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When one calculates the vector potential of a magnetic dipole (current circuit), one can arrive at an integral of the form: \begin{equation*} \vec{A} = \frac{I}{cR^3} \oint\limits_L {\vec{dl}} ( {\vec{r}'} \cdot {\vec{R}}), \end{equation*} where $\vec{R}$ --position vector of watchpoint.

Using common formula \begin{equation*} \oint\limits_L \overset{\vec{a}}{\vec{r}'}\times [\overset{\vec{b}}{d\vec{l}} \times \overset{\vec{c}}{\vec{R}}] = \oint\limits_L \overset{\vec{b}}{\vec{dl}} ( \overset{\vec{a}}{\vec{r}'} \cdot \overset{\vec{c}}{\vec{R}}) - \oint\limits_L \overset{\vec{c}}{\vec{R}} (\overset{\vec{a}}{\vec{r}'}\cdot \overset{\vec{b}}{d\vec{l}}) \end{equation*}

So \begin{equation*} \vec{A} = \frac{I}{cR^3} \oint\limits_L \vec{r}'\times [d\vec{l} \times \vec{R}]. \end{equation*}

The common answer is \begin{equation*} \vec{A} = \frac{I}{cR^3} \left( \frac12 \oint\limits_L \vec{r}'\times d\vec{l}\right) \times \vec{R} = \frac{\vec{m} \times \vec{R}}{R^3}, \end{equation*} where $$ \vec{m} = \frac{I}{2c} \oint\limits_L \vec{r}'\times d\vec{l} $$

Where does the ratio

\begin{equation*} \oint\limits_L \vec{r}'\times [d\vec{l} \times \vec{R}] = \underset{\text{area}}{\left( \frac12 \oint\limits_L \vec{r}'\times d\vec{l}\right)} \times \vec{R} \end{equation*}

come from? How to prove the equation?

Sergio
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  • For vector products $a\times b \times c$ is ambiguous. Do you mean $(a\times b)\times c$ or $a\times (b\times c)$. These are not the same thing. – mike stone Nov 18 '22 at 14:28
  • @mikestone I changed formulae – Sergio Nov 18 '22 at 14:34
  • the area of a triangle spanned by the vectors $\mathbf r'$ and $d\ell$ is $ d\mathbf S = \frac{1}{2} \mathbf r' \times d\ell$ – hyportnex Nov 18 '22 at 14:43
  • @hyportnex that's what it says, but that's not the point. – Sergio Nov 18 '22 at 14:50
  • in that case, neither I have any idea what you are asking about, sorry. – hyportnex Nov 18 '22 at 14:55
  • @hyportnex I asking about how to prove equation $\oint\limits_L \vec{r}'\times [d\vec{l} \times \vec{R}] = \left( \frac12 \oint\limits_L \vec{r}'\times d\vec{l}\right) \times \vec{R} $ – Sergio Nov 18 '22 at 15:00
  • write out the coordinates $\mathbf r' =p(cos \theta, sin\theta, 0)$ and $\mathbf R = (cos\alpha, 0, sin\alpha)$ and $d\mathbf {\ell} = p(-sin\theta, cos\theta, 0)d\theta$ and integrate over the circle. Your 2nd integral in the first eq is zero and you get the result you wish to prove for a planar circular disk. – hyportnex Nov 18 '22 at 15:25
  • @hyportnex Of course, but this is a special case. I need more general case – Sergio Nov 18 '22 at 15:40
  • span a surface over your nonplanar current loop and break it up into small circular loops planar loops, a la Cauchy and sum. – hyportnex Nov 18 '22 at 16:02

1 Answers1

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Ok, as I mention above \begin{equation*} \oint\limits_L (\vec{r}'\cdot \vec{R}) d\vec{l} = \oint\limits_L \vec{r}'\times [d\vec{l} \times \vec{R}] . \end{equation*}

Let's consider first integral $\oint\limits_L (\vec{r}'\cdot \vec{R}) d\vec{l} $. Multiply (by dot product) it by an arbitrary constant vector $\vec{a}$: \begin{equation*} \vec{a} \cdot \oint\limits_L (\vec{r}'\cdot \vec{R}) d\vec{l} = \oint\limits_L (\vec{r}'\cdot \vec{R}) \vec{a} \cdot d\vec{l}. \end{equation*}

Now use a Stokes theorem \begin{equation*} \oint\limits_L (\vec{r}'\cdot \vec{R}) \vec{a} \cdot d\vec{l} = \iint\limits_S \vec{\nabla} \times \left( (\vec{r}'\cdot \vec{R}) \vec{a}\right) \cdot d\vec{S}. \end{equation*}

Simplify the expression $\vec{\nabla} \times \left( (\vec{r}'\cdot \vec{R}) \vec{a}\right)$: \begin{equation*} \vec{\nabla} \times \left( (\vec{r}'\cdot \vec{R}) \vec{a} \right) = (\vec{r}'\cdot \vec{R}) \vec{\nabla}\times\vec{a} + (\nabla{ (\vec{r}'\cdot \vec{R})) \times \vec{a} }. \end{equation*}

Still simplification $\nabla{ (\vec{r}'\cdot \vec{R})} = \vec{R}$

Because the $\vec{\nabla}\times\vec{a} = 0$, we have \begin{equation*} \iint\limits_S \vec{\nabla} \times \left( (\vec{r}'\cdot \vec{R}) \vec{a}\right) \cdot d\vec{S} = \iint\limits_S \vec{R} \times \vec{a} \cdot d\vec{S}. \end{equation*} Swapping vectros (using triple product properties), we get \begin{equation*} \iint\limits_S \vec{R} \times \vec{a} \cdot d\vec{S} = \vec{a} \cdot \left( \iint\limits_S d\vec{S} \right) \times \vec{R}. \end{equation*}

We finally received an equation \begin{equation*} \vec{a} \cdot \oint\limits_L (\vec{r}'\cdot \vec{R}) d\vec{l} = \vec{a} \cdot \left( \iint\limits_S d\vec{S} \right) \times \vec{R}. \end{equation*}

Since the vector $\vec{a}$ is arbitrary: \begin{equation*} \oint\limits_L (\vec{r}'\cdot \vec{R}) d\vec{l} = \left( \iint\limits_S d\vec{S} \right) \times \vec{R}. \end{equation*}

Maybe there is a simpler common way.

Using the proven identity, it is now easy to calculate the torque of Ampere force that rotates the loop wire in a homogeneous field:

\begin{equation*} \vec{M} = \oint\limits_L \vec{r} \times d\vec{F} = \oint\limits_L \vec{r} \times [\frac{I}{c}d\vec{l} \times \vec{B} ] = \frac{IS}{c}\times \vec{B} = \vec{m}\times\vec{B}. \end{equation*}

Sergio
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