Edits to “Statement”, “Solution”, “Answer”
en/12.1.28.md
+9 −7
| @@ -1,13 +1,11 @@ | |||
| ### Statement | |||
| − | $12.1.28.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
| − | |||
| $12.1.28$ A plane electromagnetic wave falls on a metal wall moving at velocity $v$ perpendicular | |||
| to its surface. Electric field strength of wave $E$. What pressure in | |||
| SI and GHS does the wave exert on the wall? | |||
| + | ### Solution | ||
| + | |||
| Using the Lorentz transformation for electric and magnetic fields and the relation between these fields $E = B c$, we can calculate the electric field in | |||
| the moving frame: | |||
| \begin{equation} | |||
| E_1 = \frac{E - (-v) B}{\sqrt{1-v^2/c^2}} = \frac{E - (-v) E/c}{\sqrt{1-v^2/c^2}} = E \sqrt{\frac{1+v/c}{1-v/c}} | |||
| \end{equation} | |||
| The pressure on the moving wall is: | |||
| \begin{equation} | |||
| \mathcal{P} = 2 \frac{I}{c} = 2 \frac{c \epsilon_0 E_1^2}{c} = 2 \epsilon_0 E_1^2 = 2 \epsilon_0 E^2 \frac{1+v/c}{1-v/c} | |||
| \end{equation} | |||
| The factor of two is because there is a pressure due to the incoming flow and another due to the outgoing flow. | |||
| Also, the result in the $GHS$ system of units is: | |||
| \begin{equation} | |||
| @@ -29,10 +27,14 @@Solution | |||
| \mathcal{P} = \frac{1}{8 \pi} \epsilon_0 E^2 \frac{1+v/c}{1-v/c} | |||
| \end{equation} | |||
| + | |||
| + | |||
| + | #### Answer | ||
| + | |||
| \begin{equation} | |||
| \mathcal{P} = 2 \epsilon_0 E^2 \frac{1+v/c}{1-v/c} | |||
| \end{equation} | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| + | \begin{equation} | ||
| + | \mathcal{P} = \frac{1}{8 \pi} \epsilon_0 E^2 \frac{1+v/c}{1-v/c} | ||
| + | \end{equation} | ||
| @@ -1,13 +1,11 @@ | |||
| ### Statement | ### Statement | ||
| $12.1.28.$ [Insert the problem statement] | |||
| ### Solution | |||
| $12.1.28$ A plane electromagnetic wave falls on a metal wall moving at velocity $v$ perpendicular | $12.1.28$ A plane electromagnetic wave falls on a metal wall moving at velocity $v$ perpendicular | ||
| to its surface. Electric field strength of wave $E$. What pressure in | to its surface. Electric field strength of wave $E$. What pressure in | ||
| SI and GHS does the wave exert on the wall? | SI and GHS does the wave exert on the wall? | ||
| ### Solution | |||
| Using the Lorentz transformation for electric and magnetic fields and the relation between these fields $E = B c$, we can calculate the electric field in | Using the Lorentz transformation for electric and magnetic fields and the relation between these fields $E = B c$, we can calculate the electric field in | ||
| the moving frame: | the moving frame: | ||
| \begin{equation} | \begin{equation} | ||
| E_1 = \frac{E - (-v) B}{\sqrt{1-v^2/c^2}} = \frac{E - (-v) E/c}{\sqrt{1-v^2/c^2}} = E \sqrt{\frac{1+v/c}{1-v/c}} | E_1 = \frac{E - (-v) B}{\sqrt{1-v^2/c^2}} = \frac{E - (-v) E/c}{\sqrt{1-v^2/c^2}} = E \sqrt{\frac{1+v/c}{1-v/c}} | ||
| \end{equation} | \end{equation} | ||
| The pressure on the moving wall is: | The pressure on the moving wall is: | ||
| \begin{equation} | \begin{equation} | ||
| \mathcal{P} = 2 \frac{I}{c} = 2 \frac{c \epsilon_0 E_1^2}{c} = 2 \epsilon_0 E_1^2 = 2 \epsilon_0 E^2 \frac{1+v/c}{1-v/c} | \mathcal{P} = 2 \frac{I}{c} = 2 \frac{c \epsilon_0 E_1^2}{c} = 2 \epsilon_0 E_1^2 = 2 \epsilon_0 E^2 \frac{1+v/c}{1-v/c} | ||
| \end{equation} | \end{equation} | ||
| The factor of two is because there is a pressure due to the incoming flow and another due to the outgoing flow. | The factor of two is because there is a pressure due to the incoming flow and another due to the outgoing flow. | ||
| Also, the result in the $GHS$ system of units is: | Also, the result in the $GHS$ system of units is: | ||
| \begin{equation} | \begin{equation} | ||
| @@ -29,10 +27,14 @@Solution | |||
| \mathcal{P} = \frac{1}{8 \pi} \epsilon_0 E^2 \frac{1+v/c}{1-v/c} | \mathcal{P} = \frac{1}{8 \pi} \epsilon_0 E^2 \frac{1+v/c}{1-v/c} | ||
| \end{equation} | \end{equation} | ||
| #### Answer | |||
| \begin{equation} | \begin{equation} | ||
| \mathcal{P} = 2 \epsilon_0 E^2 \frac{1+v/c}{1-v/c} | \mathcal{P} = 2 \epsilon_0 E^2 \frac{1+v/c}{1-v/c} | ||
| \end{equation} | \end{equation} | ||
| #### Answer | \begin{equation} | ||
| \mathcal{P} = \frac{1}{8 \pi} \epsilon_0 E^2 \frac{1+v/c}{1-v/c} | |||
| [Insert a concise answer or boxed result] | \end{equation} | ||