The electrons beam is usually confined by magnetics in the vacuum tube. The plasma is usually confined by magnetics in tokomaks. The motion is govend by the Newton’s 2nd law. However the cylinderical coordinates are used in both cases as the fields are axially symmetric. The common question is how to get the equations of motion with cylindrical coordinates.
For the electron beams, the radial electric field is produced by the electron beam and the magnetic flux density used to focus the beam is in the axial direction. The position of a electron is defined by cylindrical coordinates \((r(t),\theta(t),z(t))\). It is dependent on the time t. The unit vectors \(\hat{r}(t), \hat{\theta}(t)\) in the cylindrical coordinates are changing with time t which is quite different as that in the cartesion coordinates1. Let’s identify the unit vectors firstly.
It is well known that the unit vectors in the cartesian coordinates are constant,
\(\hat x=\hat i=(1,0)\,\,\,\,\hat y=\hat j=(0,1)\)
As shown in Fig.1, The unit vectors in the cylindrical coordinates are changing with time. They can be decomposed to the cartesian coordinates
\(\hat{e_r}=\cos\theta\ \hat i+\sin \theta \hat j\)
\(\hat{e_\theta}=\cos(\theta+\pi/2)\ \hat i+\sin (\theta+\pi/2) \hat j=-\sin\theta\hat i+\cos\theta\hat j\)
So, the derivative of the unit vectors are
\[\begin{eqnarray}
\frac{d\hat{e_r}}{dt}&=&\frac{d}{dt}(\cos\theta\hat i+\sin\theta\hat j)\\
&=&-\sin\theta\frac{d\theta}{dt}\hat i+\cos\theta\frac{d\theta}{dt}\hat j\\
&=&\dot\theta(-\sin\theta\hat i+\cos\theta\hat j)\\
&=&\dot\theta\hat{e_\theta}\\
\end{eqnarray}\]
And
\[\begin{eqnarray}
\frac{d\hat{e_\theta}}{dt}&=&\frac{d}{dt}(-\sin\theta\hat i+\cos\theta\hat j)\\
&=&-\cos\theta\frac{d\theta}{dt}\hat i-\sin\theta\frac{d\theta}{dt}\hat j\\
&=&-\dot\theta(\cos\theta\hat i+\sin\theta\hat j)\\
&=&-\dot\theta\hat{e_r}\\
\end{eqnarray}\]
For the motion in cylindrical coordinates, the position vector is
\[\vec r=r\hat{e_r}+z\hat{e_z}\]
The velocity is
\[\begin{eqnarray}
\vec v&=&\frac{d\vec r}{dt}\\
&=&\frac{d}{dt}(r\hat{e_r}+z\hat{e_z})\\
&=&\frac{dr}{dt}\hat{e_r}+r\frac{d\hat{e_r}}{dt}+\frac{dz}{dt}\hat{e_z}\\
&=&\dot r\hat{e_r}+r\dot\theta\hat{e_\theta}+\dot z\hat{e_z}\\
\end{eqnarray}\]
The acceleration is
\[\begin{eqnarray}
\vec a&=&\frac{d\vec v}{dt}\\
&=&\frac{d}{dt}(\dot r\hat{e_r}+r\dot\theta\hat{e_\theta}+\dot z\hat{e_z})\\
&=&\ddot r\hat{e_r}+\dot r\frac{d\hat{e_r}}{dt}+\frac{d(r\dot\theta)}{dt}\hat{e_\theta}+r\dot\theta\frac{d\hat{e_\theta}}{dt}+\ddot z\hat{e_z}\\
&=&\ddot r\hat{e_r}+\dot r\dot\theta\hat{e_\theta}+(\dot\theta\dot r+r\ddot\theta)\hat{e_\theta}-r\dot\theta^2\hat{e_r}+\ddot z\hat{e_z} \\
&=&(\ddot r-r\dot\theta^2)\hat{e_r}+(r\ddot\theta+2\dot r\dot \theta)\hat{e_\theta}+\ddot z \hat{e_z}\\
\end{eqnarray}\]
The force applied to the electrons are electrical force and lorents force
\[
\vec F=-e \vec v \times \vec B=-e \left( \begin{array}{ccc}
\hat{e_r} & \hat{e_\theta} & \hat{e_z} \\
\dot r & r\dot\theta & \dot z \\
B_r & 0 & B_z
\end{array} \right)=-e r\dot\theta B_z\hat{e_r}+e(\dot r B_z-\dot z B_r)\hat{e_\theta}-e r\dot\theta Br\hat{e_z}
\]
Finally, the motion equation could be obtained by Newton’s 2nd law
\[(\ddot r-r\dot\theta^2)\hat{e_r}+(r\ddot\theta+2\dot r\dot \theta)\hat{e_\theta}+\ddot z \hat{e_z}=-e r\dot\theta B_z\hat{e_r}+e(\dot r B_z-\dot z B_r)\hat{e_\theta}-e r\dot\theta Br\hat{e_z}\]
Reference
[1] https://ocw.mit.edu/courses/mechanical-engineering/2-003j-dynamics-and-control-i-spring-2007/lecture-notes/lec01.pdf