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For an arbitrary arrangement of current-carrying wire, I am trying to show that the Taylor expansion around $\vec{r} =0$ of a magnetic field $\vec{B}(\vec{r})$ is given by

\begin{align} \vec{B}(\vec{r}) = \frac{\mu_0I}{{4\pi}} \int d\vec{r}' \times \Big(\frac{\vec{r}-\vec{r}'}{{r}'^3} -3\vec{r}'\frac{\vec{r}\cdot\vec{r}'}{r'^5}\Big). \end{align}

From the Biot-Savart law, I know that

\begin{align} \vec{B}(\vec{r}) = \frac{{\mu _0 I}}{{4\pi }}\int\frac{{d\ell' \times {\bf{\hat {\mathfrak{r}}}}}}{{\mathfrak{r}^2}} \end{align}

where $\vec{\mathfrak{r}} = |\vec{r}-\vec{r}'|$.

Now, I know that I can change the integrand in the Biot-Savart law using the properties of cross products and the fact that $\hat{\mathfrak{r}}=\vec{\mathfrak{r}}/\mathfrak{r}$. This causes the integrand to simplify to

\begin{split} \frac{{d\ell' \times {\bf{\hat {\mathfrak{r}}}}}}{{\mathfrak{r}^2}} & = \frac{1}{\mathfrak{r}^3}({d\ell' \times ({\bf{\vec{r}-\vec{r}'}}}))\\ & =({d\ell' \times {\vec{r}}})/\mathfrak{r}^3. \end{split}

So now I am trying to do a Taylor expansion of this term so that will hopefully result in the first equation I wrote. However, I am not sure how to Taylor expand $\frac{1}{\mathfrak{r}^3}$. Looking at the Wikipedia page for expansion in Cartesian coordinates, I feel like I am on the correct track. However, I cannot follow their example.

  • How do they get the $v_\alpha(\mathbf{R}) \equiv\left( \frac{\partial v(\mathbf{r}-\mathbf{R}) }{\partial r_\alpha}\right)_{\mathbf{r}= \mathbf0}=\frac{-R_\alpha}{{R^3}}$ calculation? What is $r_\alpha$?
  • Am I on the right track, in general, for solving this problem? Does anybody have any ideas?
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