PHYS321 Derivation

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Motivating the Derivation

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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\label{fig1} The Helmholtz configuration of two coaxial current-carrying wire coils of equal radii. Figure based on Ref. [2] Fig. 5.59.

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The Biot-Savart Law

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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\label{fig2}The magnetic field due to a single current-carrying loop of wire. Figure based on Ref. [2] Fig. 5.21.

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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 \(2\pi\textit{R}\). Adding this result to Eq. (\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 N turns of wire in each coil then the magnetic field from a single coil would be 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: \[\label{eq4} B=\frac{\mu_0\textit{N}\textit{I}\textit{R}^2}{(\textit{R}^2+\textit{z}^2)^{\frac{3}{2}