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Electro dynamics, E/B fields inside wire

  1. Apr 8, 2012 #1
    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. jcsd
  3. Apr 9, 2012 #2

    tiny-tim

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    Hi Greger! :smile:

    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. :wink:
     
  4. Apr 9, 2012 #3
    hey tiny-tim,

    thanks for your answer!

    I was thinking that for the B field but wouldnt 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 havent 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 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?
     
  5. Apr 9, 2012 #4

    tiny-tim

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

    (it looks like you used the wrong I and the wrong r)
     
  6. Apr 9, 2012 #5
    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)?
     
    Last edited: Apr 10, 2012
  7. Apr 10, 2012 #6

    tiny-tim

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    Hi Greger! :smile:

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

    Try again, integrating round a circle of radius a. :smile:
     
  8. Apr 10, 2012 #7
    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
  9. Apr 10, 2012 #8

    tiny-tim

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

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

    (also, divide your answer by the correct answer … what do you get? :wink:)
     
  10. Apr 10, 2012 #9
    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?
     
    Last edited: Apr 10, 2012
  11. Apr 10, 2012 #10

    tiny-tim

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    think!! :rolleyes:

    (what is current? :smile:)​
     
  12. Apr 10, 2012 #11
    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?
     
  13. Apr 10, 2012 #12

    tiny-tim

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    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. :smile:
     
  14. Apr 10, 2012 #13
    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
  15. Apr 10, 2012 #14

    tiny-tim

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    it should be a positive E x B, shouldn't it? :confused:
     
  16. Apr 10, 2012 #15
    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
  17. Apr 10, 2012 #16

    tiny-tim

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    A current-carrying wire gets hot, and heat radiates outward. :wink:
     
  18. Apr 10, 2012 #17
    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!
     
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