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Ampere's Law

  1. Oct 28, 2009 #1
    1. The problem statement, all variables and given/known data

    Capture.JPG

    2. Relevant equations

    [tex]\int \vec B \cdot d\vecs = \mu i_{enc}[/tex]

    3. The attempt at a solution

    This should be straight forward. I'm really starting to become frustrated with this textbook.

    Amperian Loop 1 encircles both current loops.

    [tex]\int \vec B \cdot d\vecs = \mu i_{enc}[/tex]
    [tex]\int \vec B \cdot d\vecs = (1.26x10^-6) i_{enc}[/tex]

    The enclosed current is *using the right hand rule* +6A and -15A = -9Amps.

    1.26x10^-6(-9) = -1.1x10^-5 which is apparently wrong.

    What the heck is wrong?
     
    Last edited: Oct 28, 2009
  2. jcsd
  3. Oct 28, 2009 #2
    Did you remember the minus? :)
     
  4. Oct 28, 2009 #3
    That's a typo. I actually DID submit -1.1e-5 which is wrong.
     
  5. Oct 28, 2009 #4
  6. Oct 28, 2009 #5
    Yup. I doublechecked. That's what I put in, the negative.
     
  7. Oct 29, 2009 #6
    From a homework study group on Facebook. Apparently, this gives the right answer:
    Why do I use both positive 15 and postive 6 in the equation??
     
  8. Oct 29, 2009 #7
    You don't - notice it says "(-1 * first red number)" which would give a -15
     
  9. Oct 29, 2009 #8
    It's still not what we computed though. Either you use a negative 15A current, and a positive 6A current and get -9, or you get -21A as they compute. Why would you get -21? Why would you use both negative currents?
     
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