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