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6.013, Electromagnetic Fields, Forces, and Motion Lectures 6 & 7Prof. Markus Zahn Page 1 of 40

6.013, Electromagnetics and ApplicationsProf. Markus Zahn

September 27 and 29, 2005Lectures 6 and 7: Polarization, Conduction, and Magnetization

I. Experimental Observation: Dielectric Media

A. Fixed Voltage - Switch Closed ov = V

As an insulating material enters a free-space capacitor at constant voltagemore charge flows onto the electrodes; i.e., as x increases, i increases.

B. Fixed Charge - Switch open (i=0)

As an insulating material enters a free space capacitor at constant charge,the voltage decreases; i.e., as x increases, v decreases.

II. Dipole Model of Polarization

A. Polarization Vector P Np N q d ( p q d dipole moment)

N dipoles/Volume (P is dipole density) d q

q


Courtesy of Krieger Publishing. Used with permission.

inside V pS V

Q = qN d da dV

paired charge or equivalentlypolarization charge density

6.013, Electromagnetic FProf. Markus Zahn

inside V PS V V

Q = P da P dV dV (Divergence Theorem)

P = qN d Np

PP

B. Gauss’ Law

o total u P uE P

o E

oD E

uD

C. Boundary

Unpaired chargedensity; alsocalled free charge

ields, Forces, and Motion Lectures 6 & 7Page 3 of 40

uP

P Displacement Flux Density

Conditions

density


a bu u suS V

D = D da = dV n D D

a bP P spS V

P = P da = dV n P P

a bo u P o u P o su spS V

E E da dV n E E =

D. Polarization Current Density

Q qN dV qN d da P da [Amount of Charge passing through

surface area element da ]

p

Q Pdi da

t t[Current passing through surface

area element da]

pJ da

polarization current density

pP

Jt

Ampere’s law:

u p oEx H J Jt

u oP EJt t

u oJ E Pt

u 0DJ ; D E Pt


III. Equipotential Sphere in a Uniform Electric Field

o o orlim r, E r cos E z E r cos

r R, 0

3

o 2

Rr, E r cos

r

This solution is composed of the superposition of a uniform electric fieldplus the field due to a point electric dipole at the center of the sphere:

dipole 2o

p cos4 r

with 3o op 4 E R

This dipole is due to the surface charge distribution on the sphere.

3

s o r o o o 3r R r R

2Rr R, E r R, E 1 cos

r r

o o3 E cos


IV. Artificial Dielectric

sv v

E , Ed d

sA

q A vd

q A

Cv d

Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.

For spherical array of non-interacting spheres (s >> R)

z

_3 3

o o z z o oP = 4 R E i P = N p = 4 R E N

31N = s

3 3

o e o e

R RP = 4 E = E = 4

s s

e (electric susceptibility)

o o eD = E P = 1 E = E

r(relative dielectric constant)

3

r o o e o

R= = 1 = 1 4

s

V. Demonstration: Artificial Dielectric



s

v vE = = E =

d d

s

A q Aq = A = v C = =

d v d

o

s

vi = C V =

R

3o

o

A R AC = = 4

d s d

R=1.87 cm, s=8 cm, A= (0.4)2 m2, d=0.15m

=2(250 Hz), Rs=100 k , V=566 volts peak

C=1.5 pf

0 sv = C R V


=(2) (250) (1.5 x 10-12) (105) 566 = 0.135 volts

VI. Plasma Conduction Model (Classical)

pdvm = q E m v

dt n

pdvm = q E m v

dt n

p = n kT , p = n kT

k=1.38 x 10-23 joules/oK Boltzmann Constant

A. London Model of Superconductivity [ T 0 , 0 ]

dv dvm = q E , m = q Edt dt

J = q n v , J = q n v

2q E q ndJ d dvq n v = q n = q n = E

dt dt dt m m

2p

2q E q ndJ d dv= q n v = q n = q n = E

dt dt dt m m

2p

2 22 2p p

q n q n= , =

m m(p = plasma frequency)

For electrons: q-=1.6 x 10-19 Coulombs, m-=9.1 x 10-31 kg

n-=1020/m3 , 12o 8.854 x10 farads/m

211

p

q n= 5.6 x10

mrad/s

6.013, Electromagnetic Fields, Forces, aProf. Markus Zahn

p 10

pf = 9 x102

Hz

B. Drift-Diffusion Conduction [Neglect inertia]

0

n k Tdv q k T

m = q E m v V = E ndt n m m n

0

n k Tdv q k T

m = q E m v v = E ndt n m m n

2q n q k TJ = q n v = E n

m m

2q n q k TJ = q n v = E n

m m

= q n , = q n

J = E D

J = E D

q k T= , D =

m m

q k T= , D =

m m

charge mobilities

D D k T= =q

= ther

Einstein’s Relation

MolecularDiffusion

nd Motion Lectures 6 & 7Page 10 of 40

mal voltage (25 mV@ T300o K)

Coefficients


C. Drift-Diffusion Conduction Equilibrium J J 0

J = 0 = E D = D

J = 0 = E D = D

D k T= = ln

q

D k T

= = lnq

q / kTo= e

Boltzmann Distributionsq / kT

o= e

o= 0 = = 0 = [Potential is zero when system is charge neutral]

2 q /kT q /kTo o2 q= = = e e = sinhkT

(Poisson - Boltzmann Equation)

Small Potential Approximation:q

1kT

q qsinh

kT kT

2 02 q

= 0k T

2d2

od

k T= 0 ; L =2 qL

Debye Length

6.013, Electromagnetic Fields, Forces, and MotionProf. Markus Zahn

D. Case Studies

1. Planar Sheet

d d

2x /L x /L

1 22 2d

d= 0 = A e A e

dx L

B.C. x = 0

ox 0 = V

d

d

x /Lo

x /Lo

V e x 0

x

V e x 0

/0

dx LV e

0V

( )

Lectures 6 & 7Page 12 of 40

/0

dx LV e


d

d

x /Lo

d

x

x /Lo

d

V e x 0L

dE = =

dxV

e x 0L

d

d

x /Lo2

d

x

x /Lo2

d

Ve x 0

L

dE= =

dxV

e x 0L

o

s x xd

2 Vx = 0 = E x 0 E x 0 =

L

2. Point Charge (Debye Shielding)

22

d

= 0L

22

1 rr rr

2

2

1= rr r

E. Ohmic Conduction

J = E D

J = E D

If charge density gradients smal

oJ = J J = E =

= ohmJ = E (Ohm’s Law)


l, then negligible o= =

E = E

ic conductivity

d

d

2

2 2d

r /Lr /Ld1 2

r /L

d rr = 0 0

dr L

r = A e A e

Qr = e

4 r


F. p—n Junction Diode


A Dn p 2

i

N Nk T= = ln

q n

2 2A p D n

p n

qN x qN xx = 0 = =

2 2

22A pD n

n p

qN xqN x= = +

2 2

D nn p

qN x= x x

2

VII. Relationship Between Resistance and Capacitance In Uniform Media Describedby and .

u S S

L L

D da E daq

C = = =v E ds E ds

L L

S S

E ds E dsvR = = =i J da E da

v

6.013, Electromagnetic Fields, Forces,Prof. Markus Zahn

L L

S L

E ds E daRC = =

E da E ds

Check:

Parallel Plate Electrodes: l AR = , C = RC =

A l

Coaxial

bln aR = , C =b2 l ln

2

Concentric Spherica

and M

l

a

l

,

otion Lectures 6 & 7Page 16 of 40

RC =

,

6.013, Electromagnetic Fields, ForcesProf. Markus Zahn

1 2

1 1R R

R = , C =4

VIII. Charge Relaxation in U

uuJ = 0

t

uE =

uJ = E

uE

t

u

u u

e

=t

IX. Demonstration 7.7.1 –

,

, and Motion Lectures 6 & 7Page 17 of 40

1 2

RC =1 1R R

4

niform Conductors

u

u= 0 = 0t

e = = dielectric relaxation time

etu 00 = r, t = 0 e

Relaxation of Charge on Particle in Ohmic Conductor

6.013, Electromagnetic Fields, Forces, and MoProf. Markus Zahn



u

S S

qJ da = E da = =

+

+

+

++

++ +

,

++

++

tion

dqdt



ete

e

dq q= 0 q = q t = 0 e =

dt

Partially Uniformly Charged Sphere


0 1

3u 1 0

1

r R

4t = 0 = Q = R

30 r R

et0 1 e

u

1

e r R =t =

0 r R

e e

e

t t0

131

r

t

1 22

220

r e Q r e= 0 r R3 4 R

E r, t =

Q eR r R

4 rQ

r R4 r

su 2 0 r 2 r 2r = R = E r = R E r = R

et2

2

Q= 1 e

4 R

Figure 3-21 An initial volume charge distribution within an Ohmic conductor decays exponentiallytowards zero with relaxation time =/and appears as a surface charge at an interface of discontinuity.Initially uncharged regions are always uncharged with the charge transported through by the current.


X. Self-Excited Water Dynamos

A. DC High Voltage Generation (Self-Excited)

As the drops form near the rings, image charges are induced on thesurface of the water. When the drops are completely formed (andinsulated from one another) they carry a net charge, the opposite sign tothe sign of charge on the surrounding ring.

Water drops fall into the cans through cross-connected wire loops. A potential difference of more than20 kV between cans is spontaneously generated by the motion of the drops. For optimum operation thedrops should form nearer the rings than shown. This is accomplished by increasing the flow rate.


1

2

st21 2i 1 1 i

st12 1i 2 2 i

dvnC v = C v = V e nC V = C s V

dt

dvnC v = C v = V e nC V = C s V

dt

i1

i2

nC V1Cs

= 0nC

1 VCs

Det = 0

2

i inC nC= 1 s =

Cs C

root blows up

inCt

CeAny perturbation grows exponentially with time

B. AC High Voltage Self – Excited Generation

1

2

3

st2i 1 1

st3i 2 2

st1i 3 3

dvnC v = C ; v = V e

dtdv

nC v = C v = V edtdv

nC v = C v = V edt


1i

2i

3i

VnC Cs 0

0 nC Cs V = 0

Cs 0 nC V

det = 0

13 3 i 3

i

nCnC + Cs = 0 s = 1

C

1 is = nC C (exponentially decaying solution)

i2,3

nCs = 1 3 j

2C (blows up exponentially because

sreal >0 ; but alsooscillates at frequency simag 0)

XI. Conservation of Charge Boundary Condition

uuJ = 0

t

1 3 1 3j1 1,

2


u uS V

dJ da dV = 0

dt

a b su

dn J J = 0

dt

XII. Maxwell Capacitor

A. General Equations

x

x

_

a

_

b

E t i 0 x a

E =

E t i b x 0

a

x b ab

E dx = v t = E t b E t a

sua b a a b b a a b b

d dn J J = 0 E t E t E t E t = 0dt dt

b a

v t aE t = E tb b

b ba a a a a a

dE t v(t) E t a E t v(t) E t a = 0

b dt b

bb a b ba a a

v ta dE a dvE t =

b dt b b b dt

v(t)a, a

b, b


B. Step Voltage: v t = V u t

Then dv= V t

dt (an impulse)

At t=0

b a b ba

a dE dv= = V tb dt b dt b

Integrate from t=0- to t=0+

+t 0 0t 0

b a b b ba a a

t 0t 0 t 0

a dE adt = E = V t dt = V

b dt b b b

aE t 0 = 0

b b b

a a aa b

a VE t = 0 = V E t = 0 =

b b b a

For t > 0, dv= 0

dt

b a b ba a a

a dE aE t = V

b dt b b

tb a b

aa b a b

V b aE t = A e ; =

b a b a

b b b ba

a b a b a b a b

V VE t = 0 = A = A = V

b a b a b a b a

t tb b

aa b a b

V VE t = 1 e e

b a b a

b aV aE t = E tb b

V

v(t)


su a a b b a a b a

V at = E t E t = E t E t

b b

ba a b

a V= E t

b b

b a a b t

a b

V= 1 e

b a

C. Sinusoidal Steady State:

j tv t = Re V e

j ta aE t = Re E e

b

j tbE t = Re E e

Conservation of Charge Interfacial Boundary Condition

a a b b a a b b

dE t E t E t E t = 0

dt

a a a b b bˆ Ê j E j = 0

b aˆ Ê b E a = V

ab

ˆˆ E aVE =

b b

a

a a a b b

ˆˆ E aVE j j = 0

b b

a a a b b b b

â VE j j = j

b b

a b

b b a a a a b b

ˆ ˆ Ê E V= =

j j b j a j

su a a b bˆ ˆ= E Eˆ

a b b a

a a b b

= Vb j a j


D. Equivalent Circuit (Electrode Area A)

a a a b b b

ˆ ˆI = j E A = j E A

a b

a a b b

V=

R RR C j 1 R C j 1

a ba b

a bR = , R =

A A

a ba b

A AC = , C =

a b

v(t)


XIII. Magnetic Dipoles



6.013, Electromagnetic Fields, Forces, andProf. Markus Zahn

Diamagnetism

2z

e e eI = = , m = I R i =

2 2 2

2_ _2

z ze R

R i = i2

Angular Momentum _ _ _ _

2r r ze e eL = m R i v = m R R i i = m R i

r p e2m= m

e

linear momentum

L is quantized in units of 34h , h = 6.62x10 joule sec2

(Planck’s constant)

24 2

e e e

e L e h ehm = = = 9.3x10 amp m

2m 2 2 m 4 m

Bohr magneton mB

(smallest unit ofmagnetic moment)

Imagine all Bohr magnetons in sphere of radius R aligned. Net magnetic moment is

3 0B

0

A4m m R3 M

Total massof sphere

For iron: =7.86 x 103 kg/

Courtesy of Hermann A.

Motion Lectures 6 & 7Page 28 of 40

molecular weight

m3, M0=56

Haus and James R. Melcher. Used with permission.

Avogadro’s number = 6.023 x 1026 molecules per kilogrammole


For a current loop

2 3 0 0B B

0 0

A A4 4m = i R = m R i = m R3 M 3 M

For R = 10 cm

26

-24 36.023 104i = 9.3 10 .1 7.86 10

3 56

= 1.05 x 105 Amperes

Thus, an ordinary piece of iron can have the same magnetic moment as acurrent loop of radius 10 cm of 105 Amperes current.

B. Magnetic Dipole Field

_ _0

r30

mH = 2 cos i sin i4 r

(multiply top & bottom by 0 )

Electric Dipole Field

_ _

r30

pE = 2 cos i sin i4 r

Analogy

0p m

P = Np M = Nm , N = # of magnetic dipoles / volume

Polarization Magnetization


XIV. Maxwell’s EQS Equations with Magnetization

A. Analogy to Maxwell’s EQS Equations with Polarization

EQS

0 uE = P

p = P (Polarization or paired

charge density)

α b a b

0 sun E E = n P P

a b

sp = n P P

MQS

0 0H = M

m 0= M (magnetic charge

density)

α b a b

0 0n H H = n M M

a b

sm 0= n M M

H = J

0E = H Mt

6.013, Electromagnetic Fields, Forces, and Motion Lecture 6 & 7Prof. Markus Zahn Page 31 of 40

B. MQS Equations

0B = H M Magnetic flux density B has units of Teslas

(1 Tesla = 10,000 Gauss)

B 0

a an B B = 0

BE =

t

H = J

S

dv = , = B da

dt(total flux)

XV. Magnetic Field Intensity along Axis of a Uniformly Magnetized Cylinder

a b

sm 0 sm 0 0d= n M M z = = M2

sm 0 0dz = = M2

x H = J = 0 H =


20 0 m 0H = = = M

m2 m

0 V' 0

r ' dV '= r =

4 r r '

R Rsm sm

r'=0 r'=00 0

d dz = 2 r 'dr ' z = 2 r 'dr '2 2z =4 r r ' 4 r r '

R R

0 o 0 o1 1

2 22 2r'=0 r'=02 2

0 0

M 2 r 'dr ' M 2 r 'dr '=

d d4 r ' z 4 r ' z2 2

1

2 221

2 22

r 'dr ' = r ' z ar ' z a

RR1 12 22 2

2 20

r ' 0 r ' 0

M d dz = r ' + z r ' z

2 2 2

1 12 22 2

2 20M d d d d= R z z R z z

2 2 2 2 2

( )z

Lecture 6 & 7Page 32 of 40

2 2 1/ 20( / 2){ [ ] }M R d R d


01 1

2 22 22 2

dd zzM d22 z2 2

d dR z R z2 2

zH = =z

01 1

2 22 22 2

dd zzM d d22 2 z2 2 2d dR z R z

2 2

XVI. Toroidal Coil

Courtesy of Hermann A. Haus and James R. M

-d/2 d/

Hz

Lecture 6 & 7Page 33 of 40

elcher. Used with permission.

2


1 1

1C

N i N iH dl = H 2 r = N i H =

2 r 2 R

2wB

4

2

2 2w

= N = N B4


H 1 1 1 H

1

H 2 RV = i R = R V =Horizontalvoltage to oscilloscope

N

2 v2 2 2 v v 2 2

d dVv = = i R V = V R Cdt dt

If 2 v2 2 2 2 2 2 v v

2

d dV1R R C = R C V V =Verticalvoltage tooscilloscopeC dt dt

2

2

w= N B

4

2

v 22 2

1 wV = N B

R C 4


XVII. Magnetic Circuits

In iron core:H = 0

lim B = H

B finite

NiH dl = Hs = Ni H =s

0

0

Dd N= H Dd = i

s

S

B da = 0

2 20 0Dd Dd= N = N i L = = N

s i s


XVIII. Reluctance

0

lengthNi s= = =Dd permeability cross sec tionalarea

R

[Reluctance, analogous to resistance]

Series

Parallel


A. Reluctances In Series

1 2

1 21 1 2 2

s s= , =

a D a DR R

1 2

Ni=R R

1 1 2 2C

H dl = H s H s = Ni

1 1 1 2 2 2= H a D = H a D

2 2 1 11 2

1 1 2 2 2 1 1 1 2 2 2 1

a Ni a NiH = ; H =

a s a s a s a s

B. Reluctances In Parallel

1 2 1 2C

NiH dl = H s = H s = Ni H = H

s

1 21 1 1 2 2 2 1 2

1 2

Ni= H a H a D = = Ni

R RP P

R R

1 21 2

1 1= ; =P P

R R

1=P

R[Permeances, analogous to Conductance]


XIX. Transformers(Ideal)


A. Voltage/Current Relationships

1 1 2 2N i N i l= ; =

AR

R

Another way: 1 1 2 2C

H dl = Hl = N i N i

1 1 2 2N i N iH =l

1 1 2 21 1 2 2

N i N iA= HA = N i N i =

l R


21 1 1 1 1 2 2 1 1 2

A= N = N i N N i = L i Mi

l

22 2 1 2 1 2 2 1 2 2

A= N = N N i N i = Mi L i

l

2 21 1 0 2 2 0 1 2 0 0

A 1L = N L , L = N L , M = N N L , L = =l R

12

1 2M = L L

1 1 2 1 21 1 1 0 1 2

d di di di div = = L M = N L N N

dt dt dt dt dt

2 1 2 1 22 2 2 0 1 2

d di di di div = = M L = N L N N

dt dt dt dt dt

1 1

2 2

v N=v N

1 2

1 1 2 22 1

i Nlim H 0 N i = N i =

i N

1 1

2 2

v i= 1

v i

Courtesy of Hermann A. Haus andJames R. Melcher. Used with permission.

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