Can Ampere's law be applied to 3D loop?

[kelvin490]

Usually in textbooks Ampere's law is just illustrated using 2D loops which forms a plane. Can the law be applied to 3D loops which cannot form a plane surface within the loop?

In a 3D loop which surface should we count when we count the current? In 2D case it is easy because there can only be a plane and the current is just that cut through the plane but if the loop is 3D that may be more then one possible surface formed by the loop, which section of the current should be counted?

 

[Doc Al]

In 2D case it is easy because there can only be a plane

Just because the loop is in a plane does not mean that the surface bounded by the loop is.

 

[rumborak]

If I understand the OP question correctly, I think he is wondering how the law can be applicable for all surfaces that are bounded by the loop. After all, the surface could look like an inflated soap bubble!
Kelvin, the answer lies in the fact that the electromagnetic field is what's called "holomorphic". I essence it means that a surface integral has the same value, no matter the shape of the surface.
Intuitively, if you imagine changing the surface so that it now would encapsulate some more current, because the surface has to eventually come back to meet the boundary, it will inevitably also capture that new current's return, this nullifying its influence on the integral.

 

[Doc Al]

If I understand the OP question correctly, I think he is wondering how the law can be applicable for all surfaces that are bounded by the loop. After all, the surface could look like an inflated soap bubble!

Exactly. The point of my comment is that it doesn't take some 3-D loop to get to his question. Even the simplest planar circle will bound an infinite number of oddly-shaped surfaces.

 

[kelvin490]

If I understand the OP question correctly, I think he is wondering how the law can be applicable for all surfaces that are bounded by the loop. After all, the surface could look like an inflated soap bubble!
Kelvin, the answer lies in the fact that the electromagnetic field is what's called "holomorphic". I essence it means that a surface integral has the same value, no matter the shape of the surface.
Intuitively, if you imagine changing the surface so that it now would encapsulate some more current, because the surface has to eventually come back to meet the boundary, it will inevitably also capture that new current's return, this nullifying its influence on the integral.

I think you both get the point. To me the difficulty is, how to prove for any possible surfaces bounded by the loop, the net current passes through these surfaces would be exactly the same? Image there can be a lot of current carrying wires in different orientations, and there are a lot of possible surfaces bounded by planar or 3D loops.

 

[jtbell]

how to prove for any possible surfaces bounded by the loop, the net current passes through these surfaces would be exactly the same?

Any proof must start from a set of assumptions. Which ones do you want to use?

Note that if you take Maxwell's equations as the foundation of electrodynamics, Ampere's Law is itself a fundamental assumption.

See also these earlier threads which I turned up with a Google search for "ampere's law proof":

https://www.physicsforums.com/threads/ampere-circuital-law-proof.709347/
https://www.physicsforums.com/threads/amperes-law-derivation.307609/

(there are probably other similar threads here)

 

[kelvin490]

Any proof must start from a set of assumptions. Which ones do you want to use?

Note that if you take Maxwell's equations as the foundation of electrodynamics, Ampere's Law is itself a fundamental assumption.

See also this earlier thread:

https://www.physicsforums.com/threads/ampere-circuital-law-proof.709347/

(there are probably other similar threads here)

I think what confuses me is, since there can be many different possible surfaces bounded by a loop. Therefore, I don't know which portion of current should be counted if there are many electric wires orientated (and may be curved and twisted) differently in the space. It doesn't seem like an assumption itself.

 

[jtbell]

First you choose a particular surface that spans the loop. Then, every time a wire pierces that surface, you "count" its current in the total I_enclosed. The current "counts" as positive if it goes through in one direction, negative if in the other direction (with respect to the surface).

In order for the usual basic version of Ampere's law $$\int {\vec B \cdot d \vec l} = \mu_0 I_\textrm{enclosed}$$ to "work", your currents must always form closed loops, and not "appear" or "disappear" anywhere.

 

[kelvin490]

First you choose a particular surface that spans the loop. Then, every time a wire pierces that surface, you "count" its current in the total I_enclosed. The current "counts" as positive if it goes through in one direction, negative if in the other direction (with respect to the surface).

In order for the usual basic version of Ampere's law $$\int {\vec B \cdot d \vec l} = \mu_0 I_\textrm{enclosed}$$ to "work", your currents must always form closed loops, and not "appear" or "disappear" anywhere.

If we do the same thing with another surface (say S2) that spans the loop, and every time a wire pierces that surface, you "count" its current in the total I_enclosed. Can we be sure the I_enclosed counted will be the same as that you did for the previous surface S1?

 

[Doc Al]

I recommend that you read up on Stokes' theorem. That might give you some insight into how it works.

 

[kelvin490]

I recommend that you read up on Stokes' theorem. That might give you some insight into how it works.

Thanks. This reasoning may be useful, we just need to assume current works like incompressible fluid, density doesn't change so volume flow in equals to volume flow out. Imagine curve C is simple planar loop. Now consider two different shapes for S: first is a simple disc aligned with C, and second is a hemisphere whose open end is aligned with C. Any wire that cuts through the first surface must also cut through the second - otherwise the wire would have to be cut short inside the hemisphere, but if that was the case then it wouldn't carry any current. We just need to assume current works like incompressible fluid, density doesn't change so volume flow in equals to volume flow out. In case of current no charge is accumulated.

 

[Doc Al]

Here's how I like to think of it. You have to start somewhere, so I assume that Ampere's law works for a tiny square element of area. Now take any arbitrary loop and any arbitrary surface bounded by that loop. Divide the surface into a zillion little squares. Then imagine yourself doing the line integral around each square and adding them up. Note that adjacent squares will share a boundary and that the line integral over that shared boundary must cancel, since the direction will be opposite for each square. The only place where the line integral will not vanish is along the original arbitrary loop. And that is the same regardless of the surface chosen.

A bit handwavy, but I hope that helps a bit.

 

[rumborak]

We just need to assume current works like incompressible fluid, density doesn't change so volume flow in equals to volume flow out. In case of current no charge is accumulated.

Density is irrelevant, since it all comes down to the number of charged particles moving through the surface. Keep in mind, current is defined as charge moving over time (I = Q/T).
But, to your other point, yes, Ampere's law rests on a few assumptions such as non-accumulating charges. The Wikipedia article addresses these:

http://en.wikipedia.org/wiki/Ampère..._formulation_of_Amp.C3.A8re.27s_circuital_law