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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This document was uploaded on 01/17/2014.

Ask a homework question - tutors are online