3

Why doesn't Ampère's law hold for short current carrying wires? Of course, such wires should be part of a closed circuit, but that's a physical fact, and there is a numerous amount of ways to close it. Ampere's law, on the other hand, is merely a mathematical law.

The problem is Ampère's law can't really "tell" the difference between infinite wires and finite wires. Such law states: $$\oint \mathbf{B} \cdot d\mathbf{l}=\mu_0I$$ As far as I know there is no restriction on the integration curve as long as it is closed. Then I could pick a circle away from the wire, centered at its axis, oriented in its direction. The integral then only depends on $\theta$, and $\mathbf{B}$ comes out because it does not (there is azimuthal symmetry). With this logic (which is evidently flawed), $\mathbf{B}$ is the same regardless of its wire axis coordinate (which is true for an infinite wire, but not in this case). My assumptions here were:

  1. $\mathbf{B}$ doesn't depend on $\theta$
  2. $\mathbf{B}$ goes in the azimuthal direction. Biot-Savart law forces it to be perpendicular to the wire ($I$ goes in that direction) and to the radial component ($\mathbf{r}$ has a component parallel to $\mathbf{B}$ and another one to the radial component for every segment of wire), because the direction of the $\mathbf{B}$ field generated by each small segment of wire is in the direction of $\mathbf{B}\times\mathbf{r}$.
  3. The closed amperian loop can be arbitrarily chosen.
  4. Ampère's law works for finite, open, line currents.

To finish, in Griffith's Introduction to Electrodynamics (4th ed., p. 225) the author derives a formula for a finite straight current carrying wire using the Biot-Savart law:

$$B=\frac{\mu_0 I}{4\pi s}(\sin{\theta_2-\sin{\theta_1})}$$

This is only the magnitude. But above, he writes:

In the diagram, $(d\mathbf{l'}\times \mathbf{\hat{r}})$ points out of the page(...)

Image extracted from Griffith's "Introduction to electromagnetism"

Which, by the azimuthal symmetry of the problem, implies that the magnetic field goes in the azimuthal direction. That would leave assumptions 3 and 4 as the sources of the absurdity.

Emilio Pisanty
  • 132,859
  • 33
  • 351
  • 666
sbs95
  • 153
  • Subject appart, was this question ok? Was it too long? Any advice on that matter? Thank you. – sbs95 Nov 07 '16 at 20:40
  • 1
    I'm not sure I understand. For a finite piece of wire that just ends somewhere, you could use a surface for the integration that doesn't contain the wire, so the current would be 0. This means that the value of $\oint \mathbf B \mathrm d\mathbf l$ would depend on the surface that you integrate on, for the same closed circuit of integration. This wouldn't make sense. I guess my point is that you need continuity of charge for Ampère's law to be meaningful (since the right hand side refers to "currents crossing the surface"). – Andras Deak Nov 07 '16 at 20:45
  • Yes, you're right about the surface. That's exactly the problem. But what does "continuity of charge" mean, exactly? Maybe that Ampere's law only makes sense for closed or infinite currents? Do you know where exactly that hypothetical is included, mathematically? I mean, the law in differential form always applies, right? Even in this case. So maybe Stoke's theorem? – sbs95 Nov 07 '16 at 20:53
  • 1
    I believe that it only makes sense for closed or infinite currents, yes. In keeping with the continuity equation: charge can't disappear. I'd think the same issue is there with Stokes: the integral of $\mathrm B$ is a given well-defined value, but if you transform this into the surface integral of the curl, the result might depend on the surface chosen. Unfortunately I don't see the mathematical background, this is mostly heuristic to me. But to me it has to do with the conservation of charge (density). – Andras Deak Nov 07 '16 at 20:57
  • Charge isn't really disappearing, I think. In any case, it would be jumping from one end of the wire to the other. The current is steady, and even though its not physical, what does it matter? This is a mathematical model, right? Its useful to explain real world events, but it also extends to some nonphysical scenarios (such as an infinite wire). That's why you can, for instance, calculate the field of such a short wire using the Biot-Savart law. Just why can't we use Ampere's law? – sbs95 Nov 07 '16 at 21:07
  • Again, I'm mostly speculating. But it matters that "it's not physical", since Ampère's law is physical, not mathematical. Biot$\text{--}$Savart works, since that's inherently differential, which you can add up. But Ampère is inherently integral to me. Think of it like this: Gauss' law for $\mathrm B$ only works in our universe, where there are no magnetic monopoles. If there were, the closed surface integral of $\mathrm B$ wouldn't be zero independent from the shape and position of the integration surface. There's probably a formulation somewhere that derives Ampère partly from continuity. – Andras Deak Nov 07 '16 at 21:13
  • @sbs95, re "Charge isn't really disappearing, I think. In any case, it would be jumping from one end of the wire to the other": that's 100% magical thinking, and it doesn't mean anything. If it's physically moving back from one end to the other, then there's a second current involved. If it's teleporting from one spot to the other, then I'd like to borrow your teleportation machine - not least because it would allow me to break the law of local conservation of charge, which would be a huge break in known physics. – Emilio Pisanty Nov 07 '16 at 21:33
  • Ampere's law in differential form is a direct consequence of Biot-Savart. The thing is, when taking the current's flux through any surface with the same boundary, the answer cannot change for the integral form of Ampere to hold. That's purely mathematical. You're saying it's simpler to think about it heuristically, but why the fact that Ampere is integral makes it "physical" (dependent on other physical properties of the universe, such as the continuity of charge)? It sounds interesting to me, I just don't understand. – sbs95 Nov 07 '16 at 21:33
  • @EmilioPisanty What I meant was that Biot-Savart law on its own doesn't hold any information about conservation or continuity of charge. 100% thankful for the aggression. – sbs95 Nov 07 '16 at 21:47
  • @sbs95 No aggression meant - apologies if it came off like that. Indeed, the Biot-Savart law does not know about conservation of charge - which is why one needs to be extra careful to only use it on its domain of validity, which is for static situations with no accumulation of charge and with closed circuits - there is simply no evidence, either theoretical or experimental, that it holds for finite segments, because they don't make sense on static situations or as experimental widgets. For more details, see my answer. – Emilio Pisanty Nov 07 '16 at 22:00
  • Yes, your answer was extremely helpful. Thank you very much! I guess I'll try not to sound so magical next time. – sbs95 Nov 07 '16 at 22:02
  • No worries. These are subtle issues, and it takes a while to get used to them. – Emilio Pisanty Nov 07 '16 at 22:08

2 Answers2

4

There is no such thing as a "finite, open line current". If you did have such a thing, the ends of the line segment would accumulate charge at a linear rate, and you are no longer on a static situation, and you can no longer apply the magnetostatic Ampère's law.

This isn't being nit-picky, either, because you need the current to obey the continuity equation in order for Ampère's law to even make sense. This is most cleanly stated in the integral version with a volumetric current density $\mathbf J(\mathbf r)$, for which Ampère's law reads $$ \oint_\mathcal C \mathbf B\cdot\mathrm d\mathbf l = \mu_0\int_\mathcal{S} \mathbf J\cdot \mathrm d\mathbf S, $$ for any closed curve $\mathcal C$ and any surface $\mathcal S$ whose boundary is $\mathcal C$. For this to be true, you require the integral on the right to be independent of the choice of surface, and by the standard argument this is equivalent to requiring that the flux of the current density out of any closed suface be zero: $$ \oint_\mathcal{S} \mathbf J\cdot \mathrm d\mathbf S=0. $$

This requirement fails for a current-carrying finite line segment, because the current flowing into a sphere about either end of the segment is nonzero. For such a current, then, Ampère's law makes no sense.

(Since you explicitly mention the differential form of Ampère's law, here goes that version of the argument: you have $\nabla \times \mathbf B = \mu_0 \mathbf J$, so therefore $\nabla \cdot \mathbf J = \nabla \cdot(\nabla \times \mathbf B) \equiv 0$, and $\mathbf J$ obeys the differential stationary continuity equation. There is no way to reconcile this with a finite current-carrying segment.)

So where does this leave the current-carrying line segment? It's a really neat concept, and it mostly doesn't lead us astray, so what gives?

Well, mostly, it doesn't lead us astray because we use finite segments correctly: as part of a multi-part curve $\mathcal C = \mathcal C_1 \oplus \mathcal C_2 \oplus\cdots\oplus \mathcal C_N$, where the endpoints overlap correctly, pair by pair and back again from $\mathcal C_N$ to $\mathcal C_1$, and no charge is either appearing or disappearing - there is no surface with a nonzero current flux.

That's a legit circuit, and Ampère's law holds for it. The way to bring the analysis down to each individual segment is (exclusively!) through the Biot-Savart law, which is provably the solution to the magnetostatic Ampère's law, and which reads for this circuit $$ \mathbf B(\mathbf r) = \frac{\mu_0}{4\pi} \int_{\mathcal C} \frac{I\mathrm d\mathbf l \times(\mathbf r-\mathbf r')}{\|\mathbf r-\mathbf r'\|^3} = \frac{\mu_0}{4\pi} \sum_{n=1}^N \int_{\mathcal C_n} \frac{I\mathrm d\mathbf l \times(\mathbf r-\mathbf r')}{\|\mathbf r-\mathbf r'\|^3} . $$

For such a circuit, it is convenient to split the field up into components, defined as $$ \mathbf B_n(\mathbf r) = \frac{\mu_0}{4\pi} \int_{\mathcal C_n} \frac{I\mathrm d\mathbf l \times(\mathbf r-\mathbf r')}{\|\mathbf r-\mathbf r'\|^3} , $$ and which we interpret as the field produced by the individual segment $\mathcal C_n$. However, these fields are completely in our heads: there are no physical current configurations that produce the $\mathbf B_n(\mathbf r)$, and there is no way to make sense of them beyond the Biot-Savart law (and certainly no way to make sense of them using Ampère's law).

Emilio Pisanty
  • 132,859
  • 33
  • 351
  • 666
  • 1
    This answer is extremely clear, thank you very much. So Biot-Savart on its own doesn't mathematically need for the current to be steady (even though its true only for steady currents), and that's why you can use it as a "tool" to calculate the "contribution" from a single segment of wire, as long as in reality it's part of a closed loop. Ampere's law, on the other hand, is actually a theorem derived from two physical hypothesis: Biot-Savart itself and the continuity equation of current. – sbs95 Nov 07 '16 at 21:59
  • 1
    Yeah, pretty much. You can calculate the Biot-Savart integral for whatever circuit you like, but you only get a meaningful physical result if you give it a physical circuit (or, as shown, if you re-assemble the parts after the calculation). – Emilio Pisanty Nov 07 '16 at 22:06
  • 1
    The view that Ampère's law is a theorem derived from Biot-Savart is not incorrect (and you don't actually need to give it the continuity equation - that one is baked into the current in the Biot-Savart integrand). However, it's important to realize that the two are actually equivalent, and that for fully-grown electrodynamics, it's actually Ampère's law that's more fundamental, because it generalizes to the full Ampère-Maxwell law in dynamic situations. The full view of EM sees Ampère's law as the static limit of the Maxwell equations, and the Biot-Savart law as its solution. – Emilio Pisanty Nov 07 '16 at 22:06
  • 2
    It's easier for me to grasp with electricity, of course (Couloumb and Gauss). But I understand, I'll have to keep on reading then. Thank you very much for your time. – sbs95 Nov 07 '16 at 22:14
0

It cannot hold as you stated because for stationary fields $\frac{\partial E}{\partial t}=0$ and $\frac{\partial H}{\partial t}=0$ you get $ \mathrm{curl} H = J$, $\mathrm{div} H = 0$ and $\mathrm{div} E = \rho / \epsilon $ and then $\mathrm{div} J = -\frac{\partial\rho }{\partial t}=0$, so that $\mathrm{div} J = 0$, in other words the $J$ lines must be closed for stationary currents.

hyportnex
  • 18,748