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Magnetic Fields from Currents in a Wire and a Cylindrical Shell

  1. Mar 10, 2012 #1
    1. The problem statement, all variables and given/known data
    A solid cylindrical conducting shell of inner radius a = 5.3 cm and outer radius b = 7.9 cm has its axis aligned with the z-axis as shown. It carries a uniformly distributed current I2 = 7.1 A in the positive z-direction. An infinite conducting wire is located along the z-axis and carries a current I1 = 2.7 A in the negative z-direction.


    What is [itex]\int[/itex][itex]^{P}_{S}[/itex] [itex]\vec{B}[/itex] . [itex]\vec{dL}[/itex], where the integral is taken on the straight line path from point S to point P as shown?

    Link to the picture: http://i89.photobucket.com/albums/k211/diethaltao/h15_cylindersD.png

    2. Relevant equations


    3. The attempt at a solution
    I'm not even sure how to approach this problem. At first I found the difference between the values of the magnetic field at P and at S, but this was wrong.
    Then I thought to use
    ∫[itex]\vec{B}[/itex] . [itex]\vec{dL}[/itex] = μoI
    but B is not constant over the interval [S,P] so I can't pull it out of the integral.

    I was able to calculate the integral along the dotted path in the picture, which I basically did by realizing R to S was perpendicular to the field so it didn't count, and that P to R was 1/8 of a larger circle drawn around the diagram. But in this case, B was a constant distance from the centre.

    Any input is appreciate. Thanks!
     
  2. jcsd
  3. Mar 10, 2012 #2

    tiny-tim

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    hi diethaltao! :smile:

    how about doing it along a path 8 times as long, in a square? :wink:
     
  4. Mar 10, 2012 #3
    Hi tiny-tim!

    So, I would find r using the Pythagorean Theorem:
    r = [itex]\sqrt{((0.21)(0.6))^2+(0.21-(0.21)(0.6))^2}[/itex] = 0.151.
    So I would multiply the field (which I found at point P to be 4.19E-6) by the perimeter of the square?
    And the perimeter would be 8R...? Obviously I messed up somewhere.
     
  5. Mar 11, 2012 #4

    tiny-tim

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    hi diethaltao! :smile:

    (just got up :zzz:)

    why do you want to know the field?

    try using the Ampere-Maxwell law :wink:
     
  6. Mar 11, 2012 #5
    Hi tiny-tim!

    Ampere-Maxwell's law says that [itex]\vec{B}[/itex].[itex]\vec{dL}[/itex] around a closed loop is proportional to the current enclosed in that loop, or
    [itex]\oint[/itex][itex]\vec{B}[/itex].[itex]\vec{dL}[/itex] = μo*Ienc.
    But I can't just plug in the values of μo and Ienc to solve for the integral. And if I choose a square to be my enclosed area, B wouldn't be the same along the side, correct? For example, the field at point S would be stronger than at point P.
     
  7. Mar 11, 2012 #6

    tiny-tim

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    yes, but if eg you extend PS to a point T on the y axis,

    then ∫ B.dl along PS will be the same as ∫ B.dl along ST, won't it? :wink:
     
  8. Mar 11, 2012 #7
    Ok, I understand that.
    So now I have something like the picture attached, where R is the hypotenuse of a 0.21 by 0.21 triangle (not sure if I need that value, but I calculated it anyways.)
    The integral from P to S is the same as the integral of T to S.
    And therefore, the integral from P to T would be the same as it would be for the other three sides of the square.
    I tried integrating around the whole square and dividing it by 8, but that was incorrect.

    http://i89.photobucket.com/albums/k211/diethaltao/Untitled.png
     
    Last edited: Mar 11, 2012
  9. Mar 12, 2012 #8

    tiny-tim

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    hi diethaltao! :wink:

    use the Ampere-Maxwell law

    ∫ B.dl = … ? :smile:
     
  10. Mar 12, 2012 #9
    Oh, wow, I get it now.
    Can't believe I overlooked something so simple! Thanks so much! :smile:
     
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