Maybe this question sounds easy, but I would really be grateful if someone could give me an explanation if it's true or not. I'm not sure if I understand it good enough.

  • $\begingroup$ Why not use the Biot–Savart law? $\endgroup$ – Cinaed Simson Jun 5 at 1:20
  • $\begingroup$ @CinaedSimson It depends on if the OP is actually wanting to calculate the field due to this current, or if they are really just wondering if Ampere's law could be used to do it $\endgroup$ – Aaron Stevens Jun 5 at 12:58

No, you can't use Ampere's Law directly to find $\vec{B}$ around a short segment of current. But it's for a subtle reason.

Ampere's Law in integral form is $$ \oint_\text{loop} \vec{B} \cdot d\vec{l} = \mu_0 \int_\text{surface} \vec{J} \cdot d\vec{a} $$ for any loop and any surface for which the loop forms the boundary. The quantity on the right-hand side is just the amount of current passing through the surface.

If you try to do this for a finite segment of current, you run into a problem: the right-hand side varies with respect to the surface you choose! You could draw a circular loop around the middle of the segment, and count all the current passing through the flat circular disk spanned by the circle. In this case, you get all of the current in the segment. But it's equally valid to pick a surface that bulges out like a balloon and doesn't intersect the segment at all. In this case, the enclosed current is zero.

The reason that this is ill-defined is that the current can't flow along a finite segment in a "steady" fashion: what do the charges do when they gets to the end of the segment? There are two main options. The charges could disappear when they gets to the end of segment; but this violates conservation of charge. Alternately, the charges could pile up at the ends of the wire; but then you have an electric field that's changing with time, and you have to include Maxwell's "displacement current" term in Ampere's Law.

(In more mathematical terms, a segment of wire doesn't obey the continuity equation $\nabla \cdot \vec{J} = 0$. If you're familiar with the differential form of Ampere's Law, $\nabla \times \vec{B} = \mu_0 \vec{J}$, see if you can figure out why you have to have $\nabla \cdot \vec{J} = 0$ to use this equation.)


Ampere's law is always true, but it cannot always be used to determine the magnetic field of a general current distribution (this has nice parallels with using Gauss's law as well).

Ampere's law in integral form is:

$$\int\mathbf B\cdot\text d\mathbf l=\mu_0I_{\text{enc}}$$

where $I_\text{enc}$ is the current that flows through the surface bounded by the closed curve of the line integral. The issue is that the magnetic field appears inside the integral, which we can think of as a sum. In general, knowing what a sum evaluates to tells you nothing about what each of the terms in the sum is.

The analogy I like to give is the following: Let's say I tell you I have ten numbers that add up to $100$ (this is analogous to knowing the enclosed current). If this is all you know, then you would not be able to tell me what the ten numbers (i.e. the field) actually are. However if I tell you that all of the numbers are equal (this is like having certain symmetries in your system), then you can tell me that each number is equal to $10$.

So what symmetry do we need in order to determine the field from Ampere's law. There are really just four main cases (and maybe some variations on these that do not come to mind).

  1. Infinite line currents (including 1D lines or cylinders)
  2. Infinite sheet currents
  3. Infinitely long solenoids
  4. Torroidal solenoids

In each of these cases you can be smart with your choice of Amperian loop such that along sections of the loop either the field is constant and $\mathbf B\cdot\text d\mathbf l=B\ \text dl$ or $\mathbf B\cdot\text d\mathbf l=0$. This causes the integral on the left hand side of Ampere's law to become $$\int\mathbf B\cdot\text d\mathbf l=\int B\ \text d l=B\int\text dl=BL$$ where $L$ is the total length of the loop around which $\mathbf B$ and $\text d\mathbf l$ were not orthogonal. Therefore $$B=\frac{\mu_0I_\text{enc}}{L}$$ where $I_\text{enc}$ will somehow depend on $L$ so that the field does not depend on what $L$ actually is.

The problem with the short current element is that we no longer have nice symmetries to where we can easily pick a loop where the field has a constant magnitude and a "nice" direction with respect to the chosen loop. Another issue is given in Michael's answer.

I do know that Griffith's treats the problem of the field due to a current segment with the Biot-Savart Law (it is derived in the process of the example of using the Biot-Savart law for an infinite line charge). It is a fairly straightforward process, so if you are wanting to learn how to determine the field in this situation I would suggest looking there.

  • $\begingroup$ You can't even use Ampere's Law to find the field at points in the plane bisecting the wire, for a different reason. See my answer. $\endgroup$ – Michael Seifert Jun 4 at 21:14
  • $\begingroup$ @MichaelSeifert Good point. I'll remove that from my answer $\endgroup$ – Aaron Stevens Jun 4 at 23:25

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