em potentials2
TRANSCRIPT
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Time Harmonic Waves
( ) ArB = ( ) = AjrE
Representation in phasor domain
=
jALorentz Condition
=
=+
kk ,22
Phasor potential satisfies (5)
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Now consider
( ) ( )
+
=
r
v
rrtrf
v
rrtr cos,
Consider charge density ( ) ( ) ( )( )rtrftr += cos,in phasor domain ( ) ( ) ( )rjerfr =
Corresponding phasor
( ) ( )
( ) kvereerf rrjk
v
rrj
rj
===
,
i.e. time retardation corresponds to a phase shift in
the phasor domain.
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Therefore, solution of (5) in phasor domain is
( )
( )Vdrr
er
rV
rrjk
=
4
1
(6)
( )rr
errG
rrjk
=
41, Free space Greens function for
the potentials
( ) ( )
Vdrr
erJrA
V
rrjk
=
4(7)
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Hertz PotentialsIt was shown by Hertz that it is possible to combineand and the Lorentz condition in a single vector
(called Hertz vector) from which all the fieldcomponents can be derived.
A
Electric Hertz vector is defined as follows
tA
e
= e=
e
Lets show that the Lorentz condition is satisfied
( ) 0=
+
=
+ ee
tttA
(1) (2)
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Then satisfies
Lets define a vector such that
t
pJ
= p=
p
Note that the equation of continuity is satisfiedby this choice of (show as an exercise)p
e
=
p
t
ee
2
2
2 (show as an exercise)
(4)(3)
(5)
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tAB
e
==
tB
e
=
( )eettt
AE
=
=
( ) 2
2
tE
ee
=
( )
= p
E e
or
(6)
(7)
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Given
,J
Findusing (3) & (4)p
Solve (5) for e
Findusing (6) & (7)
EB,
t
pJ
=
p=
=
p
t
ee
2
2
2
tB
e
=
( ) 2
2
tE
ee
=
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(i) Consider vacuum with ,J
Corresponding Maxwells equations are
,0 t
HE
=
,0
t
EJH
+=
0=
E
0= H
(9)
tH
e
=
0
( ) 22
00t
E e
e
=
0
2
2
00
2
=
p
t
ee Where satisfiese
(8)
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(ii) Consider nonmagnetic, source free dielectric
PED +=0
HB0
= 0,0 == J
Polarization vector
Note that is not necessarilyD E
Then, corresponding Maxwells equations are
,0
t
HE
=
,0
t
P
t
EH
+
=
0
=
PE
0= H
(11)
(10)
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Compare the two sets of Maxwells equations given by (8)
and (11)
,0 t
HE
=
,0
t
EJH
+=
0
=
E
0= H
,0
t
HE
=
,0
t
P
t
EH
+
=
0
=
PE
0= H
They are in thesamemathematicalform with
t
PJ
P
Remember that
t
pJ
= p=
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Thus, the equations given in (9) are all valid for the
representation of EM fields in the dielectric when isreplaced by , polarization vector.
pP
(9)
tH e
=0
( ) 2
2
00
t
E e
e
=
0
2
2
00
2
=
p
t
ee Where satisfiese P
: electric type Hertz vectore
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(iii) Consider source free magnetic material
,00
MHB +=,0
ED= 0,0 == J
Magnetization vector
Then, corresponding Maxwells equations are
,0 t
E
H
=
,00
t
M
t
HE
=
MH =
0= E
(13)
(12)
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Comparing the two sets of Maxwells equations given by
(11) and (13), we see that replacing
MPEHHE00000
,,,,
in (11) we can get (13).Thus, the resultant representation given by (9) can betransformed using the same replacements:
tE
m
=
0
( ) 2
2
00 tH
m
m
=
Mt
mm =
2
2
00
2 Where satisfiesm
: magnetic type Hertz vector potentialm
(14)