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+### 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
+the mass of electrons moving along these tracks is N = 1000 times greater
+than me?
+
+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}
+ \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]