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A Puzzle Based on Ampere's Circuital Law

Anyone who has a basic knowledge about electrodynamics is expected to know Ampere's circuital rule. The rule states that, in free space, the circulation of magnetic field along an arbitrary closed loop enclosing a steady current is equal to the magnitude of the enclosed current divided by the magnetic permeability of free space. Its differential form is: \[\overrightarrow { \nabla } \times \overrightarrow { B } ={ \mu }_{ 0 }\overrightarrow { J } \] where \[\overrightarrow { B }\] is the magnetic field and \[\overrightarrow { J }\] is the current density vector. Using Stokes' theorem we obtain: \[\oint { \overrightarrow { B } .d\overrightarrow { l } ={ \mu }_{ 0 }{ I }_{ enclosed } } \] which is a direct mathematical equation for the aforementioned statement. Now, it is a trivial problem, that every textbook provides, to find the magnetic field intensity at a point in a magnetic field using this Ampere's circuital law and the principle of symmetry. One such problem that is commonplace is to find the field strength at a point say P, ata certain distance say 'a' from a straight wire of infinite length carrying a steady current, say 'I'. We consider a circular Amperian loop of radius a passing through P, with centre on the wire. Then arguing using the principle of symmetry, the equation \[\oint { \overrightarrow { B } .d\overrightarrow { l } ={ \mu }_{ 0 }{ I }_{ enclosed } } \] then simply becomes \[\int { Bdl={ \mu }_{ 0 }{ I } } \\ \Longrightarrow B\int { dl } ={ \mu }_{ 0 }{ I }\\ \Longrightarrow B(2\pi a)={ \mu }_{ 0 }{ I }\\ \Longrightarrow B=\frac { { \mu }_{ 0 }{ I } }{ 2\pi a } \] (since at every point on the loop, the directions of the length element and the magnetic field are same, the dot product within the closed integral in the left hand side becomes simple product and the integral becomes a simple integral. Also, from radial symmetry, magnetic field strength has equal magnitude at all points on the loop, and B can be treated as constant.) The direction of the magnetic field at P is given by the right hand thumb rule. Now consider the infinite wire to be replaced by a wire of finite length carrying the same steady current I. Then by Ampere's law and principle of symmetry, one can argue that the magnetic field strength at a distance 'a' from the wire is same as that in the case of infinite wire, since the principle of symmetry applies here also, and since the loop encloses the same current I. But if one has gone through a few applications of the Biot-Savart Law one can peremptorily say that in the second case the result will be different. The magnitude of magnetic field strength will be then given by: \[B=\frac { { \mu }_{ 0 } }{ 4\pi a } (\sin { { \theta }_{ 2 } } -\sin { { \theta }_{ 1 } } )\] Where \[{ \theta }_{ 2 } \] and \[{ { \theta }_{ 1 } } \] are the angles that P subtends with the two ends of the finite conductor. For an infinite conductor the former becomes 90 degrees while the latter becomes (-90) degrees, which then gives: \[B=\frac { { \mu }_{ 0 } }{ 4\pi a } (\sin { { 90 }^{ 0 } } -\sin { ({ -90 }^{ 0 }) } )\\ \Longrightarrow b=\frac { { \mu }_{ 0 } }{ 4\pi a } \{ 1-(-1)\} \\ \Longrightarrow b=\frac { { \mu }_{ 0 } }{ 4\pi a } \times 2\\ \Longrightarrow b=\frac { { \mu }_{ 0 } }{ 2\pi a } \] which is our Ampere's Law. Thus we are seeing that even we are using Ampere's Law and the principle of symmetry in case of a finite conductor, our result is not tallying with that obtained from the Biot-Savart Law. This arises due to a very subtle fallacy in our considerations. Can you spot it?

Note by Kuldeep Guha Mazumder
2 years, 2 months ago

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I also had this curiosity and I got a reply that in finite wire we have to connect it to a source or it would form loop, and we take the amperian loop then magnetic field on this loop will also be affected which would not be easily integrable. Even if we managed to solve it then the result would come out to be different from biot savart law.

This problem does not occur while using biot savart law as it involves the contribution only due to the part of wire considered

I hope it helps. Ayush Jain · 1 year, 2 months ago

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@Ayush Jain Well analyzed Kuldeep Guha Mazumder · 1 year, 2 months ago

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How does the principle of symmetry apply in the second case? Vishnu C · 2 years, 2 months ago

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@Vishnu C Well, try to argue how it doesn't (actually it doesn't, you have to find out why). Kuldeep Guha Mazumder · 2 years, 2 months ago

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In case of a finite open wire, there will be charge accumulation at the end surfaces of the wire if we have a stationary current flowing in the wire. To apply ampere's law one needs to first check whether the charge conservation constraint is satisfied. For an infinite wire this is automatically satisfied. But this is not the case for a finite wire. Shouvik Mukherjee · 2 years, 2 months ago

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@Shouvik Mukherjee Why so? I want an explanation.. Kuldeep Guha Mazumder · 2 years, 2 months ago

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