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and using (2.23a)
q
∂
dt
·
K
g
=−
B
/
d
a
Note that the vectors
q
B
and
d
a
are in opposite directions. Now, since
B
is
uniform in space, we can move it outside the integral and
q
r
g
/τ
g
K
⊥
/
t
=
K
g
/τ
g
=
π
(∂
B
/∂
t
)
=
(
K
⊥
/
B
)(∂
B
/∂
t
)
and finally
K
B
2
∂μ/∂
t
=
(
K
⊥
/
B
)(∂
B
/∂
t
)/
B
−
(∂
B
/∂
t
)
⊥
/
=
0
μ
and so
is conserved.
For illustrative purposes, we can apply these findings to a “magnetic bottle”
geometry similar to the earth's magnetic field as illustrated in Fig. 2.10. Suppose
a particle starts on the symmetry axis at the place where the field strength is
B
0
and the velocity vector of the particle makes an angle
(the pitch angle) with
respect to the magnetic field line. At the magnetic equator, then,
α
V
||
=
V
cos
α
,
V
⊥
=
V
sin
α
carries the particle into a region of larger field strength, as
shown in Fig. 2.10b. However, if
The velocity
V
||
increases
MV
2
2 must
also increase. Since the particle energy must also be conserved, this can only
μ
is conserved, as
|
B
|
⊥
/
V
B
B
B
0
(a)
V
II
V
'
B
(into paper)
(b)
Figure 2.10
(a) Sketch of the magnetic field lines and the corresponding magnetic field
magnitude along the axis of a magnetic bottle. (b) The gyromotion associated with a
converging magnetic field.
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