Electro dynamics, E/B fields inside wire

[Greger]

http://img850.imageshack.us/img850/3444/asdfga.jpg

i'm used to approaching these questions by using either guass' law or amperes law but in this question i don't see how either will help since there's 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

 

[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.

 

[Greger]

hey tiny-tim,

thanks for your answer!

I was thinking that for the B field but wouldn't that just be

B = [itex]\frac{μ_{0}I}{2\pi r}[/itex] 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 [itex]\frac{1}{μ_{0}}ExB[/itex] is for this particular question and I haven't been able to produce it.

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

E = -grad(V)

but integrating across the length gives

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


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


does my problem lye with the B field?

 

[tiny-tim]

Hey Greger!

I was thinking that for the B field but wouldn't that just be

B = [itex]\frac{μ_{0}I}{2\pi r}[/itex] 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 [itex]\frac{1}{μ_{0}}ExB[/itex] is for this particular question and I haven't been able to produce it.

Can you show your full calculations for B?

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

 

[Greger]

Yea sure,

I just used amperes law,

[itex]B=\frac{{\mu}_{o} I}{2\pi r}[/itex]

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)?

 

[tiny-tim]

Hi Greger!

(just got up :zzz:)

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,

What do you think "_enc" means??

Try again, integrating round a circle of radius a. ​

 

[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 I am worried about is if you pick a radius a<r you are only enclosing some of the current, there's still current outside of the loop, wouldn't that create a field itself?

 

[tiny-tim]

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

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

your equation only contains I_enc

But anyway you would get

[itex]B=\frac{{\mu}_{o} I}{2\pi a}[/itex]

and ExB with what I have so far would give

[itex]\frac{{\mu}_{o} I V}{2\pi a L}[/itex]

but the ExB I'm trying to get to is

[itex]\frac{{\mu}_{o} I V a}{2\pi r^2 L}[/itex]

but you're still not using I_enc

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

 

[Greger]

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

dividing them you get

[itex]\frac{I_{enc} r^2}{I a^2}[/itex]

right?

 

[tiny-tim]

Does this imply Ienc =[itex]\frac{I a^2}{r^2}[/itex]?

but how can you get that without knowing the answer?

think!

(what is current? )​

 

[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?

 

[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.

 

[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).

 

[tiny-tim]

ExB will give - (z x thi) = - (-r) = r, a positive ExB, but I am looking for a negative ExB …

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

 

[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

 

[tiny-tim]

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

 

[Greger]

I guess you could say that the heat flows inwards as well 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!