Edits to “Solution”, “Answer”
en/14.4.23.md
+5 −1
| ### 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} | |||
| @@ -25,10 +25,14 @@Solution | |||
| t = \frac{2 \pi R}{\beta c} = \frac{2 \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{2 \pi m_e}{e B} \gamma | ||
| + | \end{equation} | ||
| #### Answer | |||
| \begin{equation} | |||
| − | t = \frac{2 \pi m_e}{e B} | ||
| + | t_{field} = \frac{2 \pi m_e}{e B \sqrt{1-\beta^2}} | ||
| \end{equation} | |||
| ### Statement | ### Statement | ||
| $14.4.23$ An electron enters a magnetic field at a velocity $\beta c$ perpendicular to the | $14.4.23$ 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 | field boundary and to the induction vector $B$. Determine the residence time of | ||
| the electron in the magnetic field. | the electron in the magnetic field. | ||
| ### Solution | ### Solution | ||
| When the electron enters the magnetic field, it will describe a circular path of radius $R$. | 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. | To calculate the radius we need to use Newton's second law for the radial direction. | ||
| \begin{equation} | \begin{equation} | ||
| \frac{m_e v^2}{R} = \frac{m_e (\beta c)^2}{R} = F | \frac{m_e v^2}{R} = \frac{m_e (\beta c)^2}{R} = F | ||
| \end{equation} | \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 | 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: | electron we have $F = e \beta c B$. Finally: | ||
| \begin{equation} | \begin{equation} | ||
| \frac{m_e (\beta c)^2}{R} = e \beta c B \rightarrow R = \frac{m_e \beta c}{e B} | \frac{m_e (\beta c)^2}{R} = e \beta c B \rightarrow R = \frac{m_e \beta c}{e B} | ||
| \end{equation} | \end{equation} | ||
| Thus, the time the electron spends in the magnetic field is: | Thus, the time the electron spends in the magnetic field is: | ||
| \begin{equation} | \begin{equation} | ||
| @@ -25,10 +25,14 @@Solution | |||
| t = \frac{2 \pi R}{\beta c} = \frac{2 \pi m_e}{e B} | t = \frac{2 \pi R}{\beta c} = \frac{2 \pi m_e}{e B} | ||
| \end{equation} | \end{equation} | ||
| And, taking into account the dilatation of the time we have: | |||
| \begin{equation} | |||
| t_{field} = t \gamma = \frac{2 \pi m_e}{e B} \gamma | |||
| \end{equation} | |||
| #### Answer | #### Answer | ||
| \begin{equation} | \begin{equation} | ||
| t = \frac{2 \pi m_e}{e B} | t_{field} = \frac{2 \pi m_e}{e B \sqrt{1-\beta^2}} | ||
| \end{equation} | \end{equation} | ||