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Larson Lovdal edited We_can_isolate_the_components__.tex
over 8 years ago
Commit id: 869843896dd35236caf0d8474250e17ada1632d1
deletions | additions
diff --git a/We_can_isolate_the_components__.tex b/We_can_isolate_the_components__.tex
index b07e50b..92d2404 100644
--- a/We_can_isolate_the_components__.tex
+++ b/We_can_isolate_the_components__.tex
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\begin{equation}\label{eq3}
\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 our final expression for the magnetic field due to two coils with N turns:
\begin{equation}\label{eq4}
\vec{B}=\frac{\mu_0\textit{I}\textit{R}^2{(\textit{R}^2+\textit{z}^2)^{\frac{3}{2}}}
\end{equation}