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Using ampere's law with current density

  1. May 11, 2008 #1
    K sorry for the bad drawing using paint...but anyways...so the question is :A long wire is in a conducting sheath and viewed from the top.The wire and the conducting sheath each carry opposite currents I into and out of the page with current density as J.The outer sheath has a radius of 2R and he inner wire has a radius of R from the centre.Calculate the magnetic field at the centre of the wire.

    J=I/A->I=JA
    Taking the amperian loop as a circle of radius 2R
    A= pir^2....I1= pi (R^2)I and I2= 4pi(R^2)I
    therefore the net current wud be I2-I1 which is 3(pi)(R^2)
    then using ampere's law B(2(pi)r)=(mu)I -> B(2(pi)(2R))=3(pi)(R^2)
    And I get (3/4)(pi)R^2)J...but the answer is (1/2)(pi)R^2)J
    Can you please tell me wht I am doing wrong?Please help me quickly I have an AP C exam tomorrow and I want to make sure I knw how to use Ampere's law porperly.Thank You

    Btw if possible can you explain to me wht exactly is happening...if there is a current going in one direction...how come the sheath carries a current in the opposite direction?
     

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  3. May 11, 2008 #2

    dx

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    It's very difficult to understand what you've written. Also, these questions should be posted in the homework section.
     
  4. May 11, 2008 #3
    k sorry abt the homework section thing but can u please solve it for me.I have an AP exam tomorrow.Wht dont u understand...I will explain
     
  5. May 11, 2008 #4

    dx

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    Write out the question in its original form and draw a labeled diagram.
     
  6. May 11, 2008 #5
    the problem is I dont have the question...the question was on the AP practise test which I cant bring home.So I am writing the question from memory.K I will make a labelled diagram.But basically the two circles u see is one with a radius of 4R(sheath) and smaller one with radius of R(wire).They both have current density of J.Find magnetic field at centre of wire.
     
  7. May 11, 2008 #6
    hope this clarifies.here is a labelled diagram.Note:the diagram is being viewed from top.So your looking at the end of the wire from the top.
     

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  8. May 11, 2008 #7

    dx

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    I can't make sense of it the way you're formulating it. Maybe you dont remember it correctly.
     
  9. May 11, 2008 #8
    wht do u mean "formulating it"...yes the question isn't structured exactly the way it was...but the info is all the same...which part is confusing u?
     
  10. May 11, 2008 #9
    wht abt the diagram...has it helped?
     
  11. May 11, 2008 #10

    dx

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    I can't see the diagram, it must be approved first. Maybe you can host it somewhere else and give me a link.
     
  12. May 11, 2008 #11
    oh god...can I send it to ur e-mail?I dunno anywhere where I can host it....
     
  13. May 11, 2008 #12

    dx

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  14. May 11, 2008 #13
  15. May 11, 2008 #14

    dx

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    Nope, I'm sorry I'm unable to understand it. Maybe some one else here will be able to help you.
     
  16. May 11, 2008 #15
    there is no one else communicating with me.wht exactly cant u get?
     
  17. May 11, 2008 #16
    look there is juz a current going out of the page and into the page basically and the wire is a staright long one.
     
  18. May 11, 2008 #17

    dx

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    In that case, if you consider a small circular loop of radius r around the axis as the path of integration, you will get [tex] B = 2\mu_{0} J r [/tex]. As r goes to zero, B goes to zero. So, at the centre of the wire B should be zero. But you said the correct answer was something else.
     
  19. May 11, 2008 #18
    But u havent considered the current in the sheath.I think the loop has to be 4R not just R.
     
  20. May 11, 2008 #19

    dx

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    You asked for the magnetic field at the centre of the wire. No matter what you choose the initial radius as, when you take the limit of r going to zero, B will go to zero.
     
  21. May 11, 2008 #20
    oh yeah u r right...it shud be zero_Oh wait I think the question was whts the magnetic field on the surface of the sheath.sorry abt tht.:blushing:
     
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