Edits to “Statement”, “Answer”
en/12.1.6.md
+2 −4
| @@ -1,7 +1,6 @@ | |||
| ### Statement | |||
| − | $12.1.6.$ [Insert the problem 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: | |||
| \begin{equation} | |||
| E L=\frac{d\Phi}{dt} | |||
| \end{equation} | |||
| 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 | |||
| @@ -47,5 +46,4 @@Solution | |||
| \end{equation} | |||
| #### Answer | |||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $B=\frac{E}{c}$(in SI) $B=E$(in CGS) | ||
| @@ -1,7 +1,6 @@ | |||
| ### Statement | ### Statement | ||
| $12.1.6.$ [Insert the problem 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 | ### Solution | ||
| Let $ab=L$.By applying the law of electromagnetic induction to the countour $baa'b'$ we get: | Let $ab=L$.By applying the law of electromagnetic induction to the countour $baa'b'$ we get: | ||
| \begin{equation} | \begin{equation} | ||
| E L=\frac{d\Phi}{dt} | E L=\frac{d\Phi}{dt} | ||
| \end{equation} | \end{equation} | ||
| In time $dt$ the wave moves a distance $cdt$.So the change of the magnetic flux is: | In time $dt$ the wave moves a distance $cdt$.So the change of the magnetic flux is: | ||
| \begin{equation} | \begin{equation} | ||
| d\Phi=BLcdt | d\Phi=BLcdt | ||
| \end{equation} | \end{equation} | ||
| By plugging eq(2) into the first we find that(in SI units): | By plugging eq(2) into the first we find that(in SI units): | ||
| \begin{equation} | \begin{equation} | ||
| B=\frac{E}{c} | B=\frac{E}{c} | ||
| \end{equation} | \end{equation} | ||
| In Gaussian units Faraday's law for the same countour looks like: | In Gaussian units Faraday's law for the same countour looks like: | ||
| \begin{equation} | \begin{equation} | ||
| EL=\frac{1}{c}\frac{d\Phi}{dt} | EL=\frac{1}{c}\frac{d\Phi}{dt} | ||
| \end{equation} | \end{equation} | ||
| So the answer in CGS units is: | So the answer in CGS units is: | ||
| \begin{equation} | \begin{equation} | ||
| B=E | B=E | ||
| @@ -47,5 +46,4 @@Solution | |||
| \end{equation} | \end{equation} | ||
| #### Answer | #### Answer | ||
| $B=\frac{E}{c}$(in SI) $B=E$(in CGS) | |||
| [Insert a concise answer or boxed result] | |||