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Next: Ampèreの法則 Up: denjiki Previous: Ohmの法則

Biot-Savartの法則

\fbox{\parbox{14.5cm}{
\begin{itemize}
\item Biot-Savartの法則
\par
電流密度分...
...\bm{r}-\bm{r}')}{\vert\bm{r}-\bm{r}'\vert^3}
\end{eqnarray*}
\end{itemize}
}}

[ 例題1 ] 直線電流

\includegraphics[width=3.5cm]{line.eps}

$\displaystyle \bm{r}$ $\displaystyle =$ $\displaystyle (r, 0, 0)$  
$\displaystyle \bm{r}'$ $\displaystyle =$ $\displaystyle (0, 0, z)$  
$\displaystyle \vert\bm{r}-\bm{r}'\vert$ $\displaystyle =$ $\displaystyle \sqrt{r^2 + z^2}$  


$\displaystyle B_{\theta}$ $\displaystyle =$ $\displaystyle \frac{\mu_0I}{4\pi} \int_{-\infty}^{\infty} \frac{rdz}{(r^2+z^2)^{3/2}}$  
    $\displaystyle {\color{blue} \tan\theta = \frac{z}{r}, \, \cos\theta = \frac{r}{\sqrt{r^2+z^2}} とおくと}$  
    $\displaystyle {\color{blue} \frac{d\theta}{\cos^2\theta} = \frac{dz}{r}}$  
  $\displaystyle =$ $\displaystyle \frac{\mu_0I}{4\pi} \int_{-\pi/2}^{\pi/2} \frac{\cos\theta d\theta}{r}$  
  $\displaystyle =$ $\displaystyle \frac{\mu_0I}{4\pi r} \left[ \sin\theta \right]_{-\pi/2}^{\pi/2}$  
  $\displaystyle =$ $\displaystyle \frac{\mu_0I}{2\pi r}$  

[ 例題2 ] 円電流

\includegraphics[width=3.5cm]{circle.eps} \includegraphics[width=3.5cm]{circle2.eps}

$\displaystyle \bm{r}$ $\displaystyle =$ $\displaystyle (0, 0, z)$  
$\displaystyle \bm{r}'$ $\displaystyle =$ $\displaystyle (a, \theta, 0)$  
$\displaystyle \vert\bm{r}-\bm{r}'\vert$ $\displaystyle =$ $\displaystyle \sqrt{a^2 + z^2}$  


$\displaystyle B_{z}$ $\displaystyle =$ $\displaystyle \frac{\mu_0I}{4\pi} \int_{0}^{2\pi} \frac{ad\theta}{a^2+z^2} \cos\varphi$  
    $\displaystyle {\color{blue} ここで \cos\varphi = \frac{a}{a^2+z^2}}$  
  $\displaystyle =$ $\displaystyle \frac{\mu_0I}{4\pi} \int_{0}^{2\pi} \frac{a^2d\theta}{(a^2+z^2)^{3/2}}$  
  $\displaystyle =$ $\displaystyle \frac{\mu_0I}{2} \frac{a^2}{(a^2+z^2)^{3/2}}$  


next up previous
Next: Ampèreの法則 Up: denjiki Previous: Ohmの法則
Keiichi Takasugi
平成24年1月25日