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speed, at which time no further momentum transfer would occur. The ionized
particles, however, are “locked” onto the magnetic field lines instead of being
accelerated parallel to
U
and can act as a steady drag on the neutral wind. We
now show that the linear momentum lost per unit time by the neutral fluid due
to this drag is equal to the electromagnetic force
F
E
=
B
due to the wind-
generated electric current flowing in the fluid. As shown in Chapter 2, the current
due to a neutral wind
U
is given by
J
×
=
σ
·
U
B
×
J
with
⎛
⎞
σ
P
00
0
⎝
⎠
σ
0
0
σ
=
00
σ
P
being the appropriate tensor for an equatorial F-region process with
B
in the
ˆ
a
y
direction. Since
σ
is diagonal and
U
×
B
is perpendicular to
B
, the triple cross
product in
J
×
B
yields
F
E
=−
σ
P
B
2
U
(3.21)
⊥
where
U
is the projection of the wind vector onto the plane perpendicular to
B
. The plasma thus clearly acts as a drag on the neutrals, since the force acts in
the opposite direction to
U
⊥
. We have shown previously that in the F region we
⊥
σ
P
by the expression
can approximate
ne
2
v
in
/
i
2
B
2
σ
P
=
M
=
nv
in
M
/
(3.22)
where
n
is the plasma density,
M
the ion mass,
i
the ion gyrofrequency, and
v
in
the ion-neutral collision frequency. Substituting (3.22) into (3.21) yields
F
E
=−
nMv
in
U
(3.23)
⊥
which even more clearly shows the draglike nature of the
J
B
force. At first
glance this expression has the form of a frictional or drag force,
F
D
, on the wind,
but it is not quite right, since it should include the
neutral
density, not the plasma
density, if it is to be inserted into the Navier Stokes equation; that is, it should
be of the form
×
F
D
=−
n
n
Mv
ni
U
⊥
However, for equal-mass ions and neutrals, it follows that the collision frequen-
cies for momentum transfer are related by
n
n
v
ni
=
n
i
v
in
=
nv
in
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