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\textbf{}\section{The Biot-Savart Law}  The Biot-Savart Lawrelates  shows how \textit{d}$\vec{B}$ depends on each differential piece of the current-carrying loop \textit{I}\textit{d}$\vec{l}$, the position vector $\vec{\textit{r}}$ of the point where we're finding the field, and the sine of the angle \theta between these two vectors with the equation \citep{Wolfson06}: \begin{equation}\label{eq1}  \textit{d}\vec{B} = \frac{\mu_0}{4\pi} \frac{\textit{I}\textit{d}\vec{l}\times \hat{\textit{r}}} {\textit{r}^2}  \end{equation}  The contributions to the magnetic field from the loop trace out a circle (see Figure \ref{fig2}) thus the symmetrical horizontal components of the magnetic field cancel leaving a total field $\vec{B}$ pointing upward. Due to this cancellation we only want to integrate the upward components \textit{d}$B_z$. We can isolate the components using \textit{d}$\vec{B}_z = \textit{d}\vec{B}\cos{\theta}$, which comes from the right triangle in the top of figure \ref{fig2} formed by \textit{d}$\vec{B}$, $\vec{B}$, and the radius of the circle. Thus, integrating only the $\hat{z}$ components of equation \ref{eq1} to obtain the total magnetic field we have:  \begin{equation}\label{eq2}  \vec{B}=\int{\frac{\mu_0}{4\pi}\frac{\textit{I}\textit{d}\vec{l}\times\hat{\textit{r}}} {\textit{r}^2}\cos{\theta}}   \end{equation}