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6.013/ESD.013J Electromagnetics and Applications, Fall 2005
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6.013, Electromagnetics and Applications Prof. Markus Zahn September 27 and 29, 2005 Lectures 6 and 7: Polarization, Conduction, and Magnetization
I. Experimental Observation: Dielectric Media
A. Fixed Voltage - Switch Closed ( v = Vo )
As an insulating material enters a free-space capacitor at constant voltage
more 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 = N p = N q d
N dipoles/Volume ( P is dipole density)
( p = q d dipole moment)
d
+q
−q
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lectures 6 & 7
Page 1 of 40
Courtesy of Krieger Publishing. Used with permission.
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lectures 6 & 7 Page 2 of 40 Q inside V = −
∫ q N d i da = ∫ ρ
S
p
dV
V
paired charge or
equivalently
polarization
charge density
Qinside V = − ∫ P i da = − ∫ ∇ i P dV =
S
V
∫ρ
P
dV
(Divergence Theorem)
V
P = q N d = Np
∇ i P = −ρP
B. Gauss’ Law
∇i
( ε E) = ρ
o
total
= ρu + ρP = ρu − ∇ i P
Unpaired charge
density; also
called free charge
density
∇i
(ε
o
)
E + P = ρu
D = εo E + P
Displacement Flux Density
∇ i D = ρu
C. Boundary Conditions
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lectures 6 & 7
Page 3 of 40
∫ D i da = ∫ ρ
∇ i D = ρu ⇒
S
∇ i P = −ρP ⇒
dV ⇒ n i ⎡⎣D a − Db ⎤⎦ = σ su
∫ P i da = − ∫ ρ
S
(
u
V
P
V
)
∇ i ε o E = ρu + ρP ⇒
∫ε
o
dV ⇒ n i ⎡⎣P a − P b ⎤⎦ = −σ sp
E i da =
S
∫ (ρ
V
u
+ ρP ) dV ⇒ n i ε o ⎡⎣Ea − Eb ⎤⎦ = σ su + σ sp
D. Polarization Current Density
∆Q = q N dV = q N d i da = P i da
[Amount of Charge passing through
surface area element da ]
d ip =
∂∆Q
∂P
=
i da
∂t
∂t
[Current passing through surface
area element da ]
= Jp i da
polarization current density Jp =
∂P
∂t
Ampere’s law: ∇ x H = Ju + Jp +
εo
∂E
∂t
= Ju +
∂P
+
∂t
= Ju +
∂
ε o
E + P
∂t
= Ju +
∂ D
; D = ε 0E + P
∂t
(
εo
∂E
∂t
)
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lectures 6 & 7
Page 4 of 40
III. Equipotential Sphere in a Uniform Electric Field
lim Φ (r, θ ) = −E or cos θ
r→∞
⎡⎣Φ = −Eo z = −E or cos θ ⎤⎦
Φ (r = R, θ ) = 0
⎡
R3 ⎤
Φ (r, θ ) = −Eo ⎢r − 2 ⎥ cos θ
r ⎦
⎣
This solution is composed of the superposition of a uniform electric field
plus the field due to a point electric dipole at the center of the sphere:
Φ dipole =
p cos θ
4πε or 2
with p = 4 πε oE oR 3
This dipole is due to the surface charge distribution on the sphere.
σ s (r = R, θ ) =
ε oEr (r = R, θ ) = − ε o
∂Φ
∂r
=
r =R
⎡
ε oEo ⎢1 +
⎣
2R 3
r3
⎤
⎥ cos θ
r =R ⎦
= 3εoEo cos θ
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lectures 6 & 7
Page 5 of 40
IV. Artificial Dielectric
E =
v
,
d
σs =
q = σs A =
C =
εA
d
εE =
εv
d
v
q
εA
=
v
d
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
For spherical array of non-interacting spheres (s >> R)
_
P = 4 π ε o R 3 E o i z ⇒ Pz = N p z = 4 π ε o R 3 E o N
N= 1
s3
⎡
R ⎤
ε o ⎢ 4 π ⎛⎜ ⎞⎟ ⎥ E = ψ e ε o E
3
⎛
⎛R ⎞ ⎞
⎜ ψe = 4 π ⎜ ⎟ ⎟
⎜
⎝ s ⎠ ⎠⎟
⎝
3
P=
⎢⎣
⎝ s ⎠ ⎥⎦
ψ e (electric susceptibility)
D=
ε o E + P = ε o ⎡⎣1 + ψ e ⎤⎦ E = ε E
εr (relative dielectric constant)
⎛
⎛R ⎞ ⎞
ε = εr ε o = ε o ⎣⎡1 + ψ e ⎤⎦ = ε o ⎜1
+ 4π ⎜ ⎟ ⎟
⎜
⎟
3
⎝
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
⎝s⎠ ⎠
Lectures 6 & 7
Page 6 of 40
V. Demonstration: Artificial Dielectric
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lectures 6 & 7 Page 7 of 40 Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
E=
v
εv
⇒ σs = ε E =
d
d
εA
q = σs A =
d
∆i = ω ∆C
∆C =
v ⇒ C=
V =
(ε − εo ) A
d
q
ε A
=
v
d
v
o
Rs
3
⎛R ⎞ A
= 4 π εo ⎜ ⎟
⎝ s ⎠ d
R=1.87 cm, s=8 cm, A= (0.4)2 m2, d=0.15m ω =2π(250 Hz), Rs=100 k Ω , V=566 volts peak ∆ C=1.5 pf
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lectures 6 & 7
Page 8 of 40
v 0 = ω ∆C R s V
=(2π) (250) (1.5 x 10-12) (105) 566 = 0.135 volts
VI. Plasma Conduction Model (Classical)
m+
∇p +
dv +
= q + E − m + ν + v + −
dt
n+
m−
∇p −
dv −
= −q− E − m − ν − 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 , ν ± → 0 ] m+
dv +
= q+ E
dt
J+ = q+ n + v +
,
m−
,
dv −
= −q − E
dt
J− = −q − n − v
−
(
)
q+ E
q 2n
dJ+
d
dv +
q + n + v + = q+ n +
= q+ n +
= + + E
=
dt
dt
dt
m+
m+
(
)
ωp2 + ε
(
)
−q− E
q 2n
dJ−
d
dv −
=−
q − n − v − = −q− n −
= −q− n −
= − − E
dt
dt
dt
m−
m−
(
)
ωp2 − ε
ωp2 + =
q + 2 n +
,
m+ ε
ωp2 − =
q− 2 n−
m− ε
( ωp = plasma frequency)
For electrons: q-=1.6 x 10-19 Coulombs, m-=9.1 x 10-31 kg
n-=1020/m3 , ε = ε o ≈ 8.854 x 10 −12 farads/m
ωp − =
q−2 n−
≈ 5.6 x 1011 rad/s
m− ε
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lectures 6 & 7
Page 9 of 40
fp − =
ω
p −
2π
≈ 9 x 1010 Hz B. Drift-Diffusion Conduction [Neglect inertia]
0
∇ (n + k T )
q+
dv +
kT
m+
= q+ E − m + ν + v + −
⇒ v+ =
E−
∇n +
dt
n+
m+ ν +
m + ν +n+
0
m−
∇ (n − k T ) −q −
dv −
kT
= −q− E − m − ν − v − −
⇒ v− =
E−
∇n −
dt
n−
m−ν −
m − ν −n −
J+ = q + n + v + =
q + 2n +
qkT
∇n
+
E− +
m+ ν +
m + ν +
J− = −q − n − v − =
ρ + = q+ n + ,
q − 2n −
qkT
∇ n
−
E+ −
m−ν −
m − ν −
ρ − = −q− n −
J+ = ρ + µ + E − D + ∇ρ +
J− = −ρ − µ − E − D − ∇ρ −
µ+ =
q+
m+ ν +
,
µ− =
q−
m− ν −
,
charge molulities D+ =
D− =
kT
m + ν +
kT
m − ν −
Molecular
Diffusion
Coefficients
D+
D
kT
= thermal voltage (25 mV@ T ≈ 300o K)
= − =
q
µ+
µ−
Einstein’s Relation
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lectures 6 & 7
Page 10 of 40
(
C. Drift-Diffusion Conduction Equilibrium J+ = J− = 0
)
J+ = 0 = ρ + µ + E − D + ∇ρ + = −ρ + µ + ∇Φ − D + ∇ρ +
J− = 0 = −ρ − µ − E − D − ∇ρ − = ρ − µ − ∇Φ − D − ∇ρ −
∇Φ = −
∇Φ =
D+
−k T
∇ρ + =
∇ (ln ρ + )
q
ρ+ µ +
D−
kT
∇ρ − =
∇ (ln ρ − )
ρ− µ−
q
ρ + = ρ o e − qΦ / kT
Boltzmann Distributions
ρ − = −ρ o e + qΦ / kT
ρ + ( Φ = 0 ) = −ρ − ( Φ = 0 ) = ρo
∇2 Φ =
−ρ
ε
=−
( ρ+
+ ρ− )
ε
=
−ρo
⎡e
ε ⎣
[Potential is zero when system is charge neutral]
− qΦ / kT
Small Potential Approximation:
sinh
− e+ qΦ / kT ⎤⎦ =
2ρo
ε
qΦ
kT
(Poisson - Boltzmann Equation)
sinh
qΦ
<< 1
kT
qΦ qΦ
≈
kT
kT
∇2 Φ −
2ρ0q
Φ=0
εk T
∇2 Φ −
Φ
=0 ;
L d2
Ld =
εk T
2 ρo q
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Debye Length
Lectures 6 & 7
Page 11 of 40
D. Case Studies
1. Planar Sheet
d2 Φ
Φ
− 2 =0
2
dx
Ld
B.C. Φ ( x → ±∞ ) = 0
⇒
Φ = A1 ex / L d + A2 e−x / L d
⇒ Φ (x) =
Φ ( x = 0 ) = Vo
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Vo e−x / L d
x>0
Vo e+x / Ld
x<0
Lectures 6 & 7
Page 12 of 40
Vo −x / L d
e
Ld
Ex = −
dΦ
=
dx
−
−
ρ=ε
x>0
dEx
=
dx
−
Vo x / Ld
e
Ld
ε Vo
L d2
ε V
o
L
2
d
x<0
e−x / Ld
x>0
e+x / Ld
x<0
2 ε
Vo
σs ( x = 0 ) = ε ⎡⎣Ex ( x = 0+ ) − Ex ( x = 0− ) ⎤⎦ =
L d
2. Point Charge (Debye Shielding)
∇2 Φ −
Φ
=0
L d2
⇒
r Φ = A1e−r / Ld + A2 e+r / L d
1 ∂ ⎛ 2 ∂Φ ⎞
⎜r
⎟
r2 ∂r ⎝ ∂r ⎠
=
d2
rΦ
(r Φ ) − 2 = 0 0
dr 2
Ld
Φ (r ) =
Q
e−r / L d
4 πεr
1 ∂2
(r Φ )
r ∂r2
E. Ohmic Conduction
J+ = ρ + µ + E − D + ∇ρ +
J− = −ρ − µ − E − D − ∇ρ −
If charge density gradients small, then ∇ρ ± negligible ⇒ ρ + = −ρ − = ρ o
J = J+ + J− = ( ρ + µ + − ρ − µ − ) E = ρ o ( µ + + µ − ) E = σE
σ = ohmic conductivity J = σ E (Ohm’s Law) 6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lectures 6 & 7
Page 13 of 40
F. p—n Junction Diode
Transition region
P
Charge density ρ
Electric field ε
V0
Potential V
An open-circuited pn diode
Figure by MIT OpenCourseWare.
∆Φ = Φ n − Φ p =
k T N A ND
ln
q
ni2
Φ (X = 0 ) = Φp +
∆Φ = Φ n − Φ p =
=
q NA xp2
2ε
= Φn −
q ND x n2
2ε
q NA xp2
q ND x n2
+
2ε
2ε
q ND x n
( xn + xp )
2ε
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lectures 6 & 7
Page 14 of 40
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lectures 6 & 7
Page 15 of 40
VII. Relationship Between Resistance and Capacitance In Uniform Media Described
by ε and σ .
C=
ε
∫ D i da
qu
=
v
=
S
∫ E i ds
S
∫ E i ds
L
v
R= =
i
∫ E i ds
L
∫
J i da
L
∫ E i ds
=
L
σ
S
RC =
∫ E i ds
∫
S
E i da
∫ E i da
S
ε
∫ E i da
L
σ
∫ E i da
L
∫
E i ds
=
ε
σ
L
Check:
Parallel Plate Electrodes: R =
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
l
,
σA
C=
εA
l
⇒ RC =
ε
σ
Lectures 6 & 7
Page 16 of 40
Coaxial
R=
ln b
a ,
2 πσl
C=
2 πεl
ln b
⇒ RC = ε
a
σ
Concentric Spherical
1
1
−
R1 R 2
R=
,
4πσ
C=
4πε
1
1
−
R1 R 2
⇒ RC =
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
ε
σ
Lectures 6 & 7
Page 17 of 40
VIII. Charge Relaxation in Uniform Conductors
∇ i Ju +
∂ ρu
=0
∂ t
∇ iE =
ρu
ε
Ju = σ E
σ ∇ i E
ρ
u
∂ ρu
∂t
+
=0 ⇒
+
σ
ε
ρu = 0
ε
τe =
∂ ρu
∂t
∂ ρu
∂t
+
ρu
τe
ε σ = dielectric relaxation time
=0 ⇒
ρu = ρ 0
(r, t = 0 ) e − t τ
e
IX. Demonstration 7.7.1 – Relaxation of Charge on Particle in Ohmic Conductor
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lectures 6 & 7
Page 18 of 40
∫ J i da = σ ∫ E i da =
S
S
σ qu
ε
=
−dq
dt
dq
q
+
= 0 ⇒ q = q ( t = 0 ) e − t τe
τe
dt
(τ
e
=
ε
σ
)
Partially Uniformly Charged Sphere
Courtesy of Krieger Publishing. Used with permission.
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lectures 6 & 7
Page 19 of 40
ρ0
r < R1
ρu ( t = 0 ) =
QT =
0
ρu ( t ) =
r > R1
ρ 0 e − t τe
( τe
r < R1
0
Er (r, t ) =
4
π R 13 ρ 0
3
= ε σ)
r > R1
ρ 0 r e − t τe
Q r e − t τe
=
3ε
4 π ε R 13
Q e − t τe
4 π ε r2
Q
4 π ε0 r2
0 < r < R1
R1 < r < R 2
r > R2
σ su (r = R 2 ) = ε 0 Er (r = R 2 + ) − ε Er (r = R 2 − )
=
Q
1 − e − t τe
4 π R 22
(
)
X. Self-Excited Water Dynamos
A. DC High Voltage Generation (Self-Excited)
Courtesy of Herbert Woodson and James Melcher. Used with permission.
Woodson, Herbert H., and James R. Melcher. Electromechanical Dynamics,
Part 2: Fields, Forces, and Motion. Malabar, FL: Krieger Publishing Company,
1968. ISBN: 9780894644597.
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lectures 6 & 7
Page 20 of 40
From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987.
Used with permission.
Courtesy of Herbert Woodson and James Melcher. Used with permission.
Woodson, Herbert H., and James R. Melcher. Electromechanical Dynamics,
Part 2: Fields, Forces, and Motion. Malabar, FL: Krieger Publishing Company,
1968. ISBN: 9780894644597.
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lectures 6 & 7
Page 21 of 40
−n C i v1 = C
dv 2
dt
v1 = V 1 e st
⇒
−n C i v 2 = C
dv1
dt
− n Ci V 1 = C s V 2
⇒
v 2 = V 2 e st
− n Ci V 2 = C s V 1
⎡n Ci
⎢ Cs
⎢
⎢
⎢
⎢ 1
⎣⎢
⎤
1 ⎥ ⎡ V1 ⎤
⎢ ⎥
⎥ ⎢ ⎥
⎥ ⎢ ⎥=0
n Ci ⎥ ⎢ ⎥
⎥ ⎢ ⎥
Cs ⎥⎦ ⎣ V2 ⎦
Det = 0
2
n Ci
⎛ n Ci ⎞
⎜
⎟ =1 ⇒ s = ±
C
⎝ Cs ⎠
⊕ root blows up
n Ci
t
eC
Any perturbation grows exponentially with time
B. AC High Voltage Self – Excited Generation
From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.
dv 2
dt
dv 3
−n C i v 2 = C
dt
dv1
−n C i v 3 = C
dt
−n C i v1 = C
⎡n Ci
⎢
⎢
⎢0
⎢
⎢
⎢Cs
⎣
Cs
n Ci
0
;
v1 = V 1 e st
v 2 = V 2 e st
v 3 = V 3 e st
0⎤
⎥
⎥
Cs ⎥
⎥
⎥
n Ci ⎥⎦
⎡ V1 ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ V2 ⎥ = 0
⎢ ⎥
⎢ ⎥
⎢ V3 ⎥
⎣ ⎦
det = 0
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lectures 6 & 7
Page 22 of 40
(n Ci )
3
1
⎛ n Ci ⎞
3
3
+ ( Cs ) = 0 ⇒ s = ⎜
⎟ ( −1)
⎝ C ⎠
s1 = −n Ci C (exponentially decaying solution) ( −1)
13
s2, 3 =
= −1,
1 ± 3j
2
n Ci
⎡1 ± 3 j⎤ (blows up exponentially because sreal >0 ; but also
⎦
2C ⎣
oscillates at frequency simag ≠ 0)
XI. Conservation of Charge Boundary Condition
∇ i Ju +
∫J
u
S
∂ρu
=0
∂t
i da +
d
ρu dV = 0
dt ∫V
n i ⎡⎣ Ja − Jb ⎤⎦ +
d
σsu = 0
dt
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lectures 6 & 7
Page 23 of 40
XII. Maxwell Capacitor
A. General Equations
_
Ea ( t ) i x
0<x<a
E=
_
Eb ( t ) i x
a
∫E
x
−b < x < 0
dx = v ( t ) = Eb ( t ) b + Ea ( t ) a
−b
n i ⎡⎣ Ja − Jb ⎤⎦ +
Eb ( t ) =
v (t)
b
dσsu
d
⎡εaEa ( t ) − εbEb ( t ) ⎤⎦ = 0
= 0 ⇒ σa Ea ( t ) − σb Eb ( t ) +
dt
dt ⎣
− Ea ( t )
a
b
⎡ v (t)
⎤ d ⎡
⎛ v (t)
⎞⎤
σa Ea ( t ) − σb ⎢
− Ea ( t ) a⎥ +
− Ea ( t ) a ⎟ ⎥ = 0
⎢εaEa ( t ) − εb ⎜⎜
⎟
⎢⎣ b
⎥⎦ dt ⎢⎣
⎝ b
⎠ ⎥⎦
σ v ( t ) εb dv
εb a ⎞ dEa ⎛
σ a⎞
⎛
+ ⎜ σa + b ⎟ Ea ( t ) = b
+
⎜ εa +
⎟
b ⎠ dt
b ⎠
b
b dt
⎝
⎝
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lectures 6 & 7
Page 24 of 40
B. Step Voltage: v ( t ) = V u ( t )
Then
dv
= V δ ( t ) (an impulse)
dt
At t=0
εb a ⎞ dEa εb dv εb
⎛
=
=
Vδ ( t )
⎜ εa +
⎟
b ⎠ dt
b dt
b
⎝
Integrate from t=0- to t=0+
εb a ⎞ dEa
ε a⎞
⎛
⎛
dt = ⎜ εa + b ⎟ Ea
⎜ εa +
⎟
∫
b ⎠ dt
b ⎠
⎝
t = 0− ⎝
t = 0+
t = 0+
=
t = 0−
0+
∫
t = 0−
εb
b
Vδ ( t ) dt =
εb
b
V
Ea ( t = 0 − ) = 0
εb a ⎞
εb
εb V
⎛
V ⇒ Ea ( t = 0 + ) =
⎜ εa +
⎟ Ea ( t = 0 + ) =
εb b + εb a
b ⎠
b
⎝
For t > 0,
dv
=0
dt
εb a ⎞ dEa ⎛
σ a⎞
σ
⎛
+ ⎜ σ a + b ⎟ Ea ( t ) = b V
⎜ εa +
⎟
b
dt
b
b
⎠
⎠
⎝
⎝
Ea ( t ) =
σb V
ε b + εb a
+ A e − t τ
; τ = a
σ ab + σb a
σ ab + σ b a
Ea ( t = 0 ) =
σb V
+A =
σ ab + σ b a
⎡
⎤
εb V
εb
σb
⇒ A=V⎢
−
⎥
ε ab + εb a
⎣ ε ab + ε b a σ ab + σ b a ⎦
Ea ( t ) =
σb V
εb V
1 − e−t τ +
e−t τ
ε ab + εb a
σ ab + σ b a
Eb ( t ) =
V
a
− Ea ( t )
b
b
(
)
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lectures 6 & 7
Page 25 of 40
⎛V a
⎞
σ su ( t ) = ε a Ea ( t ) − εb Eb ( t ) = ε a E a ( t ) − ε b ⎜ − E a ( t ) ⎟
⎝b b
⎠
ε a⎞
V
⎛
= Ea ( t ) ⎜ ε a + b ⎟ − εb
b ⎠
b
⎝
=
V ( σb ε a − σ a εb )
( σ ab + σb a)
(1 − e )
−t τ
C. Sinusoidal Steady State: v ( t ) = R e ⎡ V e jω t ⎤
⎣
⎦
Ea ( t ) = Re ⎡Ea e jωt ⎤
⎣
⎦
Eb ( t ) = Re ⎡Eb e jωt ⎤
⎣
⎦
Conservation of Charge Interfacial Boundary Condition
σ a E a ( t ) − σ b Eb ( t ) +
d
⎡ ε a E a ( t ) − εb Eb ( t ) ⎤⎦ = 0
dt ⎣
Ea ⎡σ
⎣ a + j ω ε a ⎤⎦ − Eb ⎡σ
⎣ b + j ω εb ⎤⎦ = 0
Eb b + Ea a = V
Eb =
V Ea a
−
b
b
⎛ V Ea a ⎞
Ea = ⎡σ
⎣ a + j ω ε a ⎤⎦ − ⎜⎜ b − b ⎟⎟ ⎣⎡σ b + j ω ε b ⎤⎦ = 0
⎝
⎠
a
V
⎡
⎤
E a = ⎢ σ a + j ω ε a + ( σ b + j ω εb ) ⎥ = ⎡⎣σ b + j ω ε b ⎤⎦ = 0
b
b
⎣
⎦
Ea
Eb
V
=
=
j ω εb + σb
j ω εa + σa
⎡b
⎣ ( σ a + j ω ε a ) + a ( σ b + j ω εb ) ⎤⎦
σ su =
=
ε a Ea − εb Eb
⎡⎣b ( σ a +
( ε a σb − εb σ a )
j ω ε a ) + a ( σb +
j ω εb ) ⎤⎦
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
V
Lectures 6 & 7
Page 26 of 40
D. Equivalent Circuit (Electrode Area A)
I = ( σ a + j ω ε a ) Ea A = ( σ b + j ω ε b ) Eb A
=
V
Ra
Rb
+
R a C a j ω + 1 R b Cb j ω + 1
R a =
Ca =
a
b
, R b =
σa A
σb A
εa A
a
, Cb =
εb A
b
Courtesy of Krieger Publishing. Used with permission.
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lectures 6 & 7
Page 27 of 40
XIII.
Magnetic Dipoles
Courtesy of Krieger Publishing. Used with permission.
Courtesy of Krieger Publishing. Used with permission.
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lectures 6 & 7 Page 28 of 40 Diamagnetism
I=
e
2π
=
ω
eω
2π
m = −I π R 2 iz =
,
−e ω
2π
_
π R2 i z =
−e ω R 2 _
iz
2
_
_
⎛_ _ ⎞
Angular Momentum L = meR i r × v = me R ( ωR ) ⎜ i r × i φ ⎟ = me ωR 2 i z
⎝
⎠
2me
r ×p
=−
m
e
linear momentum
(
L is quantized in units of
)
h
,
2π
h = 6.62 x 10−34 joule − sec
(Planck’s constant)
m =
e L
2 me
=
eh
eh
≈ 9.3 x 10−24 amp − m2
=
2 π (2 ) me
4 π me
Bohr magneton mB
(smallest unit of
magnetic moment)
Imagine all Bohr magnetons in sphere of radius R aligned. Net magnetic moment is
⎛4
⎞
m = mB ⎜ π R 3 ρ ⎟
⎠
⎝3
Total mass
of sphere
A0
M0
Avogadro’s number = 6.023 x 1026 molecules per kilogram−mole
molecular weight
For iron: ρ =7.86 x 103 kg/m3, M0=56
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lectures 6 & 7
Page 29 of 40
For a current loop m = i π R 2 = mB
A
A
4
4
π R 3 ρ 0 ⇒ i = mB R ρ 0
3
M0
3
M0
(
6.023 × 1026
⎛4⎞
For R = 10 cm ⇒ i = 9.3 × 10-24 ⎜ ⎟ (.1) 7.86 ×103
56
⎝3⎠
)
= 1.05 x 105 Amperes
Thus, an ordinary piece of iron can have the same magnetic moment as a
current loop of radius 10 cm of 105 Amperes current.
B. Magnetic Dipole Field
H=
µ0 m
4 π r3 µ0
_
_
⎡
⎤
⎢2 cos θ i r + sin θ i θ ⎥ (multiply top & bottom by µ0 )
⎦
⎣
Electric Dipole Field E=
p
_
_
⎡
⎤
2 cos θ i r
+ sin θ i θ ⎥
⎢
4 π ε0 r ⎣
⎦
3
Analogy p → µ 0 m
P = Np ⇒ M = Nm ,
Polarization
N = # of magnetic dipoles / volume Magnetization
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lectures 6 & 7
Page 30 of 40
XIV. Maxwell’s EQS Equations with Magnetization
A.
Analogy to Maxwell’s EQS Equations with Polarization
EQS
MQS
(
(
)
)
(
)
∇ i ε 0 E = ρu − ∇ i P
∇ i µ 0 H = −∇ i µ 0 M
ρp = −∇ i P (Polarization or paired
ρm = −∇ i µ 0 M (magnetic charge charge density)
(
n i ⎡ε0 E − E
⎣⎢
α
b
)⎤⎦⎥ = −n i ⎡⎣⎢P
a
b
− P ⎤ + σ su
⎥⎦
a
b
σsp = −n i ⎡P − P ⎤
⎢⎣
⎦⎥
(
density)
(
)
(
)
α
b
a
b
n i ⎡µ 0 H − H ⎤ = −n i ⎡ µ 0 M − M ⎤
⎥⎦
⎣⎢
⎣⎢
⎦⎥
(
)
a
b
σ sm = −n i ⎡ µ 0 M − M ⎤
⎢⎣
⎥⎦
∇ ×H = J
∇ ×E = −
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
)
∂
µ0 H + M
∂t
(
)
Lectures 6 & 7
Page 31 of 40
B.
MQS Equations
(
)
B = µ 0 H + M Magnetic flux density B has units of Teslas
(1 Tesla = 10,000 Gauss)
∇ iB = 0
a
a
n i ⎡B − B ⎤ = 0
⎢⎣
⎥⎦
∇ ×E = −
∂ B
∂ t
∇ × H = J
v=
dλ
,
dt
λ = ∫ B i da (total flux)
S
XV. Magnetic Field Intensity along Axis of a Uniformly Magnetized Cylinder
From Electromagnetic Fields and Energy by Hermann A. Haus and James R. Melcher. Used with permission.
(
a
σ sm = −n i µ 0 M − M
b
) ⇒ σ (z = d 2 ) = µ
sm
(
σ sm z = −d
2
0
M0
) = −µ
0
M0
∇ x H = J = 0 ⇒ H = −∇ Ψ
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 6 & 7
Page 32 of 40
(
)
(
∇ i µ 0 H = −µ 0 ∇ 2 Ψ = ρ m = −∇ i µ 0 M
∇2 Ψ =
Ψ (z) =
−ρ m
R
∫
r'=0
R
=
∫
r'=0
µ0
()
⇒ Ψ r =
∫
(
σsm z = d
2
V'
M
Ψ (z) = 0
2
M
= 0
2
( )
ρ m r ' dV '
4 π µ 0
r − r '
) 2 π r ' dr ' +
4 π µ0 r − r '
(
σsm z = − d
R
∫
r'=0
R
µ0 Mo 2 π r ' dr '
⎡
d⎞ ⎤
⎛
4 π µ0 ⎢ r '2 + ⎜ z − ⎟ ⎥
2 ⎠ ⎦⎥
⎝
⎢⎣
2
∫
)
1
−
2
µ0 Mo 2 π r ' dr '
∫
r'=0
2
⎡ 2 ⎛
d⎞ ⎤
4 π µ0 ⎢ r ' + ⎜ z + ⎟ ⎥
2 ⎠ ⎥⎦
⎝
⎢⎣
r ' dr '
⎡r '2 + ( z + a)2 ⎤
⎣
⎦
⎧
1 R
2 2
⎪⎡
d⎞ ⎤
⎪ 2 ⎛
−
⎨ ⎢r ' + ⎜ z − ⎟ ⎥
2 ⎠ ⎥⎦
⎝
⎪ ⎢⎣
r '= 0
⎪⎩
1
⎧⎡
2
2
d⎞ ⎤
d
⎪ 2 ⎛
⎨ ⎢R + ⎜ z − ⎟ ⎥ −
z −
2 ⎠ ⎥⎦
2
⎝
⎪ ⎢⎣
⎩
)
2 π r ' dr '
2
4 π µ0 r − r '
1
2
2
= ⎡r '2 + ( z + a) ⎤
⎣
⎦
⎡ 2 ⎛
d⎞ ⎤
⎢r ' + ⎜ z + ⎟ ⎥
2 ⎠ ⎥⎦
⎝
⎢⎣
2
1
2
1
1
2
2
⎫
⎪
⎪
⎬
⎪
r '=0 ⎪
⎭
R
2
⎡
d⎞ ⎤
⎛
− ⎢R 2 + ⎜ z + ⎟ ⎥
2 ⎠ ⎦⎥
⎝
⎢⎣
1
2
⎫
d⎪
+ z+ ⎬
2⎪
⎭
Courtesy of Krieger Publishing. Used with permission.
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 6 & 7
Page 33 of 40
⎧
⎪
d
z−
−M0 ⎪⎪
2
−
⎨
1
2 ⎪⎡
2
2
⎤
⎡ 2
d⎞
⎛
⎪ ⎢R 2 + ⎜ z − ⎟ ⎥
⎢R
2 ⎠ ⎦⎥
⎝
⎣⎢
⎩⎪ ⎣⎢
Hz =
⎫
⎪
⎪
⎪
⎬
1
2
2
d⎞ ⎤ ⎪
⎛
+ ⎜z + ⎟ ⎥ ⎪
2 ⎠ ⎥⎦ ⎪
⎝
⎭
d⎞
⎛
⎜z + 2 ⎟
⎝
⎠
z >
d
2
−∂ Ψ
=
∂z
⎧
⎪
d
z−
−M0 ⎪⎪
2
−
⎨
1
2 ⎪⎡
2
2
⎤
⎡ 2
d⎞
⎛
⎪ ⎢R 2 + ⎜ z − ⎟ ⎥
⎢R
2 ⎠ ⎦⎥
⎝
⎣⎢
⎩⎪ ⎣⎢
d⎞
⎛
⎜z + 2 ⎟
⎝
⎠
2
d⎞ ⎤
⎛
+ ⎜z + ⎟ ⎥
2 ⎠ ⎥⎦
⎝
1
2
⎫
⎪
⎪
⎪
+ 2⎬
⎪
⎪
⎪⎭
−
d
d
<z<
2
2
XVI. Toroidal Coil
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 6 & 7
Page 34 of 40
∫ H i dl = H
φ
2 π r = N1i ⇒ Hφ =
C
Φ ≈B
N1i
N1i
≈
2 πr 2 πR
π w2
4
λ = N 2 Φ = N2 B
π w2
4
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
VH = i1 R1 = R1
v2 =
Hφ 2 π R
N1
( VH =Horizontal voltage to oscilloscope )
dλ2
dVv
= i2 R 2 + Vv = Vv + R 2 C2
dt
dt
If R 2 >>
dλ2
dVv
1
⇒
≈ R 2 C2
⇒ λ2 = R 2 C2 Vv
C2 ω
dt
dt
=
Vv =
( Vv = Vertical voltage to oscilloscope )
π w2
N2 B
4
1
π w2
N2 B
R 2 C2 4
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 6 & 7
Page 35 of 40
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 6 & 7 Page 36 of 40 XVII.
Magnetic Circuits
Courtesy of Krieger Publishing. Used with permission.
In iron core:
H=0
lim B = µ H ⇒
µ→∞
B finite
∫ H i dl = Hs = Ni ⇒ H =
Φ = µ 0 H Dd =
Ni
s
µ 0 Dd N
i
s
∫ B i da = 0
S
λ = NΦ =
µ 0 Dd 2
µ Dd 2
λ
N i⇒ L= = 0
N
s
i
s
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 6 & 7
Page 37 of 40
XVIII. Reluctance
R=
(length)
Ni
s
=
=
Φ
µ0 Dd (permeability ) ( cross − sec tional area)
[Reluctance, analogous to resistance]
Courtesy of Krieger Publishing. Used with permission.
A. Reluctances In Series
R1 =
Φ=
s1
,
µ1 a1 D
R2 =
s2
µ2 a2 D
Ni
R1 + R2
∫ H i dl = H
1
s1 + H2 s2 = Ni
C
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 6 & 7
Page 38 of 40
Φ = µ1 H1 a1 D = µ2 H2 a2 D
H1 =
µ2 a2 Ni ;
µ1 a1 s2 + µ2 a2 s1
H2 =
µ1 a1 Ni
µ1 a1 s2 + µ2 a2 s1
B. Reluctances In Parallel ∫ H i dl = H
1
s = H2 s = Ni ⇒ H1 = H2 =
C
Ni
s
Φ = ( µ1 H1 a1 + µ2 H2 a2) D =
P1 =
P=
1
;
R1
P2 =
Ni (R1 + R2 )
R1 R2
= Ni (P1 + P2 )
1
R2
1
[Permeances, analogous to Conductance]
R
XIX. Transformers
(Ideal)
Courtesy of Krieger Publishing. Used with permission.
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 6 & 7
Page 39 of 40
A. Voltage/Current Relationships
Φ=
N1 i1 − N2 i2
;
R
Another way:
R=
l
µA
∫ H i dl = Hl = N
i − N2 i2
1 1
C
H=
N1 i1 − N2 i2
l
Φ = µHA =
N i −N i
µA
(N1 i1 − N2 i2 ) = 1 1 2 2
l
R
λ1 = N1 Φ =
µA
N12 i1 − N1 N2 i2 = L1 i1 − Mi2
l
λ2 = N2 Φ =
µA
N1 N2 i1 − N22 i2 = −Mi1 + L 2 i2
l
(
)
(
)
L1 = N12 L 0 , L 2 = N22 L 0 , M = N1 N2 L 0 , L 0 =
M = ⎡⎣L1 L 2 ⎤⎦
1
µA 1
=
R
l
2
v1 =
dλ1
di
di
= L1 1 − M 2 = N1L 0
dt
dt
dt
di2
⎤
⎡ di1
⎢N1 dt − N2 dt ⎥
⎣
⎦
v2 =
dλ2
di
di
= +M 1 − L 2 2 = N2L 0
dt
dt
dt
di1
di2 ⎤
⎡
⎢ +N1 dt − N2 dt ⎥
⎣
⎦
v1
N
= 1
v2
N2
lim H ⇒ 0 ⇒ N1 i1 = N2 i2 ⇒
µ→∞
i1
N
= 2
i2
N
1
v1 i1
=1
v2 i2
Courtesy of Hermann A. Haus and James R. Melcher.
Used with permission.
6.013, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 6 & 7
Page 40 of 40