Edits to “Solution”, “Answer”

JAMF edited Jun 14, 2026, 5:38 PM

View solution History

revision #18773 parent #18766 ← older

en/14.4.23.md +3 −3

			### 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}

		@@ -22,17 +22,17 @@Solution
			Thus, the time the electron spends in the magnetic field is:

			\begin{equation}
		−	t = \frac{2 \pi R}{\beta c} = \frac{2 \pi m_e}{e B}
		+	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{2 \pi m_e}{e B} \gamma
		+	t_{field} = t \gamma = \frac{\pi m_e}{e B} \gamma
			\end{equation}

			#### Answer

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

	@@ -22,17 +22,17 @@Solution
	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}
	t = \frac{2 \pi R}{\beta c} = \frac{2 \pi m_e}{e B}		t = \frac{\pi R}{\beta c} = \frac{\pi m_e}{e B}
	\end{equation}		\end{equation}

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

	\begin{equation}		\begin{equation}
	t_{field} = t \gamma = \frac{2 \pi m_e}{e B} \gamma		t_{field} = t \gamma = \frac{\pi m_e}{e B} \gamma
	\end{equation}		\end{equation}

	#### Answer		#### Answer

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