# Ampère's circuital law not permitted?

1. Jun 15, 2010

### nonequilibrium

Hello,

So I saw Ampère's law in the following form:

$$\int \overline{B}.d \overline{s} = \mu I$$

And I was wondering why you can not use that to find the magnetic field at a certain point around a non-infinite straight piece of conducting wire (I had to solve it with Biot-Savart to get it).

Obviously, if I tried to use Ampere's law, I'd get the wrong result, and I get you'd be using a lot less information (it doesn't ask how long the wire is) so it's evident it shouldn't work, but on the other hand, I don't see where the equation itself breaks down? What prerequisite is missing? Or was that formula itself simply derived out of Biot-Savart with the assumption the current was infinitely long? Or maybe I just saw an incomplete form of the law?

Thank you,
mr. vodka

2. Jun 15, 2010

### Studiot

Have you tried the integration?

I am more used to Ampreres law in the form

$$I = \oint {H.dL}$$

This certainly integrates correctly so long as the conductor is inside the path of integration, otherwise the value = zero.

This version applies to the field inside and around a conductor and the 'I' referred to is conduction current.
In the case of a dielectric the current is displacement current and Amperes law has an additional term.

Another form of Amperes law is one of Maxwells equations.

$$I = \oint {H.dL} = \int {\int\limits_A {J.ds} }$$

Hope the exams are going well

3. Jun 15, 2010

### nasu

It's not that the law is not valid or not "permitted". It's just not so useful for specific calculation.
For infinite wire we assume B having radial symmetry so the surface integral for a cylindrical surface is just BxArea.
For finite wire there is no nice symmetric area with B constant all over it. So what area would you pick and how would you calculate the integral?

4. Jun 15, 2010

### Staff: Mentor

A finite wire segment with a steady current is impossible, because of continuity of charge and current. You can't satisfy Maxwell's equations, including Ampere's Law, with the field calculated via Biot-Savart from a wire segment. You have to use a complete current loop.

Ampere's Law works for an infinite straight wire because you can think of the current as sort of "looping back at infinity."

5. Jun 19, 2010

### nonequilibrium

Well there is symmetry: if you pick a certain spot with a distance a above the finite conductor, then the circle around the conductor through that point will have a constant B field so you get B*2*Pi*a = mu*I, but this is not the correct answer.

Studiot - I'm not too familiar with that form, but it looks really similar, yet you say your integral gives the right answer? I've seen H before and then it was equal to B/mu, and dL looks like it's the same as my ds? So your form really looks equivalent to mine, at least in this situation. How did you get the correct answer?
PS: my exams are going well, thank you; still interested in that entropy question by the way :)

Nasu - is there something wrong with the symmetry I argued in this post?

jtbell - Interesting, but is that really of the essence? I don't want to sound denigrating, I'll readily believe you know it many times better than me, but I have the feeling (at least in my introductory course...) multiple things might not really be possible according to ALL the equations, but they still give reasonable answers. For example, for a finite conductor, I can calculate the B - field completely with Biot-Savart and in the limit of the finite wire becoming an infinite wire, it gives the right limit (i.e. the result you should become when calculating B for an infinite wire), so it looks reasonable. Now, couldn't you cheat a little by saying the conductor isn't really finite, but at both endpoints the conductor actually makes a 90° bend straight downward, making for the fact you can just as well ignore their contribution to the B-field you're interested in, certainly if your 'supposedly finite' straight conductor is quite long and you're interested in the B-field around the middle? Or do you say: no, the fact you theoretically assume a constant current in a finite straight segment is the very reason why Ampere won't work here (and not due to another kind of mistake)? (Would that also make my calculation (done in class) with the use of Biot-Savart complete rubbish?)

Thank you all for the replies

6. Jun 21, 2010

### nasu

B being constant along the circle centered on the wire is not necessarily
true. When you talk about a finite wire, that usually means that the wire is part of a finite circuit loop, like one side of a rectangular loop. The field does not have cylindrical symmetry, there is a region outside the loop and one outside the loop.
The B in the Ampere's law is the total B, not just the one produced by the piece of wire that you are interested in.
When you use Biot-Savart formula you calculate just the B produced by the specific piece of wire.

As a simple example, think about two "infinite" wires with "zero" spacing. You can use Biot-Savart to calculate the field for each one of the two wires at distance r and you'll find the well known formula.
When you write Ampere's law for a circular loop around the two wires, you get B=0, which is the total field. It is also what you get when you apply Biot-Savart for ALL the wires in circuit (the two infinite ones in this example).

For the case of an infinite wire you assume that the field has cylindrical symmetry. And this is essential for using Ampere's law. You find the total field around the wire based on the cylindrical symmetry assumption. And then you assume that the rest of the circuit is somehow too far to have any influence so this is the field of just one wire.

Those considerations have nothing to do with the validity of the law. It is valid no matter what is the geometry and symmetry (or lack of symmetry) of the problem. Ampere's law is the integral form of one of Maxwell's equations (curlB=j).

Last edited: Jun 21, 2010