(Created page with "2) <math> \begin{equation*} \boxed{d\bar{H}=\frac{I(\bar{R}')d\bar{l}'\times(\bar{R}-\bar{R}')}{4\pi\abs{\bar{R}-\bar{R}'}^3}} \end{equation*} </math> <math> \begin{align*}...") |
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<math> | <math> | ||
\begin{equation*} | \begin{equation*} | ||
| − | \boxed{d\bar{H}=\frac{I(\bar{R}')d\bar{l}'\times(\bar{R}-\bar{R}')}{4\pi | + | \boxed{d\bar{H}=\frac{I(\bar{R}')d\bar{l}'\times(\bar{R}-\bar{R}')}{4\pi|\bar{R}-\bar{R}'|^3}} |
\end{equation*} | \end{equation*} | ||
</math> | </math> | ||
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<math> | <math> | ||
\begin{align*} | \begin{align*} | ||
| − | \text | + | \text{along x:}& & \bar{R}&=(0,0,0) \longrightarrow &&\bar{H}&=\int_1^{\infty}\frac{Idx\hat{x}\times(-x\hat{x})}{4\pi x^3}=0\\ |
& &\bar{R}'&=x\hat{x} &\\ | & &\bar{R}'&=x\hat{x} &\\ | ||
| − | & & | + | & &|\bar{R}-\bar{R}'|&=x && &\\ |
| − | \text | + | \text{along y:}& & \bar{R}&=(0,0,0) \longrightarrow &&\bar{H}&=\int_1^{\infty}\frac{Idy\hat{y}\times(-y\hat{y})}{4\pi y^3}=0\\ |
| − | & &\bar{R}'&=y\hat{y} &\\ | + | & &|\bar{R}'|&=y\hat{y} &\\ |
| − | & & | + | & &|\bar{R}-\bar{R}'|&=y && & |
\end{align*} | \end{align*} | ||
</math> | </math> | ||
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<math> | <math> | ||
\begin{equation*} | \begin{equation*} | ||
| − | \text | + | \text{Superposition: Total field} \qquad \boxed{\bar{H}=0} |
\end{equation*} | \end{equation*} | ||
</math> | </math> | ||
Revision as of 19:00, 18 June 2017
2)
$ \begin{equation*} \boxed{d\bar{H}=\frac{I(\bar{R}')d\bar{l}'\times(\bar{R}-\bar{R}')}{4\pi|\bar{R}-\bar{R}'|^3}} \end{equation*} $
$ \begin{align*} \text{along x:}& & \bar{R}&=(0,0,0) \longrightarrow &&\bar{H}&=\int_1^{\infty}\frac{Idx\hat{x}\times(-x\hat{x})}{4\pi x^3}=0\\ & &\bar{R}'&=x\hat{x} &\\ & &|\bar{R}-\bar{R}'|&=x && &\\ \text{along y:}& & \bar{R}&=(0,0,0) \longrightarrow &&\bar{H}&=\int_1^{\infty}\frac{Idy\hat{y}\times(-y\hat{y})}{4\pi y^3}=0\\ & &|\bar{R}'|&=y\hat{y} &\\ & &|\bar{R}-\bar{R}'|&=y && & \end{align*} $
$ \begin{equation*} \text{Superposition: Total field} \qquad \boxed{\bar{H}=0} \end{equation*} $
