deletions | additions       diff --git a/We_can_isolate_the_components__.tex b/We_can_isolate_the_components__.tex  index 4767f8e..0d87b63 100644  --- a/We_can_isolate_the_components__.tex  +++ b/We_can_isolate_the_components__.tex 
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\end{equation}  We can use Figure \ref{fig2} to simplify this integral by substituting constant values. Using the right triangle formed by \textit{z}, \textit{R}, and \textit{$\vec{r}$} we know the magnitude $\textit{r}=\sqrt{\textit{R}^2+\textit{z}^2}$. We can also determine $\cos{\theta}=\textit{R}/{\textit{r}}=\textit{R}/{\sqrt{\textit{R}^2+\textit{z}^2}}$.Re-writing equation \ref{eq2} with these values and bringing the constant current \textit{I} outside the integral yields:  \begin{equation}\label{eq3}  \vec{B}=\frac{\mu_0}{4\pi(\textit{R}^2+\textit{z}^2)}\frac{\textit{R}}{\sqrt{\textit{R}^2+\textit{z}^2}}\int{\textit{d}\vec{l}\times\hat{\textit{r}}} \vec{B}=\frac{\mu_0\textit{I}{4\pi(\textit{R}^2+\textit{z}^2)}\frac{\textit{R}}{\sqrt{\textit{R}^2+\textit{z}^2}}\int{\textit{d}\vec{l}\times\hat{\textit{r}}}  \end{equation}  Fortunately,because $\textit{d}\vec{l}$ and $\hat{\textit{r}}$ are perpendicular $\int{\textit{d}\vec{l}\times\hat{\textit{r}}}=\int{\textit{d}l}$, which is the circumference of the loop of wire given by $2\pi\textit{R}$. Adding this result to equation \ref{eq3} gives the total magnetic field due to a single loop of current-carrying wire, but we are interested in the field from two coils with many current-carrying loops. Let us say there are \textit{N} turns of wire in each coil then the magnetic field from a single coil would be \textit{N} times our result. Since we also wish to determine the field due to two coils, we must also multiply the result by two, which yields a final expression for the magnetic field due to two coils with N turns: