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en/14.3.7.md
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| + | ### Statement | ||
| + | |||
| + | $14.3.7.$ [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} | ||
| + | $14.3.7$ a. When moving at the speed $\vec{\beta c}$ of a state in which there was only an electric field, a magnetic field with induction $\vec{B}$ | ||
| + | arises , associated with the new electric field $\vec{E}$ by the relation $\vec{B} = [\vec{\beta} \times \vec{E}]$. Prove this relation in the case | ||
| + | when the old $\vec{E}$ is perpendicular to the velocity $\vec{\beta c}$\\ | ||
| + | |||
| + | b. What magnetic field occurs when the electric field of intensity $\vec{E}$ moves at | ||
| + | the speed $\beta c$ if, $\beta = 1$? | ||
| + | |||
| + | \begin{center} | ||
| + | solution | ||
| + | \end{center} | ||
| + | |||
| + | a) In the moving frame there is no magnetic field, only a electric field. But in the Earth frame there are a electric field $\vec{E}$ and a magnetic field $\vec{B}$. | ||
| + | For prove the relation we will use the loretnz transformation of the magnetic field for calculate the magnetic field in the moving field, this is zero: | ||
| + | |||
| + | \begin{equation} | ||
| + | \vec{B}' = 0 = \frac{\vec{B} - [\vec{\beta} \times \vec{E}]}{\sqrt{1-\beta^2}} \rightarrow \vec{B} = [\vec{\beta} \times \vec{E}] | ||
| + | \end{equation} | ||
| + | |||
| + | In the equations of before I set $c=1$ for simplify the calculations, fixing the equation of before: | ||
| + | |||
| + | \begin{equation} | ||
| + | \vec{B} = \frac{[\vec{\beta} \times \vec{E}]}{c} | ||
| + | \end{equation} | ||
| + | |||
| + | b) Using the equation of before we can calculate the magnetic fields in each case: | ||
| + | |||
| + | \begin{equation} | ||
| + | \vec{B} = \frac{[\vec{\beta} \times \vec{E}]}{c} | ||
| + | \end{equation} | ||
| + | |||
| + | \begin{equation} | ||
| + | \vec{B_1} = \frac{[\vec{\beta_1} \times \vec{E}]}{c} | ||
| + | \end{equation} | ||
| + | |||
| + | \begin{equation} | ||
| + | \vec{B_c} = \frac{[\hat{\beta} \times \vec{E}]}{c} | ||
| + | \end{equation} | ||
| + | |||
| + | Where $\hat{\beta}$ is the unitary vector in the direction of velocity. The magnetic field will be zero when the velocity were parallel to the electric field. | ||
| + | |||
| + | \end{document} | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
| @@ -0,0 +1,73 @@ | |||
| ### Statement | |||
| $14.3.7.$ [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} | |||
| $14.3.7$ a. When moving at the speed $\vec{\beta c}$ of a state in which there was only an electric field, a magnetic field with induction $\vec{B}$ | |||
| arises , associated with the new electric field $\vec{E}$ by the relation $\vec{B} = [\vec{\beta} \times \vec{E}]$. Prove this relation in the case | |||
| when the old $\vec{E}$ is perpendicular to the velocity $\vec{\beta c}$\\ | |||
| b. What magnetic field occurs when the electric field of intensity $\vec{E}$ moves at | |||
| the speed $\beta c$ if, $\beta = 1$? | |||
| \begin{center} | |||
| solution | |||
| \end{center} | |||
| a) In the moving frame there is no magnetic field, only a electric field. But in the Earth frame there are a electric field $\vec{E}$ and a magnetic field $\vec{B}$. | |||
| For prove the relation we will use the loretnz transformation of the magnetic field for calculate the magnetic field in the moving field, this is zero: | |||
| \begin{equation} | |||
| \vec{B}' = 0 = \frac{\vec{B} - [\vec{\beta} \times \vec{E}]}{\sqrt{1-\beta^2}} \rightarrow \vec{B} = [\vec{\beta} \times \vec{E}] | |||
| \end{equation} | |||
| In the equations of before I set $c=1$ for simplify the calculations, fixing the equation of before: | |||
| \begin{equation} | |||
| \vec{B} = \frac{[\vec{\beta} \times \vec{E}]}{c} | |||
| \end{equation} | |||
| b) Using the equation of before we can calculate the magnetic fields in each case: | |||
| \begin{equation} | |||
| \vec{B} = \frac{[\vec{\beta} \times \vec{E}]}{c} | |||
| \end{equation} | |||
| \begin{equation} | |||
| \vec{B_1} = \frac{[\vec{\beta_1} \times \vec{E}]}{c} | |||
| \end{equation} | |||
| \begin{equation} | |||
| \vec{B_c} = \frac{[\hat{\beta} \times \vec{E}]}{c} | |||
| \end{equation} | |||
| Where $\hat{\beta}$ is the unitary vector in the direction of velocity. The magnetic field will be zero when the velocity were parallel to the electric field. | |||
| \end{document} | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||