Larson Lovdal edited We_can_isolate_the_components__.tex  over 8 years ago

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\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 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}}} B=\frac{\mu_0\textit{I}\textit{R}^2}{(\textit{R}^2+\textit{z}^2)^{\frac{3}{2}}}  \end{equation}