30

prrprE rdip

Electric field of a dipole• Choosing p along the z axis

and

Clifford form:

30

20 4

1ˆ

4

1)(

r

pz

rVdip

rp

r

304

1)()(

r

zpVdipdip

rrE

)ˆˆ3(2

1

4

1)(

30

prprrE rdip

Physics BYU 76

Polarization

• P(r) polarization, vector field = = dipole moment/volume. Compare:

w/ potential due to dipole distribution:

')'(

4

1)(

0

dr

V distch r

r

')'(ˆ

4

1)(

20

dr

Vpol

rPr

r

Physics BYU 77

Equivalent “bound charges”• Integration by parts:

so that:

PrPr

rPrPrP

rr

rrr

1)'(ˆ

)'('11

')'()'(

'

2

')'(1

4

1')'(

1

4

1)(

00

dar

dr

V bbpol rrr

Physics BYU 78

where

and only if is not uniform

• surface charge:

-- +• In a dielectric charge does not migrate. It gets polarized.

PnP ˆ and bb

0 P P

Physics BYU 79

0 E

Electric Displacement field D(r)• Total charge density: free plus bound

• Defining

we get: and

• Gauss’s law for D:

Prrrr )()()()( fbf

)()()( 0 rPrErD

f D

encfQad D

Physics BYU 80

• Example: Find the electric field E produced by a uniformly polarized sphere of radius R

where

and

sph

dr

V 'ˆ

4

1)(

20

rPr

232 /ˆ3

4'

ˆ

rRd

rsph r

rr

Rr

RrV

field dipole

3

1

0

PE

Physics BYU 81

• Example of Gauss’s law for D: Long straight wire, uniform line charge l,

surrounded by insulation radius a. Find D.

Outer region, we can calculate the electric field:

hshD )2( sD ˆ2 s

ass

for ˆ2

1 0

00

sDEP

Physics BYU 82

Linear Dielectrics• D, E, and P are proportional to each other for

linear materials e 0 - permittivity spacee - permittivity materialk e / e 0 dielectric constantce - k 1 susceptibility

EED 0

EEEDP e 000 )( 0CC

Physics BYU 83

• Theorem: In a linear dielectric material with constant

susceptibility the bound charge distribution is proportional to the free charge distribution.

fe

eeb

10

DP

Physics BYU 84

Boundary conditions

• Boundary between two linear dielectrics e 1 n e 2

Boundary conditions:

1 2

and 0

/ in 1, and / in 2

f

D E

E D E D

f

f

n

V

n

V

22

11

21 n̂DD

Physics BYU 85

• Example: Homogeneous dielectric sphere, radius R in uniform external electric field E0

• Solution:

in terms of the unknown P. Total field: and polarization: eliminating P:

PE03

1 0

sphfb

0EEE sph

EP 0)1(

0 0 0

3 1 and 3

2 2

E E P E

Ε

Physics BYU 86

Energy in dielectrics• vacuum energy density:

• dielectric:

• Energy stored in dielectric system:

• energy stored in capacitor:

EEr )(2

1)( 0u

EDr 2

1)(u

dW ED2

1

20

2

22

1VCCVW

Physics BYU 87

02

20 Vd

wF e

Fringing field in capacitors l F for fixed charge Q: x

where

so

dx

CdV

xCdx

dQ

dx

dWF 22

2

1

)(

1

2

1

)()]([)( 00 xl

d

wxlx

d

wxC e

Physics BYU 88

Magnetostaticsduality in Clifford space

• Electric• Field lines: from + to -• E(r): vector field• pos. &neg. charge• r (r): scalar source• V (r): scalar potential• units [E] = V/m

• Magnetic• Field lines: closed• i B: bivector field• no magnetic monopoles• J(r): vector source• A (r): vector potential• units [B] = Vs/m2

0

1E JB 0i

Physics BYU 89

Lorentz force• Point charge q in an external magnetic field B

• velocity dependent (but no-friction)• perpendicular to motion: no work done

Example: uniform magnetic field B define cyclotron (angular) frequency:

BvF q

0)( tdqd vBvlF

m

Bq

v

Physics BYU 90

equation of motion:

solution:

with and radius:

B

Bnvnv ˆ whereˆ

)0(ˆ)sin()0()cos()0(

)0()ˆexp()(

//

vnvv

vnv

tt

tt

.const)0()( 22 vv t

v

RR

vmBqv so

2

Physics BYU 91

Add uniform electric field E perpendicular to B:

a particular solution is given by:

and

Applications:• cyclotron motion• velocity selector

( )q F E v B ˆq

m

Ev n v

( ) given that 0tB

E n

v E n

2ˆ( ) exp( ) (0)t t

B

E B

v n v

Physics BYU 92

Physics BYU 93

Charge current densities• current [ I ] = Amp = C/s flow of charge per unit time• current density

• magnetic force

dqI

dt

2 volume (A/m )

surface (A/m)

wire (A)

q

J v

v K v

v

d

dq da

dl

J B

F v B K B

Ι B

Physics BYU 94

• Relation between I and J

• a) consider cylinder, radius R, with uniform steady current I

• b) for

2

IJ

R

I k s

2 3

0

2( ) 2

3

R

I d ks sdsd ks ds kR

J a

Physics BYU 95

Charge conservation

• current across any closed surface

• local (differential) form:

• steady currents:

dI d d d

dt J a J

0u

J

0 J

Physics BYU 96

Solving Magnetostatic Equation• Two Maxwell equations rewritten as

formally solved as:

where Explicitly:

0i B J

0 01 1 ( ) ( ) ( ') '

4 4d

r r

A r J r J r

0 0 02 2 2

1 1 1i i i

B J J J

B A 0 2

1

A J

Physics BYU 97

Using

the magnetic field is given in terms of the vector source as:

For current on a wire (Biot-Savart expression):

2

ˆ( ') 1 ( ')( ')

r r r

J r J r r

J r

0 02 3

ˆ( ') ( ') ( ')( ) ' '

4 4 | ' |d d

r

J r r J r r r

B rr r

0 02 2

ˆ ˆ( ') '( ) '

4 4

I ddl

r r

I r r l rB r

98

0 0

0

ˆ ˆ( ) sin4 2

I Id

s s

B r

Calculate the magnetic field due to a uniform steady current I

Trig:

Biot-Savart:

Integration:

Physics BYU

dl’

f

s

I

q

in

2

2 2 2

1 sin' cot '

sin

sdz s dz

r s

ˆˆ' 'sind dz l r2

0 02 2 2

ˆ' sin ˆsin4 4 sin

I Id sdd

r s

l rB

0 sin ˆ4

Id d

s

B

Physics BYU 99

I2

a

I1

in

f

Force between two // currents

at wire 2:

attractive force per unit length:

0 11 2

IB

a

2 1I B 1 2 2 1F I B l

1 2 0 1 22ˆ( )

2

I Is

l a

F

f

Physics BYU 100

0 encd I B l

• using Stokes’ theorem

we get over “Amperian loop”. Symmetric case:

Ampere’s Law

0 0enc

surf

L I d B J a

( )S S

d d

B a B l

Physics BYU 101

0 ˆ(sgn )2

z K j

B

0 ˆ2

I

s

B

Applications of Ampere’s law B tangent to Amperian loop

Current in x direction

0 where 2BL I L s

0ˆ so (2 ) 0Ki B l Kl K

Physics BYU 102

• Solenoid: n turns/length• current I,• magnetic field in z direction• Outer loop:• Straddling loop:

• Torus: Amperian loop inside the torus N = TOTAL number of turns

ˆnIK

( ) ( ) 0a b B B

0BL nIL

0ˆ inside; 0 outsidenI k B B

0(2 )B s I 0 ˆ inside; 0 outside2

N I

s

B B

Physics BYU 103

Vector Potential A • Poisson equation:

• Solution:

• Disadvantages: integral diverges it is a vector field (not scalar)• Advantage: in the same direction

20 A J

0 ( ')( ) '

4d

r

J r

A r

d dA J

Physics BYU 104

Examples:• Long thin wire with current I

so

2 20 0

2 20

0

'2 ln( )

4 4'

22ln

4

LL

L

I IdzA z z s

z s

I L

s

0 0

0

ˆˆ( ) ln and ( )2 2

I Is

s s

A r k B r

Physics BYU 105

• Example: Find the vector field corresponding to a uniform magnetic field.

• Using Stokes’ theorem: magnetic flux• Amperian loop FOR A

• For a solenoid inside, and

( )d d d A l A a B a2(2 ) ( )A s B s

1 1 1ˆ ˆ( ) ( )2 2 2Bs B A r k s B r

0B nI2

022

1 ˆ ˆ( )/2 2

ss nIB

a ss a

A r

Physics BYU 106

Boundary conditions for B • Integrating Maxwell’s equation for B

or, equivalently • is CONTINUOUS while

0

1

02ˆ

i

i

B J

nB K

1 2 0 0ˆ ˆi B B nK K n

1

2

B/ / / /1 2 0 B B K

Physics BYU 107

Multipole expansion: magnetic moment• Applying the expansion

to the vector potential for a CLOSED loop:

...'ˆ1

)'(cos'11

20

rrP

r

r

rr l

l

l

rr

0

02

1( ) '

41 1

ˆ' ' ' ...4

Id

rI

d dr r

A r l

l r r l

Physics BYU 108

• The dipole term is:

• Using the fundamental theorem in 2-d

and

02

1ˆ( ) ( ') '

4dipole

Id

r

A r r r l

, ( ) a scalar field S S

i d d

a l r C r

ˆ ˆ( ) k r k ˆ ˆ' ( ') '( ') r r r r

Physics BYU 109

• Finally:

where the magnetic dipole moment is:

Curl:

02

ˆ( )

4dipole r

m rA r

AreaI d I m a

m

Area

ˆ ˆ ˆ( ') ' ' 'd i d d r r l a r a r

3 3 4

1 3ˆ( ) ( )

r r r

m r

m r r m r

Physics BYU 110

Magnetic field for a dipole

compared to the electric dipole field:

03

1ˆ ˆ[3( ) ]

4 r

B A m r r m

30

1 1ˆ ˆ[3( ) ]

4 r E p r r p

Physics BYU 111

Magnetism in Matter• Paramagnets M // B• Diamagnets M // -B• Ferromagnets nonlinear - permanent M

Dipoles

0 uniform

torque

( ) point dipole

q

W

p s

F Ε

N p Ε

p Ε

F p Ε

0 uniform

torque

( ) point dipole

I

W

m a

F B

N m B

m B

F m B

Physics BYU 112

Atomic diamagnetism

B add B m

Energy:

-e -e

2

20

2

ˆ2

cent

evI

aev

aa

m vF

a

m k

20 02m v m

F v v evBa a

0

ˆ2 2

eva eW

m m B m B L B

02

eBav

m

2 2

0

ˆ2 4

e e aa v

m m k B

Physics BYU 113

Magnetization• Magnetization = (dipole moment)/volume• Vector potential: 0

2

ˆ( ')'

4

dd

r

m r rA

0 02

ˆ 1( ) ( ') ' ( ') ' '

4 4d d

r r

rA r M r M r

0 0

0 0

' ( ') ( ') '( )

4 4( ') ( ') '

4 4b b

d

r rda

r r

M r M r aA r

J r K r

Physics BYU 114

Sphere - constant magnetization• sphere radius R - uniform M

with magnetic moment 34

3Rm M

0

2 inside

3(dipole @ center) outside

MB A

0 0 32

2

insideˆ

( ) '4 3 ˆ outside

d Rrr

r

rA r M M

r

Physics BYU 115

A useful integral over the sphere

corresponds to electric field w/ constant density:

using Gauss’s law:

Applications: constant charge density, polarization, and magnetization

0 2

ˆ( ) '

sph

dr

r

I r

30

2

inside4

( )3 ˆ outside

R

r

r

I rr

04

Physics BYU 116

• Bound current densities:

• Auxiliary field H

• Defining

ˆ( ) ( ) and ( ) ( )b b J r M r K r M r n

0

1f b f

B J J J J M

0

1 we get f

H B M H J

Physics BYU 117

• An equivalent problem: sphere with uniform surface charge s spinning with constant angular velocity .w

Tangential velocity & surface current

equivalent to uniform M w/bound Kb

inside spinning sphere

R

w

v ω s ω R

K v ω R

ˆ, so b M R K M r

0

2

3R B ω

Physics BYU 118

• Ampere’s law for magnetic materials

• Experimentally: I & V H & E Example: copper rod of radius R carrying a

uniform free current I Amperian loop radius s: outside M = 0 and

f encd I H l

2 2/ inside(2 )

1 outside

s RH s I

0

0ˆ ( )

2

Is R

s

B H

Physics BYU 119

Boundary conditions for B & H

• Integrating Maxwell’s equations:

or, equivalently

1 1

2 2

0

ˆ ˆ 0

f

f

H J B

n H K n B

/ / / /1 2 1 2ˆ and f

H H K n B B

Physics BYU 120

Linear Magnetic Materials• Susceptibility and permeability:

• diamagnetic ~ -10-5

paramagnetic ~ +10-5

Gadolinium ~ +0.5

0

0

and ( )

1 magnetic permeabilitym

m

M H B H M H

B H

Physics BYU 121

• Theorem: In a linear homogeneous magnetic material

(constant susceptibility) the volume bound current density is proportional to the free current density :

Bound surface current:

example: solenoid

( )b m m f J M H J

ˆ ˆb m K M n H n

ˆb mn I K

Physics BYU 122

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