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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 $\vec{F_B)=q\vec{v}\times\vec{B}$ $\vec{F_B}=q\vec{v}\times\vec{B}$  \citep{Wolfson1}. In our set up 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.