Geoscience Reference
In-Depth Information
The boundary condition on the top of the ionosphere follows from the
continuity equation for the total current:
∂
∂t
−
j
(
x, s, t
)=
Σ
P
sin
I
∂
2
∂x∂t
I
xd
(
x, t
)
∂
∂s
c
A
−
|
x
=0
,
(15.31)
where
Σ
P
=
Σ
P
/Σ
A
and
Σ
A
=
c
2
/
4
πc
A
.
Let the Alfven velocity,
c
A
, be constant along the field-lines. Then, by
substituting the solution of (15.28) in the form of a traveling wave
j
(
t
−
s/c
A
)
into (15.31), we obtain
∂
∂x
I
xd
(
x, t
−
s/c
A
)
j
=
−
.
(15.32)
Σ
P
/
sin
I
1+
There is no
Σ
H
in this relation because, as it was shown in Chapter 7,
Σ
H
has
no influence on the Alfven wave for not so large horizontal scales of
100 km.
However, the external ionospheric current
I
xd
is defined by
Σ
H
.
The current induced by the pulse in the conducting layer is then chan-
neled into the flux tube about that of the fore front of the pulse. This
field-aligned current produces a magnetic pulse. Let us substitute (15.30) in
(15.32). Integration of the obtained relation gives magnetic field of the current
pulse
b
y
B
0
Σ
H
Σ
P
1
1+sin
I/ Σ
P
v
n
(
x, t
−
s/c
A
)
=
−
.
(15.33)
c
A
The transversal distribution of the magnetic field
b
y
in the Alfven pulse is
proportional to the horizontal distribution of macroscopic velocity of the
ionospheric neutral gas in the pulse. The current is defined by its spatial
derivative.
Since
Σ
P
1 in day then (15.33) reduces to
b
y
B
0
≈−
Σ
H
Σ
P
v
n
c
A
.
Σ
P
and in night
1
v
n
c
A
.
Let us estimate the magnetic field above the ionosphere. Let
v
n
=5
b
y
B
0
≈−
Σ
H
Σ
A
×
10
2
cm/s,
c
A
=1
10
8
cm/s,
Σ
P
≈
×
Σ
H
, then in day
10
−
6
b
=0
.
25
nT
≈
5
×
G=0
.
5nT
and in night
10
−
6
b
=0
.
025
nT
≈
0
.
5
×
G=0
.
05 nT
.
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