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Simple line integral:confused with directions

  1. Nov 2, 2013 #1
    5-P-E-P-F%20(10).PNG

    I am supposed to do the integral ∫(1/Ω)dl, where dl is the vector pointing along the loop and Ω is the distance from the origin to a point on the loop.

    Let region 1 be from x=a to x=b
    region 2: outer semicircle of radius b
    region 3: x=-b to x=-a
    region 4: inner semicircle of radius a.

    Let's just look at regions 1 and 3:
    in these regions, dl=dx x, x is the unit vector. dx x is a positive quantity, since it is parallel (not antiparallel) to the arrows in the above figure drawn in these regions (my path of integration). Also in regions 1 and 3, Ω=x.
    Thus i set up the integral in region 1 as:

    x∫(1/x)dx , with limits from a->b

    and in region 3:

    x∫(1/x)dx , with limits from -b->-a.

    However, when I add these two results, they cancel, but the solution says they should add. I am having same problem with regions 2 and 4 (they should cancel but in my case they add). Thus I must probably be confused with signs/directions, but i do not see where my error is.

    Help is much appreciated!
     
  2. jcsd
  3. Nov 2, 2013 #2

    tiny-tim

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    Hi Aziza! :smile:
    nooo :wink:
     
  4. Nov 2, 2013 #3
    huh?

    if you agree with my region 1 set up, then region 3 is same...except for the limits
     
  5. Nov 2, 2013 #4

    tiny-tim

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  6. Nov 2, 2013 #5
    ohhh ;DDDD yes you are right, thank you!

    However i am still having trouble with regions 2 and 4:

    region 2 i set up as:
    (1/b)∫dl2 from 0->pi

    and region 4:
    (1/a)∫dl4 from pi->0

    Now,
    dl2 = bθdθ, where θ is unit vector and positive if pointing in direction of increasing θ (ie counterclockwise)

    So since in region 4, θ is pointing clockwise:
    dl4 = -aθ

    The sum of regions 2 and 4 however then gives me 2∫dθ from 0->pi, but it should give zero...
     
    Last edited: Nov 2, 2013
  7. Nov 2, 2013 #6

    tiny-tim

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    that (dθ) doesn't make any sense :redface:

    the RHS needs to be a vector, but dθ isn't a vector
     
  8. Nov 2, 2013 #7
    oops you are right, that is not wat i meant. i edited the post above,
     
  9. Nov 2, 2013 #8

    tiny-tim

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    why the minus? :confused:

    i think you're being too analytical about this whole problem …

    just use common-sense: look at the diagram, and remember that ∫dl is always the length (and is always positive) :wink:
     
  10. Nov 2, 2013 #9
    but in the end what i am calculating is the magnetic vector potential, so the result of that integral must be a vector.

    lol i have no common sense, thus i really need the formal mathematical way of doing this
     
  11. Nov 2, 2013 #10

    tiny-tim

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    sorry, i got carried away, i meant to leave out the "∫" :redface:

    dl is always positive (ie if you use l instead of θ, you shouldn't get confused)
     
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