physics 122 Lecture 02 October 1st 2013_final

# In magnitude this drift is e v m m ve b tesla sec ve

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Unformatted text preview: ly imposed fields Finite E We may choose E to lie in the x-z plane so that Ey = 0. As before, the z-component 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. (2-11) 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 &gt; 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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