Правка разделов «Solution», «Answer»
en/14.4.22.md
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| ### Statement | |||
| $14.4.22$ What is the magnetic field induction on storage tracks of radius R = 6 m, if | |||
| the mass of electrons moving along these tracks is N = 1000 times greater | |||
| than me? | |||
| ### Solution | |||
| In this problem it is necessary to use the concept of relativistic mass. The relativistic mass of each electron in this case is: | |||
| \begin{equation} | |||
| M = m_e N = \frac{m_e}{\sqrt{1-\beta^2}} \rightarrow \beta = \frac{\sqrt{N^2-1}}{N} | |||
| \end{equation} | |||
| Thus, the velocity of the electrons inside the storage is: | |||
| \begin{equation} | |||
| v = \beta c = c \frac{\sqrt{N^2 - 1}}{N} | |||
| \end{equation} | |||
| On the other hand, the radial force is equal to the magnetic force acting on the electron $F = e v B = e c B \frac{\sqrt{N^2 - 1}}{N}$. Using Newton's second law | |||
| @@ -22,11 +22,11 @@Solution | |||
| we can calculate the magnetic field induction inside the storage: | |||
| \begin{equation} | |||
| − | \frac{N m_e v^2}{R} = e v B \rightarrow B = \frac{N m_e v}{e R} = \frac{m_e c}{e R | ||
| + | \frac{N m_e v^2}{R} = e v B \rightarrow B = \frac{N m_e v}{e R} = \frac{m_e c}{e R \sqrt{N^2 - 1}} = \frac{m_e c}{e \cdot 6 \cdot \sqrt{1000^2 - 1}} \approx 0.28 \mu\text{T} | ||
| \end{equation} | |||
| #### Answer | |||
| \begin{equation} | |||
| − | B \approx | ||
| + | B \approx 0.28 \mu\text{T} | ||
| \end{equation} | |||
| ### Statement | ### Statement | ||
| $14.4.22$ What is the magnetic field induction on storage tracks of radius R = 6 m, if | $14.4.22$ What is the magnetic field induction on storage tracks of radius R = 6 m, if | ||
| the mass of electrons moving along these tracks is N = 1000 times greater | the mass of electrons moving along these tracks is N = 1000 times greater | ||
| than me? | than me? | ||
| ### Solution | ### Solution | ||
| In this problem it is necessary to use the concept of relativistic mass. The relativistic mass of each electron in this case is: | In this problem it is necessary to use the concept of relativistic mass. The relativistic mass of each electron in this case is: | ||
| \begin{equation} | \begin{equation} | ||
| M = m_e N = \frac{m_e}{\sqrt{1-\beta^2}} \rightarrow \beta = \frac{\sqrt{N^2-1}}{N} | M = m_e N = \frac{m_e}{\sqrt{1-\beta^2}} \rightarrow \beta = \frac{\sqrt{N^2-1}}{N} | ||
| \end{equation} | \end{equation} | ||
| Thus, the velocity of the electrons inside the storage is: | Thus, the velocity of the electrons inside the storage is: | ||
| \begin{equation} | \begin{equation} | ||
| v = \beta c = c \frac{\sqrt{N^2 - 1}}{N} | v = \beta c = c \frac{\sqrt{N^2 - 1}}{N} | ||
| \end{equation} | \end{equation} | ||
| On the other hand, the radial force is equal to the magnetic force acting on the electron $F = e v B = e c B \frac{\sqrt{N^2 - 1}}{N}$. Using Newton's second law | On the other hand, the radial force is equal to the magnetic force acting on the electron $F = e v B = e c B \frac{\sqrt{N^2 - 1}}{N}$. Using Newton's second law | ||
| @@ -22,11 +22,11 @@Solution | |||
| we can calculate the magnetic field induction inside the storage: | we can calculate the magnetic field induction inside the storage: | ||
| \begin{equation} | \begin{equation} | ||
| \frac{N m_e v^2}{R} = e v B \rightarrow B = \frac{N m_e v}{e R} = \frac{m_e c}{e R |
\frac{N m_e v^2}{R} = e v B \rightarrow B = \frac{N m_e v}{e R} = \frac{m_e c}{e R \sqrt{N^2 - 1}} = \frac{m_e c}{e \cdot 6 \cdot \sqrt{1000^2 - 1}} \approx 0.28 \mu\text{T} | ||
| \end{equation} | \end{equation} | ||
| #### Answer | #### Answer | ||
| \begin{equation} | \begin{equation} | ||
| B \approx |
B \approx 0.28 \mu\text{T} | ||
| \end{equation} | \end{equation} | ||