Module 3 : MAGNETIC FIELD
Lecture 15 : Biot- Savarts' Law

Once the integration is carried out, the expression above can be shown to be symmetrical between the two circuits. To show, we express the vector triple product

\begin{displaymath}\vec{dl_2}\times(\vec{dl_1}\times\vec{r_{12}}) = \vec{dl_1}(... ..._2}\cdot\vec{r_{12}}) - \vec{r_{12}}(\vec{dl_1}\cdot\vec{dl_2})\end{displaymath}

so that

\begin{displaymath}\vec F = \frac{\mu_0}{4\pi}I_1I_2\left[ \oint\oint\frac{\vec{... ...\frac{\vec{r_{12}}(\vec{dl_1}\cdot\vec{dl_2})}{r_{12}^3}\right]\end{displaymath}

The integrand in the first integral is an exact differential with respect to the integral over $\vec{dl_2}$as \begin{displaymath}\oint\oint\frac{\vec{dl_1}(\vec{dl_2}\cdot\vec{r_{12}})}{r_{1... ...} = \oint\vec{dl_1}\oint\nabla(\frac{1}{r_{12}})\cdot\vec{dl_2}\end{displaymath}

The integral above being an integral of a gradient over a closed path vanishes. Thus

\begin{displaymath}\vec F = \frac{\mu_0}{4\pi}I_1I_2\oint\oint\frac{\vec{r_{12}}(\vec{dl_1}\cdot\vec{dl_2})}{r_{12}^3}\end{displaymath}

which is explicitly symmetric between the two circuits, confirming the validity of Newton 's third law.

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