Правка разделов «Statement», «Solution», «Answer»
en/14.4.22.md
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| ### Statement | |||
| − | $14.4.22.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
| − | |||
| − | \documentclass[12pt,a4paper]{article} | ||
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| − | \begin{document} | ||
| − | |||
| $14.4.22$ 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 | |||
| we can calculate the magnetic field induction inside the storage: | |||
| \begin{equation} | |||
| @@ -50,8 +25,8 @@Solution | |||
| \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} \sqrt{1000^2 - 1} \approx 1.7 \, \text{T} | |||
| \end{equation} | |||
| − | \end{document} | ||
| − | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | \begin{equation} | ||
| + | B \approx 1.7\,\text{T} | ||
| + | \end{equation} | ||
| @@ -1,36 +1,11 @@ | |||
| ### Statement | ### Statement | ||
| $14.4.22.$ [Insert the problem statement] | |||
| ### Solution | |||
| \documentclass[12pt,a4paper]{article} | |||
| \usepackage[english]{babel} | |||
| \usepackage{float} | |||
| \usepackage{wrapfig} | |||
| \usepackage{lmodern} | |||
| \usepackage[T1]{fontenc} | |||
| \usepackage[utf8]{inputenc} | |||
| \usepackage{microtype} | |||
| \usepackage{graphicx} | |||
| \usepackage{booktabs} | |||
| \usepackage{amsmath,amssymb} | |||
| \usepackage{hyperref} | |||
| \usepackage{csquotes} | |||
| \usepackage{geometry} | |||
| \usepackage{subcaption} | |||
| \usepackage{tikz} | |||
| \usepackage{array} | |||
| \usepackage{pgfplots} | |||
| \usepackage{wrapfig} | |||
| \usepackage{subcaption} | |||
| \begin{document} | |||
| $14.4.22$ 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 | |||
| 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 | ||
| we can calculate the magnetic field induction inside the storage: | we can calculate the magnetic field induction inside the storage: | ||
| \begin{equation} | \begin{equation} | ||
| @@ -50,8 +25,8 @@Solution | |||
| \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} \sqrt{1000^2 - 1} \approx 1.7 \, \text{T} | \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} \sqrt{1000^2 - 1} \approx 1.7 \, \text{T} | ||
| \end{equation} | \end{equation} | ||
| \end{document} | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | \begin{equation} | ||
| B \approx 1.7\,\text{T} | |||
| \end{equation} | |||