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In pursuit of an empirically determined value for the electron charge-to-mass \(e/m\) ratio we must guide electrons in a variable circular path using a homogeneous magnetic field. The Helmholtz configuration of two current-carrying coils with radius *R* separated by a distance *d* as shown in Fig. \ref{fig1} provides the necessary field. To determine the magnetic field \(\vec{\textit{B}}\) we will first find the magnetic field due to a single current-carrying loop using the Biot-Savart Law.

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The Biot-Savart Law shows how *d*\(\vec{B}\) depends on each differential piece of the current-carrying loop *Id*\(\vec{l}\), 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 [1]: \[\label{eq1}
\textit{d}\vec{B} = \frac{\mu_0}{4\pi} \frac{\textit{I}\textit{d}\vec{l}\times\hat{\textit{r}}} {\textit{r}^2}\] Where \(\vec{\textit{r}}\) has been decomposed into its magnitude *r* and its unit vector \(\hat{r}\). The contributions to the magnetic field from the loop trace out a circle (see Fig. \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 *d*\(B_z\).

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We can isolate the components using *d*\(\vec{B}_z = \textit{d}\vec{B}\cos{\theta}\), which comes from the right triangle in the top of Fig. \ref{fig2} formed by *d*\(\vec{B}\), \(\vec{B}\), and the radius of the circle. Thus, integrating only the \(\hat{z}\) components of Eq. (\ref{eq1}) to obtain the total magnetic field we have: \[\label{eq2}
\vec{B}=\frac{\mu_0}{4\pi}\int{\frac{\textit{I}\textit{d}\vec{l}\times\hat{\textit{r}}} {\textit{r}^2}\cos{\theta}}\] We can use Fig. \ref{fig2} to simplify this integral by substituting constant values. Using the right triangle formed by *z*, *R*, and *\(\vec{r}\)* we know the magnitude \(\textit{r}=\sqrt{\textit{R}^2+\textit{z}^2}\) and we can also determine \(\cos{\theta}=\textit{R}/{\textit{r}}=\textit{R}/{\sqrt{\textit{R}^2+\textit{z}^2}}\). Re-writing Eq. (\ref{eq2}) with these values and bringing the constant current *I* outside the integral yields: \[\label{eq3}
\vec{B}=\frac{\mu_0\textit{I}}{4\pi(\textit{R}^2+\textit{z}^2)}\frac{\textit{R}}{\sqrt{\textit{R}^2+\textit{z}^2}}\int{\textit{d}\vec{l}\times\hat{\textit{r}}}\] Fortunately,because \(\textit{d}\vec{l}\) and \(\hat{\textit{r}}\) are perpendicular \(\int{\textit{d}\vec{l}\times\hat{\textit{r}}}\) becomes \(\int{\textit{d}l}\), which is the circumference of the wire loop given by \(2\pi\textit{R}\