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Bio-Savart Law, current through wire in semi circle

  1. Feb 4, 2013 #1
    1. The problem statement, all variables and given/known data

    I solved this problem. I just have a general question at the very end.


    A current flows in a wire that has straight sections on either side of a semicircular loop of radius b. Find the mag and direction of the magnetic field B at point P(center of loop).

    ............................................... periods = mag field pointing out
    .................._____.................... x's = mag field pointing in
    .............../xxxxxxx\................. radius = "b" from point P to any part in the semi circle.
    ............../xxxxxxxxx\..............................y
    -->--->--| xxxxPxxxx|--->---->--..............|
    xxxxxxxxxxxxxxxxxxxxxxxxxxxxx......z(out)|___x

    2. Relevant equations

    Bio-Savart

    3. The attempt at a solution

    B = μ/4pi ∫(I(dl x r ))/r^2

    I started out by pulling out the constants and everything I knew. Since dl is in the same direction as the current, the magnetic field from the straight pieces of the wire does not contribute to the magnetic field at point P due to the angle between dl and r:

    B = μI/(4pi*b^2) ∫(dl x r )

    As the current goes around the semi circle, a perfect 90 degrees is maintained between dl and r so r cancels out( sin(90)= 1 )

    B = μI/(4pi*b^2) ∫ dl

    The next step is finding dl which is the sum of which dl travels around the semi circle. Since it is half of a circle it would be pi multiplied by the radius which is b:

    B = μI*pi*b/(4pi*b^2) >>>>>> B = -μI/4b in the negative z direction due to the magnetic field.


    --------------
    My Question
    --------------

    What if the semicircle was not a semicircle, but a square that was cut in half? How would I deal with finding dl then?
     
  2. jcsd
  3. Feb 6, 2013 #2
    It sounds like your confusion has to do with computing the cross product [itex]d\mathbf{\vec{l}}\times \mathbf{\vec{r}}[/itex]. I would suggest breaking the wire up into the three separate sections, for example, the integral for the longest part of the half-square would have [itex]d\mathbf{\vec{l}}= dx\mathbf{\hat{x}}[/itex], since the current of that segment lies parallel to the [itex]\hat{\mathbf{x}}[/itex] axis. Then you just need to compute the cross product. Think about what [itex]\mathbf{\vec{r}}[/itex] could be; you are integrating along the current in the wire that lies at [itex]y=b/2[/itex] and [itex]z=0[/itex], from [itex]x=\left[-b/2,b/2\right][/itex]. You just need to compute the cross-product of the two quantities [itex]d\mathbf{\vec{l}}\times \mathbf{\vec{r}}[/itex] once you have this.
     
  4. Feb 6, 2013 #3

    rude man

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    Homework Helper
    Gold Member



    You would have to use the Biot-Savart law using vectors.

    Set up an x-y coord. system with P at the origin. The square is bounded by (-a,0), (-a,2a), (a,2a) and (a,0). I use i and j for unit vectors.
    Biot-Savart: dB = kI(dl x r)/r^3

    For the left vertical part of your square (x = -a): dl = dy j and r = a i - y j. r = |r|.
    Integrate from y = 0 to y = 2a.

    For the horizontal stretch y = 2a, dl = dx i and r = -x i - 2a j. Integrate from x = -a to +a.
    Etc. Get the idea?
     
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