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\begin{equation}\label{eq6}  \frac{q}{m}=\frac{v}{Br}  \end{equation}  We now have an expression for the electron $e/m$ however it depends on the speed $v$, which we must determine using kinetic energy and the accelerating potential. A particle of charge \textit{q} starting from rest and accelerating through a given electric potential \textit{V} to some speed $v$ gains all of its kinetic energy \textit{qV} from this acceleration. We can set this equal to the classical expression $\frac{1}{2}mv^2$ and solve for $v$ to find $v=\sqrt{\frac{2q\textit{V}}{m}}$. Substituting this expression into Eq. (\ref{eq6}) and solving we find $\frac{q}{m}=\frac{2v}{B^2r^2}$. Replacing \textit{B} with \textit{B}with  Eq. (\ref{eq5}) our expression for the electron $e/m$ is: \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}=0.33m