Difference between revisions of "Forest UCM PnCP"

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:<math>v_x +i v_y = A \left ( \cos(\omega t) - i \sin ( \omega t) \right )</math>
 
:<math>v_x +i v_y = A \left ( \cos(\omega t) - i \sin ( \omega t) \right )</math>
  
 +
The above expression indicates that <math>v_x</math> and <math>v_y</math> oscillate at the same frequency but are 90 degrees out of phase.  This is characteristic of circular motion with a magnitude of <math>v_{\perp}</math> such that
 +
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:<math>v^* = v_{\perp}e^{-i\omega t}</math>
  
 
==Determine the position as a function of time==
 
==Determine the position as a function of time==

Revision as of 11:28, 27 August 2014

Charged Particle in uniform B-Field

Consider a charged particle moving the x-y plane in the presence of a uniform magnetic field with field lines in the z-dierection.

[math]\vec{v} = v_x \hat i + v_y \hat j[/math]
[math]\vec{B} = B \hat k[/math]


Lorentz Force
[math]\vec{F} = q \vec{E} + q\vec{v} \times \vec{B}[/math]
Note
the work done by a magnetic field is zero if the particle's kinetic energy (mass and velocity) don't change.
[math]W = \Delta K.E.[/math]

No work is done on a charged particle force to move in a fixed circular orbit by a magnetic field (cyclotron)


[math]\vec{F} = m \vec{a} = q \vec{v} \times \vec{B} = q\left ( \begin{matrix} \hat i & \hat j & \hat k \\ v_x & v_y &0 \\ 0 &0 & B \end{matrix} \right )[/math]
[math]\vec{F} = q \left (v_y B \hat i - v_x B \hat j \right )[/math]

Apply Newton's 2nd Law

[math]ma_x = qv_yB[/math]
[math]ma_y = -qv_x B[/math]
[math]ma_z = 0[/math]


Motion in the z-direction has no acceleration and therefor constant (zero) velocity.
Motion in the x-y plane is circular

Let

[math]\omega=\frac{qB}{m}[/math] = fundamental cyclotron frequency

Then we have two coupled equations

[math]\dot{v}_x = \omega v_y[/math]
[math]\dot{v}_y = - \omega v_x[/math]

determine the velocity as a function of time

let

[math]v^* = v_x + i v_y[/math] = complex variable used to change variables
[math]\dot{v}^* = \dot{v}_x + i \dot{v}_y[/math]
[math]= \omega v_y + i (-\omega v_x)[/math]
[math]= -i \omega \left ( \omega v_x +i\omega v_y \right )[/math]
[math]= -i \omega v^*[/math]
[math]\Rightarrow[/math]
[math]v^* = Ae^{-i\omega t}[/math]

the complex variable solution may be written in terms of [math]\sin[/math] and [math]\cos[/math]

[math]v_x +i v_y = A \left ( \cos(\omega t) - i \sin ( \omega t) \right )[/math]

The above expression indicates that [math]v_x[/math] and [math]v_y[/math] oscillate at the same frequency but are 90 degrees out of phase. This is characteristic of circular motion with a magnitude of [math]v_{\perp}[/math] such that

[math]v^* = v_{\perp}e^{-i\omega t}[/math]

Determine the position as a function of time

http://hep.physics.wayne.edu/~harr/courses/5200/f07/lecture10.htm


http://www.physics.sfsu.edu/~lea/courses/grad/motion.PDF

http://physics.ucsd.edu/students/courses/summer2009/session1/physics2b/CH29.pdf

http://cnx.org/contents/77faa148-866e-4e96-8d6e-1858487a520f@9

Forest_Ugrad_ClassicalMechanics