1
$\begingroup$

I'm using the Barnes Hut method to calculate the magnetic vector potential induced by an applied current. Given as:

$\begin{equation} A(r) = \frac{\mu_0}{4\pi} \int_V\frac{\bf{J(r')}}{|r-r'|}dV(r') \end{equation}$

In general this equation can be written as:

$\begin{equation} \Phi_i = \sum^N_{j=0} \frac{m_j}{r_{ij}} \end{equation}$

By Taylor expanding the general form $\frac{m_j}{r_{ij}}$ at a cell center I can calculate my multipole which are only dependet on $j$:

$\begin{equation}\begin{split} \frac{m_j}{r_{ij}} & \approx \frac{m_j}{R} + m_j(x_c-x_j)\left(-\frac{x_i-x_c}{R^3}\right) + m_j(y_c-y_j)\left(-\frac{y_i-y_c}{R^3}\right) \\ & + m_j(z_c-z_j)\left(-\frac{z_i-z_c}{R^3}\right) + \frac{m_j(x_c-x_j)^2}{2}\left(\frac{3(x_i-x_c)^2}{R^5} - \frac{1}{R^3}\right) \\ & + \frac{m_j(y_c-y_j)^2}{2}\left(\frac{3(y_i-y_c)^2}{R^5} - \frac{1}{R^3}\right) + \frac{m_j(z_c-z_j)^2}{2}\left(\frac{3(z_i-z_c)^2}{R^5} - \frac{1}{R^3}\right) \\ & + \frac{m_j(x_c-x_j)(y_c-y_j)}{2}\frac{3(x_i-x_c)(y_i-y_c)}{R^5} \\ & + \frac{m_j(y_c-y_j)(z_c-z_j)}{2}\frac{3(y_i-y_c)(z_i-z_c)}{R^5} \\ & + \frac{m_j(z_c-z_j)(x_c-x_j)}{2}\frac{3(z_i-z_c)(x_i-x_c)}{R^5} \end{split}\end{equation}$\

$\begin{equation} \Phi_i = \sum^N_{j=0} \frac{m_j}{r_{ij}} = \sum^N_{j=0} A_j B_i \end{equation}$

My question now is: can I do a multipole expansion for calculating the magnetic field intesity with consideration of the magnetization:

$\begin{equation} H(r) = H_i(r) -\frac{1}{4\pi} \int_V [\frac{M(r')}{|r-r'|^3} - \frac{3(r-r')}{|r-r'|^5}(r-r') \cdot M(r')]dV(r') \end{equation}$

Here $M$ is the magnetization of a ferromagnetic material which actually depends on the magnetic field intesity. The magnetic field intensity $H_i$ is given by Biot-Savarts law from all external electric currents. If this is possible what would be a good strategy to get independent $A_j$ and $B_i$? I actually try to Taylor expanded the equation but the variables are not independet as you can see for the first derivative in x-direction (without ($H_i$)):

\begin{equation} H_{x_j|c} = \newcommand{\Bold}[1]{\mathbf{#1}}\left(-\frac{6 \, M_{\mathit{xj}} {\left(x_{c} - x_{i}\right)}}{{\left({\left| -x_{c} + x_{i} \right|}^{2} + {\left| -y_{c} + y_{i} \right|}^{2} + {\left| -z_{c} + z_{i} \right|}^{2}\right)}^{\frac{5}{2}}} + \frac{15 \, {\left(M_{\mathit{xj}} {\left(x_{c} - x_{i}\right)} + M_{\mathit{yj}} {\left(y_{c} - y_{i}\right)} + M_{\mathit{zj}} {\left(z_{c} - z_{i}\right)}\right)} {\left(x_{c} - x_{i}\right)}^{2}}{{\left({\left| -x_{c} + x_{i} \right|}^{2} + {\left| -y_{c} + y_{i} \right|}^{2} + {\left| -z_{c} + z_{i} \right|}^{2}\right)}^{\frac{7}{2}}} - \frac{3 \, {\left(M_{\mathit{xj}} {\left(x_{c} - x_{i}\right)} + M_{\mathit{yj}} {\left(y_{c} - y_{i}\right)} + M_{\mathit{zj}} {\left(z_{c} - z_{i}\right)}\right)}}{{\left({\left| -x_{c} + x_{i} \right|}^{2} + {\left| -y_{c} + y_{i} \right|}^{2} + {\left| -z_{c} + z_{i} \right|}^{2}\right)}^{\frac{5}{2}}},\,-\frac{3 \, M_{\mathit{yj}} {\left(x_{c} - x_{i}\right)}}{{\left({\left| -x_{c} + x_{i} \right|}^{2} + {\left| -y_{c} + y_{i} \right|}^{2} + {\left| -z_{c} + z_{i} \right|}^{2}\right)}^{\frac{5}{2}}} - \frac{3 \, M_{\mathit{xj}} {\left(y_{c} - y_{i}\right)}}{{\left({\left| -x_{c} + x_{i} \right|}^{2} + {\left| -y_{c} + y_{i} \right|}^{2} + {\left| -z_{c} + z_{i} \right|}^{2}\right)}^{\frac{5}{2}}} + \frac{15 \, {\left(M_{\mathit{xj}} {\left(x_{c} - x_{i}\right)} + M_{\mathit{yj}} {\left(y_{c} - y_{i}\right)} + M_{\mathit{zj}} {\left(z_{c} - z_{i}\right)}\right)} {\left(x_{c} - x_{i}\right)} {\left(y_{c} - y_{i}\right)}}{{\left({\left| -x_{c} + x_{i} \right|}^{2} + {\left| -y_{c} + y_{i} \right|}^{2} + {\left| -z_{c} + z_{i} \right|}^{2}\right)}^{\frac{7}{2}}},\,-\frac{3 \, M_{\mathit{zj}} {\left(x_{c} - x_{i}\right)}}{{\left({\left| -x_{c} + x_{i} \right|}^{2} + {\left| -y_{c} + y_{i} \right|}^{2} + {\left| -z_{c} + z_{i} \right|}^{2}\right)}^{\frac{5}{2}}} - \frac{3 \, M_{\mathit{xj}} {\left(z_{c} - z_{i}\right)}}{{\left({\left| -x_{c} + x_{i} \right|}^{2} + {\left| -y_{c} + y_{i} \right|}^{2} + {\left| -z_{c} + z_{i} \right|}^{2}\right)}^{\frac{5}{2}}} + \frac{15 \, {\left(M_{\mathit{xj}} {\left(x_{c} - x_{i}\right)} + M_{\mathit{yj}} {\left(y_{c} - y_{i}\right)} + M_{\mathit{zj}} {\left(z_{c} - z_{i}\right)}\right)} {\left(x_{c} - x_{i}\right)} {\left(z_{c} - z_{i}\right)}}{{\left({\left| -x_{c} + x_{i} \right|}^{2} + {\left| -y_{c} + y_{i} \right|}^{2} + {\left| -z_{c} + z_{i} \right|}^{2}\right)}^{\frac{7}{2}}}\right) \end{equation}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.