Geoscience Reference
In-Depth Information
conductivity to be constant as a function of
y
.) The current parallel to
B
now is
nonzero and must be supplied by the dynamo and completed in the E layer. The
full divergence equation must be used. First, as discussed in Chapter 2, we note
that for large-scale dynamo sources, the magnetic field lines are nearly equipo-
tentials due to the high ratio of
10
5
. The electric field is thus mapped
down to the E-region altitudes, where we have assumed that the neutral wind
vanishes in order to study just the F-region dynamo. The
σ
0
to
σ
P
(
≥
)
∇·
J
=
0 condition in
the F region now yields
=
−
dJ
y
dy
d
[
σ
P
(
E
z
+
uB
)
]
dz
y
1
,
which corresponds to integrating from the equatorial plane to the base of the F
layer in the Northern Hemisphere. By symmetry the contribution to the integral
from the Southern Hemisphere is identical. Symmetry also requires that the field-
aligned current vanish directly at the equator
We now integrate this equation along the
y
direction from
y
=
0to
y
=
[
J
y
(
0
)
=
0
]
, and the integral yields
y
1
d
dz
[
σ
P
(
E
z
+
uB
)
]
dy
=−
J
y
(
y
1
)
(3.8)
0
Due to the low perpendicular conductivity in the “valley” between the E and
F regions, the current we have just calculated, which leaves the F-region dynamo
at
y
=
y
1
, must be equal to the field-aligned current that enters the E region at
y
y
2
. In the E layer we have only the (mapped) electric field,
E
z
, to contend
with because we have set the wind speed as
u
=
=
0 for now. Applying
∇·
J
=
0
yields
d
(σ
P
E
z
)/
dz
=−
dJ
y
/
dy
Integrating from
y
2
to
y
3
and requiring
J
y
(
y
3
)
=
0—that is, an insulating atmo-
sphere below
y
3
—we have
y
3
d
dz
(σ
P
E
z
)
dy
=
J
y
(
y
2
)
=
J
y
(
y
1
)
(3.9)
y
2
and, finally, using (3.8) and (3.9),
y
1
y
3
dz
dy
dz
PN
E
z
dy
d
d
F
E
σ
P
(
E
z
+
uB
)
=−
σ
(3.10)
y
2
0
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