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?

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    $\begingroup$ 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. $\endgroup$ Jul 14, 2015 at 12:58

1 Answer 1


Although Ampere's law and the Biot-Savart law are magnetostatic approximations, they will still approximately hold when the charge is moving slowly enough.

The problem is you're just naively using an Ampere's law result derived for a steady current to a situation with a moving charge.

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 $$ \vec{B} = \frac{\mu_0}{4\pi}q\frac{{\vec{v}\times\hat{r}}}{r^2}.$$

  • $\begingroup$ A school textbook describes ampere's law something like this: A.C.L assumes an open surface with a boundary. The surface has current passing through it. We consider the boundary to be made up of a number of small line elements, dl. we take the scalar products of the magnetic feilds with all such dl's and add them together in an integral. this integral is equal to u.I. The textbook never mentioned a steady current. Even an electron moving is a current, isn't it? Also, the book did not mention infinite wire. So is the book incorrect? Or will what it says hold for finite wire too? It should not $\endgroup$
    – Prem
    Jul 14, 2015 at 16:06
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    $\begingroup$ There is as assumption of steady current, otherwise the book's explanation doesn't make sense. What if you try to use Ampere's law to calculate the magnetic field before the electron enters the surface? The electron hasn't passed through, so no current has passed through the surface, and we'd find zero magnetic field. In general when E&M books first talk about magnetism induced by current, they'll be looking at steady state or nearly steady state systems. $\endgroup$ Jul 14, 2015 at 16:37
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    $\begingroup$ I was also wrong, Ampere's law does not need an infinite wire to hold. However, Ampere's law isn't very useful for any arbitrary configurations of current because a line integral is just a magnitude. Once you take the line integral of the B field, you lose information about direction of the magnetic field or the magnitude of the field at a specific point in your loop. $\endgroup$ Jul 14, 2015 at 16:38

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