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8/11/2019 AK Lecture5
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PHL 100
Instructor: Arun KumarLecture - 5
Electromagnetic Waves and Quantum Mechanics
1
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, 1 1
2r ik k
= +
K is complex
0
0
exp( z) exp [ ( )]exp( z) exp [ ( )]
i r
i r
E E k i k z tB B k i k z t
= =
Real part of k determines the wavelength, phase velocityand refractive index:
2r p r
r p
c
k v n k k v
= = =
EM Waves in conductors:Recap:
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Poor conductor >
,
2
r ik k
1 1
2i r
d
k k
= = =and
1
id k
=
Skin Depth
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4
Skin depth for silver at optical frequencies
0 0
7 1 15
,
10 ( ) , ~ 10m Hz
= >>
21 2 10 o
i
d Ak
=
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Skin depth for salt water at radio frequencies
21d cm
0
10 2 2
1
6 10 /
5 ( ) , 10
C N m
m GHz
= =
Acoustic communication or direct cable link to sea
surface is required
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0
0
exp( z) exp [ ( )]
exp( z) exp [ ( )]
i r
i r
E x E k i k z t
kB y E k i k z t
=
=
Relation between E and B fields:
Sincek
is complex
1tan i
r
k
k
=
i
k k e
=
This means E and B are not in phase B E =
0
0
exp( z) cos [ ( )]
exp( z) cos [ ( )]
i r E
i r E
E x E k i k z t
kB y E k i k z t
= +
= + +
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E and B are not in phase
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Reflection/ Transmission at conducting surface
iE
1 1,
2 2, ,
tE
rE
0 1
0 11
exp[ ( ) ]
exp[ ( ) ]
i i
ii
E x E i k z t
EB y i k z tv
=
=
x y
zIncident beam
Reflected beam
Transmitted beam
0 1
01
1
exp[ ( ) ]
exp[ ( ) ]
r r
rr
E x E i k z t
EB y i k z t
v
=
=
0 2
20 2
exp[ ( ) ]
exp[ ( ) ]
t t
t t
E x E i k z t
kB y E i k z t
=
=
8
k2 is complex valued
Dielectric
Conductor
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Boundary conditions
E-parallel condition at boundary z = 0
0 0 0i r tE E E+ =
H-parallel condition at boundary z = 0
20 0 0
1 1 2
1( )i r t
kE E E
v =
Solve simultaneous equations to get reflected and
transmitted field amplitude in terms of incident field.
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Solve simultaneous equations
1 12
2
vk
=Define
0 0 0 0
1 2,
1 1r i t iE E E E
= =
+ +
2 2,
2
r ik k
is large for
good conductor
0 0 0, 0r i tE E E Almost all the energy
is reflected
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Energy Density:
( )
( )
20
2
2202
2 20
exp( 2 z)4
11exp( 2 z)
4
1 exp( 2 z)4
E i
B i
i
U E k
U E k
E k
=
+=
= +
For good conductors, magnetic contribution dominates:
20
1exp( 2 z)
4B iU E k
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0
2
1( )
exp( 2 z) cos ( ) cos ( )i r r
S E B
kz E k k z t k z t
=
= +
Poynting Vector in Metals:
2 cos ( ) cos ( )cos{2 ( ) } cos ( )
r r
r
k z t k z t k z t
+= + +
Using
0
0
2
2
1 ( ) exp( 2 z) cos2
exp( 2 z) cos2
i
r
i r
kS E B z E k
kS z E k where k k
= =
= =
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Problem: As the wave propagates in a metal, the energy
density decreases because of the attenuation factorexp( 2 z)ik
Where does this lost energy go? This energy is equal to theJoule Heating in the medium due to the current density
generated by the electric field of the EM Wave.
The above was also discussed in the class.
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Free electrons in metals and plasma
2
02 exp ( )q Ed x d x i td t md t
+ =
2
2
total driving dampingF F F
d x d x
m q E m d td t
= +
=
Steady state solution 0( ) exp ( )x t x i t=
2 0
0 0
q E
x i x m =
0
0 2
/q E mx
i =
+
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0( ) exp ( )x t x i t=
00
2
/q E mx
i =
+
Current Density:
( )d x
J N f q
d t
=
2 /N f q mJ E
i
=
J and E are
not in phase
0 00
/ /
exp ( )
q E m q E md xi x i t id t i i = = =+
2 /N f q m
i
=
Conductivity
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