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Looking at a worked example in Chengs Field and Wave Electromagnetics (newest ed, p 236) about Biot-Savarts law.

A conducting wire of length $2L$ is carrying the current $I$. Find $\textbf{B}$ by first finding the magnetic vector potential.

The book then applies the following equation

$$\textbf{A} = \frac{\mu_0 I}{4\pi}\oint_c \frac{dl}{R}$$

and integrates from $-L$ to $L$. But how can we do that? It's not a closed loop. I guess this is due to some sloppy notation somewhere?

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    $\begingroup$ This looks sloppy all round. What is $R$? Where is the unit vector on the RHS? What assumptions are being made? I don't think this should be a closed integral. $\endgroup$ – Rob Jeffries Nov 29 '14 at 10:26
  • $\begingroup$ The way this was presented to me in my undergraduate E&M course (Wangsness' Electromagnetic Fields) is that you can treat the straight segment as a closed loop if you consider the parallel segments to be infinitely far apart such that their contributions to the line integral fail to contribute at all field points in your calculation/region of interest. I don't care for the reasoning much myself though, it seemed a bit crude to me then and now even moreso that I'm remembering it. I asked the same question as I recall. $\endgroup$ – kbh Jan 11 '15 at 6:01
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It is true. A straight line segment is not a closed loop. But, it is like an important mathematical building block if the closed integration path $c$ can be decomposed into straight line segments $c_k$ $k=1,\ldots,n$. In this case you compute $$ \textbf{A} = \frac{\mu_0 I}{4\pi}\oint_c \frac{dl}{R} = \sum_{k=1}^n\frac{\mu_0 I}{4\pi}\int_{c_k} \frac{dl}{R}. $$ You can compute the values for the sum terms with the formula for a straight line segment (after appropriate coordinate transformations).


Rob Jeffries critized the sloppy notation in the above formula. I do not like that notation either. So, I give a derivation of the vector potential as base:

Ampere's law for the stationary electro-magnetic field is $$ \newcommand{\mylabel}[1]{\tag{#1}\label{#1}} \newcommand{\rot}{\mathrm{rot}} \newcommand{\div}{\mathrm{div}} \newcommand{\grad}{\mathrm{grad}} \rot \vec{H} = \vec{J}\mylabel{Ampere} $$ where $\vec{J}$ is the current density in the conducting region $C$. It is extented to the free space by continuation with zero value.

The lack of magnetic sources is described by the magnetic vector potential $\vec{B}=\rot\vec{A}$ with the Coulomb gauge $\div\vec{A}=0$. Substituting this with $\vec{B}=\mu\vec{H}$ gives \begin{align} \rot\rot\vec{A} &= \mu_0\vec{J}\\ \grad\underbrace{\div\vec{A}}_{=0} - \Delta \vec{A} &= \mu_0\vec{J}\\ \Delta\vec{A} &= -\mu_0\vec{J}\mylabel{VecPotPDE} \end{align} With Green's function $-\frac{1}{4\pi|\vec{r}-\vec{r}_c|}$ for the Laplacian in $\mathbb{R}^3$ we can write the free space solution of \ref{VecPotPDE} as \begin{align} \vec{A}(\vec{r}) &= \frac{\mu_0}{4\pi}\int_{C} \frac{\vec{J}dV_C}{|\vec{r}-\vec{r}_C|} \end{align} In the limit case of a thin long conductor one replaces $\vec{J}d V_C$ with $I d\vec{r}_C$ and the three-dimensional domain $C$ with an oriented curve ("current thread") and obtains \begin{align} \vec{A}(\vec{r}) &= \frac{\mu_0 I}{4\pi}\int_{C} \frac{d\vec{r}_C}{|\vec{r}-\vec{r}_C|}.\mylabel{VecPot} \end{align} You are right that from the integral form of \ref{Ampere} \begin{align} \oint_{\partial \Omega} \vec{H}(\vec{r})d\vec{r} &= \int_{\Omega} \vec{J}\cdot d\vec{A} \end{align} with a closed surface $\Omega = \partial G$ (meaning $\partial\Omega=\emptyset$) we get the lack of current sources \begin{align} \oint_{\partial G} \vec{J}\cdot d\vec{A} = 0 \end{align} (From this point of view technical stationary current sources would better be named "current drives".) This implies that all current threads should be closed.

Nevertheless, you can decompose the closed current thread $C$ into a sequence of partial paths $C_k$ $k=1,\ldots,n$ and obtain the overall vector potential as a sum of partial integrals \begin{align} \vec{A}(\vec{r}) &= \sum_{k=1}^n \frac{\mu_0 I}{4\pi}\int_{C_k}\frac{d\vec{r}_C}{|\vec{r}-\vec{r}_C|}. \end{align}

This is especially useful if the closed path can be decomposed into a kind of standard paths such as straight line segments where you can provide the integrals in pre-calculated form.

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  • $\begingroup$ How does decomposing it make it more of a closed integral? I don't get it. $\endgroup$ – Benjamin Lindqvist Apr 9 '14 at 16:25
  • $\begingroup$ We can discuss it at chat.stackexchange.com. $\endgroup$ – Tobias Apr 9 '14 at 16:29
  • $\begingroup$ Explain it in your answer. $\endgroup$ – Rob Jeffries Nov 29 '14 at 10:23

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