# Electro dynamics, E/B fields inside wire

1. Apr 8, 2012

### Greger

http://img850.imageshack.us/img850/3444/asdfga.jpg [Broken]

i'm used to approaching these questions by using either guass' law or amperes law but in this question i dont see how either will help since theres no enclosed charge its a current, and also since the whole wire carries the current and you need to find the B field inside the wire.

could anyone please give me a push in the right direction

Last edited by a moderator: May 5, 2017
2. Apr 9, 2012

### tiny-tim

Hi Greger!

The wire is a cylinder of radius r, and you can assume the current (the moving electrons) to be uniformly distributed across the cross-section of the cylinder.

To find the B field at distance a from the centre, consider a circle of radius a.

3. Apr 9, 2012

### Greger

hey tiny-tim,

I was thinking that for the B field but wouldnt that just be

B = $\frac{μ_{0}I}{2\pi r}$ for a inside the cylinder? And that would be in the ti direction (in cylindrical co-ords)

The reason I didn't think this was right was because I know what the $\frac{1}{μ_{0}}ExB$ is for this particular question and I havent been able to produce it.

For the E field i've been trying to get it in terms of V,

but integrating across the length gives

E = -V/L in the z direction (since thats the direction of current flow)

using these im a factor of a/r out from ExB (where a is some internal radius)

does my problem lye with the B field?

4. Apr 9, 2012

### tiny-tim

Hey Greger!
Can you show your full calculations for B?

(it looks like you used the wrong I and the wrong r)

5. Apr 9, 2012

### Greger

Yea sure,

I just used amperes law,

$B=\frac{{\mu}_{o} I}{2\pi r}$

I only evaluated at r since if i picked some a less then r, wouldn't there also be a contribution to the field by whatever is outside that a like,

Picking 0<a<r then doing the same as i did above only using a instead of r would get the I enclosed, but wouldn't there also still left over current flowing through the ring a<r (whatever was left over after selecting a)?

Last edited: Apr 10, 2012
6. Apr 10, 2012

### tiny-tim

Hi Greger!

(just got up :zzz:)
What do you think "enc" means??

Try again, integrating round a circle of radius a.

7. Apr 10, 2012

### Greger

haha good morning!

enc means the current enclosed by the loop right?

but if you pick a circle of radius a<r that will only be the current enclosed by that loop so it would be Ienc and not just I right? If you take a circle at radius r it will enclose all of the current so you can write I?

But what im worried about is if you pick a radius a<r you are only enclosing some of the current, theres still current outside of the loop, wouldn't that create a field itself?

Last edited: Apr 10, 2012
8. Apr 10, 2012

### tiny-tim

yes of course, but it wouldn't be relevant to your equation …

your equation only contains Ienc
but you're still not using Ienc

(also, divide your answer by the correct answer … what do you get? )

9. Apr 10, 2012

### Greger

oooo whoops sorry I forgot about Ienc for the a, I'll fix it now,

dividing them you get

$\frac{I_{enc} r^2}{I a^2}$

right?

Last edited: Apr 10, 2012
10. Apr 10, 2012

### tiny-tim

think!!

(what is current? )​

11. Apr 10, 2012

### Greger

I get that for a<r, there is less charge moving through the loop of radius a in comparison with the loop of radius r per unit time. So I get that the Ienc has to be lower then I, so in this case its compensated for by a^2/r^2, which is the ratio between the area of each of the circles right?

So is that it? It's enough to say that since a<r Ienc is just a factor less then I and that factor is just the ratio between the areas because for a small area a smaller amount of charge will flow through?

12. Apr 10, 2012

### tiny-tim

Yes.

Since we're not told otherwise, we can assume that the moving electrons are equally distributed through the wire, and so the number of them is proportional to the area.

13. Apr 10, 2012

### Greger

Thanks tiny-tim!

I just noticed that this means the poynting vector will point radially in, which is kind of weird cause that means the energy is flowing radially inwards right?

I guess this is a little off the question now but how can the energy by flowing radially inwards? I would guess it to be flowing opposite to the direction of the current ( the direction of the electron flow).

Last edited: Apr 10, 2012
14. Apr 10, 2012

### tiny-tim

it should be a positive E x B, shouldn't it?

15. Apr 10, 2012

### Greger

i'm still thinking about what it means to have the the energy flow radially inwards,

like in a circuit i would imagine the energy to flow through the wire in some direction, not radially out of the wire ahah

Last edited: Apr 10, 2012
16. Apr 10, 2012

### tiny-tim

A current-carrying wire gets hot, and heat radiates outward.

17. Apr 10, 2012

### Greger

I guess you could say that the heat flows inwards aswell since the vector is in the -s hat direction right?

I found similar, more numerical example online and they get a similar result (the energy vector in the -s hat direction).

I guess its hard to imagine the only energy being represented is the resistence, but if R=0 then S=0 since V=IR,

Thanks tiny-tim!