$$\vec{F}_\text{mag} = \frac{\mu_0 I_1 I_2}{4\pi } \oint\oint\frac{\mathrm{d}\vec{l}_1\times(\mathrm{d}\vec{l}_2\times\hat{r})}{r^2}$$

Is there any derivation for Ampère's force law or is it just derived empirically?

If there is any derivation please include it in your answer.

But if it is derived empirically then Ampère must have made this formula from the following facts:

  1. $F_\text{mag} \propto I_1$

  2. $F_\text{mag} \propto I_2$

  3. $F_\text{mag} \propto \frac{1}{r^2}$

  4. $F_\text{mag} \propto \oint\oint[\mathrm{d}\vec{l}_1\times(\mathrm{d}\vec{l}_2\times\hat{r})]$

The first three can easily be seen experimentally:

  1. The current in the first circuit is directly proportional to magnetic force.

  2. The current in the second circuit is directly proportional to magnetic force.

  3. The square of the distance is inversely proportional to magnetic force.

But what about the last one?

How did Ampere came to the conclusion that this complicated math stuff $\oint\oint[\mathrm{d}\vec{l}_1\times(\mathrm{d}\vec{l}_2\times\hat{r})]$ is directly proportional to magnetic force?

What if the magnetic force is larger for a smaller value of $\oint\oint[\mathrm{d}\vec{l}_1\times(\mathrm{d}\vec{l}_2\times\hat{r})]$ and the magnetic force is smaller for a larger value of $\oint\oint[\mathrm{d}\vec{l}_1\times(\mathrm{d}\vec{l}_2\times\hat{r})]$?

  • 1
    $\begingroup$ For a derivation from Special Relativity see Purcell or Schwartz. $\endgroup$
    – jinawee
    Commented Feb 18, 2014 at 13:57
  • $\begingroup$ I've attempted to convert the math into MathJax, our standard math format for this site, but some of the characters in your original post were unreadable. If I omitted anything, please point it out in a comment, or make another edit to fix the MathJax yourself. $\endgroup$
    – David Z
    Commented Feb 18, 2014 at 14:52
  • $\begingroup$ Thanks for the edit.Sorry I didnt know how to use MathJax.I just copied and pasted the formulas and everything was messed up. You havent missed anything.The question is exactly as I want it to be. $\endgroup$
    – user40833
    Commented Feb 18, 2014 at 16:34
  • $\begingroup$ I am waiting for answers.I dont want anybody to relate it to special relativity as it is the extension of lorentz transformation which itself arises due to the above puzzle in the question $\endgroup$
    – user40833
    Commented Feb 18, 2014 at 16:43
  • $\begingroup$ I changed the question title since 'doubt' and 'confusion' could mean anything. $\endgroup$
    – Wouter
    Commented Feb 18, 2014 at 16:46

2 Answers 2


This is a combination of the law of Biot-Savart $$ \vec{B}(\vec{r}_2) = \frac{\mu_0 I_1}{4\pi}\oint_{\vec{r}_1\in{\rm Wire}_1} \frac{d \vec{r}_1\times (\vec{r}_2-\vec{r}_1)}{|\vec{r}_2-\vec{r}_1|^3} $$ and the formula for the Lorentz force $$ d\vec{F}_2 = I_2\cdot d\vec{r}_2 \times \vec{B}(\vec{r}_2) $$ where $\vec{r}_2$ is a point on the second wire and $d\vec{r}_2$ a corresponding path element. The path integral over the second wire gives your formula.

Biot-Savart's law

From $\def\div{\operatorname{div}}\def\rot{\operatorname{rot}}\def\grad{\operatorname{grad}}\div(\vec{B})=0$ follows the existence of some vector potential $\vec{A}$ with $\vec{B}=\rot\vec{A}$. Substituting this into Ampere's law (for steady state)

$ \rot(\vec{H}) = \vec{S} $

$ \rot(\vec{B}) = \mu_0\vec{S} $


$ \rot(\rot \vec{A}) = \mu_0 \vec{S} $

With the formula $\rot\rot\vec{A}=\vec{\nabla}\times(\vec{\nabla}\times\vec{A})=\vec{\nabla}(\vec{\nabla}\cdot\vec{A})-(\vec{\nabla}\cdot\vec{\nabla})\vec{A}=\grad\div\vec{A}-\Delta\vec{A}$ and the Coulomb gauge condition $\div\vec{A}=0$ one obtains

$ \Delta\vec{A} = -\mu_0\vec{S} $

For the free-space problem this equation can be solved with the help of Green's function of the Laplacian

$$ \vec{A}(\vec{r}) = -\frac{\mu_0}{4\pi}\int_{\vec{r}_1\mathbb{R}^3} \frac{\vec{S}(\vec{r_1})}{|\vec{r}_1-\vec{r}|}d V_1 $$ Using $\vec{B}=\rot\vec{A}$ one gets the flux density

$\displaystyle \vec{B}(\vec{r}) = -\frac{\mu_0}{4\pi}\int_{\vec{r}_1\mathbb{R}^3} \rot_{\vec{r}}\left(\frac{\vec{S}(\vec{r_1})}{|\vec{r}_1-\vec{r}|}\right)d V_1 $

$\displaystyle \phantom{\vec{B}(\vec{r})} = -\frac{\mu_0}{4\pi}\int_{\vec{r}_1\mathbb{R}^3} \grad_{\vec{r}}\left(\frac{1}{|\vec{r}_1-\vec{r}|}\right)\times \vec{S}(\vec{r_1})d V_1 $

$\displaystyle \phantom{\vec{B}(\vec{r})} = \frac{\mu_0}{4\pi}\int_{\vec{r}_1\mathbb{R}^3} \frac{(\vec{r}-\vec{r}_1)\times \vec{S}(\vec{r_1})}{|\vec{r}-\vec{r}_1|^3}d V_1 $

For the integral over the cross-section area of the wire the changes in $r_1$ are neglected and $\int_{A_{\rm cross}}\vec{S}d V$ is set to $I_1 d \vec{r}_1$.

This gives Biot-Savart's law

$\displaystyle \vec{B}(\vec{r}) = \frac{\mu_0 I_1}{4\pi}\int_{\vec{r}_1\in \rm Wire_1} \frac{(\vec{r}-\vec{r}_1)\times d \vec{r}_1}{|\vec{r}-\vec{r}_1|^3}. $


We know that the Coulomb inverse square law between static charges is correct, because it was checked experimentally to the highest possible accuracy. When charges move however, the force is modified due to the limited propagation speed of the action of one charge on the other- another experimental fact. If you take this fact into account and put the Coulomb potential in the retarded potential integral, you get all of Maxwell equations- including Ampere, Biot-Savart and the rest. These are included in additional terms appearing in the integral representing; magnetism(proportional to the speed and normal to it, and inversely with the separation distance) and radiation(proportional to acceleration and inversely, not inverse square, with the separation distance. See literature on the retarded potential for the complete derivation.


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