Magnetostatics and Ampere's law on a finite length wire

[Telemachus]

Homework Statement

This doubt arised a few days, as a teaching assistant. A student asked about ampere's law in the context of this specific problem.

I had presented this problem: Calculate the magnetic field for a point localized at a distance "r" from the midpoint of a straight wire of length "2L" on which a current "i" is circulating.

So, the wire is along, let's say the z axis, and goes from z=-L to z=L.

Relevant Equations

We have Ampere's law in integral form: $\oint \vec{B}\cdot d\vec{l}=\mu_0 i_{enc}$.

And the Biot Savart law: $\vec{B}=\frac{\mu_0 i}{4\pi}\int \frac{d\vec{s}\times \hat{r}}{r^2}$


Using Biot-Savart law, it is easy to obtain in cylindrical coordinates, with $\vec{r}=(r,\phi,z)$ that
$\vec{B}(\vec{r})=\frac{\mu_0 i L}{2\pi r \sqrt{L^2+r^2}} \hat \phi$.

Then the question, for which I couldn't give a satisfactory answer, was that why can't we apply Ampere's law at the point z=0 in this problem? over that plane, the field is symmetric, and goes only along the $\hat \phi$ direction, so, in principle one could make a loop and obtain the field. But if we do this, the field obtained is the one for an infinite wire.

I tried to think why Ampere's law seems to fail in this case. For me it was clear that there is no symmetry in the z direction, there is no translational symmetry because of the finiteness of the wire. On the other hand, I know that Ampere's law is independent of the loop we take. This also poses some sort of contradiction, because if instead of a circle in the plane z=0 we use a different loop which doesn't live in the plane z=0, B is not constant anymore outside the plane. Something similar happens if we think of the surface which is enclosed by this loop. By Stokes theorem, one can deform that surface, until the point outside of the wire, and no current would be flowing through this "Stokes" surface, so there will be no current, and again a contradiction.

And then, the finite wire itself with a current that begins at a point, and ends at another point doesn't seem very physical neither.

So, I was looking for some discussion on this physical problem, formally, how to explain why Ampere's law can't be applied? is Ampere's law violated in this case? and why?

Thanks in advance.

 

[Delta2]

Its not that Ampere's law fails, it is that we use the law in a situation that is purely theoretical and does not correspond to a real world situation.

The current ends suddenly at points z=+-L of the wire and this is what makes the situation non real world. This also creates a discontinuity in the current density (more specifically the divergence of the current density becomes infinite at the points z=+-L) which makes Biot Savart law to not be equivalent to the differential form of Ampere's law (the divergence of the current density must be equal to zero everywhere in order for Biot Savart and Ampere's law to be equivalent). Hence we can get different results from Biot Savart law and from the differential (or integral form) of Ampere's law.

You can read this link of wikipedia if you want to know why we need the divergence of the current density to be identically zero in order for the two laws to be equivalent.
https://en.wikipedia.org/wiki/Biot–..._circuital_law,_and_Gauss's_law_for_magnetism

 

[PeroK]

So, I was looking for some discussion on this physical problem, formally, how to explain why Ampere's law can't be applied? is Ampere's law violated in this case? and why?

Thanks in advance.

Ampere's law requires either a closed loop of current or an idealised infinite wire of some description. A finite length of straight wire does not meet these criteria.

In this case, as you have tacitly assumed that there is no current in the opposite direction - no matter how large you make $r$ you never pick up any current in the opposite direction - which reduces to the case of an infinite wire, as far as Ampere is Law is concerned.

 

[rude man]

If the wire is of finite length there must be other wiring completing the circuit. Those other wires also have their own magnetic fields.

Ampere's law is correct when you include those fields in any contour enclosing your current.. However, due to lack of symmetry, that law is inapplicable to solving your problem.

 

[vanhees71]

You can calculate the integral, but it has no physical meaning, because the magnetic field, you'd calculate with Biot-Savart's Law is not a solution of the magnetostatic Maxwell equations since the continuity equation must hold $\vec{\nabla} \cdot \vec{j}=0$, which is not the case here. See:

https://dx.doi.org/10.1088/0143-0807/35/5/058001https://itp.uni-frankfurt.de/~hees/publ/ampere-law-discussion-ver2.pdf

 

[Telemachus]

Great answers. Thank you all.