Правка разделов «Statement», «Solution», «Answer»
en/12.1.6.md
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| @@ -1,49 +1,25 @@ | |||
| ### Statement | |||
| $12.1.6.$ An electromagnetic wave occupies the space between two parallel infinite planes $AB$ and $A'B'$. The illustrated segment of the electromagnetic field moves at the speed of light $c$ in a direction perpendicular to the plane $AB$. The electric field strength of the wave is $E$. By applying the law of electromagnetic induction to the rectangular loop (contour) $baa'b'$, determine the magnetic induction of the wave in SI and CGS units. | |||
| − | ### Solution | ||
| + |  | ||
| − | Let $ab=L$.By applying the law of electromagnetic induction to the countour $baa'b'$ we get: | ||
| + | ### Solution | ||
| + | Let $ab = L$. By applying Faraday's law of electromagnetic induction to the contour $baa'b'$ we get: | ||
| + | $$EL = \frac{d\Phi}{dt}$$ | ||
| + | In time $dt$, the wave moves a distance $c dt$. So the change of the magnetic flux is: | ||
| + | $$d\Phi = B L c dt$$ | ||
| − | \begin{equation} | ||
| − | |||
| − | \end{equation} | ||
| + | By plugging the second equation into the first, we find the induction in SI units: | ||
| + | $$EL = B L c \implies B = \frac{E}{c}$$ | ||
| + | In Gaussian units (CGS), Faraday's law for the same contour looks like: | ||
| + | $$EL = \frac{1}{c} \frac{d\Phi}{dt}$$ | ||
| + | Substituting the flux expression, the answer in CGS units is: | ||
| + | $$EL = \frac{1}{c} B L c \implies B = E$$ | ||
| − | In time $dt$ the wave moves a distance $cdt$.So the change of the magnetic flux is: | ||
| − | |||
| − | \begin{equation} | ||
| − | d\Phi=BLcdt | ||
| − | \end{equation} | ||
| − | |||
| − | |||
| − | |||
| − | By plugging eq(2) into the first we find that(in SI units): | ||
| − | |||
| − | |||
| − | \begin{equation} | ||
| − | B=\frac{E}{c} | ||
| − | \end{equation} | ||
| − | |||
| − | |||
| − | |||
| − | In Gaussian units Faraday's law for the same countour looks like: | ||
| − | |||
| − | \begin{equation} | ||
| − | EL=\frac{1}{c}\frac{d\Phi}{dt} | ||
| − | \end{equation} | ||
| − | |||
| − | |||
| − | |||
| − | So the answer in CGS units is: | ||
| − | |||
| − | \begin{equation} | ||
| − | B=E | ||
| − | \end{equation} | ||
| − | |||
| #### Answer | |||
| − | $B | ||
| + | $B = \frac{E}{c}$ (in SI); $B = E$ (in CGS) | ||
| @@ -1,49 +1,25 @@ | |||
| ### Statement | ### Statement | ||
| $12.1.6.$ An electromagnetic wave occupies the space between two parallel infinite planes $AB$ and $A'B'$. The illustrated segment of the electromagnetic field moves at the speed of light $c$ in a direction perpendicular to the plane $AB$. The electric field strength of the wave is $E$. By applying the law of electromagnetic induction to the rectangular loop (contour) $baa'b'$, determine the magnetic induction of the wave in SI and CGS units. | $12.1.6.$ An electromagnetic wave occupies the space between two parallel infinite planes $AB$ and $A'B'$. The illustrated segment of the electromagnetic field moves at the speed of light $c$ in a direction perpendicular to the plane $AB$. The electric field strength of the wave is $E$. By applying the law of electromagnetic induction to the rectangular loop (contour) $baa'b'$, determine the magnetic induction of the wave in SI and CGS units. | ||
| ### Solution | |||
|  | |||
| Let $ab=L$.By applying the law of electromagnetic induction to the countour $baa'b'$ we get: | ### Solution | ||
| Let $ab = L$. By applying Faraday's law of electromagnetic induction to the contour $baa'b'$ we get: | |||
| $$EL = \frac{d\Phi}{dt}$$ | |||
| In time $dt$, the wave moves a distance $c dt$. So the change of the magnetic flux is: | |||
| $$d\Phi = B L c dt$$ | |||
| \begin{equation} | By plugging the second equation into the first, we find the induction in SI units: | ||
| $$EL = B L c \implies B = \frac{E}{c}$$ | |||
| \end{equation} | |||
| In Gaussian units (CGS), Faraday's law for the same contour looks like: | |||
| $$EL = \frac{1}{c} \frac{d\Phi}{dt}$$ | |||
| Substituting the flux expression, the answer in CGS units is: | |||
| $$EL = \frac{1}{c} B L c \implies B = E$$ | |||
| In time $dt$ the wave moves a distance $cdt$.So the change of the magnetic flux is: | |||
| \begin{equation} | |||
| d\Phi=BLcdt | |||
| \end{equation} | |||
| By plugging eq(2) into the first we find that(in SI units): | |||
| \begin{equation} | |||
| B=\frac{E}{c} | |||
| \end{equation} | |||
| In Gaussian units Faraday's law for the same countour looks like: | |||
| \begin{equation} | |||
| EL=\frac{1}{c}\frac{d\Phi}{dt} | |||
| \end{equation} | |||
| So the answer in CGS units is: | |||
| \begin{equation} | |||
| B=E | |||
| \end{equation} | |||
| #### Answer | #### Answer | ||
| $B |
$B = \frac{E}{c}$ (in SI); $B = E$ (in CGS) | ||