Suppose we have some wire of length $2b$ with a current $I$ flowing through it and we want to evaluate the field at some distance $a$ radially outwards from the centre of the wire (so a distance b parallel to the wire from either end).

If we try to approach this using Ampere's law (yes I know you can't do that but we'll get to that), we obtain:

$\int \vec B \cdot \vec dl = \mu_0 I$ (the integral should be closed but I don't think MathJax has that).

For simplicity, we can choose a circular Amperian loop such that $\vec B$ is always parallel to the path. From here, the system appears symmetrical in both the directions parallel and perpendicular to the wire. So surely, by symmetry, $\lVert B\rVert$ will be the same everywhere along the loop and will always act in the same direction relative to the path? If it were different, that would imply some asymmetry rotationally along the axis of the wire which we clearly don't have for the simple case of a straight wire. So $\vec B \cdot \vec dl$ should simplify to just $\lVert B\rVert \lVert dl\rVert$, where $\lVert B\rVert$ should not depend on the point along the Amperian loop.

Well I must have made a mistake somewhere in the above logic because this would imply that I can factor out $\lVert B\rVert$ from the integral and just get:

$\lVert B\rVert \int dl = \mu_0 I$

$\lVert B\rVert 2 \pi a = \mu_0 I$

$\lVert B\rVert = \frac{\mu_0 I}{2 \pi a}$

And I know this can't be true because if we solve the problem using Biot-Savart, we instead obtain:

$\lVert B\rVert = \frac{\mu_0 I}{2 \pi a} \frac{1}{\frac{a^2}{b^2}+1}$

Could someone please point out where I've gone wrong?



EDIT: Is it something to do with Ampere's law implicitly assuming that the wire is infinitely long to start with? I've googled why this fails and the explanations all invoke symmetry arguments but the system looks pretty symmetrical to me...

EDIT 2: In this post I'm assuming that Ampere's law is still applicable for finite wires. My confusion stems from this post Ampere's circuital law for finite current carrying wire in which the answer implies that the only reason Ampere's law fails is due to a lack of symmetry.

  • 1
    $\begingroup$ I am trying to visualize what you are describing but I have not been able to do so. Maybe a drawing would help. How does a divergence-free current start and end $2b$ distance apart? $\endgroup$
    – hyportnex
    Feb 5, 2023 at 18:51
  • $\begingroup$ It isn't divergence free, I'm thinking that's the problem here. $\endgroup$
    – David
    Feb 5, 2023 at 18:53
  • $\begingroup$ In this post there's an explanation that implies that Ampere's law would still work for wires that seemingly break physics and just generate charge out of nowhere at the ends: physics.stackexchange.com/questions/131471/… $\endgroup$
    – David
    Feb 5, 2023 at 18:56
  • $\begingroup$ In the stationary (time-independent) case $\text{div}\mathbf J=0$ and is compatible with Ampere's law $\text{curl}\mathbf H=\mathbf J$, otherwise it is not. $\endgroup$
    – hyportnex
    Feb 5, 2023 at 18:57
  • $\begingroup$ You can use the symbol $\oint$ with \oint $\endgroup$
    – FGSUZ
    Feb 5, 2023 at 19:43

1 Answer 1


Your wire-carrying current is of finite length, and thus current has to start at one end and stop at the other end. This is possible, but then necessarily negative charge accumulates on one end and positive on the other, so we have a polarized charge distribution that grows in time its electric field changes in time. This makes the system break Ampere's law.

James Maxwell noticed this deficiency of the Ampere law and fixed it by a minimal change: by adding an additional term due to time-dependent electric field to the right-hand side of the Ampere equation. The resulting Ampere-Maxwell equation due to Maxwell is

$$ \oint_{\partial S} \mathbf B \cdot d\mathbf s = \int_S \mu_0 \mathbf j \cdot d\mathbf S + \int_S \mu_0\epsilon_0 \frac{\partial \mathbf E}{\partial t} \cdot d\mathbf S. $$

This has been verified to work well ever since (besides fixing the law of magnetostatics in presence of charging/discharging charge distributions it predicted EM waves too). The new term is traditionally called "displacement current", even though in vacuum displacement current needs no real current being present.

The Biot-Savart formula is more general and is applicable to your example, as long as electric field is potential field (gradient of some function of position). As long as the current does not change in time, or changes only very slowly, electric field is the Coulomb field of the charge distribution on the wire, and the Biot-Savart formula gives accurately the magnetic field.


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