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Ampere's circuital law

  1. Aug 8, 2012 #1
    1. The problem statement, all variables and given/known data

    State Ampere's circuital law. The figure shows a region A in which the magnetic field [itex]H[/itex] and of value 20Am^-1 to the right and a region B in which the magnetic field [itex]H[/itex] is uniform and of value 30Am^-1 to the left. Find the current enclosed by the dashed rectangular loop, which has dimensions 8cm x 5cm.

    2. Relevant equations

    [itex]\int_c H\cdot dl[/itex][itex]= I_{enc}[/itex]

    3. The attempt at a solution

    So we already have a constant value for H for both segments and they are constant so I think its adding together two different values. So..

    [itex] H \int \cdot dl = I_{enc} = H * (2x + y) [/itex] for x = 4cm and y = 5cm?

    Or am I missing something?
  2. jcsd
  3. Aug 8, 2012 #2


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    hi zhillyz! :smile:

    (try using the X2 button just above the Reply box :wink:)
    i don't understand :redface:

    where is the electric wire (for the current)?

    what shapes are A and B, and where are they in relation to each other and the wire, and is anything moving? :confused:
  4. Aug 8, 2012 #3
    Apologies perhaps the image itself will clarify and yes I have been that busy brushing up my Latex I forgot the more basic tools were there lol.

  5. Aug 8, 2012 #4


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    ah! :smile:
    what does the "dot" do to the "vertical" bits? :wink:

    (and you'll need an H1 and an H2)
  6. Aug 9, 2012 #5
    For each half of the rectangles vertical sides the magnetic field moves in equal an opposite directions and therefore would cancel each other out? So it is just..

    [itex]I_{1enc} = H_1 \cdot y = 20\cdot 0.05 = 1A[/itex]
    [itex]I_{2enc} = H_2 \cdot y = 30\cdot 0.05 = 1.5A[/itex]

    Seem correct?
  7. Aug 9, 2012 #6


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    hi zhillyz! :smile:
    no, you're missing the point (or the dot) …

    H.dl is 0 because H is perpendicular to that part of the circuit! :rolleyes:

    otherwise correct :smile:
  8. Aug 9, 2012 #7
    Thank you very kindly :).
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