NE528 Fall 2009 HW#3 (total 80 points)
Due Wednesday, October 7 by class time
1. (40 points)
We derived the expression for the current density “the diamagnetic current” at equilibrium
condition and have shown that it is given by
2
B
p
J
B
×∇
=
.
Now, consider a plasma where both the density and temperature have radial profiles and vary
as
(
)
0
cos
/
n
n
r
a
=
,
, and
(
cos
/
e
eo
T
T
r
a
=
)
4
e
i
T
T
a
=
, where
a
is the plasma radius and
r
is the
radial distance.
Obtain the expression of the diamagnetic current density at equilibrium condition under the above
conditions and discuss your result. Use the single fluid equation
d u
E
J
B
dt
ρ
σ
p
=
+
×
− ∇
and
keep the term
E
σ
in the equation for a finite charge density, i.e.
0
0
and hence
E
σ
σ
≠
≠
Helpful vector relation
:
(
)
(
)
.
.
A
B
C
AC B
A B C
×
×
=
−
Where
A
,
B
and
C
are vectors.
Worked Solution:
Single fluid equation is:
p
B
J
E
dt
u
d
∇
−
×
+
σ
=
ρ
The steady-state equilibrium condition is:
0
+
E
J
B
p
σ
=
×
−∇
Taking the cross product of
p
J
B
E
σ
∇
=
×
+
with B and using the vector relation
(
)
(
)
A
B
C
A.C
.
B
A B C
×
×
=
−
(
)
(
)
.
.
B
p
B
J
B
B
E
B B J
B J
B
B
σ
σ
×∇
=
×
×
+
×
=
−
+
×
E
With B and J perpendicular to each other the second term
(
)
.
B J
B
vanishes and solving for the
current gives
2
2
2
2
B
p
B
E
B
p
E
J
B
B
B
B
B
σ
σ
×∇
×
×∇
×
=
−
=
+
The term
2
E
B
B
σ
×
represents a drift current because
σ
is the charge density of the fluid, and
2
E
B
B
×
is
the drift velocity, thus this term is
,
E
B drift
J
×

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