Правка разделов «Statement», «Answer»
en/12.1.7.md
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| ### Statement | |||
| − | $12.1.7.$ [Insert the problem statement] | ||
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| − | ### Solution | ||
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| − | ### Statement | ||
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| $12.1.7^\ast$. Solve problem 12.1.6 in the case where the wave propagates in a medium with a dielectric permittivity $\varepsilon$. The velocity of the wave in the medium is $c/\sqrt{\varepsilon}$. | |||
| ### Solution | |||
| The method of solution is identical to that in problem 12.1.6. In the rectangular contour, let's denote the sides as $ab = L$ and $aa' = x$. Faraday's law of electromagnetic induction in magnitude for the contour is: | |||
| $$\mathscr{E}_i = \frac{d\Phi}{dt}$$ | |||
| where $\mathscr{E}_i$ is the induced electromotive force (EMF) created by the electric field $E$, and $\Phi$ is the magnetic flux through the contour. | |||
| From the geometry of the setup: | |||
| $$\mathscr{E}_i = E L$$ | |||
| $$d\Phi = B L dx$$ | |||
| Substituting these two formulas into the induction law, we get after canceling $L$: | |||
| $$E = B \frac{dx}{dt}$$ | |||
| Here, the rate of change of distance $dx/dt$ is simply the propagation velocity of the wave in the medium, $v = c/\sqrt{\varepsilon}$. Therefore: | |||
| $$E = B \frac{c}{\sqrt{\varepsilon}} \implies B = \frac{E\sqrt{\varepsilon}}{c}$$ | |||
| This is the magnetic induction in the SI system. | |||
| In the CGS (Gaussian) system, Faraday's law of electromagnetic induction contains an additional factor of $1/c$ on the right side: | |||
| $$EL = \frac{1}{c} \frac{d\Phi}{dt} = \frac{1}{c} B L \frac{dx}{dt}$$ | |||
| Substituting the velocity $dx/dt = c/\sqrt{\varepsilon}$: | |||
| $$EL = \frac{1}{c} B L \frac{c}{\sqrt{\varepsilon}} = B L \frac{1}{\sqrt{\varepsilon}}$$ | |||
| Canceling $L$, we find the magnetic induction in the CGS system: | |||
| $$E = B \frac{1}{\sqrt{\varepsilon}} \implies B = E\sqrt{\varepsilon}$$ | |||
| @@ -34,7 +28,3 @@Solution | |||
| #### Answer | |||
| $B = \frac{E\sqrt{\varepsilon}}{c}$ (in SI); $B = E\sqrt{\varepsilon}$ (in CGS) | |||
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| − | #### Answer | ||
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| − | [Insert a concise answer or boxed result] | ||
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| ### Statement | ### Statement | ||
| $12.1.7.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $12.1.7^\ast$. Solve problem 12.1.6 in the case where the wave propagates in a medium with a dielectric permittivity $\varepsilon$. The velocity of the wave in the medium is $c/\sqrt{\varepsilon}$. | $12.1.7^\ast$. Solve problem 12.1.6 in the case where the wave propagates in a medium with a dielectric permittivity $\varepsilon$. The velocity of the wave in the medium is $c/\sqrt{\varepsilon}$. | ||
| ### Solution | ### Solution | ||
| The method of solution is identical to that in problem 12.1.6. In the rectangular contour, let's denote the sides as $ab = L$ and $aa' = x$. Faraday's law of electromagnetic induction in magnitude for the contour is: | The method of solution is identical to that in problem 12.1.6. In the rectangular contour, let's denote the sides as $ab = L$ and $aa' = x$. Faraday's law of electromagnetic induction in magnitude for the contour is: | ||
| $$\mathscr{E}_i = \frac{d\Phi}{dt}$$ | $$\mathscr{E}_i = \frac{d\Phi}{dt}$$ | ||
| where $\mathscr{E}_i$ is the induced electromotive force (EMF) created by the electric field $E$, and $\Phi$ is the magnetic flux through the contour. | where $\mathscr{E}_i$ is the induced electromotive force (EMF) created by the electric field $E$, and $\Phi$ is the magnetic flux through the contour. | ||
| From the geometry of the setup: | From the geometry of the setup: | ||
| $$\mathscr{E}_i = E L$$ | $$\mathscr{E}_i = E L$$ | ||
| $$d\Phi = B L dx$$ | $$d\Phi = B L dx$$ | ||
| Substituting these two formulas into the induction law, we get after canceling $L$: | Substituting these two formulas into the induction law, we get after canceling $L$: | ||
| $$E = B \frac{dx}{dt}$$ | $$E = B \frac{dx}{dt}$$ | ||
| Here, the rate of change of distance $dx/dt$ is simply the propagation velocity of the wave in the medium, $v = c/\sqrt{\varepsilon}$. Therefore: | Here, the rate of change of distance $dx/dt$ is simply the propagation velocity of the wave in the medium, $v = c/\sqrt{\varepsilon}$. Therefore: | ||
| $$E = B \frac{c}{\sqrt{\varepsilon}} \implies B = \frac{E\sqrt{\varepsilon}}{c}$$ | $$E = B \frac{c}{\sqrt{\varepsilon}} \implies B = \frac{E\sqrt{\varepsilon}}{c}$$ | ||
| This is the magnetic induction in the SI system. | This is the magnetic induction in the SI system. | ||
| In the CGS (Gaussian) system, Faraday's law of electromagnetic induction contains an additional factor of $1/c$ on the right side: | In the CGS (Gaussian) system, Faraday's law of electromagnetic induction contains an additional factor of $1/c$ on the right side: | ||
| $$EL = \frac{1}{c} \frac{d\Phi}{dt} = \frac{1}{c} B L \frac{dx}{dt}$$ | $$EL = \frac{1}{c} \frac{d\Phi}{dt} = \frac{1}{c} B L \frac{dx}{dt}$$ | ||
| Substituting the velocity $dx/dt = c/\sqrt{\varepsilon}$: | Substituting the velocity $dx/dt = c/\sqrt{\varepsilon}$: | ||
| $$EL = \frac{1}{c} B L \frac{c}{\sqrt{\varepsilon}} = B L \frac{1}{\sqrt{\varepsilon}}$$ | $$EL = \frac{1}{c} B L \frac{c}{\sqrt{\varepsilon}} = B L \frac{1}{\sqrt{\varepsilon}}$$ | ||
| Canceling $L$, we find the magnetic induction in the CGS system: | Canceling $L$, we find the magnetic induction in the CGS system: | ||
| $$E = B \frac{1}{\sqrt{\varepsilon}} \implies B = E\sqrt{\varepsilon}$$ | $$E = B \frac{1}{\sqrt{\varepsilon}} \implies B = E\sqrt{\varepsilon}$$ | ||
| @@ -34,7 +28,3 @@Solution | |||
| #### Answer | #### Answer | ||
| $B = \frac{E\sqrt{\varepsilon}}{c}$ (in SI); $B = E\sqrt{\varepsilon}$ (in CGS) | $B = \frac{E\sqrt{\varepsilon}}{c}$ (in SI); $B = E\sqrt{\varepsilon}$ (in CGS) | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||