Subfield: Electromagnetism

A beam of electrons emitted by a point source $\mathrm{P}$ enters the magnetic field $\vec{B}$ of a toroidal coil (toroid) in the direction of the lines of force. The angle of the aperture of the beam $2 \cdot \alpha_{0}$ is assumed to be small $\left(2 \cdot \alpha_{0}<<1\right)$. The injection of the electrons occurs on the mean radius $\mathrm{R}$ of the toroid with acceleration voltage $\mathrm{V}_{0}$.

Neglect any interaction between the electrons. The magnitude of $\vec{B}, B$, is assumed to be constant.

<img_4363>


Data:

$\frac{\mathrm{e}}{\mathrm{m}}=1.76 \cdot 10^{11} \mathrm{C} \cdot \mathrm{kg}^{-1} ; \quad \mathrm{V}_{0}=3 \mathrm{kV} ; \quad \mathrm{R}=50 \mathrm{~mm}$
Context question:
1. To guide the electron in the toroidal field a homogeneous magnetic deflection field $\vec{B}_{1}$ is required. Calculate $\vec{B}_{1}$ for an electron moving on a circular orbit of radius $R$ in the torus.
Context answer:
\boxed{$1.48 \cdot 10^{-2}$}


Context question:
2. Determine the value of $\vec{B}$ which gives four focussing points separated by $\pi / 2$ as indicated in the diagram.

Note: When considering the electron paths you may disregard the curvature of the magnetic field.
Context answer:
\boxed{$0.37 \cdot 10^{-2}$}


Extra Supplementary Reading Materials:

3. The electron beam cannot stay in the toroid without a deflection field $\vec{B}_{1}$, but will leave it with a systematic motion (drift) perpendicular to the plane of the toroid.

Note: The angle of aperture of the electron beam can be neglected. Use the laws of conservation of energy and of angular momentum.

Question: a) Show that the radial deviation of the electrons from the injection radius is finite.

Solution:
3. Finiteness of $r_{1}$ and direction of the drift velocity

In the magnetic field the lines of force are circles with their centres on the symmetry axis (z-axis) of the toroid.

In accordance with the symmetry of the problem, polar coordinates $r$ and $\varphi$ are introduced into the plane perpendicular to the z-axis (see figure below) and the occurring vector quantities (velocity, magnetic field $\overrightarrow{\mathrm{B}}$, Lorentz force) are divided into the corresponding components.

<img_4485>

Since the angle of aperture of the beam can be neglected examine a single electron injected tangentially into the toroid with velocity $\mathrm{v}_{0}$ on radius $\mathrm{R}$.

In a static magnetic field the kinetic energy is conserved, thus

$$
\mathrm{E}=\frac{\mathrm{m}}{2}\left(\mathrm{v}_{\mathrm{r}}^{2}+\mathrm{v}_{\varphi}^{2}+\mathrm{v}_{\mathrm{z}}^{2}\right)=\frac{\mathrm{m}}{2} \mathrm{v}_{0}^{2}
\tag{12}
$$

The radial points of inversion of the electron are defined by the condition

$$
\mathrm{v}_{\mathrm{r}}=0
$$

Using eq. (12) one obtains

$$
\mathrm{v}_{0}^{2}=\mathrm{v}_{\varphi}^{2}+\mathrm{v}_{\mathrm{z}}^{2}
\tag{13}
$$

Such an inversion point is obviously given by

$$
\mathrm{r}=\mathrm{R} \cdot\left(\mathrm{v}_{\varphi}=\mathrm{v}_{0}, \mathrm{v}_{\mathrm{r}}=0, \mathrm{v}_{\mathrm{z}}=0\right)
$$

To find further inversion points and thus the maximum radial deviation of the electron the components of velocity $\mathrm{v}_{\varphi}$ and $\mathrm{v}_{\mathrm{Z}}$ in eq. (13) have to be expressed by the radius.

$\mathrm{V}_{\varphi}$ will be determined by the law of conservation of angular momentum. The Lorentz force obviously has no component in the $\varphi$ - direction (parallel to the magnetic field). Therefore it cannot produce a torque around the z-axis. From this follows that the $\mathrm{z}$-component of the angular momentum is a constant, i.e. $\mathrm{L}_{\mathrm{z}}=\mathrm{m} \cdot \mathrm{v}_{\varphi} \cdot \mathrm{r}=\mathrm{m} \cdot \mathrm{v}_{0} \cdot \mathrm{R}$ and therefore 
$$
\mathrm{v}_{\varphi}=\mathrm{v}_{0} \cdot \frac{\mathrm{R}}{\mathrm{r}}
\tag{14}
$$

$\mathrm{v}_{\mathrm{z}}$ will be determined from the equation of motion in the z-direction. The z-component of the Lorentz force is $\mathrm{F}_{\mathrm{z}}=-\mathrm{e} \cdot \mathrm{B} \cdot \mathrm{v}_{\mathrm{r}}$. Thus the acceleration in the $\mathrm{z}$-direction is

$$
\mathrm{a}_{\mathrm{z}}=-\frac{\mathrm{e}}{\mathrm{m}} \cdot \mathrm{B} \cdot \mathrm{v}_{\mathrm{r}} .
\tag{15}
$$

That means, since B is assumed to be constant, a change of $v_{z}$ is related to a change of $r$ as follows:

$$
\Delta \mathrm{v}_{\mathrm{z}}=-\frac{\mathrm{e}}{\mathrm{m}} \cdot \mathrm{B} \cdot \Delta \mathrm{r}
$$

Because of $\Delta \mathrm{r}=\mathrm{r}-\mathrm{R}$ and $\Delta \mathrm{v}_{\mathrm{z}}=\mathrm{v}_{\mathrm{z}}$ one finds

$$
\mathrm{v}_{\mathrm{z}}=-\frac{\mathrm{e}}{\mathrm{m}} \cdot \mathrm{B} \cdot(\mathrm{r}-\mathrm{R})
\tag{16}
$$

Using eq. (14) and eq. (15) one obtains for eq. (13)

$$
1=\left(\frac{\mathrm{R}}{\mathrm{r}}\right)^{2}+\mathrm{A}^{2} \cdot\left(\frac{\mathrm{r}}{\mathrm{R}}-1\right)^{2}
\tag{17}
$$

where $\mathrm{A}=\frac{\mathrm{e}}{\mathrm{m}} \cdot \mathrm{B} \cdot \frac{\mathrm{R}}{\mathrm{v}_{0}}$

Discussion of the curve of the right side of eq. (17) gives the qualitative result shown in the following diagram:

<img_4532>

Hence $\mathrm{r}_{1}$ is finite. Since $\mathrm{R} \leq \mathrm{r} \leq \mathrm{r}_{1}$ eq. (16) yields $\mathrm{v}_{\mathrm{z}}<0$. Hence the drift is in the direction of the negative $z$-axis.