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\section{Solving for the $e/m$ Ratio}  We begin with the force on a charged particle in a magnetic field given by $\vec{F_B}=q\vec{v}\times\vec{B}$ \citep{Wolfson1} [1]  where $q$ is a charge with a velocity $\vec{v}$ traveling in a magnetic field $\vec{B}$. In our arrangement where electrons are being accelerated through a potential perpendicular to $\vec{B}$ the force becomes $qvB$. We can set this equal to Newton's Second Law with acceleration expressed as $\frac{v^2}{r}$ for uniform circular motion where \textit{r} is the radius of the circular path. Solving for the charge-to-mass ratio we find: \begin{equation}\label{eq6}  \frac{q}{m}=\frac{v}{Br}  \end{equation} 
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\begin{equation}\label{eq7}  \frac{q}{m}=\frac{125\textit{V}\textit{R}^2}{{32\mu_0^2\textit{I}^2\textit{r}^2\textit{N}^2}}  \end{equation}   Using the numerical values from our experimental set up $\textit{R} = 0.33$ (m) $\textit{N} = 72$ turns and $\mu_0=4\pi\times 10^{-7} (\frac{\text{N}}{\text{A}^2})$ [1]  the equation becomes: \begin{equation}\label{eq8}  \frac{q}{m}=5.196\times10^{7}\frac{\textit{V}}{\textit{I}^2\textit{r}^2}(\frac{\text{A}^2\text{m}}{\text{N}})  \end{equation}