# Evaluation and Explanation of the Experiment

## Observations:

In the experiment the following relationships are visible:
• Rotation $\phi$ ∝ Coil Current I ∝ Magnetic Field B
• Rotation $\phi$ ∝ $\frac{1}{\sqrt{V_a}}$

## General Considerations:

In contrast to the experiment with the cathode ray tube, where a fine electron beam enters a magnetic field perpendicular, here a divergent electron beam from an electron gun (assumed as point P) is used. The beam enters the uniform magnetic field in parallel direction. By the pitch angle $\delta$ the electrons inherit two velocity components - $v_{\perp}$ perpendicular and $v_{\parallel}$ parallel to the magnetic field B. So the electrons move on a helix with the height h. The initial velocity $v_0$ can be seperated in two components. The component parallel to the magnetic field is: $$\begin{equation}v_{\parallel}=v_0\cdot \cos(\delta).\end{equation}$$ The component perpendicular to the magnetic field is: $$\begin{equation}v_{\perp}=v_0\cdot \sin(\delta).\end{equation}$$
Motion perpendicular to the magnetic field:
Here the same equations as in the experiment with the cathode ray tubes can be used: $$\begin{equation}F_{\rm{Lorentz}}=F_{\rm{Zentripetal}}\qquad \Rightarrow\qquad e\cdot v_{\perp}\cdot B=m_e\frac{v_{\perp}^2}{r}\end{equation}$$ So the radius of the helix is: $$\begin{equation}r=\frac{v_{\perp}\cdot m_e}{e\cdot B}\end{equation}$$ With $v_{\perp}=\omega\cdot r$ the angular velocity $\omega$ is: $$\begin{equation}\omega=\frac{e\cdot B}{m_e}\end{equation}$$ Equation (5) shows, that the angular velocity $\omega$ is independend from the initial speed $v_0$ and from the divergence angle $\delta$. So $\omega$ is equal for all electrons. In a time t all electrons perform a rotation by the same angle $\phi$.

Motion parallel to the magnetic field:
Here no forces act on the electrons - they move with constant speed. So the time tdurchlauf until the impact of the electrons on the screen can be calculated with: $$\begin{equation}t_{\text{durchlauf}}=\frac{l_x}{v_{\parallel}}\end{equation}$$ To get tdurchlauf for all electrons independend from their divergence angle $\delta$, the luominicend screen is spherical instead of flat. So the length lx is shorter for electrons with big devergence angle (smaller$v_{\parallel}$) and tdurchlauf is constant.

## Apply on the experiment:

Equation (5) gives the angular velocity of the electron on theit circular orbits. Multiplied by tdurchlauf the rotation $\phi$ is: $$\begin{equation}\phi=\omega\cdot t_{durchlauf}=\frac{e\cdot B}{m_e}\cdot t_{durchlauf}\end{equation}$$ Then one has as seen in the experiment $\phi$ ∝ BI.
Moreover with (6) is $$\begin{equation}\phi=\frac{e\cdot B\cdot l_x}{m_e\cdot v_{\parallel}}\end{equation}$$ So one has $\phi$ ∝ $\frac{1}{v_{\parallel}}$. As shown here $v$ ∝ $\sqrt{U_b}$ and so is also $\phi$ ∝ $\frac{1}{\sqrt{U_b}}$. This matches with the observations in the experiment.