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Biot-savat integral

  1. May 14, 2006 #1
    so trying to understand the derivation of magentic field due to a long straight conductor:

    so biot-savat says in this case:

    B= u0I/4pi * integrate over L and -L for: xdy/(x^2+y^2)^(3/2)

    and i should turn up with the answear:

    B= (u0I/4pi)*(2L/xsqrt(x^2+y^2))

    but can't do the integration.. so well i understand biot-savat but yeah can't do the integration, would anybody maybe show me how to do this?
  2. jcsd
  3. May 14, 2006 #2


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    Fundamentally, that's
    [tex]\int_{-L}^L (a^2+ y^2)^{-\frac{3}{2}}dy[/itex]
    and should make you think of a trig substitution (any time you have a square inside a square root and cannot simply substitute for the function in the square root itself, think of a trig substitutin). If I remember correctly, 1+ tan2= sec2 so try [itex]y= x tan(\theta)[/itex]. Then [itex]x^2+ y^2= x^2+ x^2 tan^2(\theta)= x^2(1+ tan^2(\theta))= x^2sec^2(\theta)[/itex] and
    [tex](x^2+ y^2)^{-\frac{3}{2}}= x^{-3}sec^{-3}(\theta)=x^{-3}sin^3(\theta)[/tex]
    Also, [itex]dy= x sec^2(\theta)[/itex] so your integral becomes
    which, except possibly for evaluating at the limits of integration, is easy.

    Added in edit: I just noticed you say

    "and i should turn up with the answear:
    B= (u0I/4pi)*(2L/xsqrt(x^2+y^2))"

    Well, that's obviously wrong, isn't it? If you are integrating, with respect to y, from -L to L, the result will not involve y.
    Last edited: May 14, 2006
  4. May 14, 2006 #3
    Not unless he means y for a
  5. May 15, 2006 #4
    yes ment L^2 not y^2 at the bottom..

    thanks for pointing out that.

    so nobdy who knows exatcly how one goes from the firs tpoint of the bio savart law to the later (stated in the first post)

    yeah so let's sat that we can use the trig but then how do i get it out to "normal" form so i can evaluate?

    how did the people who found this in the first place do it?
  6. May 15, 2006 #5


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    Well, integral of sin(y) is -cos(y) so, evaluating at the limits of integration, we have [itex]-2cos(tan^{-1}(\frac{L}{x})[/itex].
    To evaluate that, either use trig identities or imagine right triangle with "opposite side" of length L and "near side" of length x (i.e. the tangent of the angle is L/x). It's hypotenuse has length (Pythagorean theorem) [itex]\sqrt{L^2+ x^2}[/itex] and so the cosine of that angle is [itex]\frac{x}{\sqrt{L^2+ x^2}}[/itex]. That should be simple enough.
  7. May 16, 2006 #6


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    Also a hyperbolic substitution ([itex] y=x \sinh t [/itex]) might work.


    P.S. And it's Jean Baptiste Biot and Felix Savart (sic!)
    Last edited: May 16, 2006
  8. May 16, 2006 #7


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    Here's a character description of Biot:

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