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Next: この文書について... Up: denjiki Previous: Faradayの電磁誘導の法則

Maxwell方程式

\fbox{\parbox{14.5cm}{
\begin{itemize}
\item Maxwell方程式
\begin{eqnarray*}
...
...{3mm} = \hspace{3mm} \mu\bm{H} \hspace{62mm}
\end{eqnarray*}
\end{itemize}
}}

[ 例題1 ] 電磁波

電荷なし$ \rho = 0$ , 電流なし $ \bm{J} = 0$

\begin{displaymath}
\left\{
\begin{array}{ccl}
\vspace{1mm}
\nabla\times\bm{...
...on\mu \frac{\partial\bm{E}}{\partial t}
\end{array}
\right.
\end{displaymath}


$\displaystyle \nabla\times (\nabla\times\bm{E}) = -\frac{\partial}{\partial t} (\nabla\times\bm{B})$ $\displaystyle =$ $\displaystyle -\epsilon\mu \frac{\partial^2\bm{E}}{\partial t^2}$  
$\displaystyle \rotatebox{90}{=} \hspace{37mm}$      
$\displaystyle \nabla(\nabla\cdot\bm{E}) - \nabla^2\bm{E} \hspace{21mm}$      
$\displaystyle \rotatebox{90}{$\leftarrow$} \hspace{39mm}$      
$\displaystyle 0 \hspace{39mm}$      

$\displaystyle \nabla^2\bm{E} = \epsilon\mu \frac{\partial^2\bm{E}}{\partial t^2}
$

$\displaystyle \bm{E} = \bm{E}_0 \exp i(\bm{k}\cdot\bm{r} - \omega t) \hspace{5mm} とおくと
$

\begin{displaymath}
\left\{
\begin{array}{ccl}
\vspace{1mm}
\nabla^2\bm{E} &...
...}_0\exp i(\bm{k}\cdot\bm{r} - \omega t)
\end{array}
\right.
\end{displaymath}


$\displaystyle k^2$ $\displaystyle =$ $\displaystyle \epsilon\mu\omega^2$  
$\displaystyle \frac{\omega}{k}$ $\displaystyle =$ $\displaystyle \pm \frac{1}{\sqrt{\epsilon\mu}} \hspace{5mm} : \ 位相速度$  
$\displaystyle \frac{1}{\sqrt{\epsilon_0\mu_0}}$ $\displaystyle =$ $\displaystyle c = 3.00\times10^8 \ [{\rm m/s}]$  

\includegraphics[width=4cm]{wave.eps}

$\displaystyle \bm{B} = \bm{B}_0 \exp i(\bm{k}\cdot\bm{r} - \omega t) \hspace{5mm} とおくと
$

\begin{displaymath}
\left\{
\begin{array}{ccl}
\vspace{1mm}
\nabla\times\bm{...
...}_0\exp i(\bm{k}\cdot\bm{r} - \omega t)
\end{array}
\right.
\end{displaymath}


$\displaystyle \bm{B}_0$ $\displaystyle =$ $\displaystyle \frac{\bm{k}\times\bm{E}_0}{\omega}$  
$\displaystyle \frac{E_0}{H_0}$ $\displaystyle =$ $\displaystyle \frac{E_0}{B_0/\mu} = \sqrt{\frac{\mu}{\epsilon}} \hspace{5mm} : \ インピーダンス$  
$\displaystyle \sqrt{\frac{\mu_0}{\epsilon_0}}$ $\displaystyle =$ $\displaystyle \frac{4\pi c}{10^7} = 377 \ [\Omega]$  

\fbox{\parbox{14.5cm}{
\begin{itemize}
\item Poyntingベクトル
\begin{displaym...
...s\bm{H} = \frac{1}{\mu} \bm{E}\times\bm{B}
\end{displaymath}
\end{itemize}
}}

[ 例題2 ] 抵抗体のエネルギーの流れ

\includegraphics[width=4cm]{poynting.eps}

半径$ a$ , 長さ$ \ell$ , 抵抗率$ \eta$

$\displaystyle \bm{E} = \eta\bm{J} = \eta J \bm{e}_z
$

$ r<a$

$\displaystyle 2\pi rH$ $\displaystyle =$ $\displaystyle \pi r^2J$  
$\displaystyle \bm{H}$ $\displaystyle =$ $\displaystyle \frac{Jr}{2} \bm{e}_{\theta}$  
$\displaystyle \bm{S}$ $\displaystyle =$ $\displaystyle \bm{E}\times\bm{H} = -\frac{\eta J^2}{2}r \bm{e}_r$  
$\displaystyle 2\pi rS$ $\displaystyle =$ $\displaystyle \eta J^2\pi r^2 \hspace{5mm} \leftarrow \ rとともに減少し吸収される$  

$ r>a$

$\displaystyle 2\pi rH$ $\displaystyle =$ $\displaystyle \pi a^2J$  
$\displaystyle \bm{H}$ $\displaystyle =$ $\displaystyle \frac{Ja^2}{2r} \bm{e}_{\theta}$  
$\displaystyle \bm{S}$ $\displaystyle =$ $\displaystyle \bm{E}\times\bm{H} = -\frac{\eta J^2}{2} \frac{a^2}{r} \bm{e}_r$  
$\displaystyle 2\pi rS$ $\displaystyle =$ $\displaystyle \eta J^2\pi a^2 \hspace{5mm} \leftarrow \ 一定値をとり吸収されない$  


next up previous
Next: この文書について... Up: denjiki Previous: Faradayの電磁誘導の法則
Keiichi Takasugi
平成24年1月25日