1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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.

    PastPaper.jpg
     
  5. Aug 8, 2012 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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 :).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Ampere's circuital law
  1. Ampere's Circuital Law (Replies: 4)

  2. Ampere's Law (Replies: 1)

Loading...