Edits to “Statement”, “Solution”, “Answer”
en/14.3.20.md
+10 −6
| @@ -1,13 +1,11 @@ | |||
| ### Statement | |||
| − | $14.3.20.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
| − | |||
| $14.3.20$ In a straight wire, the current density is $j$. How will this density change when | |||
| the wire moves with the speed $\beta_1 c$ in the longitudinal direction? What is the | |||
| − | volume charge that appears in the wire? | ||
| + | volume charge that appears in the wire? | ||
| + | ### Solution | ||
| + | |||
| To solve this problem we can use the Lorentz transformation for current density and charge density. | |||
| The transformations are: | |||
| \begin{equation} | |||
| j_{\text{Earth}} = \gamma (j + \rho \beta c) \qquad \text{and} \qquad \rho_{\text{Earth}} = \gamma (\rho + \frac{j \beta_1}{c}) \quad \text{where} \quad \gamma = \frac{1}{\sqrt{1 - \beta_1^2}} | |||
| \end{equation} | |||
| It is also necessary to understand that in the frame where the wire is at rest, there is no charge density. This idea may seem strange, but | |||
| you need to think of the current as two currents: one of positive charge and one of negative charge moving in opposite directions. | |||
| The difference between these current densities is $j$. You can choose an arbitrary volume around the wire and the sum of the charges inside it is zero at every moment. | |||
| Using this idea in the Lorentz transformations we have: | |||
| \begin{equation} | |||
| j_{\text{Earth}} = \gamma j = \frac{j}{\sqrt{1-\beta_1^2}} | |||
| \end{equation} | |||
| And: | |||
| \begin{equation} | |||
| \rho_{\text{Earth}} = \frac{\gamma j \beta_1}{c} = \frac{j \beta_1}{c \sqrt{1-\beta_1^2}} | |||
| \end{equation} | |||
| @@ -34,4 +32,10 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | \begin{equation} | ||
| + | j_{\text{Earth}} = \gamma j = \frac{j}{\sqrt{1-\beta_1^2}} | ||
| + | \end{equation} | ||
| + | |||
| + | \begin{equation} | ||
| + | \rho_{\text{Earth}} = \frac{\gamma j \beta_1}{c} = \frac{j \beta_1}{c \sqrt{1-\beta_1^2}} | ||
| + | \end{equation} | ||
| @@ -1,13 +1,11 @@ | |||
| ### Statement | ### Statement | ||
| $14.3.20.$ [Insert the problem statement] | |||
| ### Solution | |||
| $14.3.20$ In a straight wire, the current density is $j$. How will this density change when | $14.3.20$ In a straight wire, the current density is $j$. How will this density change when | ||
| the wire moves with the speed $\beta_1 c$ in the longitudinal direction? What is the | the wire moves with the speed $\beta_1 c$ in the longitudinal direction? What is the | ||
| volume charge that appears in the wire? |
volume charge that appears in the wire? | ||
| ### Solution | |||
| To solve this problem we can use the Lorentz transformation for current density and charge density. | To solve this problem we can use the Lorentz transformation for current density and charge density. | ||
| The transformations are: | The transformations are: | ||
| \begin{equation} | \begin{equation} | ||
| j_{\text{Earth}} = \gamma (j + \rho \beta c) \qquad \text{and} \qquad \rho_{\text{Earth}} = \gamma (\rho + \frac{j \beta_1}{c}) \quad \text{where} \quad \gamma = \frac{1}{\sqrt{1 - \beta_1^2}} | j_{\text{Earth}} = \gamma (j + \rho \beta c) \qquad \text{and} \qquad \rho_{\text{Earth}} = \gamma (\rho + \frac{j \beta_1}{c}) \quad \text{where} \quad \gamma = \frac{1}{\sqrt{1 - \beta_1^2}} | ||
| \end{equation} | \end{equation} | ||
| It is also necessary to understand that in the frame where the wire is at rest, there is no charge density. This idea may seem strange, but | It is also necessary to understand that in the frame where the wire is at rest, there is no charge density. This idea may seem strange, but | ||
| you need to think of the current as two currents: one of positive charge and one of negative charge moving in opposite directions. | you need to think of the current as two currents: one of positive charge and one of negative charge moving in opposite directions. | ||
| The difference between these current densities is $j$. You can choose an arbitrary volume around the wire and the sum of the charges inside it is zero at every moment. | The difference between these current densities is $j$. You can choose an arbitrary volume around the wire and the sum of the charges inside it is zero at every moment. | ||
| Using this idea in the Lorentz transformations we have: | Using this idea in the Lorentz transformations we have: | ||
| \begin{equation} | \begin{equation} | ||
| j_{\text{Earth}} = \gamma j = \frac{j}{\sqrt{1-\beta_1^2}} | j_{\text{Earth}} = \gamma j = \frac{j}{\sqrt{1-\beta_1^2}} | ||
| \end{equation} | \end{equation} | ||
| And: | And: | ||
| \begin{equation} | \begin{equation} | ||
| \rho_{\text{Earth}} = \frac{\gamma j \beta_1}{c} = \frac{j \beta_1}{c \sqrt{1-\beta_1^2}} | \rho_{\text{Earth}} = \frac{\gamma j \beta_1}{c} = \frac{j \beta_1}{c \sqrt{1-\beta_1^2}} | ||
| \end{equation} | \end{equation} | ||
| @@ -34,4 +32,10 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | \begin{equation} | ||
| j_{\text{Earth}} = \gamma j = \frac{j}{\sqrt{1-\beta_1^2}} | |||
| \end{equation} | |||
| \begin{equation} | |||
| \rho_{\text{Earth}} = \frac{\gamma j \beta_1}{c} = \frac{j \beta_1}{c \sqrt{1-\beta_1^2}} | |||
| \end{equation} | |||