# Ampere's Law for a finite wire

1. May 10, 2015

### Septim

Greetings,

I am working as a TA and I encountered a particular question which asks the student to use the Ampere's Law in order to get the magnetic field created by a semi-infinite wire. I know that there will be charge accumulation a time-dependent electric-field, hence a displacement current but could not formulate the problem.

The semi-infinite wire which extends to infinity only in one direction. There are no other circuit elements at the other end(finite end) of the wire and the current does not loop. The magnetic field obviously has cylindrical symmetry when the Amperian contour is taken as a circle with its center on the wire.

However, due to charge accumulation there is a time-dependent electric field; hence a displacement current as mentioned before. How can I formulate the Ampere's law and show that the magnetic field is the half of that the infinite wire at the finite end of the wire? Do you think treating charge accumulation as a point charge with changing amount of charge right at the finite end of the wire will suffice? What should be the geometry and size of the charge accumulation, should a capacitor be used instead to represent it for example?

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2. May 10, 2015

### Staff: Mentor

I'm not sure if that is true.
For the cylindrical component around the end of the wire: sure. But the magnetic field could have components in radial direction and along the wire direction.

For the cylindrical component, you can use symmetry: add another wire for the other side that accumulates the opposite type of charge at the connection point. If you can show that both wire parts including their point charges give the same contribution, and together they form a regular infinite wire, you are done.

3. May 10, 2015

### Septim

Well mfb thanks for the answer. From Biot-Savart law you can solve this problem exactly and the field has no radial component or a component along the wire as you can infer by the differential magnetic field elements which all point in the same direction. If I add another wire which will accumulate opposite charge at the finite end of the semi-infinite wire in question then I will have an infinite wire which is not my intention. I asked this question elsewhere (physics.stackexchange) and they told me to put a spherical capacitor of finite size at the end but even with that I think one still runs into singularities.

4. May 10, 2015

### Staff: Mentor

How do you handle the charge accumulation with Biot-Savart?

It helps to find a solution for a specific region in space.

5. May 10, 2015

### Septim

Actually I do not account for that charge accumulation with regular Biot-Savart law, yet it gives you a result due entirely to the constrained current. The answer accepted in this question is half of the value of the magnetic field of an infinite wire which makes sense when you think of the integration limits in the Biot-Savart case(minus infinity to zero vs. minus infinity to plus infinity).

6. May 10, 2015

### Staff: Mentor

You cannot ignore some part of the setup in the calculation just because it is inconvenient.

7. May 11, 2015

### Septim

I know that you cannot ignore the displacement current or charge accumulation but wonder if applying Biot-Savart law blindly just ignores them. In some university websites and lecture notes it is accepted that the B-field of a semi-infinite wire is half of that of an infinite wire when the B-field is measured on an axis perpendicular to the wire and which intersects the semi-infinite wire at the finite end. For now I think we can think of this as a hypothetical setup not a real physical one.

8. May 11, 2015

### Staff: Mentor

Yes it does.
That is fine for the phi component, and I guess you can hand-wave the other components away, but "it works in this special case" is not an argument for the general case.

9. May 11, 2015

### Delta²

The magnetic field B , is the curl of magnetic vector potential A. But A depends only on the current density (as long as we choose the lorentz gauge) therefore the magnetic field will depend only on the current in the wire and not on the displacement current.

I am not sure though if Lorentz Gauge is applicable to this problem.

10. May 12, 2015

### Septim

Well thanks for the answer, actually in this question it particularly asks the student to use Ampere's law since they are freshman engineering major mostly. Is it possible to use Ampere's law and maybe a basic symmetry argument to prove that the field of the semi-infinite wire is half of that of the infinite wire?

11. May 12, 2015

### Septim

I think we do not have any other component except phi component. This is a special case since we are directly looking above the finite end and we can exploit the symmetry I guess. Furthermore, I think application of Ampere's law only is not sufficient, one also has to use symmetry arguments and/or displacement current. This course is for freshman engineering majors and they only knew Ampere's law, Biot-Savart and displacement current when this question was asked.

12. May 12, 2015

### Delta²

Ok so, using Ampere's law and the symmetry that $B_{\phi}$ will be the same (in magnitude) along the circumference of a circle that passes from point A and the circle's plane is perpendicular to the axis of the wire, we can conclude that $B_{\phi}=\frac{\mu_0}{2\pi y}I_D$ ,w here $I_D$ is the displacement current that passes through the surface of the circle towards to the right(the regular current that passes through is zero). Now all we have to do is to prove that $I_D=I/2$.

Well this seems to hold because the charge will be accumulated on the thin surface of the circle with radius R i.e the cross section at the finite end of the wire. It will be $I=dq/dt$ and the electric field from a thin surface $S=\pi R^2$ that has charge q is $E=q/2S$ on each side of the surface. So the electric field will be $E=q/2S$ and from that you can find the displacement current $\frac{dE}{dt}S$ will be equal to I/2.

13. May 12, 2015

### Septim

Thanks Delta you helped me to finally figure it out. Actually I began to do the same analysis but my mistakes were assuming that the wire penetrated the surface to some extend (ohmic current flow through it) and I did not consider the back surface . The capacitor analogy that was offered on Physics Stack Exchange had really got into me such that I was visioning two parallel plates of opposite charges.

14. May 14, 2015

### Septim

Actually I have one more question: Why is the displacement current and hence the charge distribution confined to our are of interest? That is can it not extend beyond the contour of Amperian loop, what hinders it?

Last edited: May 14, 2015
15. May 14, 2015

### Delta²

Well we have to make this simplifying assumption in order to be able to solve the problem. You said that this problem is for freshman engineering and i know that lots of simplifying assumptions and approximations schemes are considered for engineering classes.

The problem becomes more complex if we assume a charge distribution along the whole surface of the current carrying conductor that extents from the finite end along all the way to infinity.

16. May 14, 2015

### Septim

Okay agreed but how can one proceed if the charge distribution is not confined? I think the result should be independent of the selected charge configuration.

17. May 14, 2015

### Delta²

Well back to my original post in this thread, if we treat Maxwell's equations in their differential form and introduce the vector potential $A(\vec{r},t)$ and the scalar potential $\phi(\vec{r},t)$ and use the lorentz gauge $\nabla\cdot A+\frac{1}{c}\frac{\partial\phi}{\partial t}=0$ (1) the resulting equation after some math involved is

$\nabla^2A-\frac{1}{c^2}\frac{\partial^2A}{\partial^2t}=\mu_0J$

where $J$ is the regular (ohmic) current density. If the current density does not vary with time then this last equation has the solution

$A(\vec{r})=\int \frac{J(\vec{r'})}{|r-r'|}d^3r'$ (2)

So from (2) we can see that A depends only on the ohmic current density and not on the charge density. The magnetix field B is the curl of A, that is

$B=\nabla \times A$
The fact that the divergence of A ($\nabla\cdot A$) depends (due to (1)) to the scalar potential $\phi$ which depends on the charge density doesnt have an effect on the magnetic field B, because for any vector field A the curl of A can be totally independent from the divergence of A (this is called the Helmholtz decomposition theorem in vector calculus).

18. May 15, 2015

### Septim

Got it thanks, I will look into Helmholtz's theorem which is this one I think.

19. May 15, 2015

### vanhees71

I've not read the entire thread, but just let me warn against the use of Ampere's Law or Biot Savart's Law for finite non-closed wires (for the time-dependent problem it's different, but than Ampere's Law must be generalized to the Ampere-Maxwell Law anyway). The application of Ampere's law to finite (or even semifinite) wires is always wrong on very basic principles of electrodynamics. For details, see

H. van Hees, Comment about "Didactical formulation of the Ampere law"
Comment about D. Barchiesi, Eur. J. Phys. 35, 038001 (2014)
http://dx.doi.org/10.1088/0143-0807/35/5/058001
http://fias.uni-frankfurt.de/~hees/publ/ampere-law-discussion-ver2.pdf

20. May 15, 2015

### Septim

Thanks for the answer, I will look at the articles when I have the time. Furthermore, although this is not a physical case I think one should be able to solve it via careful application of corrected for of Ampere's Law(Maxwell-Ampere's Law). The ironic thing is this question is asked to freshman engineering student who encounter electromagnetism for the first time.