this is for holding javascript data
Larson Lovdal edited section_Solving_for_the_e__1.tex
over 8 years ago
Commit id: 9ed4f52cda83038c38c32e21cc920268e2376a85
deletions | additions
diff --git a/section_Solving_for_the_e__1.tex b/section_Solving_for_the_e__1.tex
index 3456392..7673572 100644
--- a/section_Solving_for_the_e__1.tex
+++ b/section_Solving_for_the_e__1.tex
...
\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} \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}