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Magnetic field for coaxial cable

  1. Dec 11, 2014 #1

    NYK

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    1. The problem statement, all variables and given/known data
    https://scontent-b.xx.fbcdn.net/hphotos-xpa1/v/t1.0-9/10846336_1517298108546407_4211362504714893270_n.jpg?oh=2f11c165933554a0e4003b398c877004&oe=5542789C

    2. Relevant equations
    ∫B⋅nda = Iencμo

    Ienc=∫J⋅da

    3. The attempt at a solution
    For r < a:

    Ienc = I

    ∫B⋅nda = 2πr = Ienc μo

    B = (Ienc μo)/(2πr)

    For r > b:

    Ienc = I - I

    B =0

    For a < r < b:

    dI = J⋅nda = IJIda da = 2πrdr

    dI = Jo(b/r)(2πrdr)

    Ienc = ∫Jo(b/r)(2πrdr) = Jo2πb∫dr (integrating from b to a)

    Ienc = Jo2πb(b-a)

    ∫B⋅nda = B2πr = Ienc μo

    B2πr = μo( Jo2πb(b-a))

    B = (μo/r)( Job(b-a))



    I think I did this correctly just looking for a double check
     
  2. jcsd
  3. Dec 11, 2014 #2

    TSny

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    The integral on the left should be an integral along a path rather than an integral over an area. Also, your middle expression is incomplete.

    This looks correct for r < a.

    Correct.

    OK

    You need to find B at some arbitrary value of r that lies between a and b. To find B at r, what will be the choice of the path of integration? How does this affect the integration limits?

    Also, will the path of integration enclose the thin wire at the center of the cable? If so, should the current in the wire also be included as part of Ienc?
     
  4. Dec 11, 2014 #3

    NYK

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    That makes sense the current in the middle wire is for sure having an effect on the total current so;

    Ienc = Io - Imeat

    and Imeat = ∫Jo(b/r)(2πrdr) = Jo(2πb)∫dr (integrating from a to r)

    Imeat = Jo(2πb)(r-a)

    Then; Ienc = Io - Jo(2πb)(r-a)

    So, B = (μo/r)(Io - Jo(2πb)(r-a))

    That makes more sense to me because that is saying that that current is the current running through the meat subtracted from the current running through the wires. Which it should be since the currents are going in opposite directions,
     
  5. Dec 12, 2014 #4

    TSny

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    OK, that all looks good. There is one more thing you should probably do. The problem states that the metal carries the same current ##I_o## as the wire. This means that you can express ##J_o## in terms of ##I_o## and simplify your result.
     
  6. Dec 12, 2014 #5

    NYK

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    Could I define Jo = Io

    So for a < r < b

    the B field would be simplified to:


    B = (μo/r)(Io(1-b(r-a)))
     
  7. Dec 12, 2014 #6

    TSny

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    No. ##J_o## has the units of current density while ##I_o## has the units of current. So, they cannot be equal.

    What would you get if you integrated ##J## over the area between r = a and r = b?
     
  8. Dec 12, 2014 #7

    NYK

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    I integrated over that area to begin with and came up with:

    Imeat = ∫ Jo(b/r)(2πrdr) = Jo2πb∫dr (integrating from a to b)

    Imeat = Jo2πb(b-a)

    Not sure I am understanding what you are saying though
     
  9. Dec 12, 2014 #8

    TSny

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    OK.

    Now, how is Imeat related to Io?
     
  10. Dec 12, 2014 #9

    NYK

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    Well their reciprocals of one another but the shaded portion has a current density while the middle portion does not.

    So,

    Ienc = Io - Imeat = Io - Jo2πb(b-a)

    am I on the right track?
     
  11. Dec 12, 2014 #10

    TSny

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    No. Note that you got ##I_{meat}## by integrating the current density ##J## over the entire cross-section of the metal. Therefore, ##I_{meat}## is the total current carried by the metal between r = a and r = b. The problem statement tells you what this total current is equal to.

    Therefore you know what Jo2πb(b-a) is equal to. Hence, you can solve for Jo. Then substitute into your final expression for B in post #3.
     
  12. Dec 12, 2014 #11

    NYK

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    Pretty sure I got it:

    If Imeat = Jo2πb(b-a)

    then by the same method of integrating

    Io = Jo2πb(r-a)

    putting htis back into the equation:


    B = (μ0/r)(Jo2πb((b-a)(r-a))

    How does that look?
     
  13. Dec 13, 2014 #12

    TSny

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    OK. What I would do here is solve for ##J_o## in terms of ##I_o## and then substitute for ##J_o## in the result of post #3. This way, you will get B expressed in terms of ##I_o##.

    This doesn't look correct. Note that B should go to zero when r = b since you know that B = 0 for r>b. Also you can check that the units on the right side of your result are not the correct units for B. Check the part that is in red.
     
  14. Dec 14, 2014 #13

    NYK

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    Your right,

    I re worked the problem and came up with:

    for a < r < b:

    B(r) = (μo/2πr)Io(1-(b/r)((r2-a2)/(b2-a2)))

    so now when r meets b, B = 0
     
  15. Dec 14, 2014 #14

    TSny

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    But this expression also implies that B = 0 when r = a. So something's wrong.

    What did you get for the relationship between ##J_o## and ##I_o##?
     
  16. Dec 14, 2014 #15

    NYK

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    I said in the equation the Jo = Io/π(b2-a2)

    But i just tried redefining that to: Jo = Io/π(r2-a2)

    So then Imeat = Io/π(r2-a2)(b/r)(π(r2-a2) = Io(b/r)

    so then my Ienc = I(1-(b/r))

    so, B(r) = (μo/2πr)(I(1-(b/r))
     
  17. Dec 14, 2014 #16

    TSny

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    The current density ##J(r)## is not uniform. It is given by the equation ##J(r) = J_ob/r##. To find the relation between the constant ##J_o## and the total current in the outer conductor, ##I_o##, integrate ##J(r)## over the cross-section of the outer conductor.

    This goes to zero for r = b, but does it give the correct answer for r = a?
     
  18. Dec 14, 2014 #17

    NYK

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    I integrated J(r) from a to r and came up with:

    J(r) = Jobln(r/a)

    Then for a < r < b

    Ienc = Io - Job(ln(r/a))

    So, B(r) =(μo/2πr)(Io - Job(ln(r/a))
     
  19. Dec 15, 2014 #18

    TSny

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    In your first post you had

    Use this to relate ##J_o## to ##I_o##.

    Then use this in your result for B in post #11 to express B in terms of ##I_o##.
     
  20. Dec 15, 2014 #19

    NYK

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    So using Io = Jo2πb(b-a)

    and solving for Jo

    I come up with Jo = (Io/2πb(b-a))

    Then plugging this into the B field equation quoted above I come up with:

    B = (μo/r)(Io(r-a))
     
  21. Dec 15, 2014 #20

    TSny

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    OK. (I modified the parentheses a bit to make sure the (b-a) factor is in the denominator.)

    Oops! My fault. You should substitute for ##J_o## in your expression that you obtained in post #3. Your expression in post #11 was incorrect. Sorry.
     
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