Larson Lovdal edited section_Solving_for_the_e__1.tex  over 8 years ago

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\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}}) \frac{q}{m}\frac{\text{kg}}{\text{C}}=5.196\times10^{7}\frac{\textit{V}}{\textit{I}^2\textit{r}^2}(\frac{\text{A}^2\text{m}^2}{\text{V}})  \end{equation}