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Flux, how to find n (normal) and derivation of formula?

  1. Aug 8, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the flux of the following fields:
    F_1 = xi + yj
    F_2 = -yi + xj

    across the following curve: The circle r(t) = (cost) i + (sint) j
    t is from [0,2pi]


    2. Relevant equations

    Flux = ∫F dot n ds = ∫M dy - N dx

    3. The attempt at a solution

    For F_1 I got:
    M = x = cos t
    N = y = sin t
    dy = cost t dt
    dx = -sin t dt

    Flux for F_1 = ∫[0,2pi] cos^2 t + sin^2 t dt = ∫[0,2pi] dt => 2pi

    For F_2 I got:
    M = -y = -sin t
    N = x = cos t
    dy = cos t dt
    dx = - sin t dt

    Flux for F_2 = ∫[0, 2pi] -costsint + costsint dt = ∫0 = 0

    -------
    My answers are correct (can someone verify if method was used correctly?), but the answer key used a different method:
    It said "n = <cost , sin t>, and proceeded to dot that with each field.

    My concern was how n was found, I thought n = T' / |T'|, where T = r' / |r'| :
    so: r' = <-sint, cos t> = T (because |r'| = 1)
    T' => <-cost, -sint> = n (because |T'| = 1 as well), so how did they get their n?

    Also, I'm a bit unsure on how Flux = ∫M dy - N dx is derived -- in the textbook they did:
    n = T x K = (dx / ds i + dy/ ds j) x k = (dy / ds i - dx/ ds j), but if someone can spell that out for me I'd appreciate it.

    Thanks
     
  2. jcsd
  3. Aug 8, 2012 #2
    Formula [itex]n=\frac{T'}{|T'|}[/itex] is incorrect, try parabola [itex]r(t)=[t,t^2][/itex]. Vector product [itex]T\times k[/itex] is perpendicular to [itex]k[/itex] which means it lies in [itex]XY[/itex] plane. It is also perpendicular to [itex]T[/itex] so it is normal to the curve.
     
  4. Aug 8, 2012 #3
    Is the normal vector here in the flux case, different from the "unit vector normal" ? If so, how is n for flux calculated then?
     
  5. Aug 10, 2012 #4
    No, it is not different, except [itex]T\times k[/itex] is not unit vector, so [itex]n=\frac{T\times k}{|T\times k|}[/itex]
     
  6. Aug 10, 2012 #5

    LCKurtz

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    They just observed that the curve was a circle and the position vector to the circle is perpendicular to it.
     
  7. Aug 10, 2012 #6
    Oh that makes sense, thanks for the clarifications szy and LC.
     
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