problem4.tex

\begin{flushleft}
\textit{Problem 4}
\end{flushleft}
 In gas units, the force experienced by He particle is:
 \begin{equation*}
  \overrightarrow{F}=q(\overrightarrow{E}+\frac{\overrightarrow{v}}{c}\times\overrightarrow{B})
 \end{equation*}
 
Let's consider:

$$\overrightarrow{v}=v\widehat{z},\quad \overrightarrow{E}=E\widehat{x} and \overrightarrow{B}=B \widehat{y} $$

then $$\overrightarrow{F}=q(\overrightarrow{E}+\frac{\overrightarrow{v}}{c}\times\overrightarrow{B})=qE(1-\frac{v}{c}\frac{B}{E})\widehat{x}$$

Following suggestion let $$u\equiv\frac{cE}{B}$$, then 
\begin{equation}
 \overrightarrow{F}=qE(1-\frac{v}{u})\widehat{x}
\end{equation}

so if $$v=u$$ the particle is not at all detected; contributions from $\overrightarrow{E}$ and $\overrightarrow{B}$ in the Lorentz force compensate each other.
If $v=u+\Delta u$ then $\overrightarrow{F}=qE(1-\frac{u+\Delta u}{u})\widehat{x}=-qE\frac{\Delta u}{u}\widehat{x}= \texttt{F}\widehat{x}$.
The equation of motion is:
\begin{equation}
 \gamma m \ddot{x}=\texttt{F}\Rightarrow\quad\ddot{x}=-\frac{qE}{\gamma mu}\Delta u \widehat{x}
\end{equation}

or upon integration 

\begin{equation}
 x=-\frac{qE}{\gamma mu}\Delta u \frac{t^2}{2}+x_0
\end{equation}

$t=\frac{L}{u}$ and let $\Delta x =x-x_0$ then the particle displacement is:
\begin{equation*}
 |\Delta x|=\frac{qEL^2}{2\gamma m u^3} \Delta u
\end{equation*}

so $\Delta u =|\Delta x|\frac{2\gamma m u^3}{qEL^2}$