Exercise 5

As shown in the textbook, for a particle in a uniform magnetic field the following equations of motion apply: \[ a_x = \frac{eb}{mc} v_y \] \[ a_y = - \frac{eb}{mc} v_x \] \[ a_z = 0 \]

We are looking for the equations of motion in the \(xy\) plane.

To simplify notation, let’s define a constant \(\omega\) \[ \omega := \frac{eb}{mc} \]

and rewrite the equations of motion: \[ \dot{v}_x = \omega v_y \] \[ \dot{v}_y = -\omega v_x \]

It is easy to prove that the following functions are solutions to these differential equations: \[ v_x(t) = v_0 \sin \omega t \] \[ v_y(t) = v_0 \cos \omega t \]

By integration we get the position vector: \[ \dot{x}(t) = v_x(t) = v_0 \sin \omega t \Rightarrow x(t) = - \frac{v_0}{\omega} \cos \omega t + x_0 \] \[ \dot{y}(t) = v_y(t) = v_0 \cos \omega t \Rightarrow y(t) = \frac{v_0}{\omega} \sin \omega t + y_0 \]

These equations describe a circular motion with orgin \((x_0, y_0)\) and radius \(r\): \[ r = \frac{v_0}{\omega} = v_0 \frac{m c }{e b} \]

The centripetal force \(F_z\) is nothing else than the Lorentz force \(F_L\): \[ F_z = \frac{m v^2_0}{r} = \frac{m v^2_0 \omega}{v_0} = \frac{e}{c} v_0 b = F_L \]