24. electromagnetic wavesphome.postech.ac.kr/user/genphys/download/phy102-24.pdf · production of...
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24. Electromagnetic Waves24. Electromagnetic Waves24-1. Displacement Current and the Ampere’s Law
• Ampere’s Law
I = 0
IsdB 0µ=⋅∫rr
• Induction
dtd BΦ
−=ε
dtdsdE BΦ
−=⋅∫rr
Faraday’s Law
• Similar form for B-field
dEE I
dtd
dtdsdB 000 µ=
Φεµ=
Φ∝⋅∫
rr
dtdI E
dΦ
ε≡ 0 : Displacement current
+ −
Er
ii
C
ε
R⊗ ⊗
Br
?
Q
Cε
t
c/te τ−−1
( ) RC,er
ieCQ C/t/t cc =τ=→−= τ−τ− εε 1
riBIsdBπµ
=→µ=⋅∫ 20
0rr
( )c/teCQV τ−−== ε 1 ( )c/te
ddVE τ−−==
ε 1
( )c/tE e
dA τ−−=Φ ε 1
d/t
/t
/t
c
E
ieR
ed
ARC
ed
Adt
drBsdB
c
c
c
00
00
00
00
1
1
2
µ=µ=
εµ=
τεµ=
Φεµ=π⋅=⋅
τ−
τ−
τ−
ε
ε
ε∫
rr
dAC 0ε=
c/td e
Ri τ−ε=
Er
r
Br
R
dirBsdB 02 µ=π⋅=⋅∫rr
CtEd e
Rdtdi τε
=Φ
ε= 0
dirB
πµ2
0=
riR
B d20
2πµ
=
for r > R
for r < R
dtdiiisdB E
dΦ
+=+=⋅∫ 00000 εµµµµrr id : displacement current
0εQAEE =⋅=Φ Gauss’s Law
ε
=ε=⋅
==A
QdA
dQ
CdQE,
CQV
00
dE i
dtdQ
dtd
==Φ
0ε
• Magnetic fields are produced both by conduction currents and by changing electric fields.
24-2. Maxwell’s Equation
0εQAdE =⋅∫
rrGauss’s Law
0=⋅∫ AdBrr
No magnetic monopole
dtdsdE BΦ
−=⋅∫rr
dtdisdB EΦ
µε+µ=⋅∫ 000rr
Faraday’s Law (Induction)
Ampere-Maxwell’s Law
∫∫∫ ερ
=⋅∇=⋅ dvdvEAdE0
rrrrCf)
0ερ
=⋅∇ Err
⇒
0=⋅∇=⋅ ∫∫ dvBAdBrrrr
∫∫∫ ⋅−=⋅×∇=⋅ AdBdtdAdEsdE
rrrrrrr
∫∫
∫∫
⋅εµ+⋅µ=
Φεµ+µ=⋅×∇=⋅
AdEdtdAdj
dtdiAdBsdB E
rrrr
rrrrr
000
000
tEjB∂∂
εµ+µ=×∇r
rrr000
tBE∂∂
−=×∇r
rr⇒
⇒
0=⋅∇ Brr
⇒
djtE rr
=∂∂
ε0 ( )djjBrrrr
+µ=×∇ 0⇒
Lorentz force: force law for a particle of charge q
BvqEqFrrrr
×+=
24-3. Electromagnetic WaveMaxwell’s equations
dtdsdE
dtd BB Φ
−=⋅⇒Φ
−= ∫ε rr
tBE∂∂
−=×∇r
rr
dtdildB E
incΦ
+=⋅∫ 000 εµµrr
dtdldB EΦ
=⋅∫ 00εµrr
tEB∂∂
=×∇r
rr00εµ
iinc = 0
( )
∂∂
−∂∂
=×∇∂∂
=×∇×∇tB
tE
tB
rrrrrr
0000 εµεµ
( ) BBrrrr
2−∇=×∇×∇
kz
jy
ix
ˆˆˆ∂∂
+∂∂
+∂∂
=∇r
( ) ( ) BBBBrrrrrrrr
22 −∇=∇−⋅∇∇=×∇×∇
( ) ( ) ( )CBABCACBArrrrrrrrr
⋅−⋅=××
2
2
002
tBB
∂∂
=∇r
rεµ
2
2
002
tEE
∂∂
=∇r
rεµ
02
2
002
2
=∂∂
−∂∂
tB
xB εµ
02
2
002
2
=∂∂
−∂∂
tE
xE εµ
Wave equations
02
2
002
2
=∂∂
−∂∂
tB
xB εµ 02
2
002
2
=∂∂
−∂∂
tE
xE εµ
)cos(0 tkxBB ω−=
)cos(0 tkxEE ω−=
0200
2 =ωεµ−k
cvk
≡==00
1εµ
ω
BEtEB
rrr
rr⊥⇒
∂∂
εµ−=×∇ 00
Speed of Light
EkB ωεµ 00=
ckk
kkBE
==
==
ωωωεµω
2
00
1
smmc /103sec/1099792.2 88 ×≈×=
Electromagnetic Wave
24-4. Electromagnetic Oscillators and Alternating Current
LC-Circuit -- Oscillator
CQU E 2
20=i Energy
+ + +− − − CQ0
L
S
0=+−CQ
dtdiL
02
2
=+⇒CQ
dtQdL
dtdQi −=
02
2
=+LCQ
dtQd
General solution
LC12 =ω)cos(0 φω += tQQ
tcosQQQQ ω=⇒= 00
tsinQdtdQi ωω=−= 0
CQ
tsinC
QtcosC
Q
tsinQLtcosC
Q
LiC
QUUU BE
2
22
21
2
21
2
20
2202
20
220
2220
22
=
ω+ω=
ωω+ω=
+=+=
Energy conserved
At t = 0,
Total Energy
QUE UB
CQ2
20
i
• Alternating Currenttsinω= εε 0
iR=εtsin
Ri ω=ε0
tsinR
RiP ω==ε 2
202
21sinsin
2
0
22 == ∫π
θθθ dave
rmsrmsrmsrms
ave
iRiR
RiR
P
εε
ε
===
==
22
20
20
21
21
20εε =rms 2
0iirms =
Power
Power
( Root mean square )
ε
i
• Hertz’s Apparatus
−
+
−
+
Transmitter Receiver
LCf
π21
= Resonance frequency
24-5. Production of Electromagnetic Wave by an Antenna
A charged particle undergoes an acceleration, and it must radiate energy via electromagnetic radiation.
Magnetic field lines are always perpendicular to Electric field lines.
24-6. Energy carried by Electromagnetic WavesPoynting Vector : Intensity of an electromagnetic wave
BESrrr
×=0
1µ
2
0
2
0
0
1
1
BcEc
EBS
µ=
µ=
µ=
(Watt/m2)
= c
EB
2m
rmsEE = 2
0
2
0
12
1rms
m Ec
Ec
Sµµ
==
202
1 EuE ε=Energy density associated with an Electric field :
2
021 BuB µ
=Energy density associated with a Magnetic field :
For an electromagnetic wave
BE uBEu =µ
=ε=0
22
0 221
Total instantaneous energy density :
0
22
0 µ=ε=+=
BEuuu BE aveave cuSI ==
00
2 1εµ
=c( )0
22
0212
0 2µ=ε=ε= m
maveaveBEEu
Example 24.2 Fields due to a Point Source
Average power of the source : Pave
Intensity at a distance r
24 rPI ave
π=
2
00
2
22 maxmax
ave Bcc
ESIµ
=µ
==
At a distance r
cE
crP
cIB
rcPcIE
maxavemax
avemax
=π
µ=
µ=
πµ
=µ=
200
20
0
22
22
For Pave = 800 W and r = 3.5 m
Emax = 62.6 V/mBmax = 2.09 × 10-7 T
24-7. Momentum and Radiation Pressure
reflectedincident Total absorption caseMomentum change:
cUp =∆ U: incident energy
Total reflection caseMomentum change:
cUp 2
=∆
tpF∆∆
= tIAU ∆=
Total absorption case
cIAF =⇒ctIAp /∆=∆
cI
AFP .A.T ==Pressure
Total reflection case
cIAF 2
=⇒c/tIAp ∆=∆ 2
cI
AFP .R.T
2==Pressure
24-8. The Spectrum of Electromagnetic Wave
f : frquencyc = f λ
mmfc 300
101103
6
8
=××
==λ1 MHz :
24-9. Polarization
0I 021 II = θ= 2
021 cosII
θθ= 2202
1 sincosII
θcosEEy = θ∝ 2cosI