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imposed fields Finite E
We may choose E to lie in the xz plane so that Ey = 0.
As before, the zcomponent of velocity is unrelated to
the transverse component. dv
m = q ( E + v × B)
dt
whose z component is dvz q
= Ez
dt m
or qEz
vz =
t + vz 0
m (2.8) This is a straightforward acceleration along B. The transverse
components of eq. (2.8) are dv x q
= E x ± ωc v y
dt m
dv y = 0 ωc v x dt
Differentiating, we have (for constant E)
2 v x = −ω c v x
⎛ྎ q
⎞ྏ v y = ω c ⎜ྎ E x ± ω c v y ⎟ྏ
⎝ྎ m
⎠ྏ
⎞ྏ
2 ⎛ྎ E x
= −ω c ⎜ྎ
+ v y ⎟ྏ
⎝ྎ B
⎠ྏ (2.11) We can write this as E x ⎞ྏ
E x ⎞ྏ
d 2 ⎛ྎ
2 ⎛ྎ
v + ⎟ྏ = −ωc ⎜ྎ v y + ⎟ྏ
2 ⎜ྎ y
B ⎠ྏ
B ⎠ྏ
⎝ྎ
dt ⎝ྎ
wo that eq. (211) is reduced to the previous case if we replace vy by ⎛ྎ E x ⎞ྏ
v y + ⎜ྎ ⎟ྏ . Eq. (2.4) is therefore replaced
⎝ྎ B ⎠ྏ v x = v⊥eiωct
v y = iv⊥ e iω c t Ex
−
B The Larmor motion is the same as before, but there is superimposed
a drift vgc of the guiding center of the –y direction (for Ex > 0). Figure 2.3 Particle drifts in crossed electric and magnetic fields. To obtain a general formula for vE, we can solve eq. (2.8) in
vector form. We may omit the m dv/dt term in eq (2.8) since this
term gives only the circular motion at wc, which we already know
about. Then eq. (2.8) becomes E+...
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 Winter '14

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