Edits to “Statement”, “Answer”
en/12.1.27.md
+11 −2
| @@ -1,7 +1,16 @@ | |||
| ### Statement | |||
| − | $12.1.27.$ | ||
| + | $12.1.27.$ | ||
| + | The fictitious wave method (see the problem 12.1.19) can also be used to solve | ||
| + | the problem of reflection of an electromagnetic wave from a metal surface | ||
| + | moving at speed v. To solve this problem, it is necessary to choose a fictitious | ||
| + | wave so that it enters the region outside the metal and becomes real, and | ||
| + | when superimposed on the incident wave, it gives an electric field strength | ||
| + | in the GHS that isv | ||
| + | times less than the magnetic induction. Explain this | ||
| + | c | ||
| + | condition | ||
| ### Solution | |||
| The method of the fictitious wave for a moving reflector requires adapting the boundary condition. For a perfect metal at rest, the fictitious wave is chosen so that the total tangential electric field vanishes at the surface. If the metal moves with velocity v , the perfect conductor condition must be satisfied in the reference frame where the metal is at rest. In that frame, the tangential electric field vanishes: | |||
| $\mathbf{E}'_{\text{tan}} = 0$ | |||
| By transforming this condition to the laboratory system (CGS) using the field transformation laws to first order in $v/c$ one obtains: | |||
| $\mathbf{E}_{\text{tan}} + \frac{\mathbf{v}}{c} \times \mathbf{B} = \mathbf{0}$ | |||
| For a plane wave incident normally, $\mathbf{v}, \mathbf{E}, and \mathbf{B}$ are mutually perpendicular. At the surface, the superposition of the incident wave and the fictitious wave (which will become the real reflected wave upon exiting the metal) must satisfy the above relation. From it, it follows that the amplitudes of the combined fields satisfy: | |||
| $E = \frac{v}{c}\,B$ | |||
| This is precisely the condition mentioned: the intensity of the resulting electric field is$ v/c $times smaller than the magnetic induction. | |||
| @@ -20,4 +29,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $\boxed{E = \frac{v}{c}\,B}$ | ||
| @@ -1,7 +1,16 @@ | |||
| ### Statement | ### Statement | ||
| $12.1.27.$ |
$12.1.27.$ | ||
| The fictitious wave method (see the problem 12.1.19) can also be used to solve | |||
| the problem of reflection of an electromagnetic wave from a metal surface | |||
| moving at speed v. To solve this problem, it is necessary to choose a fictitious | |||
| wave so that it enters the region outside the metal and becomes real, and | |||
| when superimposed on the incident wave, it gives an electric field strength | |||
| in the GHS that isv | |||
| times less than the magnetic induction. Explain this | |||
| c | |||
| condition | |||
| ### Solution | ### Solution | ||
| The method of the fictitious wave for a moving reflector requires adapting the boundary condition. For a perfect metal at rest, the fictitious wave is chosen so that the total tangential electric field vanishes at the surface. If the metal moves with velocity v , the perfect conductor condition must be satisfied in the reference frame where the metal is at rest. In that frame, the tangential electric field vanishes: | The method of the fictitious wave for a moving reflector requires adapting the boundary condition. For a perfect metal at rest, the fictitious wave is chosen so that the total tangential electric field vanishes at the surface. If the metal moves with velocity v , the perfect conductor condition must be satisfied in the reference frame where the metal is at rest. In that frame, the tangential electric field vanishes: | ||
| $\mathbf{E}'_{\text{tan}} = 0$ | $\mathbf{E}'_{\text{tan}} = 0$ | ||
| By transforming this condition to the laboratory system (CGS) using the field transformation laws to first order in $v/c$ one obtains: | By transforming this condition to the laboratory system (CGS) using the field transformation laws to first order in $v/c$ one obtains: | ||
| $\mathbf{E}_{\text{tan}} + \frac{\mathbf{v}}{c} \times \mathbf{B} = \mathbf{0}$ | $\mathbf{E}_{\text{tan}} + \frac{\mathbf{v}}{c} \times \mathbf{B} = \mathbf{0}$ | ||
| For a plane wave incident normally, $\mathbf{v}, \mathbf{E}, and \mathbf{B}$ are mutually perpendicular. At the surface, the superposition of the incident wave and the fictitious wave (which will become the real reflected wave upon exiting the metal) must satisfy the above relation. From it, it follows that the amplitudes of the combined fields satisfy: | For a plane wave incident normally, $\mathbf{v}, \mathbf{E}, and \mathbf{B}$ are mutually perpendicular. At the surface, the superposition of the incident wave and the fictitious wave (which will become the real reflected wave upon exiting the metal) must satisfy the above relation. From it, it follows that the amplitudes of the combined fields satisfy: | ||
| $E = \frac{v}{c}\,B$ | $E = \frac{v}{c}\,B$ | ||
| This is precisely the condition mentioned: the intensity of the resulting electric field is$ v/c $times smaller than the magnetic induction. | This is precisely the condition mentioned: the intensity of the resulting electric field is$ v/c $times smaller than the magnetic induction. | ||
| @@ -20,4 +29,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $\boxed{E = \frac{v}{c}\,B}$ | ||