Ampere's equation

In physics, more particularly in electrodynamics, Ampère's equation describes the force between two infinitesimal elements of electric-current-carrying wires. The equation is named for the early nineteenth century French physicist and mathematician André-Marie Ampère.

Rather than giving Ampère's original infinitesimal equation, which is not without problems, we will describe two common cases obtained by integration: a system consisting of two straight wires and a system of two closed loops. Since the integrals over disputed terms in Ampère's infinitesimal equation vanish, the equations for these integrated systems are generally accepted and, moreover, are in full agreement with experiment.

Electromagnetic units
Equations will be given in two common systems of  electromagnetic units (SI and  Gaussian units) and to that end we define the  constant k as follows,
 * $$ k = \begin{cases} {\displaystyle \frac{\mu_0}{4\pi}} & \hbox{for SI units}\\ \\ {\displaystyle \frac{1}{c^2}} & \hbox{for Gaussian units}. \end{cases} $$

Here &mu;0 is the magnetic constant (also known as vacuum permeability). The quantity c is the velocity of light in the vacuum (299&thinsp;792&thinsp;458&thinsp;m&thinsp;s&minus;1 exactly).

Two straight, infinite, and parallel wires
Consider two wires, one carrying an electric current i1, the other i2. Both currents are constant in time; the wires are infinite, straight, and parallel. If the wires are a distance r apart, by Ampère's law the force (per length l of wire) between them is,
 * $$ \frac{F}{l} = 2 k \frac{i_1 i_2}{r}.\, $$

The force F is attractive if the currents run in the same direction and repulsive if they flow in opposite direction.

This equation is used to define the SI unit A (ampere) of current. Take two infinitely long wires in vacuum at a distance r = 1 m, consider the force that one meter of these wires exert on one another (l = 1 m) and let this force be F = 2&sdot;10&minus;7 N (newton). Then for i1 = i2  the current strengths are by definition both equal to one ampere (1 A). In SI units this implies that k = 10&minus;7  N/A2 and hence that the magnetic constant is:
 * &mu;0 = 4&pi; k = 4&pi;&sdot;10&minus;7 N/A2.



Two loops
Let $$\scriptstyle i_1$$ and $$\scriptstyle i_2$$ be electric currents constant in time. They run in separate loops (closed curves) C1 and C2, see figure on the right, where all quantities are defined. The total force between two loops is given by the double path integral over the loops
 * $$ \mathbf{F}_{12} = -k i_1 i_2 \oint_{C_1}\oint_{C_2} \frac{ (d\mathbf{l}_1 \cdot d\mathbf{l}_2)\, \mathbf{r}_{12} }{|\mathbf{r}_{12}|^3}. \qquad\qquad\qquad(1) $$

Alternative expression
One often finds (see, e.g., Ref. ) the following expression for the force between two electric-current-carrying loops:
 * $$ \mathbf{F}_{12} = k i_1 i_2 \oint_{C_1}\oint_{C_2} \frac{ d\mathbf{l}_1 \times( d\mathbf{l}_2\,  \times \mathbf{r}_{12} )}{|\mathbf{r}_{12}|^3}, \qquad\qquad\qquad\qquad\qquad\qquad\qquad (2) $$

instead of the simpler expression in Eq. (1). Here the multiplication signs (&times;) indicate vector products. The integrand [expression under the integral of Eq. (2)] follows from the Biot-Savart-Laplace expression for the magnetic induction B(r1) due to a segment of the second loop. Insertion of B(r1) into the Lorentz force that acts on the current in segment dl1 gives the integrand of Eq. (2).

The labeling of the segments being arbitrary, one would expect the same force (in absolute value) when the labels 1 and 2 are interchanged, or in other words, one would expect Newton's third law  $$\scriptstyle \mathbf{F}_{12} = -\mathbf{F}_{21}$$ (action is minus reaction) to hold. This is not the case, the integrand in equation (2) is non-symmetric under interchange of labels 1 and 2 and hence the integral also appears to be non-symmetric. However, after integration the expression becomes antisymmetric (changes sign under interchange of 1 and 2) and hence satisfies Newton's third law.

To see this we note that the force in Eq. (2) has in fact two contributions, as follows from a result well-known in vector analysis,
 * $$ \frac{ d\mathbf{l}_1 \times( d\mathbf{l}_2\,  \times \mathbf{r}_{12} )} {|\mathbf{r}_{12}|^3} = \frac{(  d\mathbf{l}_1 \cdot  \mathbf{r}_{12} )\, d\mathbf{l}_2}{|\mathbf{r}_{12}|^3}  - \frac{(d\mathbf{l}_1 \cdot d\mathbf{l}_2)\, \mathbf{r}_{12}}{|\mathbf{r}_{12}|^3} . $$

The second contribution gives an integral that is obviously equal to Eq. (1), which is manifestly antisymmetric under interchange of labels (the dot product does not change, the vector r12 changes sign). We will show that the first contribution vanishes after integration over a closed curve.

Write
 * $$ \frac{\mathbf{r}_{12}}{|\mathbf{r}_{12}|^3} = -\boldsymbol{\nabla}_1 \left(\frac{1}{|\mathbf{r}_{12}|}\right) \equiv -\mathbf{A}, $$

where A is a short hand notation. Applying Stokes' theorem, we find that the first contribution becomes
 * $$ -k i_1 i_2 \oint_{C_2}\left[ \oint_{C_1} (d\mathbf{l}_1 \cdot \mathbf{A})\,\right] d\mathbf{l}_2 = - k i_1 i_2 \oint_{C_2} \left[\iint_S (\boldsymbol{\nabla}_1 \times \mathbf{A}) \cdot d\mathbf{S}_1 \, \right]d\mathbf{l}_2. $$

It is well-known in vector analysis that
 * $$ \boldsymbol{\nabla}_1 \times \boldsymbol{\nabla}_1 \Phi = 0 $$

for any scalar function &Phi; and in particular for &Phi; &equiv; 1/|r12|. Hence the first contribution to the force of Eq. (2) vanishes.