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}\