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Homework Help: Magnetic Fields and Two circular loops of wire

  1. Feb 24, 2010 #1
    1. The problem statement, all variables and given/known data

    Two circular loops of wire, each containing a single turn, have the same radius of 3.6 cm and a common center. The planes of the loops are perpendicular. Each carries a current of 2.0 A. What is the magnitude of the net magnetic field at the common center?

    I have been using this equation: B = uI/2(3.14)r

    Where u = 4(3.14) x 10^-7, I= 2.0 Amperes, and r = .036 meters.

    When I work through the equation this way, I end up with:

    B = 1.111 x 10^-5

    It tells me that this answer is wrong and I only have one more attempt at this problem.

    thanks.
     
  2. jcsd
  3. Feb 24, 2010 #2

    tiny-tim

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    Hi PhilCam! :smile:

    (have a pi : π and a mu: µ and try using the X2 tag just above the Reply box :wink:)
    Is that the magnitude for just one loop?

    You need to add the vectors for the two fields, and then take the magnitude of the resultant vector.
     
  4. Feb 24, 2010 #3
    Thanks for the reply!

    Would the magnitude for the loops be the same?

    I assume so, so would my final answer be:

    B Final = Square root of B2 + B2..?

    Thanks
     
  5. Feb 24, 2010 #4

    tiny-tim

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    Looks good! :smile:
     
  6. Feb 24, 2010 #5
    By doing the math, I finished with an answer of 1.571 x 10-5, but it says this is incorrect.

    I thought maybe I had done the calculations incorrectly, but double-checked and got the same thing.

    Is there something wrong with my formula?
     
  7. Feb 24, 2010 #6

    tiny-tim

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    ah, I think you may have had the wrong number of πs in your original formula …

    isn't it µ0 (which contains a 4π), divided by 4π, and then multiplied by the length of the wire, which also contains a 2π?

    (see http://en.wikipedia.org/wiki/Biot–Savart_law" [Broken])
     
    Last edited by a moderator: May 4, 2017
  8. Feb 24, 2010 #7
    Thanks, I'll check that out when I have a few more minutes.

    Thanks again.
     
  9. Feb 24, 2010 #8
    I'm having a difficult time applying the law as it is on the wiki page.

    It says that the r should be squared. And times by four pi. That four pi would cancel out with the four pi from µ.

    So I would just end up with r squared on the denominator?
     
  10. Feb 24, 2010 #9

    tiny-tim

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    Hi PhilCam! :smile:

    Yes (I think that's something to do with why µ0 is defined with that extra 4π).
    Yes, but it's partly balanced by a single r in the ∫dl on top (which comes out as the length of the wire, 2πr). :wink:
     
  11. Feb 24, 2010 #10
    I'm not sure what my final equation would look like. The original looks like

    B= uI/2(pi)r

    The pi would cancel out from the constant in the "u." I used this and still had the wrong answer. Any other ideas?
     
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