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Biot-Savart Law, right angle

  1. Mar 28, 2012 #1
    1. The problem statement, all variables and given/known data

    http://i.minus.com/1333003834/Q661ScjxBUxkrL2FfmVFPQ/iTEfM3UTAVtTa.png [Broken]


    2. Relevant equations

    B = ∫ ([μ0 / 4pi] * I * ds-vector x r-hat) / r^2

    3. The attempt at a solution

    I know the horizontal line will not add anything to the magnetic field (B), so focusing on the vertical line.

    I take a little bit of length (ds) which I will call dy.

    dy x r-hat = dy sin θ
    r = sqrt(x^2+y^2)
    sin θ = x / r

    do all your substitutions and get:

    B = ([μ0 / 4pi] * I * x ) ∫ dy / (x^2+y^2)^(3/2)

    At this point I am confused on my limits of integration, I know for an infinite long straight wire I use -∞ to ∞.

    In my notes I have an example where it goes from -y1 to y2 and comes out with

    B = ([μ0 / 4pi] * I ) / x * (cos θ1 - cos θ2)

    where θ1 is the angle between -y1 and the point i am finding, and θ2 is 180 - θ1.
    This whole θ thing is tripping me up, how did it get there ( I am assuming trig subsitutation). Further more how can I get a grasp on what θ1 would be?

    I guess -y1 in my situation is just y, and y2 is 0.

    so θ1 = inverse-tan (x/y) and thus θ2 = 180 - inverse-tan (x/y)?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Mar 29, 2012 #2

    tiny-tim

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    Hi ParoXsitiC! :smile:

    (try using the X2 and X2 buttons just above the Reply box :wink:)

    (and write arctan or tan-1, not inverse-tan)


    (You're rambling a bit :redface:, so I won't answer point-by-point)

    θ isn't a substitutuion, it's the actual angle, between the current and the line from the point to P :smile:

    You can either do ∫ dy, in which case your limits are the endvalues of y, in this case 0 and ∞

    or you can do ∫ dθ, in which case your limits are the endvalues of θ, in this case 0 and π/2.

    (btw, can't you just say it's half the value for a whole line?)
     
  4. Mar 29, 2012 #3
    That makes sense but how did they get to the two cosines mathematically?

    I guess my understand is that the formula is a simplified equation that does not include an integral, but I dont understand how it was derived, nor do I understand which angle is taken for θ2
     
  5. Mar 29, 2012 #4

    tiny-tim

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    oh, cos = adj/hyp (adjacent/hypotenuse)

    = y/√(x2 + y2) :smile:

    (and θ2 is the angle PYO, where Y = (0,y), as y -> ∞)
     
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