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In the realm of magnetostatics, consider the integral form of Ampere's law:

$$ \oint_C \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{enclosed}$$

What I realized is when asked the question "what is the enclosed current enclosed by?"

The most common answer I get is "enclosed by the Amperian loop of course!"

I think this is a huge misconception, because if we look at how the integral form of Ampere's law is derived (in quasistatic situations): $$\nabla \times \mathbf{B} = \mu_0 \mathbf{J} \longrightarrow \iint_S (\nabla \times \mathbf{B}) \cdot d\mathbf{a} = \mu_0 \iint_S \mathbf{J} \cdot d\mathbf{a} \longrightarrow \oint_C\mathbf{B} \cdot d\mathbf{a}= \mu_0 \iint_S \mathbf{J} \cdot d\mathbf{a}$$

In other words, the answer should be that the current is enclosed by the surface BOUNDED by the Amperian loop, because of the surface integral.

However, I notice that this definition of enclosed current is not without issues, because if we consider the situation below:

enter image description here

Both surfaces $S_1$ and $S_2$ are enclosed by the same Amperian loop, however, one may argue that the surface $S_2$ "encloses" more current than the surface $S_1$. But we know this is not true because the magnetic field for both cases should be the same, since it is the same line integral.

To resolve this, we may argue that for surface $S_2$, the current outside the Amperian loop is "not really enclosed", since it penetrates from outside the surface and exits, so the net contribution to the surface integral is zero.

But all I need to do is shade the Amperian loop to make it a closed surface, and the same argument can be applied, that the current passing through inside the Amperian loop is "not really enclosed" as well.

I think I am hugely misunderstanding something but I am not sure what it is.

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You have highlighted the fact that you can choose *any (well belaved) surface as long as it is bounded by the Amperian loop which means that $\displaystyle \mu_0 \iint_{S_1} \mathbf{J} \cdot d\mathbf{a}=\mu_0 \iint_{S_2} \mathbf{J} \cdot d\mathbf{a} = . . . . . =\mu_0 \iint_{S_{\rm n}} \mathbf{J} \cdot d\mathbf{a} = \, . . . . .$

The analogy which is often used is that the Amperian loop and the surface are equivalent to a butterfly net.

enter image description here

Once the direction of integration has been chosen, clockwise in this case, the direction of the normals to the surface is defined by the right-hand rule, so in the diagram above the normals are pointing "outwards, from the surface.

Consider the surfaces defined in your diagram with normals to the surfaces being shown.

enter image description here

Surface $S_1$ has all the contributions from $\mathbf{J} \cdot d\mathbf{a}$ being positive.

For surface $S_2$ there are positive (blue normal) and negative (red normal) to the integral. The negative contributions cancelling out some of the positive contributions to make the integral the same as for surface $S_1$.
One way to visualise this is to imagine areas projected onto a plane perpendicular to $\mathbf J$.

Often the simplest surface to consider is the plane defined by the Amperian loop $S_0$ where the normals are all parallel to one another and to $\mathbf{J}$ which makes the integration easier to do with $\displaystyle \mu_0 \iint_{S_{\rm n}} \mathbf{J} \cdot d\mathbf{a} =\mu_0 \iint_{S_{\rm 0}} \mathbf{J} \cdot d\mathbf{a}$.

If you think about it in simple terms then the term $\mathbf{J} \cdot d\mathbf{a}$ is the same as $J\,da\,\cos \theta$ where $da\,\cos \theta$ is the projected area onto a plane and the sum of the areas will be the same for positive and negative contributions to the integral. I have tried to illustrate this below.

enter image description here

The term $\mathbf{J} \cdot d\mathbf{a}$ relates to a flux of charge through an area.
If no charge accumulates in the volume bounded by areas $S_0$ and $S_2$ then the flux of charge through area $S_0$ into the volume must be the same as the flux through area $S_2$ out of the volume.

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  • $\begingroup$ Very nice. How do we know with certainty that after the negative contributions cancel out with the positive, regardless of surface chose, then we will always obtain that of $S_0$? $\endgroup$
    – D. Soul
    Aug 6 '20 at 8:54
  • $\begingroup$ I am sure that a Mathematician can provide a general proof but if you think about it in simple terms then the term $\mathbf{J} \cdot d\mathbf{a}$ is the same as $J\,da\, \cos (\theta)$ where $da\,\cos(\theta)$ is the projected area onto a plane and the sum of the areas will be the same for positive and negative contributions to the integral. The term $\mathbf{J} \cdot d\mathbf{a}$ relates to a flux of charge trough an area. If no charge accumulates in the volume bounded by areas $S_0$ and $S_2$ then the flux of charge through area $S_0$ must be the same as the flux through area $S_2$. $\endgroup$
    – Farcher
    Aug 6 '20 at 9:50
  • $\begingroup$ @D.Soul I have added to my original answer. $\endgroup$
    – Farcher
    Aug 6 '20 at 13:41
  • $\begingroup$ Lovely! Accepted your answer! $\endgroup$
    – D. Soul
    Aug 7 '20 at 3:09
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"But all I need to do is shade the Amperian loop to make it a closed surface," That doesn't work. The surface bounded by the closed loop always has to be an open surface. What you have produced is two surfaces for the current to go through, so you are just doing it Ampere's law twice.

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You can think of the currents enclosed by the Amperian loop $C$ as the currents that go through whatever surface $S$ -no matter how you deform it, as long as you don't rip holes in it- that is enclosed by $C$.

From a topological point of view, the Amperian loop $C$ along which you are computing the integral and the loop along which an enclosed current is passing are concatenated, like two adjacent links of a chain: you cannot move them apart without them intersecting one another.

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  • $\begingroup$ Isn't the loop $C$ and the loop in which then enclosed current is passing the exact same loop? $\endgroup$
    – D. Soul
    Aug 6 '20 at 9:04
  • $\begingroup$ No they are two distinct loops: the latter is the physical circuit along which current is flowing, while $C$ is the loop along which you are computing the integral $\oint_C \bf{B} \cdot d\bf{l}$: it doesn't need to correspond to a physical loop, it is just a path of integration $\endgroup$
    – Luca M
    Aug 7 '20 at 12:28
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Electricity is always an open system, so enclosed does not really exist. Your electric field is either endothermic or exothermic. This means which direction force is in regard to the wire. Do you understand me? If not, ask

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  • $\begingroup$ While I did not downvote your answer, I don't think this is the right answer, enclose DOES exist, you can enclose it by a surface bounded by an amperian loop. What I meant is the "contradiction" between different surfaces $\endgroup$
    – D. Soul
    Aug 6 '20 at 9:06
  • $\begingroup$ The evidence of my answer is here. Use lussacs gas law to show a change in pressure in the electric field charge during the endothermic charge of the electric field ntrs.nasa.gov/citations/20070032054 $\endgroup$ Aug 8 at 12:43

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