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Under the right circumstances, Ampere's law $\oint \vec H\cdot d\vec \ell=I_{encl}$ can be used to deduce the field $\vec H$ at a point from the current enclosed by the circuit which produces $\vec H$. This can be done when one can find a current-enclosing contour on which the field is constant in magnitude, something that can occur only in highly symmetrical situations: the symmetries of the current distribution are reflected in the symmetries of $\vec H$, meaning that the geometry of the Amperian loop enclosing the current is usually closely related to the symmetry of the source current distribution.

All textbook examples use cylindrical or planar current distributions (or modifications thereof, such as the infinite solenoid or the toroid, or even semi-infinite cylinders), resulting in circular or rectangular loops.

Can people provide examples of other non-trivial current distributions, coordinate systems and contours for which one can put Ampere's law to good use to find the field $\vec H$?

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    $\begingroup$ You can use the differential form $\vec\nabla\times{}\vec{B}=\mu_0\vec{J}$ in any geometry, and you probably do every time you use a FEM method to solve a magnetics problem. $\endgroup$
    – The Photon
    Mar 12, 2017 at 16:44
  • $\begingroup$ @ThePhoton I can't see your suggestion being correct. For an infinitely long wire carrying current $I$ the $\vec B$ field at distance $\rho$ is just $\vec B=\frac{mu_0I}{2\pi \rho}\hat \phi$ and one easily verifies that $\nabla\times \vec B=0$ near that point, consistent with $\vec J=0$ near that point. Of course this is expected since $\vec\nabla\times B$ is local whereas the integral form of Ampere's law is global. $\endgroup$ Mar 12, 2017 at 17:45
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    $\begingroup$ The two forms are mathematically equivalent according to Stokes' theorem. $\endgroup$
    – The Photon
    Mar 13, 2017 at 15:39
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    $\begingroup$ The differential form works at every point not on contours and surfaces. So if you are using the differential form, you will have to work out the curl of B at every point. $\endgroup$ Mar 16, 2017 at 11:57
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    $\begingroup$ Ampere's law is used to derive the continuity conditions for H. Hence Ampere's law is used extensively whenever there is an interface between different media. $\endgroup$
    – ProfRob
    Nov 25, 2018 at 23:03

2 Answers 2

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Ampere's Law is useful only for finding the magnetic intensity or magnetic magnetic field in those electrical distributions where the current is steady and there is a high degree of symmetry. Elsewhere, this law is valid but the mathematical equations become extremely complicated and hence it is not useful for finding the magnetic intensity or magnetic magnetic field in non-symmetrical electrical distributions.

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    $\begingroup$ In what sense do you think your answer adds anything informative to whatever @ZeroTheHero was asking? Unless I have missed it, you are repeating what he wrote. $\endgroup$
    – hyportnex
    Mar 12, 2017 at 15:48
  • $\begingroup$ @hyportnex Presumably there ought to be situations where elliptical or hyperbolic coordinates would be appropriate, but I've never seen this done anywhere. I will take anything above and beyond cylindrical or planar. $\endgroup$ Mar 12, 2017 at 15:54
  • $\begingroup$ @ZeroTheHero neither have I, and I think the reason for that is (warning: here comes the vigorous handwaving) that cylindrical symmetry has really one free variable (the radial distance) and in the integral formulation you have one current against which on the other side you have a scalar product of two vectors, and now try to find the 3 components from one equation. $\endgroup$
    – hyportnex
    Mar 12, 2017 at 15:59
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The question behind your question may be: is there a 3D version of the relevant law and its associated theorem? Yes, there is. The equivalences go like this: $$\int_{βˆ‚S} d𝐫(β‹―) = \int_S dπ’Γ—βˆ‡(β‹―), \quad \int_{βˆ‚V} d𝐒(β‹―) = \int_V dVβˆ‡(β‹―),$$ where $S$ and $V$ are, respectively, a 2D surface and 3D volume, where $βˆ‚S$ and $βˆ‚V$ are their respective 1D and 2D boundaries, where $d𝐫 = (dx, dy, dz)$ is a line element on $βˆ‚S$, $d𝐒 = (dy∧dz, dz∧dx, dx∧dy)$ is surface element on $S$ and on $βˆ‚V$, and $dV = dx∧dy∧dz$ a volume element in $V$.

Thus, for the magnetic field $𝐇$, one has the usual 2D version of the law: $$\int_{βˆ‚S} d𝐫·𝐇 = \int_S dπ’Γ—βˆ‡Β·π‡ = \int_S dπ’Β·βˆ‡Γ—π‡ = \int_S d𝐒·𝐂 = I_S,$$ as the integral version of $βˆ‡Γ—π‡ = 𝐂$, where $$𝐂 = 𝐉 + \frac{βˆ‚πƒ}{βˆ‚t}$$ is the "total" current density; but also a 3D version: $$\quad \int_{βˆ‚V} d𝐒×𝐇 = \int_V dVβˆ‡Γ—π‡ = \int_V dV𝐂.$$

As to the question, per se: there are no restrictions on the shapes of the 2D region $S$, just so long as it has a well-defined boundary $βˆ‚S$. (That means: no fractal surfaces or the like). So the surface $S$ for the current $I_S$ is passing through can be other shapes, not just solid circles or rectangles. The 3D volume $V$ doesn't have to be a cylinder. It can be any shape, as well, just so long as it has a well-defined boundary $βˆ‚V$.

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