mitschrift physik iii
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8/7/2019 Mitschrift Physik III
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Physik III
Vorlesungs-Skript
Prof. Dr. Simon Lilly
Mitschrift
Marc Maetz
HS 2010
8/7/2019 Mitschrift Physik III
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Contents
Contents 10.1 General Information . . . . . . . . . . . . . . . . . . . . . . . . 3
0.1.1 Testat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.1.2 Exam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.1.3 books . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.1.4 webpage . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1 Electrostatics: Electric charge 5
1.1 empirical fact: charges exist . . . . . . . . . . . . . . . . . . . . 51.2 Coulomb’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Potential energy of a system of charges . . . . . . . . . . . . . 61.4 Concept of electric field . . . . . . . . . . . . . . . . . . . . . . 61.5 Charge distributions . . . . . . . . . . . . . . . . . . . . . . . . 71.6 Concept of Flux and Gauss’ Law . . . . . . . . . . . . . . . . . 71.7 Applications of Gauss’ Law . . . . . . . . . . . . . . . . . . . . 81.8 Energy associated with E-field . . . . . . . . . . . . . . . . . . 91.9 Concept of electrical Potential . . . . . . . . . . . . . . . . . . . 101.10 Experiment: charging two spheres equally . . . . . . . . . . . 101.11 Other useful operators . . . . . . . . . . . . . . . . . . . . . . . 11
1.12 summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2 Conductors 13
2.1 Conductors + insulators . . . . . . . . . . . . . . . . . . . . . . 132.2 Conditions for conductor . . . . . . . . . . . . . . . . . . . . . 142.3 The general electrostatic problem . . . . . . . . . . . . . . . . . 142.4 Uniqueness Theorem . . . . . . . . . . . . . . . . . . . . . . . . 152.5 Some interesting cases . . . . . . . . . . . . . . . . . . . . . . . 15
2.5.1 Enclosed cavity . . . . . . . . . . . . . . . . . . . . . . . 152.5.2 Faraday cage . . . . . . . . . . . . . . . . . . . . . . . . 16
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Contents
2.6 Some tricks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.6.1 Mirror charges . . . . . . . . . . . . . . . . . . . . . . . 172.7 Capacitance and capacitors . . . . . . . . . . . . . . . . . . . . 172.8 Energy stored in capacitor . . . . . . . . . . . . . . . . . . . . . 18
2.8.1 Parallel plate capacitor . . . . . . . . . . . . . . . . . . . 192.9 General System of conductors . . . . . . . . . . . . . . . . . . . 19
3 Elecric Currents 21
3.1 Electric current . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.2 Charge conservation . . . . . . . . . . . . . . . . . . . . . . . . 223.3 Ohm’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.4 Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
3.4.1 Kirchoff’s Rules . . . . . . . . . . . . . . . . . . . . . . . 233.5 Energy dissipation in a resistor . . . . . . . . . . . . . . . . . . 243.6 Sources of energy . . . . . . . . . . . . . . . . . . . . . . . . . . 253.7 Circuits with capacitors . . . . . . . . . . . . . . . . . . . . . . 26
4 Fields of moving charges 28
4.0 Magnetic Fields or Forces . . . . . . . . . . . . . . . . . . . . . 284.1 Reminder about special Relativity . . . . . . . . . . . . . . . . 294.2 Invariance of charge . . . . . . . . . . . . . . . . . . . . . . . . 294.3 Transformation of E-fields . . . . . . . . . . . . . . . . . . . . . 304.4 Field of accelerated charge . . . . . . . . . . . . . . . . . . . . . 31
4.5 Force on moving charge . . . . . . . . . . . . . . . . . . . . . . 334.6 Interaction of moving charge with other moning charges . . . 33
5 Magnetic Fields 35
5.1 Simplest case . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355.1.1 Force between two ∞ long wires . . . . . . . . . . . . . 36
5.2 Properties of magnetic fields . . . . . . . . . . . . . . . . . . . 365.2.1 Uniqueness theorem . . . . . . . . . . . . . . . . . . . . 37
5.3 Vector Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . 375.4 Fields of coils and solenoids . . . . . . . . . . . . . . . . . . . . 38
5.4.1 Biot-Savart law . . . . . . . . . . . . . . . . . . . . . . . 39
5.5 summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405.6 Change in B-field across a current sheet . . . . . . . . . . . . . 405.7 How do E and B transform . . . . . . . . . . . . . . . . . . . . 425.8 Hall Effect (1879) . . . . . . . . . . . . . . . . . . . . . . . . . . 44
6 Magnetic Induction 45
6.1 Magnetic Induction . . . . . . . . . . . . . . . . . . . . . . . . . 456.1.1 Wire in circuit . . . . . . . . . . . . . . . . . . . . . . . . 47
6.2 Faraday’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 486.3 Lenz’ Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
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Contents
6.4 Mutual inductance . . . . . . . . . . . . . . . . . . . . . . . . . 50
6.5 Self-inductance . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
7 Alternating current circuits 53
7.1 The RLC circuit as damped oscillator . . . . . . . . . . . . . . 537.2 Circuits driven by alternating voltage . . . . . . . . . . . . . . 54
7.2.1 Solenoid pushing . . . . . . . . . . . . . . . . . . . . . . 55
7.2.2 Some circuits . . . . . . . . . . . . . . . . . . . . . . . . 567.3 Power Consumption . . . . . . . . . . . . . . . . . . . . . . . . 60
8 Maxwell’s Equations 61
8.1 Wave equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
8.2 Superposition of two opposite directions . . . . . . . . . . . . 628.3 Standing wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
8.4 Energy Transport of E and M waves . . . . . . . . . . . . . . . 638.5 Lorentz transformation of waves . . . . . . . . . . . . . . . . . 638.6 summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
9 Dielectric materials 65
9.1 Introdiuction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 659.2 Electric dipoles . . . . . . . . . . . . . . . . . . . . . . . . . . . 669.3 Atomic and molecular dipoles . . . . . . . . . . . . . . . . . . 67
9.3.1 Permanent dipoles . . . . . . . . . . . . . . . . . . . . . 67
9.3.2 Induced dipoles . . . . . . . . . . . . . . . . . . . . . . . 679.4 Electric fields from polarized matter . . . . . . . . . . . . . . . 68
9.4.1 Gauss’ Law in medium and vector field D . . . . . . . 709.5 Currents in dielectrics and Maxwell’s equations . . . . . . . . 71
9.6 Eloctromagnetic waves in dielectric . . . . . . . . . . . . . . . 719.7 Example: Electric field around dielectric sphere . . . . . . . . 72
10 Magnetic phenomena in matter 76
10.1 Phenomenology . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
10.2 Magnetic dipoles . . . . . . . . . . . . . . . . . . . . . . . . . . 7610.3 Force on m in external field . . . . . . . . . . . . . . . . . . . . 77
10.4 Current in loop in atom . . . . . . . . . . . . . . . . . . . . . . 7810.5 Electron spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7910.6 Magnetic fields of magnetized matter . . . . . . . . . . . . . . 79
10.7 Maxwell’s equations . . . . . . . . . . . . . . . . . . . . . . . . 8010.8 Ferromagnetism (Fe,Ni) . . . . . . . . . . . . . . . . . . . . . . 82
11 Generation of electromagnetic waves 84
11.1 Potentials and potential wave equations . . . . . . . . . . . . . 84
11.2 Delayed potentials . . . . . . . . . . . . . . . . . . . . . . . . . 8511.2.1 Hertzian Dipole . . . . . . . . . . . . . . . . . . . . . . . 86
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Contents
0.1 General Information
If you find any mistake, feel free to send me an e-mail to ’mmaetz@student.ethz.ch’.
0.1.1 Testat
To receive the Testat, you must make a serious attempt and hand-in at least3/4 of the questions on at least 3/4 of the Exercise Sheets.
0.1.2 Exam
• 2-hour written
• calculator, ”translation” dictionary English-German and 10 sides of notes in their original own handwriting
• all questions in German and English and can be answered in Germanand English
0.1.3 books
english deutsch
Purcell Electricity and Magnetism Elektrizitat and Magnetismus
Jackson Classical Electrodynamics Klassische Elektrodynamik
Tipler & Mosca Physik Physics for Scientists and Engine
Kanzig Elektrizitat und Magnetismus (v—d—f)
0.1.4 webpage
http://www.exp-astro.phys.ethz.ch/PhysikIII
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Chapter 1
Electrostatics: Electric
charge
1.1 empirical fact: charges exist
t0 explain observed forces
a) + and − charges - like repel opposite attract
b) Net charge∑ q is conserved, and is relativistically invariant
c) Charge is apparently quantized qe = 1.602 · 10−19C
d) Charges appear ∼ pointlike r < 10−15m
1.2 Coulomb’s Law
Fig.1
F21 = kq1q2
r221
r21
cgs: F =q1q2
r2SI: F =
1
4πε 0
q1q2
r2
ε0 = permittivity of free space = 8.854 · 10−12Fm−1
a)
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1. Electrostatics: Electric charge
b)
c) effect of charges is additive
Fx = ∑ i=x
1
4πε 0
qxqi
r2xi
rX I → superposition Principle
”Force experienced by one charge from another is not affected by thepresence of a third ”
1.3 Potential energy of a system of charges
Work done on q2 W 2 =
F21 · ds. If radial path:
W 2 = − r2
∞
1
4πε 0
q1q2
r2dr =
1
4πε 0
q1q2
r12= W 1
Independent of path3rd charge brought up
F3 = F31 + F32
W 3 = W 31 + W 32
Total of all charges W 31 + W 32 + W 21
W j =1
4πε 0
j−1
∑ i=1
q jqi
rij
Total W = ∑ j
W j =1
4πε 0∑
j
j−1
∑ i=1
qiq j
rij=
1
4πε 0∑
i∑ i= j
qiq j
rij· 1
2
e.g. NaCl
W energy per ion
=e2
8πε 0
−1.748
a
1.4 Concept of electric field
Fq =1
4πε 0q∑
qi
r2i
ri
Can consider this due to a field
Fq = qE(x, y, z) with E(x, y, z) =1
4πε 0∑
i
qi
r2i
ri
key concept: E describes locally the effects of distant charges
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1.5. Charge distributions
1.5 Charge distributions
Often useful to introduce charge density, ρ(x, y, z)
dq = ρdV per unit volume
→ E(x, y, z) =1
4πε 0
all
space
ρ(x, y, z)r2
r dx dydz r =
x − x
y − y
z − z
also:
σ -charge per unit area in 2-d system dq = σ da (1.1)
λ-charge per unit system in 1-d system dq = λdb (1.2)
1.6 Concept of Flux and Gauss’ Law
da is ⊥ to surface and oriented outwards. Define Flux Φ
dΦ = E · da.
Φ =1
4πε 0
q
r2· 4π r2 =
q
ε0
→ Φ =∑ i qi
ε0for any closed surface =
1
ε0
V
ρdV
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1. Electrostatics: Electric charge
1.7 Applications of Gauss’ Law
a) Uniformly charged spherical surface
+
+
++
+
+
+
+
++
+
+
S1
σ
E = 0
Φ = 0
E = Q4πε0r2 as if all Q at center!
R
Q = 4π R2δ (1.3)
Inside surface E = 0 (1.4)
Outside surface 4π r2E(r) =1
ε04π R2σ =
1
ε0Q → E(r) =
Q
4π r2ε0=
σ
ε0at the surface R = r
(1.5)
b) Infinite line charge, constant λ
missing picture
2π r lE(r) =lλ
ε0(1.6)
E(r) =λ
2πε 0r(1.7)
c) Infinite sheet charge
2π a2E(r) =π a2σ
ε0(1.8)
E(r) =σ
2ε0(1.9)
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1.8. Energy associated with E-field
1.8 Energy associated with E-field
+
+
++
+
+
+
+
++
+
+
E =σ
ε0
R2
r2(1.10)
→P.E. of charge at surface dq =
1
4πε 0
Q
r
d (1.11)
Total P.E. W =1
2
Q2
4πε 0r2(1.12)
dW
dr= −1
2
Q2
4πε 0r2(1.13)
dW
dr= −UEdV = −UE4π r2dr (1.14)
E =Q
4πε 0r2→ UE =
ε0
2E2 (1.15)
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1. Electrostatics: Electric charge
1.9 Concept of electrical Potential
Charge moves in E-field from A to B. Work by E-field = q B
A Eds Indepen-dent of path
C Eds = 0
Define electric Petential ϕ so that ∆ϕ AB = − B A E · ds → ∆ϕ(x, y, z) +
fix zero-point→ ϕ(x, y, z) usually ϕ(∞) = 0→ P.E of individual charge = qϕ(x, y, z) e.g E-field from single charge
ϕ =
1
4πε 0
1
R
+superposition
ϕ =1
4πε 0∑
i
qi
ri=
1
4πε 0
all
space
ρ(x, y, z)r
dV
P.E of ” last charge” = qϕ P.E. of system of charges = 12 ∑ i qi ϕi
dϕ = −E · ds = − Exdx + E ydy + Ezdz
(1.16)
always dϕ =∂ϕ
∂xdx +
∂ϕ
∂ ydy +
∂ϕ
∂zdz (1.17)
l.e.Ex = −∂ϕ
∂x, . . . (1.18)
E = −
ϕ (1.19)
A =
∂ A
∂xx +
∂ A
∂ yy . . . (1.20)
→E is always ⊥ to surfaces of constant ϕ (1.21)
1.10 Experiment: charging two spheres equally
r force 1 force 2
10 55 44
14 31 25
20 14 13
40 7
charge 1 charge 2
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1.11. Other useful operators
Gauss’ Law
S
E · da =
V
ρ
ε0dV = Φs (1.22)
for general F(x, y, z)
S
S · da =
S1
F · a +
S2
F · da (1.23)
= ∑ i
Si
Fda = ∑ i
V i
S
Fda
V i(1.24)
(1.25)
take limit V i → 0 ∑ i V i → dV
S
Fda =
V
limV →0
S
F · da
V
divE
dV
SF · da =
FdV Gauss Theorem
(1.26)
⇒ ρ
ε0= divE Gauss’ Law
(1.27)
→ divF =∂Fx
∂x+
∂F y
∂ y+
∂Fz
∂z(1.28)
1.11 Other useful operators
•
E =
−
ϕ (1.29)
divE =ρε0
(1.30)
→ div
ϕ
= − ρ
ε0(1.31)
2ϕ = − ρ
ε0Poisson’s equation (1.32)
2=
∂2
∂x2+
∂2
∂ y2+
∂2
∂z2
Laplacian operator (1.33)
(1.34)
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1. Electrostatics: Electric charge
• curl Remember ”circulation”
Γ =
cF · ds = ∑
i
Γi = ∑ i
ai
ci
F · ds
ai(1.35)
curl F = lima→0
C
F · ds
a(1.36)
→
CF · ds =
A
curlF · da Stokes’ The
(1.37)
curl F =
∂Fz
∂ y− ∂F y
∂z
x +
∂Fz
∂z− ∂Fz
∂x
y +
∂F y
∂x− ∂Fx
∂ y
z =
×F
(1.38) E · ds = 0 in Electrostatics → curl E = 0 + . . . (1.39)
1.12 summary
E = −
ϕ (1.40)
ϕ = − Eds (1.41)
ρ = −ε0
2ϕ (1.42)
ϕ =
ρ
4πε 0rdV (1.43)
ρ = ε div E (1.44)
E =
ρ
4πε 0r2 rdV (1.45)
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Chapter 2
Conductors
A conductor can’t have an electric field inside because the charge can move.
+
2.1 Conductors + insulators
Difference is mobility of charges. ≈ 1020
→ Key point: E (x, y, z) inside conductor = 0
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2. Conductors
E
2.2 Conditions for conductor
a) E = 0 inside conductor
b) π = constant on surface and throughout conductor. ϕ:
c) At surface, E is perpendicular to surface, E = σ ε0
(Gauss’ Law)
d) Total charge on conductor Q =
S σ da = ε0
S Eda
2.3 The general electrostatic problem
General set of conductors k in Vacuum. ϕK, Qk
→ Solve2 ϕ = 0 subject to .b.
define ϕK Dirichlet problem (2.1)
Qk Neumann bonday conditions (2.2)
mixture of ϕK + QK but don’t overconstrain (2.3)
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2.4. Uniqueness Theorem
2 ϕ ≡ 0
2.4 Uniqueness Theorem
There is at most one unique solution.
Proof: Let ϕ1 (x, y, z) be a solution to2 ϕ = 0 + ϕK . . .
ϕ2 (x, y, z) be another solution to 2 ϕ = 0 + ϕK . . .
→ ψ = ϕ1 − ϕ2 is a solution to
2 ϕ = 0 and ϕ = 0 on all surfaces→ ψ = 0 everywhere ⇒ ϕ1 = ϕ2
2.5 Some interesting cases
2.5.1 Enclosed cavity
2 ϕ = 0
ϕ ≡constant
⇒ ϕ = constant throughout cavity
→ E = 0
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2. Conductors
2.5.2 Faraday cage
Farraday cage
+
+
+
+++
+
+
+
+
+
+
++ +
+
+
+
−
−
−−
−−−
−
−
−
−−
−− −
−−
−
+Q
E = 0
−Q
+Q
+
+
+
+++
+
+
+
+
+
+
++ +
+
+
+
+Q
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2.6. Some tricks
2.6 Some tricks
2.6.1 Mirror charges
ϕ
r
ϕ ≡constant
The both following cases match the same conditions. (Mirror trick)q q
−q
2.7 Capacitance and capacitors
Single conductor carrying a charge Q and some potential ϕ
Q
ϕ
from superposition Q&ϕ Q = Cϕ
C = ”capacitance” = Qϕ units
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2. Conductors
Coulombs Volt−1 = ”FARAD” Normally, two conductors close together,
+Q on one (2.4)
−Q on the other (2.5)
V = ∆ϕ between them (2.6)
+Q
−Q
ϕ1
ϕ2s
Area A
Uniform E between plates
E =ϕ1 − ϕ2
s=
V
SV = ϕ1 − ϕ2 (2.7)
σ = ε0E = ε0V
S(2.8)
Q = Aσ =Aε0V
S(2.9)
→ C =A
Sε0 (2.10)
2.8 Energy stored in capacitor
+Q −QE
V = ϕ1 − ϕ2
V = QC
dW = Vdq (2.11)
W = Q
0
q
cdq =
1
2
Q2
C=
1
2CV 2 =
1
2QV (2.12)
(2.13)
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2.9. General System of conductors
2.8.1 Parallel plate capacitor
Note:
C =A
Sε0 (2.14)
Q = Aδ = AEε0 δ = Eε0 (2.15)
W = A2E2ε20 · 1
2
S
Aε0=
1
2ε0E2 (Volume) (2.16)
2.9 General System of conductors
Q1ϕ1
∞
ϕ∞ = 0
Q3
ϕ3
Q2
ϕ2
2 ϕ = 0
If we define all ϕi → E (x, y, z) → σ (x, y, z) (2.17)
→ Qi (2.18)
or define all Qi → E (x, y, z) → ϕi (2.19)
(2.20)
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2. Conductors
Set ϕ = 0 on all except i = 1 (2.21)
Q1 = C11 ϕ1 (2.22)
Q1 = C21 ϕ1 (2.23)
Q1 = C31 ϕ1 (2.24)
Now ϕ2 varies, all other ϕ = 0
Q1 = C12 ϕ2 (2.25)
Q2 = C22 ϕ2 (2.26)
Q3 = C32 ϕ2 (2.27). . . (2.28)
→ By superposition for general ϕi
Q1 = C11 ϕ1 + C12 ϕ2 + C13 ϕ3 + . . . (2.29)
Q2 = C21 ϕ1 + C22 ϕ22 . . . (2.30)
→ Qi = Cij ϕ j with Cij matrix capacitance (2.31)
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Chapter 3
Elecric Currents
3.1 Electric current
Net non-zero charge in net motion
a
a
Charges q
number density n
mean velocity u
Define current through a, I a=rate of net motion of charges through a.
I a = nqu · a (3.1)
For multiple species I a = ∑ i
(niqiui) J : current density
·a (3.2)
J = ∑ Ji J = ρu 3-d
I = λv 1-d
(3.3)
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3. Elecric Currents
3.2 Charge conservation
S
net flow of charge rate of charge into surface of enclosed charge
S J · da = − d
dt V ρdV (3.4)
applying Gauss’ Theorem
div J = −d ρ
dt(3.5)
3.3 Ohm’s Law
Motion of charge due to E-field in non-perfect conductor. Conductors are
not perfect. The charges are not infinitely mobile.Expect J (x, y, z) = σ E Ohm’s Lawσ is the conductivity. In practice the σ can be a Tensor dependend on itsposition. That is just an approximation. It’s not an as fundamental Law asCoulomb’s law or Gauss’ law.Consider a simple cylindrical ”component”.
A J
e−
L
I =
Jda = JA (3.6)
Potential Voltage differenc =
Eds = EL (3.7)
Define Resistance R =V
I =
EL
J A=
1
σ = ρ =
1
δ⇒ R = ρ
resistivity
L
A
(3.8)
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3.4. Circuits
Aside: In steady state
div J = − dϕ
dt= 0 (3.9)
→ div E = 0 (Ohm’s Law) (3.10)
Charge density ρ = 0
div E =
ρ
ε0
(3.11)
3.4 Circuits
Circuit = discrete ”component” linked together by ideal conducting wires,may join in ”nodes”
eg
A B
C D
E F
ϕ A = ϕB = ϕC = ϕD
ϕC > ϕE
ϕD > ϕF
ϕ
For each component I i current through it, V i potential difference acress it.For any arbitrary circuit, we want to know I i, V i for all component:→ Commonsense rules (≡ )Kirchoff’s Rules
3.4.1 Kirchoff’s Rules
• For every element i I i = V iR (Ohm’s Law)
• ∑ I in = ∑ I out or ∑ all I in = 0 since I out = −I in(charge conservation in steady state)
• Around any ”loop” ∑ V = 0→ Can usually break down a circuit into simple equivalent elements
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3. Elecric Currents
eg series connection
R1 R2
I 1 = I 2 = I (3.12)
V 1 + V 2 = V (3.13)
R =V
I =
V 1I
+V 2I
=V 1I
1
+V 2I
2
= R1 + R2 (3.14)
In parallel
V 1 = V 2 = V
I = I 1 + I 2
→ 1
R=
1
R1+
1
R2(3.15)
R1
R2
3.5 Energy dissipation in a resistor
ϕ1 ϕ2IV = ϕ1 − ϕ2
P.E. released when Q flows through V = QV .
Energy per second = V I = V 2
R = I 2R appear as heat.
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3.6. Sources of energy
3.6 Sources of energy
A
B
I Rgives energy to charges
”electro-motiveforce” e.m.f -units of Volt
eg chemical reactions:
H +
H 2
SO
HSO−4
PB
PbO2
2e− 2e−
+ −
1. Pb + HSO−4 → PbSO4 + H + + 2e− + energy
2. PbO2 + HSO−4 + 3 H + + 2e− → PbSO4 + 2 H 2O + energy
3. combined effect Pb + PbO2 + 2 HSO−4 − 2 H + → 2PbSO4 + 2 H 2O +
energy+2e− used on LHS reapon on RHS, +2 H + cross from right to left
Some energy wasted as heat
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3. Elecric Currents
$I$
$R$
$+\phi$
$-\phi$$R i$
internal resistanc
I = εR+Ri
useful V = ε
−IRi Thevenin’s Theorem any system of emf &
resistors equivalent to a single emf and intend resistance.
3.7 Circuits with capacitors
Q = CV (3.16)
I =V
R= −dQ
dt(3.17)
−dQ
dt=
Q
CR(3.18)
dQ
dQ= − dt
CR(3.19)
In Q = −t
RC + constant (3.20)
Q = Q0e− tRC (3.21)
I =V 0R
e−t tRC (3.22)
RC time
FΩv
cs−1· cv (3.23)
Also charging of CPic 3.2
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3.7. Circuits with capacitors
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Chapter 4
Fields of moving charges
4.0 Magnetic Fields or Forces
F = q E + v × B Lorentz Force
I
B
q v
F
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4.1. Reminder about special Relativity
4.1 Reminder about special Relativity
F F
∆x = γ∆x − βγc∆t Lenght/Lorentz contraction (4.1)
∆ y = y β =
v
c , γ =
1 1 − β2 (4.2)
∆z = ∆z (4.3)
c∆t = γc∆t − βγ∆x time dilation (4.4)
(4.5)
ux =
ux − v
1 − uxvc2
ux =u
x + v
1 + uxvc2
(4.6)
u y =
u y
γ 1 −uxv
c2 u y =u
y
γ 1 +uxv
c2 (4.7)
cpx = γcpx − βrE (4.8)
cp y = cp y (4.9)
cpz = cpz (4.10)
E = γE − βγcpx (4.11)
Forces? If particle at rest in F
f x = f x (4.12)
f y = γ f y (4.13) f z = γ−1 f z (4.14)
4.2 Invariance of charge
S(t)
E(t)da =
S(t)Eda =
Q
ε0(4.15)
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4. Fields of moving charges
4.3 Transformation of E-fields
V
+σ −σ
E = σ ε0
At rest in F. In F, moving related to F ⊥to plates
σ = σ (4.16)
E =σ
ε0=
σ
ε0= E (4.17)
Now v plates
V
⇒ must be general result:
E = E
E⊥ = γE⊥
Now consider point charge moving with velocity v
qvx
In F, rest-frame of particle
Ex =q
4πε 0
x
(x2 + z2)32
(4.18)
Ez =q
4πε 0
z
(x2 + z2)32
(4.19)
sety = 0 xz plane (4.20)
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4.4. Field of accelerated charge
Now consider F moving −vx (our lab!) swap primes and use
x = γx − γβct (4.21)
z = z (4.22)
t = γt − γβx
c2(4.23)
Set t = 0 when x = 0, look at field at t = 0
(4.24)
Note:
a) Field is radial: Ex
Ez
= xz , field points at origin
x
at origin
at t = 0
radial but not isotropic
q v v
→ E · ds = 0 ⇒ curl E = 0
4.4 Field of accelerated charge
Start at rest, then accelerate
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4. Fields of moving charges
q
Look at deceleration
q 0v0
12 v0t
a = v0τ
Look at this time T later. Assume 12 v0τ v0T cT
origin Where it would have been
θv0T sin θ
ErcT Eθ
Er
cτ
Er
→ Eθ
Er=
v0T sin θ
cτ (4.25)
Er =q
4πε 0R2=
q
4πε 0c2T 2(4.26)
⇒ Eθ =v0T sin θ
cτ · q
4πε 0c2T 2=
q
4πε 0
v0
τ a
·sin θ
Rc2(4.27)
→ Eθ =qa sin θ
4πε 0c2R(4.28)
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4.5. Force on moving charge
4.5 Force on moving charge
q vxF our lab, Fq at rest.
Ex = Ex
E y = γE y
→dp
x
dt= E
xq = Exq
dp y
dt= E
yq = γE yq
(4.29)
• Particle is at rest in F, so flip primes in above eqn for F, F
dpx
dt= dpx
dt(4.30)
dp y
dt=
1
γ
dp y
dt(4.31)
back in F,
dpx
dt= Exq and
dp y
dt= E yq = E yq (4.32)
→ i.e. Force on q due to E does not change with vq
4.6 Interaction of moving charge with other moning
charges
q vx
I Infinite wire is uncharged
λ+0 = −λ−
0
−ve → v0
+ve 0
Negative charge moves, positive doesn’t → I = −λ0v0
Transform to F = rest-frame of q
λ+= γλ0 (4.33)
λ−= γ
γ due tovelocity less
than V
λ0 (4.34)
⇒wire looks positively charged in F (4.35)
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4. Fields of moving charges
Algebra work out λ− in rest of negative charges
net λ = γβ F→F
β0 v0
λ0
q
E
λ(> 0)
Er = λ
2πε 0r = γββ0λ0
2πε 0r (4.36)
New force is radial from wire and perpendicular to motion.
F y = qE y = −qγββ0λ0
4πε 0r for y = −r (4.37)
Transform to F y =F
y
γ= − q ββ0λ0
2πε 0r(4.38)
but I = −λ0v0 = −λ0 β0c (4.39)
F y = I 2πε 0c2
qvr
looklike magnetic force
∝ v
∝ I in wire
⊥ to velocity
(4.40)
v0
I
+ + + + + + + + +− − − − − − − − −
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Chapter 5
Magnetic Fields
Total force on electric charge
F = qE + qv × B Lorentzforce (5.1)
B come from current and act on currentE come from charge and act on charge
dF = dqv × B = (λdl)v × B (5.2)
dF = I × Bdl (5.3)
5.1 Simplest case
B =µ0 I
2π r
µ0 = Permeability of Force Space (5.4)
Units of B held is Tesla. 1 Tesla excerts a force of 1Nm−1 on a current of 1Amp.
I
B
vF
⊗I
B
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5. Magnetic Fields
5.1.1 Force between two ∞ long wires
B1
FI1
B2
FI2
d
B1 =µ0 I 12π d B2 =
µ0 I 22π d (5.5)
F1 =µ0 I 22π d
I 1 = F2 (5.6)
µ0 = 4π × 10−7Tm−1 A−1 (5.7)
F = 2 × 10−7 Nm−1 S.I. convention for defining an amp (5.8)
5.2 Properties of magnetic fields
Consider simple case of ∞ wire
B =µ0 I
2π r(5.9)
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5.3. Vector Potential
B · ds = 0
B · ds = µ0 I
2π r · 2π r = µ0 I
General rule: Amperes Law B · ds = µ0 I = µ0
A
J0da (5.10)
5.2.1 Uniqueness theorem
For a given J(x, y, z), there is a unique B(x, y, z).
Proof:Suupose there are two solutions B1&B2 for J(x, y, z).→ D = B1 − B2 is a solution
div D = div B1 − div B2 = 0 (5.11)
curl D = curl B1 − curl B2 = µ0 J − µ0 J = 0 (5.12)
5.3 Vector Potential
ϕ = 14πε 0
ρr
dV (5.13)
⇒ E = −
ϕ (5.14)
Imagine
B =
×A (5.15)
div B = 0 = div(curl A) = 0 (5.16)
curl B = µ0 J ⇒ curl(curl A) = µ0 J (5.17)
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5. Magnetic Fields
Expand× (
×A)
5.4 Fields of coils and solenoids
x-component:
∂
∂ y
∂ A y
∂x− ∂ Ax
∂ y
− ∂
∂z
∂ Ax
∂z− ∂ Az
∂x
= µ J x (5.18)
→∂2 AY
∂ y∂x− ∂2 Ax
∂ y2− ∂2 Ax
∂z2+ ∂2 Az
∂x∂z= µ J x
(5.19)
−∂2 Az
∂ y2− ∂2 Ax
∂z2 −∂2 Ax
∂x2 2 Ax
+∂2 A y
∂ y∂x+
∂ Az
∂x∂z +
∂2 Ax
∂x2 ∂
∂x(div A)=µ0 J x
= µ0 J x (5.20)
Trick: div A = f (x, y, z)
Then we can find another F, div F = f and curl F = 0Can add −F to A ⇒ div F = 0 and the same B
2 Ax = −µ0 J x
c.f.
ϕ = − ρ
ε0
for y, z − comp.analogous (5.21)
We had
ϕ =1
4πε 0
ρ
rdV (5.22)
A =µ0
4π
J
rdV Vectorpotential (5.23)
Apply this to general current carrying loop.
dA =µ2
4π r JdV =
µ0
4π rIdldB =
×dA =
I µ0
4π
×1
rdl
(5.24)
=I µ0
4π
−dl ×
1
r
=
I µ0
4π dl × r
r2(5.25)
(5.26)
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5.4. Fields of coils and solenoids
5.4.1 Biot-Savart law
dB = Idl × r · µ0
4π r2= JdV × r
µ0
4π r2Biot-Savart law (5.27)
c.f
dE = ρ r1
4πε 2r2dV (5.28)
dB = ( J × r)µ0
4π r2dV (5.29)
usually Biot-Savart is easier to work with then calculating A and the takingits curl.on axis Bx = B y = 0
Bz =µ0
4π (b2 + z2)· I · 2π b
sin θ
b
(b2 + z2) 12
(5.30)
@z = 0 (θ = 90).
Bz =µ0 I
2b(5.31)
Bz = µ0 I 2
b2
(b2 + z2) 32
(5.32)
z
x
Bz
rθ
I
b
Now consider long Solenoid n turns per unit lenght.
current =rdθ
sin θ· n · I (5.33)
On the axis
dBz =µ0
2
rdϑ nI
sin ϑ
b sin ϑ
r2=
µ0
2nI sin ϑ (5.34)
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5. Magnetic Fields
Integrating over ϑ
Bz =µ0
2nI
ϑ 2
ϑ 1sin ϑ 2dϑ =
µ0
2nI [cos ϑ 2 − cos ϑ 1] (5.35)
For an infinitely long solenoid
Bz = µ0nI (5.36)
Applying Ampere’s Law, it is clear that B cannot depend on position withinthe solenoid, for an infinite solenoid.
5.5 summary
GaussLaw AmperesLaw
S E · da = Q
ε0= 1
ε0
ρdV
B · ds = µ0 I = µ0 Jdacurl B = µ0 J
div E = ρε0
div B = 0
ϕ = 14πε0
ρr dV A = µ0
4π
Jr dV
E = − ϕ B = curl A
dE = 14πε0
ρdV rr2 dB = µ0 Idl× r
4π r2 = JdV × rµ0
4π r2
5.6 Change in B-field across a current sheet
σ
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5.6. Change in B-field across a current sheet
∆E⊥ = σ ε0
Gauss’ Law (5.37)
∆E = 0
E · ds = 0 (5.38)
→ Pressure on sheet = σ
E1 + E2
2
= σ E⊥ (= 0 if no extend E-fields)
(5.39)
Now look at the current sheet
J = σ v Am−1 (5.40)
x
z
y
σ
J
B1B2 ∆B = (B1 − B2)
B · ds = µ0 J xl
I l
(5.41)
∆Bz = µ0J x (5.42)
∆Bx = 0 (5.43)
∆B y = 0 (div B = 0) (5.44)
∆B = µ0J (5.45)
∆B = 0 (5.46)
∆B⊥ = 0 (5.47)
∆B parallel to sheet perpendicular to current∆B parallel to sheet and parallel to current.Exactly as in electrostatics
Pressure =1
2µ
B2
2 − B21
(5.48)
+energy density in B-field =1
2µ0
c f
1
2εE2
(5.49)
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5. Magnetic Fields
5.7 How do E and B transform
x
z
y
F
v
F
v0 v0
−σ σ
E y =σ
ε0(5.50)
Bz = µ0J = µ0σ v0 (5.51)
v0 = v0 − v
1 − vv0
c2
= c ( β0 − β)(1 − ββ0)
(5.52)
σ = ϕ0
charge densityof charges
in F
σ
γ0
charge densityof charges
in rest-frame
(5.53)
γ0 =
1 − v
02
c2
(5.54)
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5.7. How do E and B transform
σ = σγ (1 − ββ0) (5.55)
and J = σ v0 = σγ (1 − ββ0) c ( β0 − β)
(1 − ββ0)= σγ (v0 − v) (5.56)
E y =
σ
ε0=
σγ
ε0− σγββ0
ε0(σ V 0µ0 = B2) (5.57)
= γE y − γβ
ε0µ0cBz (5.58)
εµ0= 1c2
= γ
E y − βc
=v
Bz
= γ
E y − vBz
(5.59)
Bz = µ0
J (5.60)
= µ0σγv0 − µ0σγv (5.61)
= γ
Bz − εγ0 βE y
→ Bz = γ
Bz − β
cE y
(5.62)
Other components
Ex = Ex (5.63)
E y = γ
E y − βcBz
(5.64)
Ez = γ
Ez + βcB y
(5.65)
Bx = Bx (5.66)
B y = γ
B y +
β
cEz
(5.67)
Bz = γ
Bz − β
cE y
(5.68)
If represent + ⊥ components to v (motion of frames)
E = E B
= B (5.69)
E⊥ = γ E⊥ + cfi × B⊥ B
⊥ = γ
B⊥ − 1
cfi × E⊥
(5.70)
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5. Magnetic Fields
5.8 Hall Effect (1879)
x
z
y
J
+
−
B
F = qE + qv × B (5.71)
If +ve charge carrier
If −ve charge carrier
+
− E
Eint = vBSign of Ez tells us +ve or −ve charge carriers.Answer: −ve!
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Chapter 6
Magnetic Induction
6.1 Magnetic Induction
F = qE + qv × B (6.1)
Charges in moveing conductor in a B-field experience Lorentz force.
x
z
y
Bz
v
F−
F+
E
Eint =
−v
×B
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6. Magnetic Induction
Eint = −v × B B
− − − − − − −
+ + + + + + +
E
−
+
F−
F+
Aside: Look is rest-frame of conductor
E y = γ
E y
=0
− βcBz
(6.2)
yz
x
Eint
E y
Eint, y = γvBz (6.3)
Eint = vBz (6.4)
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6.1. Magnetic Induction
6.1.1 Wire in circuit
A
B
lRF+
F−
B
+ charges moving A → Bgains energy (6.5)
+ charges moving B→
Agains energy (6.6)
different Situations
B1 B2 B(t)
Energy gain = B
Aq (v × B) ds = qvlB = qE
(6.7)
i.e.E = vlB = rate of charge of ” enclosed b” =Φ
t(6.8)
In general:E =
(v × B) ds (6.9)
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6. Magnetic Induction
6.2 Faraday’s Law
Φ(t) =
SB · da (6.10)
Φ (t + dt) =
S+dSB · da = Φ(t) +
dS
B · da dΦ
(6.11)
On rim of loop
da = vdt
×ds (6.12)
⇒ dΦ =
dSB · da =
C
B · (vd × ds) (6.13)
dΦ
dr=
CB · (v × ds) = −
c
(v × B) · ds = −E from before (6.14)
i.e.E = −dΦ
dtFaraday’s Law (6.15)
Note
Φ = S
B · da (6.16)
div B = 0 (6.17)
What if loop fixed and B is changing
→ look in rest-frame of loop
B1 B2
v v fixed
E⊥ = γ E⊥ + v × B⊥ (6.18)
E = γv × B (6.19)
E = γv × B1 (6.20)
E2 = γv × B2 (6.21)
energy gain per q = E = γv (B1 − B2) l (6.22)
= −γdΦ
dt= −dΦ
dt(6.23)
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6.3. Lenz’ Law
Consider a static loop Faraday’s Law
E = − d
dt Ψ = − d
dt
S
B · da (6.24)
moving loop E =
(v × B) · ds (6.25)
E =
Eds = − d
dt
S
Bda (6.26) S
curl Eda =? (6.27)
(6.28)
curl E = −∂B
∂t(6.29)
note: this does not uniquely specify E (x, y, z), - can add any E (x, y, z) fromstatic ρ (x, y, z), which has curl E = 0 as before.
6.3 Lenz’ Law
B1 B2
v
I
F1+ F2+F−
Binducted
I B
E
emf drives current around loop
B1 > B2, F1 > F2 anddΦ
dtis negative
→I is in direction to reduce change in Φ
(6.30)
also force on loop ← F resists motion
dF = I (ds × B) E =
(v × B) · ds (6.31)
dW = dF · v I =E R
(6.32)
W =
I (ds × B) · v (6.33)
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6. Magnetic Induction
6.4 Mutual inductance
Change in IU in one circuit may produce a change in Φ in a second circuit(mutual ind.) or a change in Φ in itself (self ind.)
I 1C1
Let
Φ21 = flux though C2due to I , in C1 (6.34)
=
S2
B1 · da = kI 1 (6.35)
E 21 = emf in C2due to change in I 1 (6.36)
= − ddtΦ21 (6.37)
E 21 = −kdI 1dt
k = “mutual inductance“ M21
(6.38)
units = VA−1s = Ωs = “Henry“ (6.39)
⇒ M21 =Φ21
I 1(6.40)
Use A
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6.5. Self-inductance
Φ =
SB · da (6.41)
=
Scurl A · da (6.42)
Stoke’s=
C
A · ds (6.43)
A1from Chap 4?
=µ0
4π
all
space
J
rdV =
µ0
4π I 1
C
ds1
r(6.44)
Φ21 = C2
A1 · ds2 = C2
µ04π I 1
C1ds1
r ds2 (6.45)
Φ12 =
C1
µ0
4π I 2
C2
ds2
rds1 (6.46)
Since
C1 C2
ds2 · ds1
r= C2 C1
ds1 · ds2
r(6.47)
→ Φ12
I 2=Φ21
I 1(6.48)
6.5 Self-inductance
Pic 6.1
Changing I → change in Φ → E to oppose change in I Clase switch at t = 0
I =ε0
R
1 − e− R
C t
(6.49)
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6. Magnetic Induction
I
t
ε0R
Pic 6.3
UC =1
2
CV 2 (6.50)
UL =1
2LI 20 (6.51)
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Chapter 7
Alternating current circuits
7.1 The RLC circuit as damped oscillator
Pic 7.1R
LC+Q
−Q
Q = CV (7.1)
I = −dQ
dt= −C
dV
dt(7.2)
V − LdI
dt− RI = 0 Kirchoff nr.2∑ v = 0 (7.3)
d2V
dt2 +R
L
dV
dt +1
LC V = 0 → damped harmonic oscillator (7.4)
Try solution V ∝ eλt
λ2 +R
Lλ +
1
LC= 0 (7.5)
λ =− R
L ±
R2
L24
LC
2=
−R
2L±
R2
4L2− 1
LC(7.6)
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7. Alternating current circuits
a) R2<
4Lc
V (t) = Ae−R2L t cos ωt ω2 =
1
LC− R2
4L2
(7.7)
pic7.2 (7.8)
b) R2>
4LC
V (t) = Ae− β1t + Be− β2t β =R
2L±
R2
4L2− 1
LC(7.9)
pic 7.3
c) R2 = 4LC
V (t) = Ae−R2L t (1 + Bt) (7.10)
pic 7.4
7.2 Circuits driven by alternating voltage
Easy to generate voltage source that varies sinusordiallyPic 7.5
R
LVAC
Calculate amplitude and phase of I ,which must have same ω
I = I 0 cos (ωt + α) (7.11)
Apply Kirchoff’s nr 2
ε0 cos ωt = LdI
dt+ RI (7.12)
Try solution as above
ε0 cos ωt = −LI 0ω sin (ωt + α) + RI 0 cos (ωt + α) (7.13)
= −LI 0ω sin ωt cos α + cos ωt sin α + RI 0 cos ωt cos α − sin ωt sin α(7.14)
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7.2. Circuits driven by alternating voltage
Balance sin ωt, cos ωt
−I 0 Lω cos α − RI 0 sin α = 0 → tan α =−ωL
R(7.15)
→I peaks after V , I “lags“ V − LI 0ω sin α + RI 0 cos α = ε0
(7.16)
→ I 0 =ε0
R cos α − ωL sin α(7.17)
=ε0
R (cos α + sin α tan α)(7.18)
=ε0
Rcos α (7.19)
tan α =ωL
R→ cos α =
R
(R2 + ω2L2)12
→ I 0 =ε0√
R2 + ω2C2I is reduced in amplitude
(7.20)
E
current in loop lags
I in solenoid
I s, I lis negative (7.21)
E = −dΦ
dt(7.22)
Φin ring (7.23)
7.2.1 Solenoid pushing
Some description would be nice. . .
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7. Alternating current circuits
7.2.2 Some circuits
pic 7.7
V = ε0 cos ωt (7.24)
ε0 cos ωt = − Q
C+ IR (7.25)
I = I 0 cos(ωt + ϕ0) (7.26)
I = −dQ
dt(7.27)
Q = −
Idt = − I
ωsin(ωt + ϕ) (7.28)
ε0 cos ωt =I 0
ωCsin(ωt + ϕ) + RI 0 cos(ωt + ϕ) (7.29)
(7.30)
steady-state
tan ϕ =1
RωC, I 0 =
ε0 R2 +
1
ωC
2(7.31)
I = I 0 cos ωt (7.32)
V L = LdI
dt= −LI 0ω sin (ωt + ϕ) (7.33)
V C = − Q
C=
1
C
Idt =
I 0ωC
sin (ωt + ϕ) (7.34)
V = V L + V C = −
ωL − 1
ωC
I 0 sin (ωt + ϕ) (7.35)
ωL = ωL − 1
ωC(7.36)
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7.2. Circuits driven by alternating voltage
pic 7.9
ωL > 1ωC ; tan ϕ = − ωL
R − 1ωRC I 0 =
ε0 R2 +
ωL − 1
ωC
2 (7.37)
ωL =1
ωC⇒ ω ≡ ω0 =
1√LC
(7.38)
I = I 0 =ε0
R; (7.39)
P(ω0) =ε2
0
R2; ω; P(ω) =
1
2P(ω0) (7.40)
I
ω/ω0
RR> R
Q f = ω
energy stored
energy power dissipated =
ωL
R (7.41)
ω = ω0 + ∆ω, ω20 =
1
LC⇒ 1
ω0C= ω0 L (7.42)
ωL − 1
ωC= (ω0 + ∆ω) L − 1
ω0 + ∆ω
1
C= ω0L
1 +
∆ω
ω0
−
1 + ∆ωω0
−1
ω0C(7.43)
= ω0L
1 +
∆ω
ω0−
1 − ∆ω
ω0
= ω0L
2∆ω
ω0
+ O
∆ω
ω0
(7.44)
pic 7.10
I 2 = I 02 cos (ωt + ϕ2) (7.45)
V 2 = V 02 cos (ωt + ϑ 2) (7.46)
eiϑ = cos ϑ + i sin ϑ , i2 = −1 (7.47)
1) I = I 0 cos (ωt + ϕ) → I 0eiϕ (7.48)
2) z = x + iy Re
zeiωt
(7.49)
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7. Alternating current circuits
I 1 = I 01 cos (ωtϕ1) (7.50)
I 2 = I 12 cos (ωt + ϕ2) (7.51)
I 1 = I 01eiϕ1 (7.52)
I 2 = I 02eiϕ2 (7.53)
I 1 + I 2 = Re
I 01eiϕ1 + I 02eiϕ2
eiωt
(7.54)
Kirchoff’s rules
a) ∑ I in = 0
b) ∑ ∆V = ∑ emt
V = ε0 (7.55)
I = I 0eiϕ (7.56)
admittance (7.57)
Y =eiϕ
(R2 + ω2 L2);tan =
−ωL
R(7.58)
impendance (7.59)
z =1
y(7.60)
I = YV , V = ZI (7.61)
(7.62)
R → 0
C → 0, ϕ = −π
2; y =
−i
ωL, V = IR (7.63)
Im
Re
V
I
π 2
L → 0, ϕ = 0; ϕ =1
R; z = R, C = 0, V = IR (7.64)
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7.2. Circuits driven by alternating voltage
Im
Re
V
L → 0, R → 0,tan ϕ =1
ωRC, ϕ =
π
2; y = iωC; z =
−i
ωC(7.65)
Im
Re
V
I
π 2
pic 7.11
I = YV (7.66)
V = ε0 (7.67)Y = YC + YR + YL = iωC +
1
R− i
1
ωL(7.68)
I = ε0
1
R+ i
ωC − 1
ωL
= I 0eiϕ (7.69)
I 0 = ε0
1
R2+
ωC − 1
ωL
2
(7.70)
tan ϕ =
RωC − R
ωL
(7.71)
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7. Alternating current circuits
7.3 Power Consumption
P = V I = I 2R =V 20R
cos2 ωt (7.72)
V = V 0 cos ωt (7.73)
P =v2
0
R
cos2 ωt
=
1
2
V 20R
=1
2I 2R (7.74)
P = V I = V 0 cos ωtI 0 cos (ωt + ϕ) (7.75)
= V 0 I 0 cos ωt (cos ωt cos ϕ−
sin ωt sin ϕ) (7.76)
= V 0 I 0
cos2 ωt cos ϕ − /cosωt sin ωt sin ϕ
(7.77)
(7.78)
P = V 0 I 01
2cos ϕ (7.79)
cos ωt sin ωt =1
2sin2ωt (7.80)
V rms =
1
√2 V 0 (7.81)
I rms =1√
2I 0 (7.82)
P = V rms I rms cos ϕ (7.83)
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Chapter 8
Maxwell’s Equations
8.1 Wave equation
E & B
×E =
∂B
∂t(8.1)
×B = µ0ε0
∂E
∂t(8.2)
·E =
·B = 0 (8.3)
×
×B
= ε0µ0
×∂E
∂t(8.4)
×B
0
−2
B = −ε0µ0∂2B
∂t2(8.5)
⇒2
B − ε0µ0∂2B
∂t2= 0 (8.6)
2E − ε0µ0
∂2E
∂t2= 0 (8.7)
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8. Maxwell’s Equations
x
z
y
E
B
v
Direction of travel = E × B and E · B = 0
Note: Wave doesn’t have to be a sine wave. Any function of (K · r − ωt) will
do so long spatial and temporal derivatives match and E, B, v are perpen-dicular.
8.2 Superposition of two opposite directions
Maxwell equations linear in E and B → Sum should also be a solution.evt ein Fehler hier. . .
E = z · E0 (sin (Ky − ωt) + sin (Kyωt)) (8.8)
= z · 2E0 sin (Ky) cos (ωt) (8.9)
B = xB0 (sin (Ky − ωt) − sin (Ky − ωt)) (8.10)
= x2B0 cos (Ky) sin(ωt) (8.11)
(?Standing wave)
8.3 Standing wave
E = 0 at certain points at all times.Satisfies the boundary conditions for conducting surface.⇒ conductor to reflect lightB is changing rapidly at the surface and is zero inside the conductor.⇒ surface currents
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8.4. Energy Transport of E and M waves
8.4 Energy Transport of E and M waves
Compute the energy density for aur travelling wave.
dU =
ε0E2
2+
1
2µ0B2
dV = ε0E2dV =
1
µ0B2dV (8.12)
=1
cµ0EBdv (8.13)
from before: ε0E2 = B2
µ0(8.14)
Energy transport per unit area
S =1
µ0EB (8.15)
Pointing Flux
S =1
µ0E × B (8.16)
8.5 Lorentz transformation of waves
Ex = Ex E
y = γ
E y − βcBz
E
z = γ (Ez + βcBy) (8.17)
Bx = Bx B
y = γ
By +
β
cEz
B
z = γ
Bz − β
cE y
(8.18)
E · B = 0 (8.19)
ε0E2 − B2
µ0= 0 (8.20)
E · B = Ex B
x + E y B
y + EzB
z (8.21)
= Ex Bx + γ2 (Ey − βcBz)
B y +
β
cEz
+ γ2
Ez + βcB y
Bz − β
cE y
(8.22)
= Ex Bx + γ2
1 − β2
1
E y B y + EzBz
= E · B (8.23)
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8. Maxwell’s Equations
ε0E2 − B2
µ0= ε0E2 − B2
µ0(8.24)
→ em wave in one Frame appear as an em in all frame including ωK = c
What about a frame moving at c?Consider E y = E0 and Bz = E0
c
E y = γ
E0 − βc
E0
c
= E0γ (1 − β) (8.25)
Bz = γE0
c −β
c E0 =
E0
c γ (1 − β) (8.26)
as v → c, β → 1, E&B → 0
8.6 summary
f (k · r − ωt) curl E = − ∂B∂t I = YV = 1
Z V
ωk = v = 1√ε0µ0
= c curl B = µ0ε0 ∂E∂t + µ0 J
B0 = E0c div E = ρ
ε0YR = 1
R ZR = R
E · B = 0 div B = 0 YL = −iωL ZL = iωL
E × B = direction of travel div J = − ∂ρ∂t YC = iωC ZC = −i
ωC
(8.27)
⇒ (10)Φ =
S B · da = BS cos θ
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Chapter 9
Dielectric materials
9.1 Introdiuction
When I have a material in a capacitor, capacitance changes C = QV → C =
εCVac , ε ≥ 1
For fixed V (= ∆ϕ) Q = εQVac (9.1)
For fixed Q V =V Vac
ε(9.2)
(9.3)
ε = 1.0 for vacuum (9.4)
≈ 1.0006 for air (9.5)
1.01 typical gases (9.6)
≈ 2 − 10 typical solids (9.7)
≈ 20 − 100 typical liquids (9.8)
empirically, related to density or mobility
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9. Dielectric materials
9.2 Electric dipoles
Recall p = ql r from − to +
p =
allspace
r ρdV = “first moment“ of p (9.9)
→ p in E
a) Torque N = p × E → align p with E
b) Work to align dipole with E ⊥→ ω = pE dω = − ρE sin θdθ = averageω for randomly oriented dipole
c) potential energy: E pot = −p · E
d) Net force in inhomogeneous E-field
Fx = p ·
Ex (9.10)
F y = p ·
E y (9.11)
Fz = p ·
Ez (9.12)
(9.13)
V ≡ ∆ϕ
Q
−Q−σ
σ
V =Q
εCVacε ≤ 1.0 (9.14)
E =EVac
ε(9.15)
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9.3. Atomic and molecular dipoles
9.3 Atomic and molecular dipoles
9.3.1 Permanent dipoles
H
+
Cl
−
H +
H +
O−−
→ external E-field will pref-erentially align permanent dipoles.
9.3.2 Induced dipoles
+
−
F+F−
E
⇒
−
+p
Expect p ∝ E (9.16)
= E4πε 0α α : dimensions = Volume (9.17)
Compare E-field insied atom to external one
Inside E =1
4πε 0
e
a20
for Hydrogenexpect∆z
a0≈ E pot
14πε0
ea2
0
(9.18)
p = e∆z (9.19)
α ∝ a30 (9.20)
actually α =q
2a3
0 for H (9.21)
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9. Dielectric materials
9.4 Electric fields from polarized matter
Consider material composed of dipoles number density N .
Total dipole moment
in volume dV = p · N · dV
P
P = density of polarization (9.22)
P
dz
da
−σ
+σ
E p
i.e.σ = P (9.23)
E p =σ
ε0= − P
ε0(9.24)
H P is result of external Eext
→ E p is opposite to Eext
→ E final = Eext + E p is recuced → look at capacitor
+Q
−Q −σ
σ
Evac = σ ε0
+Q
−Q−σ
σ
P E p
E final ≡ Evac − pε0
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9.4. Electric fields from polarized matter
1 =Evac
E final− p
E finalε0(9.25)
! E =Evac
E final= 1 +
p
ε0E final(9.26)
→ E = 1 +p
E finalε0 X =“permittivity“
(9.27)
Note: Final E-field makes sence! Will drop “final“ from now on.
E = 1 +p
ε0E final X susceptibility
(9.28)
Look again at Capacitor with E (≡ ∆ϕ) constant
V E
+ + + + + + + + +
− − − − − − − − −
+ + + + + + +
− − − − − − −
E from po-larization P = (E − 1) Eε0 from →
top surface has σ = −P (9.29)
= (1 − ε) ε0E (9.30)
(9.31)
top plate of capacitor needs extra charge
total charge = ε0E (9.32)
ε0E = δC + (1 − ε) ε0 top dielectric
E (9.33)
→ σ C = εε0∃ = εσ 0 σ : capacitance ↑ ×ε (9.34)
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9. Dielectric materials
9.4.1 Gauss’ Law in medium and vector field D
q
E
dilectric ε
E reduced by factor 1ε
E =q
4πεε0r2(9.35)
Now introduc concept ρ free (what we control)and ρbound what nature controls due to polarization Always true
div E =1
ε0 ρ =
1
ε0
ρ free + ρbound
(2) (9.36)
div EVac =ρ free
ε0 = ε div E +ρbound
ε0(9.37)
→ (ε − 1) div E = − ρbound
ε0(3) (9.38)
Aside: we had before
P
E= (ε − 1) ε0 (9.39)
P = ε0 (ε − 1) E (9.40)
(3) → ρbound = − div PReturn to (3)
div E = ε div E − 1
ε0div P (9.41)
div
E +
P
ε0
D/ε0
=ρ free
ε0(9.42)
div D = ρ free (9.43)
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9.5. Currents in dielectrics and Maxwell’s equations
D = E1 +p
Eε0 ε0 ε
Electric Displacement (9.44)
D = εε0E (9.45)
i.e If we deal with D, things are as in vacuum.→ get E from D
9.5 Currents in dielectrics and Maxwell’s equations
ρbound = − div P (9.46)
div J = −∂ρ
∂t(9.47)
→ Jbound =∂P
∂t(9.48)
Maxwell
curl B = µ0ε0
∂E
∂t + µ0 J free +
∂P
∂t (9.49)
and E +P
ε0=
D
ε0(9.50)
“boxed“
curl B = µ0∂D
∂t+ µ0 J free (9.51)
= µεε 0∂E
∂t+ µ0 J free (9.52)
9.6 Eloctromagnetic waves in dielectric
ρ free = 0 J free = 0 (9.53)
curl E = −∂B
∂tdiv E =
1
εε0div D = 0 (9.54)
curl B = εε0µ0∂E
∂tdiv = 0 (9.55)
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9. Dielectric materials
→ Same wave-like solutions but now
ω
k=
1√εε0µ0
→ c =c
ε12
(9.56)
ε12 = refractive index n
B0 = ε12
E0
c=
E0
c (9.57)
9.7 Example: Electric field around dielectric sphere
Consider a polarized sphere
r 0 p− ⇔
outside looks like
+
−
+
−
p
Total dipole is the same
p0 =4
3
π pr30 (9.58)
→ use standard expressions for ϕ ∝ E outside of sphere
ϕ =1
4πε 0 p0
cos θ
r2=
1
3ε0
r30
r2p cos θ (9.59)
On surface sphere, r = r0
ϕ =1
3ε0r0 P cos θ
z
=P
3ε0z (9.60)
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9.7. Example: Electric field around dielectric sphere
Inside2 ϕ = 0. By inspection and uniqueness
Ez = − p3ε0
(9.61)
ϕ =pz
3ε0(9.62)
p
Field at poles:
outside E = − ∂
∂r
Pr30 cos θ
3ε0r2=
2P
3ε0
(r = r0,cos θ = 1) (9.63)
D ≡ 2P3
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9. Dielectric materials
Eout = 2P3ε0
(9.64)
Ein =−P
3ε0(9.65)
∆E⊥ =P
ε0(9.66)
∆D⊥ = 0 (9.67)
What about E?
→ General bondary conditions for dielectric surface without dfree charges
∆E = 0 (9.68)
∆D⊥ = 0 (9.69)
What if polarization produced by extend E-field (E0)?
E = E0 + E p (9.70)
also P = (ε − 1) χ
ε0Eint (9.71)
Eint = E0 − P
3ε0(9.72)
= E0 − (ε − 1)
3Eint (9.73)
Eint =3
2 + εE0 (9.74)
P =3 (ε − 1)
ε + 2ε0E0 (9.75)
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9.7. Example: Electric field around dielectric sphere
vacuum dielectric
curl E = 0 curl E = 0
div E = ρε0
div D = ρ free
E = − ϕ D = εε0E2 ϕ = 0
∆E
= 0 ∆⊥
= 0 (9.76)
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Chapter 10
Magnetic phenomena in
matter
10.1 Phenomenology
• Ferromagneticsm (Fe,Ni, permanent magnets)
• Para-magnetism (few, eg Bi)
• Dia-magnetism (almost all meterials)
PARA DIA
Force ∝ B dBdx
10.2 Magnetic dipoles
B-field from current loop
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10.3. Force on m in external field
I
is (for from loop) exactly same as E-field from dipole.
+
−
f
p = ql
E
I M
Some response in external fields: eg
N = m × B c.f.N = p × E (10.1)
10.3 Force on m in external field
As with electric dipole
Fx = m ·
Bx (10.2)
m
BrBr
I
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10. Magnetic phenomena in matter
F = l × B (10.3)
Fz = −2π b Oa of loop
IBr (10.4)
Using div B = 0
→ π b2
dBz
dz
∆z + 2π b∆zBr = 0 (10.5)
→ Fz = 2π bI ·b
2dBz
dz (10.6)
= π b2 I ∂Bz
∂z(10.7)
= m · ∂Bz
∂z= m ·
Bz
(10.8)
10.4 Current in loop in atom
mev2
r=
Ze2
4πε 0r2(10.9)
e−
+Ze
Change B → emf in loop → E-field
E
·ds
2π rE
=
−dΦ
dt
= π r2 ∂B
∂t
(10.10)
E =r
2
dB
dt(10.11)
medv
dt= e
r
2
dB
dt
(10.12)
∆v =e
me· r
2∆B (10.13)
Change in v → change in I = ev2π r
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10.5. Electron spin
dI =e
2π rdv =
e2
me4π (10.14)
dm = − e2r2
4medB from Lenz’s Law (10.15)
• not dependent on sense of v.Applying external B, all electron orbitals aqcuire opposite ∆m. Because ∆mis anti-parallel to B → diamagnetism
10.5 Electron spin
“Spin“ of electron → mangetic moment (Quantum mechanical effect)→ External B can align m of electron spin → paramagnetismNot in all atoms because electrons paired with opposite spin, ∑ m = 0 -noeffect
10.6 Magnetic fields of magnetized matter
As before for p
M = N m (10.16)
J
M dz
da
MdV = (J dz) da = M (dadz) dV
(10.17)
M = J c.f. P = σ (10.18)
curl M = Jbound (10.19)
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10. Magnetic phenomena in matter
10.7 Maxwell’s equations
curl B = µ0ε0∂E
∂t+ µ0
J free + J bound
curl M
(10.20)
curl (B − µM)
H
= µε0∂E
∂t+ µ0 J free (10.21)
B = µ (H + M) (10.22)
i.e. Amperes Law:
C
H · ds = I enclosed free
(10.23)
cf
SD · da = ρ free (10.24)
Usually M ∝ H (10.25)
B = µµ0H (10.26)D = εε0E (10.27)
B = µµ0H (10.28)
D = εε0E (10.29)
¡++¿
∆E = 0 curl E = 0 always when∂
∂t= 0 (10.30)
∆D⊥ = 0 div D = 0 when ρ free = 0 (10.31)
∆ H = 0 curl H = 0 when J free = 0 (10.32)
∆B⊥ = 0 div B = 0 always (10.33)
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10.7. Maxwell’s equations
[surfaces with ρ free = J free = 0]
I q
filed lines will bend at surfaces
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10. Magnetic phenomena in matter
E
θ
E
θ
E⊥
E⊥
E
E⊥ =
E⊥ε
E = E (10.34)
θ> θ E
= E⊥
2 + E2
< E (10.35)
¡++¿
¡++¿
10.8 Ferromagnetism (Fe,Ni)
• magnetic moment m even in absence of Bext
• suggests that M due to stable alignment of atomic m
Iron permanent magnet M ≈ 1.8 · 106 JT −1m−3
M ≡ every atom with 2 aligned electron spins. QM effect, enerycally favor-able to align spins
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10.8. Ferromagnetism (Fe,Ni)
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Chapter 11
Generation ofelectromagnetic waves
11.1 Potentials and potential wave equations
We had before B = curl A+Maxwell-equation
curl E = −∂B
∂t= − curl
∂ A
∂t(11.1)
E = −∂ A
∂t+ anything with curl = 0 (11.2)
E = −∂ A
∂t−
ϕ (11.3)
Remember
div E =ρ
ε0curl E = −∂B
∂t(11.4)
div B = 0 curl B = µ0 J + ε0µ0∂E
∂t(11.5)
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11.2. Delayed potentials
div E = ρε0
(11.6)
+E=− ∂ A∂t − ϕ−−−−−−−−→ div
−∂A
∂t−
ϕ
=
ρ
ε0(11.7)
−2
ϕ − ∂
∂t(div A) =
ρ
ε0(11.8)
Earlier, we imposed div A = 0 because we can find a field, F, such that
curl F = 0 div F = anything (11.9)
Now, generalize this to force div A = −ε0µ0∂ϕ∂t “Lorentz condition“
−2
ϕ + ε0µ0∂2 ϕ
∂t2=
ρ
ε0in vacuum (11.10)
curl B = µ0 J + ε0µ0∂E
∂t(11.11)
+ fact E = −∂A
∂t−
ϕ (11.12)
+ fact B = curl A (11.13)
curl (curl A) div A−2A
−ε0µ0
∂2 A
∂t2− ∂ϕ
∂t div A
ε0µ0
= µ0 J (11.14)
−2
A + ε0µ0∂2A
∂t2= µ0 J = 0 in vacuum (11.15)
11.2 Delayed potentials
Solution to −2 ϕ = ρε0
is from earlier
ϕ =1
4πε 0
V
ρ(r)|r − r|dV (11.16)
Solution to full equation
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11. Generation of electromagnetic waves
ϕ(r, t) =1
4πε 0
V
ρ
r, t − |r−r|c
|r − r| dV (11.17)
“delayed“ or “retarded“ potentials. Likewise:
A(r, t) =m0
4π
V
J r, t
−|r−r |
c |r − r| (11.18)
11.2.1 Hertzian Dipole
+
− Hertzian dipolelI
q = q0 sin ωt (11.19)
I =dq
dt= I 0 cos ωt I 0 = ωq0 (11.20)
p = ql = p0 sin ωt p0 = l · q0 =I 0l
ω(11.21)
Calculate A
Az(r, t) =µ0
4π
l2
− l2
dzI
z, t − |r − z|
c
· 1
|r − z| =µ0
4π
I
t − rc
l
r
(11.22)
Ax = A y = 0 because I z(11.23)
Now use
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11.2. Delayed potentials
div A = −ε0µ0∂ϕ∂t
Lorentz condition (11.24)
∂ A
∂t= . . . (11.25)
−ε0µ0∂ϕ
∂t=
µ0l
4π
∂
∂z
I
t − rc
r
(11.26)
=µ0l
4π
− I
t − r
c z
r3− ∂I
t − r
c
∂
t − rc
τ
cr2
(11.27)
∂ϕ
∂t=
l
4πε 0 z
r3
I t−
r
c +z
cr2
∂I t − rc∂ t − rc (11.28)
→ ϕ =l
4πε 0
z
r3q
t − r
c
Delayed potentialfrom ϕ
→ ignore at large r
+z
cr2I
t − r
c
∂B∂t effect
(11.29)
→ E&B from ϕ&A
Er = 0 Eϕ = 0 Eθ = −ωl I 0 sin θ sin ω t −r
c4πε 0c2 · r
(11.30)
Br = 0 Bθ = 0 Bϕ =−µ0ωl I 0 sin ω
t − r
c
r
(11.31)
rθ Eθ
Bϕ
v = c
ϕ
Amplitude of E&B
|E|
θ → maximum in plane of dipole
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11. Generation of electromagnetic waves
Instanteous power
W =
S
1
µ0(E × B) da =
ω2l2 I 206πε 0c3
sin2
ω
t − r
c
(11.32)
W =ω2l2 I 20
12πε 0c3=
p20ω4
12πε 0c3not related to r (11.33)
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