Special theory of relativity > Motion of relativistic particles in electric and magnetic fields > Problem 14.4.23

Statement

$14.4.23$ An electron enters a magnetic field at a velocity $\beta c$ perpendicular to the
field boundary and to the induction vector $B$. Determine the residence time of
the electron in the magnetic field.

Solution

When the electron enters the magnetic field, it will describe a circular path of radius $R$.
To calculate the radius we need to use Newton's second law for the radial direction.

\begin{equation}
\frac{m_e v^2}{R} = \frac{m_e (\beta c)^2}{R} = F
\end{equation}

The force acting on the electron is the magnetic force $F = e [\vec{v} \times \vec{B}]$; because the magnetic field is perpendicular to the velocity of the
electron we have $F = e \beta c B$. Finally:

\begin{equation}
\frac{m_e (\beta c)^2}{R} = e \beta c B \rightarrow R = \frac{m_e \beta c}{e B}
\end{equation}

Thus, the time the electron spends in the magnetic field is:

\begin{equation}
t = \frac{\pi R}{\beta c} = \frac{\pi m_e}{e B}
\end{equation}

And, taking into account the dilatation of the time we have:

\begin{equation}
t_{field} = t \gamma = \frac{\pi m_e}{e B} \gamma
\end{equation}

Answer

\begin{equation}
t_{field} = \frac{\pi m_e}{e B \sqrt{1-\beta^2}}
\end{equation}