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$A_{1}=\frac{2\mu_{1}I_{2}}{c}\ln\frac{r_{1}}{a}$

$A_{2}=\frac{2\mu_{2}}{c}(I\ln\frac{r_{1}}{a}+I_{1}\ln\frac{r_{2}}{a}-I_{1}\ln\frac{r}{a})$

$r_{1}^{2}=r^{2}+b^{2}-2rb\cos\theta$

$r_{2}^{2}=r^{2}+(b')^{2}-2rb'\cos\theta=r^{2}+\frac{a^{4}}{b^{2}}-2r\frac{a^{2}}{b}\cos\theta=\frac{a^{2}}{b^{2}}r_{1}^{2}$

$b'=\frac{a^{2}}{b}$

$B_{1n}=(\nabla\times A_{1})_{r}=\frac{1}{r}\frac{\partial A_{1}}{\partial\theta}=-\frac{\mu_{1}2I_{2}}{c}\frac{b\sin\theta}{r_{1}^{2}}$

$B_{2n}=(\nabla\times A_{2})_{r}=\frac{1}{r}\frac{\partial A_{2}}{\partial\theta}=-\frac{2\mu_{2}}{c}\left(I\frac{b\sin\theta}{r_{1}^{2}}+I_{1}\frac{a^{2}\sin\theta}{br_{2}^{2}}\right)$

$\left.B_{2n}\right|_{a}=-\frac{2\mu_{2}\sin\theta}{c}\left(I\frac{b}{r_{1}^{2}}+I_{1}\frac{a^{2}}{br_{2}^{2}}\right)=-\frac{2\mu_{2}\sin\theta}{c}\frac{b}{r_{1}^{2}}\left(I+I_{1}\right)$

$\mu_{1}I_{2}=\mu_{2}\left(I+I_{1}\right)$

$H_{1\tau}=(\nabla\times A_{1})_{\theta}=-\frac{1}{\mu_{1}}\frac{\partial A_{1}}{\partial r}=\frac{2I_{2}}{c}\frac{\left(r-b\cos\theta\right)}{r_{1}^{2}}$

$\left.H_{1\tau}\right|_{a}=\frac{2I_{2}}{c}\frac{\left(a-b\cos\theta\right)}{r_{1}^{2}}$

$H_{2\tau}=(\nabla\times A_{2})_{\theta}=-\frac{1}{\mu_{2}}\frac{\partial A_{2}}{\partial r}=\frac{2}{c}\left(I\frac{\left(r-b\cos\theta\right)}{r_{1}^{2}}+I_{1}\frac{\left(r-\frac{a^{2}}{b}\cos\theta\right)}{r_{2}^{2}}-I_{1}\frac{1}{r}\right)$

$\left.H_{2\tau}\right|_{a}=\frac{2}{c}\left(I\frac{\left(a-b\cos\theta\right)}{r_{1}^{2}}+I_{1}\frac{\left(a-\frac{a^{2}}{b}\cos\theta\right)}{\frac{a^{2}}{b^{2}}r_{1}^{2}}-I_{1}\frac{1}{a}\right)=\frac{2}{c}\left(I\frac{\left(a-b\cos\theta\right)}{r_{1}^{2}}+I_{1}\frac{\left(\frac{b^{2}}{a}-b\cos\theta\right)}{r_{1}^{2}}-I_{1}\frac{1}{a}\right)=$

$\frac{2}{car_{1}^{2}}\left(I\left(a^{2}-ab\cos\theta\right)+I_{1}\left(b^{2}-ab\cos\theta\right)-I_{1}\left(a^{2}+b^{2}-2ab\cos\theta\right)\right)=$

$\frac{2}{cr_{1}^{2}}\left(I\left(a-b\cos\theta\right)-I_{1}\left(a-b\cos\theta\right)\right)$

$I-I_{1}=I_{2}$

\textcyr{\CYRT\cyro\cyrg\cyrd\cyra}

$\mu_{1}\left(I-I_{1}\right)=\mu_{2}\left(I+I_{1}\right)$ \textcyr{\cyri} $I_{1}=I\frac{\mu_{1}-\mu_{2}}{\mu_{1}+\mu_{2}}$, $I_{2}=I\frac{2\mu_{2}}{\mu_{1}+\mu_{2}}$