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Homework Help: Biot-Savart law My proof includes a negative sign

  1. Jan 1, 2012 #1
    1. The problem statement, all variables and given/known data
    A square coil of wire of side length 2a lies in the yz plane. A current I flows through the loop. The x axis is defined such that it passes through the centre of the loop, with the loop boundaries being at z±a and y=±a respectively. Given that I flows in a direction anticlockwise with respect to the x axis use the Biot Savart law to show that the B field at a point P along the positive x axis due to the side of the loot at z=a is given by

    B=[itex]\frac{I*u_{0}}{4* pi}[/itex]∫(x[itex]\hat{k}[/itex] + a[itex]\hat{i}[/itex])dy/(y[itex]^{2}[/itex] + a[itex]^{2}[/itex] + x[itex]^{2}[/itex])[itex]^{3/2}[/itex] . This integral goes from negative to positive a.

    2. Relevant equations
    Biot Savart law, written in the solution. Note that dl has direction, same as current.

    3. The attempt at a solution

    Shown in the image

    My issue is that I get a negative sign in my solution, thus I believe that the negative sign was forgotten in the given answer of the problem.

    http://img263.imageshack.us/img263/8818/solutiondj.jpg [Broken]
    1. The problem statement, all variables and given/known data
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 1, 2012 #2
    I should remove the sign in dl=-dly, that will fix things, and also makes the magnetic field at point p be appropriate.
  4. Jan 2, 2012 #3

    rude man

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    The given answer is correct:
    r = x i - y j - a k
    dl = -dy j
    dl x r = a*dl i + x*dl k
    dB = (μ0/4π)(a*dy i + x*dy k)
    B = (μ0/4π)(a i + x k){∫dy/r^3}

    where r^2 = x^2 + a^2 + z^2
    and the integration is from y= -a to y = +a.

    Note that in the answer the term a i + x k can be taken outside the integration sign.
    Last edited: Jan 2, 2012
  5. Jan 2, 2012 #4

    rude man

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    Your error was integrating from a to -a. Why did you do that?
  6. Jan 2, 2012 #5
    Hi! thanks for you're response. I integrated from a to negative a because it is a line integral and the curve is from a to negative a (because the current flows in that direction)
    therefore I integrated along the direction of the current ( also because my problem is set up to be integrated like that)

    However, i believe the problem lies in dl and the integration. If i choose that dl=-dyj then i have to integrate form negative
    a to a to get it correct. But if I choose dl=dyj and i integrate it from negative a to a then it becomes correct.

    I read about the current density which the biot-savart law comes from. The current density has a direction along the current,
    by changing a volume integral into line integral, dl and changing current density to current, dl is in direction of current density, current, which gives me -dyj.
    but i believe i should have dyj and the direction of the integral takes care of the sign for dyj.
    Last edited: Jan 2, 2012
  7. Jan 2, 2012 #6

    rude man

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    Well, you have to be careful with your integration.

    You're integrating wrt y, not l. So you have to integrate from - to +. There is no choice.

    The direction of current is taken care of when you defined dl = - dy. From then on it's strictly math; the physics is left in the dust!

    Anyway - all's well ....
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