deletions | additions       diff --git a/We_can_isolate_the_components__.tex b/We_can_isolate_the_components__.tex  index 41d79e7..a61910b 100644  --- a/We_can_isolate_the_components__.tex  +++ b/We_can_isolate_the_components__.tex 
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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 our final expression for the magnetic field due to two coils with N turns:   \begin{equation}\label{eq4}  B=\frac{\mu_0\textit{N}\textit{I}\textit{R}^2}{(\textit{R}^2+\textit{z}^2)^{\frac{3}{2}}}  \end{equation} We can further simplify equation \ref{eq4} by making two careful choices about the experimental setup. First, we choose to investigate the point exactly halfway between the coils, as shown in Figure \ref{fig1}, such that $\textit{z}=\frac{\textit{d}}{2}$. Second, we set the coils such that the separation distance \textit{d} is the same as the radius \textit{R}. The denominator of equation \ref{eq4} then becomes $(\textit{R}^2 + (\frac{\textit{R}}{2})^2)^\frac{3}{2}$.