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ik
R
V
0
1
q
2
q
m
b
y
D
f
a
2
exp .p
z
/
3
exp .qk
R
z
/
g
;
(7.116)
a
3
exp .p
z
/
C
s
exp .sk
R
z
/
;
k
R
V
0
m
i
1
q
exp .qk
R
z
/
C
i
2
b
z
D
(7.117)
where a
1
, a
2
, and a
3
are undermined coefficients while the coefficient
1
and
2
are
determined by Eq. (
7.68
) and the coefficient
3
is given by
3
D
B
0y
!
2
1
i!
m
C
l
:
(7.118)
Besides the coefficients q and s are given by Eq. (
7.61
) and p
D
k
R
i!=
m
1=2
is defined in such a way that Rep>0.
In a similar manner one can find the solutions of Eq. (
7.66
) for the atmosphere
.
z
>0/
b
j
D
d
j
exp .
k
R
z
/;j
D
x;y;
z
;
(7.119)
where d
j
denote the undefined coefficients.
It is easy to show that the solutions (
7.115
)-(
7.117
) and (
7.119
) satisfy Maxwell
equation
r
ı
B
D
0 under the requirements that ik
R
a
1
D
pa
3
and d
3
D
id
1
.
All the components of electromagnetic perturbations must be continuous at the
boundary
z
D
0. In addition, the normal component of the conduction current is
equal to zero at
z
D
0 because the atmosphere is supposed to be an insulator. It
follows from these boundary conditions that
.1
C
q/
1
;
p
k
R
C
p
q
C
.1
C
s/
2
a
1
D
(7.120)
s
.qk
R
p/
1
;
k
R
V
0
m
q
C
.sk
R
p/
2
d
1
D
(7.121)
s
a
2
D
3
;d
2
D
0:
(7.122)
We thus have just found all the coefficients in Eqs. (
7.115
)-(
7.117
)forthe
amplitudes b
x
, b
y
, and b
z
of magnetic perturbations. Substituting these amplitudes
into Eq. (
7.62
) gives the vector ı
B
.x;
z
;t/, that is, the solution of problem. The
electric field
E
is related to ı
B
through Eq. (
7.11
).
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