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Appendix A5: Solution to Equation 11.42
Equation 11.42 is reproduced and renumbered as below:
d x
dt
e
m
V
d
t B
dy
dt
2
2
0
0
1= − − + +
( cos )α ω
(A5.1a)
d y
dt
e
m
B
dx
dt
2
2
0
= − −
(A5.1b)
and
d z
dt
2
2
0=
(A5.1c)
Equation A5.1c gives that z = 0 for all times.
Let
ω
0
= eB
0
/m or B
0
= m
ω
0
/e (A5.2a)
and
eV md k
0
/( ) =
(A5.2b)
d x
dt
k t
dy
dt
2
2
0
1= + −( cos )α ω ω
(A5.3a)
d y
dt
dx
dt
2
2
0
= ω
(A5.3b)
dU
dt
k t U
x
y
= + −( cos )1
0
α ω ω
(A5.4a)
dU
U
y
x
= ω
0
(A5.4b)
From Equations A5.4a and A5.4b
U
dU
dt
x
y
=
1
0
ω
(A5.5a)