Forest UCM PnCP

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