physics 122 Lecture 02 October 1st 2013_final

# 8 are dv x q e x c v y dt m dv y 0 c v x dt

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Unformatted text preview: simple cyclotron gyration. The equation of motion dv m = qv × B dt Taking z to be the direction of B = Bz mvx = qBv y mv y = − qBvx mvz = 0 2 qB ⎛ྎ qB ⎞ྏ v vx = v y = −⎜ྎ ⎟ྏ x m ⎝ྎ m ⎠ྏ (2.1) 2 − qB ⎛ྎ qB ⎞ྏ v vy = vx = −⎜ྎ ⎟ྏ y m ⎝ྎ m ⎠ྏ (2.2) ± The denoting the sign of q. The solution to this equation vx = v⊥eiωct = x (2.4a) where v⊥ is a positive constant denoting the speed in the plane perpendicular to B. then m vy = vx qB = ±iv ⊥ e iωct =y Integrating once again, we have v x − x0 = −i ⊥ eiωct ωc y − y0 = ± v⊥ ωc e iωct (2.4b) We define the Larmor radius to be v⊥ mv⊥ rL ≡ = ωc q B Taking the real part of eq. (2.5) we have x − x0 = rL sin ωct y − y0 = rL cos ωct x B - + electron Larmor orbits in a magnetic field ion Figure 2.2 This describes a circular orbit in a guiding center (x0, y0) which is fixed. The direction of the gyration is always such that the magnetic field generated by the charged particle gyration is opposite to the external...
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## This document was uploaded on 01/17/2014.

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