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Display information for equation id:math.2095.13 on revision:2095

* Page found: Magnetostatische Feldgleichungen (eq math.2095.13)

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\begin{align}
& \Rightarrow \nabla \cdot \bar{A}(\bar{r})=-\frac{{{\mu }_{0}}}{4\pi }\int_{{{R}^{3}}}^{{}}{{}}{{d}^{3}}r\acute{\ }{{\nabla }_{r\acute{\ }}}\cdot \left( \frac{\bar{j}(\bar{r}\acute{\ })}{\left| \bar{r}-\bar{r}\acute{\ } \right|} \right)-\frac{\partial }{\partial t}\frac{{{\mu }_{0}}}{4\pi }\int_{{{R}^{3}}}^{{}}{{}}{{d}^{3}}r\acute{\ }\frac{\rho (\bar{r}\acute{\ },t)}{\left| \bar{r}-\bar{r}\acute{\ } \right|} \\
& \frac{{{\mu }_{0}}}{4\pi }\int_{{{R}^{3}}}^{{}}{{}}{{d}^{3}}r\acute{\ }\frac{\rho (\bar{r}\acute{\ },t)}{\left| \bar{r}-\bar{r}\acute{\ } \right|}={{\mu }_{0}}{{\varepsilon }_{0}}\Phi (\bar{r},t) \\
& \Rightarrow \nabla \cdot \bar{A}(\bar{r})=-\frac{{{\mu }_{0}}}{4\pi }\oint\limits_{S\infty }{{}}{{d}^{3}}\bar{f}\acute{\ }\left( \frac{\bar{j}(\bar{r}\acute{\ })}{\left| \bar{r}-\bar{r}\acute{\ } \right|} \right)-{{\mu }_{0}}{{\varepsilon }_{0}}\frac{\partial }{\partial t}\Phi (\bar{r},t) \\
\end{align}

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A ¯ ( r ¯ ) = - μ 0 4 π R 3 d 3 r ´ r ´ ( j ¯ ( r ¯ ´ ) | r ¯ - r ¯ ´ | ) - t μ 0 4 π R 3 d 3 r ´ ρ ( r ¯ ´ , t ) | r ¯ - r ¯ ´ | μ 0 4 π R 3 d 3 r ´ ρ ( r ¯ ´ , t ) | r ¯ - r ¯ ´ | = μ 0 ε 0 Φ ( r ¯ , t ) A ¯ ( r ¯ ) = - μ 0 4 π S d 3 f ¯ ´ ( j ¯ ( r ¯ ´ ) | r ¯ - r ¯ ´ | ) - μ 0 ε 0 t Φ ( r ¯ , t ) missing-subexpression absent ¯ 𝐴 ¯ 𝑟 subscript 𝜇 0 4 𝜋 subscript superscript 𝑅 3 superscript 𝑑 3 𝑟 ´ absent subscript 𝑟 ´ absent ¯ 𝑗 ¯ 𝑟 ´ absent ¯ 𝑟 ¯ 𝑟 ´ absent 𝑡 subscript 𝜇 0 4 𝜋 subscript superscript 𝑅 3 superscript 𝑑 3 𝑟 ´ absent 𝜌 ¯ 𝑟 ´ absent 𝑡 ¯ 𝑟 ¯ 𝑟 ´ absent missing-subexpression subscript 𝜇 0 4 𝜋 subscript superscript 𝑅 3 superscript 𝑑 3 𝑟 ´ absent 𝜌 ¯ 𝑟 ´ absent 𝑡 ¯ 𝑟 ¯ 𝑟 ´ absent subscript 𝜇 0 subscript 𝜀 0 Φ ¯ 𝑟 𝑡 missing-subexpression absent ¯ 𝐴 ¯ 𝑟 subscript 𝜇 0 4 𝜋 subscript contour-integral 𝑆 superscript 𝑑 3 ¯ 𝑓 ´ absent ¯ 𝑗 ¯ 𝑟 ´ absent ¯ 𝑟 ¯ 𝑟 ´ absent subscript 𝜇 0 subscript 𝜀 0 𝑡 Φ ¯ 𝑟 𝑡 {\displaystyle{\displaystyle\begin{aligned} &\displaystyle\Rightarrow\nabla% \cdot\bar{A}(\bar{r})=-\frac{{{\mu}_{0}}}{4\pi}\int_{{{R}^{3}}}{{}}{{d}^{3}}r% \acute{\ }{{\nabla}_{r\acute{\ }}}\cdot\left(\frac{\bar{j}(\bar{r}\acute{\ })}% {\left|\bar{r}-\bar{r}\acute{\ }\right|}\right)-\frac{\partial}{\partial t}% \frac{{{\mu}_{0}}}{4\pi}\int_{{{R}^{3}}}{{}}{{d}^{3}}r\acute{\ }\frac{\rho(% \bar{r}\acute{\ },t)}{\left|\bar{r}-\bar{r}\acute{\ }\right|}\\ &\displaystyle\frac{{{\mu}_{0}}}{4\pi}\int_{{{R}^{3}}}{{}}{{d}^{3}}r\acute{\ }% \frac{\rho(\bar{r}\acute{\ },t)}{\left|\bar{r}-\bar{r}\acute{\ }\right|}={{\mu% }_{0}}{{\varepsilon}_{0}}\Phi(\bar{r},t)\\ &\displaystyle\Rightarrow\nabla\cdot\bar{A}(\bar{r})=-\frac{{{\mu}_{0}}}{4\pi}% \oint\limits_{S\infty}{{}}{{d}^{3}}\bar{f}\acute{\ }\left(\frac{\bar{j}(\bar{r% }\acute{\ })}{\left|\bar{r}-\bar{r}\acute{\ }\right|}\right)-{{\mu}_{0}}{{% \varepsilon}_{0}}\frac{\partial}{\partial t}\Phi(\bar{r},t)\\ \end{aligned}}}

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A¯(r¯)=μ04πR3d3r´r´(j¯(r¯´)|r¯r¯´|)tμ04πR3d3r´ρ(r¯´,t)|r¯r¯´|μ04πR3d3r´ρ(r¯´,t)|r¯r¯´|=μ0ε0Φ(r¯,t)A¯(r¯)=μ04πSd3f¯´(j¯(r¯´)|r¯r¯´|)μ0ε0tΦ(r¯,t)

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