# Use of Ampere's law in case of a finite wire

I myself tried computing magnetic field due to a finite current carrying wire using Ampere's law and I found the expression comes similar to the case of infinite wire. Obviously, there must be something wrong...

So I checked this on internet and found this

A solution giving physical sense to the finite wire is to put a source and a sink of electrical charge +q and −q at each extremity. It is then possible to have an electrical current I flowing along the wire (figure 2). The major point is now that both charges q(t) and −q(t) are time dependent and create an electrical field E(r) which is time dependent! Hence this problem is no longer a magnetostatics problem: we have to treat it in the more general framework of electromagnetism.

Then it says that as field is time dependent, therefore we need to use Maxwell Ampere's law.

But, what I don't understand is that how come the electric field is time dependent? Why is that necessary? And why are the charges time dependent? Is it because charges are flowing and thus need to be maintained continuously? I am not sure and hence asking here.

Also I want you to note that I'm studying this for the very first time and I'm not comfortable with complex higher maths. So, it would be nice if you restrict mathematical rigor in your answer as much as possible.

Please help me in this and I would be highly grateful.

Thanks.

## 1 Answer

Assume you have a straight conducting wire.

At one end there is an excess of electrons and at the other end a deficit of electrons.
That separation of charges produces an electric field.

Now the electrons are allowed to flow from the excess end to the deficit end of the wire.
An electric current is flowing along the wire but the number of electrons at one end of the wire is decreasing whilst the number of electrons at the end of the wire is increasing. The electric field produced by the separation of charges is changing ie time dependent.

• But can't we supply fresh charge continuously at the ends of wire to establish a steady field? – Abhinav Dhawan Jul 2 '18 at 13:55
• @AbhinavDhawan How would you do that? Remember that you are trying to find the magnetic field due to a wire of finite length and so cannot connect extra bits of wire to the wire under consideration. – Farcher Jul 2 '18 at 13:57
• So, how come we got the charge q and -q initially? – Abhinav Dhawan Jul 2 '18 at 14:03
• I don't know whether this would work... but, can't we put charges on the both side without connecting it with the wire and then this would lead to a steady electric field setting up? – Abhinav Dhawan Jul 2 '18 at 14:09
• @AbhinavDhawan There must still be charges moving along the wire to produce the electric current. – Farcher Jul 2 '18 at 14:22