What are the interpretations of ampere's circuital law for a single moving electron?

I understand from Ampere's circuital law that when a current passes though any open surface with a boundary (a circular boundary, for simplicity), then limit of summation of the scalar products of magnetic field and a small length 'dl' on the circumference yields an integral which is equal to '$\mu$I'.

Or, $$\int B.dl = \mu.I$$ Do I understand it right?

If yes, then it seems to contradict the Bio-savart law, which says that $$B= \frac{\mu} {4\pi}.\frac{I X dl}{r^3}$$ Because, supposing there is just one electron or a small heavily charged fat globule causing the magnetic field, what does ampere's circuital law predict about the magnetic field around it in this case, and how? To me, Bio-savart law and ampere's law seem to contradict here.

If we assume an circular surface of radius r through which the charged body just passed, then this implies that at the boundary of the surface, at any single point, $$B=\frac{\mu.I}{2\pi.r}$$ ,which is, I know, incorrect. So what actually does ampere's circuital law says, because I am assuming it does not makes a false statement?

• Raja, I'm a little confused, so I'll comment rather than answer. See Wikipedia for Ampère's circuital law and the Biot-Savart law. Also remember that the electron has an electromagnetic field, and motion is relative. If you were motionless with respect to the electron, you might claim it had an electric field. If you moved around it in a circular fashion, you might claim it had a magnetic field. But your motion doesn't change the electron's field. Just the way you see it. Jul 14, 2015 at 12:58

Ampere's circuit law comes from $$\nabla \times \vec{B} = \mu_0 \vec{J} .$$ You then simply take a surface integral on both sides, and using Stokes theorem you find $$\oint \vec{B}\cdot d\vec{l} = \int\int_S \mu_0\vec{J}\cdot dS.$$ For the familiar case of a steady current, the surface integral of $\vec{J}$ is just $I_{enc}$. But in this case, you're dealing with a point charge moving in space, and your current density has to be written as a dirac delta function $\vec{J}=q\delta(x-vt)\delta(y)\delta(z)v\hat{x}$. Integrating this term is difficult, but in the end it should give you a result consistent with what you would get from the Biot-Savart law, which gives the field of a slow-moving point charge as $$\vec{B} = \frac{\mu_0}{4\pi}q\frac{{\vec{v}\times\hat{r}}}{r^2}.$$