physics 122 Lecture 02 October 1st 2013_final

2 the denoting the sign of q the solution to this

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Unformatted text preview: qB ⎞ྏ v vx = v y = −⎜ྎ ⎟ྏ x m ⎝ྎ m ⎠ྏ (2.1) 2 − qB ⎛ྎ qB ⎞ྏ v vy = vx = −⎜ྎ ⎟ྏ y m ⎝ྎ m ⎠ྏ (2.2) This describes a simple harmonic oscillator at the cyclotron frequency, which we define to be qB ωc = m So for a field into the black board (x) proton will gyrate anticlockwise and electrons will gyrate clockwise An<clockwise Clockwise proton electron If q >0 particle gyrate in left hand sense If q<0 particle gyrate in right hand sense Radius of circle (rc) -cyclotron radius or Larmor radius v⊥ = rcωc mv⊥ rc = qB The gyroradius (rc) is function of energy (v⊥) For static magnetic field, the magnetic force does no work on the particle ⎛ྎ 1 mv 2 ⎞ྏ d ⎜ྎ ⎟ྏ dv ⎝ྎ 2 ⎠ྏ F ⋅v = m ⋅v = dt dt F ⋅ v = (qv × B )⋅ v = 0 Only the electric field can energize particles Guiding Center Uniform E and B fields, E = 0 In this case, a charged particle has a...
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This document was uploaded on 01/17/2014.

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