Revision as of 13:50, 25 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.

$\vec{v} = v_x \hat i + v_y \hat j$

$\vec{B} = B \hat k$

$\vec{F} = q \vec{E} + q\vec{v} \times \vec{B}$

Note: the work done by a magnetic field is zero if the particle's kinetic energy (mass and velocity) don't change.; $W = \Delta K.E.$

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

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

$\vec{F} = q \left (v_y B \hat i - v_x B \hat j \right )$

$ma_x = qv_yB$

$ma_y = -qv_x B$

$ma_z = 0$

Motion in the z-direction has no acceleration and therefor constant (zero) velocity.

Let

$\omega=\frac{qB}{m}$ = fundamental cyclotron frequency

Then we have two coupled equations

$\dot{v}_x = \omega v_y$

$\dot{v}_y = - \omega v_x$

let

$v^* = v_x + i v_y$ = complex variable used to change variables

$\dot{v}^* = \dot{v}_x + i \dot{v}_y$

$= \omega v_y + i (-\omega v_x)$

$= -i \omega \left ( \omega v_x +i\omega v_y \right )$

$= -i \omega v^*$

@@ Line 46: / Line 46: @@
 :<math>\dot{v}^* = \dot{v}_x + i \dot{v}_y</math>
-:: = \omega v_y + i (-\omega v_x)
+:: <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>
 http://hep.physics.wayne.edu/~harr/courses/5200/f07/lecture10.htm