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Magnetic field on an axis

  1. Nov 1, 2009 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    I must calculate the magnetic field for any point [tex]P[/tex] on an axis passing through the middle of a square wire loop (whose radius is [tex]a[/tex] and current is [tex]I[/tex]).
    The situation is such that the square wire is on a plane that is perpendicular to the axis in which I must calculate the magnetic field. If the situation is unclear, please tell it to me so that I scan a sketch of the situation.


    2. Relevant equations

    Biot-Savart, despite the symmetry, I don't think that Ampère's law can be useful.

    3. The attempt at a solution

    I drew a sketch as follow : a horizontal axis (called x-axis) passing in the center of a square with radius [tex]a[/tex]. I drew the square such that its sides coincide with up-bottom-left-right directions, for the sake of simplicity.
    Let [tex]\theta (x)[/tex] be the angle measured from the point [tex]P[/tex], between the x-axis and the upper side of the square. I have that [tex]\theta (x)= \frac{a}{2} \cdot \frac{1}{x} \Rightarrow \theta = \arctan \left ( \frac{a}{2x} \right )[/tex].
    I've found this angle because I wanted to know the angle within [tex]\vec l[/tex] and [tex]\vec r[/tex] in Biot-Savart law. This is where I'm stuck. Am I in the right direction?

    From B-S law : [tex]\vec B =\frac{\mu _0}{4 \pi} I \int d\vec l \times \frac{d \vec r}{r^3}[/tex].

    So I wanted to calculate the contribution of each side of the square on the magnetic field situated at point [tex]P[/tex].
     
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  3. Nov 1, 2009 #2

    rl.bhat

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    Consider a current element dl at a distance l from the center of the conductor.
    Can you find the distance r from dl to P?
    If O is the center of the loop, draw a parallel line through O to the upper wire.If A is the position of dl from the center of the wire, Draw AB perpendicular to the parallel line. Now you can find the angle BAP which is needed in Biot-Savart Law
     
  4. Nov 2, 2009 #3

    fluidistic

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    Thank you for helping!


    Hmm, on the axis passing through the center of the loop and by point P?
    I guess yes, but I'm unsure where to place dl.
    Waiting for a little clarification.

    Anyway, here comes my sketch, in case of confusion :
     

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  5. Nov 2, 2009 #4

    rl.bhat

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    dl is a small element of any one wire at a distance l from the center of the wire.
     
  6. Nov 2, 2009 #5

    fluidistic

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    Ah then [tex]r=\sqrt{x^2+l^2}[/tex] where [tex]x[/tex] is the distance from the center of the conductor (see figure).
    I will try to follow your instructions.

    Edit: I reached [tex]\theta= \arccos \left ( \frac{a}{2 \sqrt{x^2+l^2}} \right )[/tex]. Where [tex]\theta[/tex] is the angle BAP.
     
    Last edited: Nov 2, 2009
  7. Nov 2, 2009 #6

    rl.bhat

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    The distance AP = sqrt[l^2 + x^2 + (a/2)^2]
    And sin( BAP) = l/sqrt[l^2 + x^2 + (a/2)^2]
    Direction of th field is not in the direction of the x-axis. You have to resolve them to get the net field along x-axis.
     
    Last edited: Nov 2, 2009
  8. Nov 2, 2009 #7

    fluidistic

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    Ok thanks a lot. I will try to understand my error about the distance AP.
     
  9. Nov 4, 2009 #8

    fluidistic

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    Sorry to bother again, I still don't reach the angle BAP.
    Here's the picture of how I understood how to plot everything. Please tell me if there's a confusion somewhere.
    Which gave me [tex]\sin \theta = \sqrt{r^2- \frac{a^2}{4}} \cdot \frac{1}{r}[/tex] where [tex]r= \sqrt{l^2+x^2}[/tex].
    I don't really know why you've searched for [tex]\sin \theta[/tex] instead of [tex]\cos \theta[/tex] since [tex] \vec l \times \vec r = lr\cos \theta[/tex].
    At last I obtained [tex]B= \frac{\mu _0Ia^2}{8 \pi r^3}[/tex] which I almost know it's wrong. I should probably get something over [tex]r[/tex], not [tex]r^3[/tex].
     

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  10. Nov 4, 2009 #9

    rl.bhat

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    In the figure OB = l, OP = x and BP = sqrt(l^2 + x^2)
    BA = a/2. So AP = r = sqrt[l^2 + x^2 + (a/2)^2]
    AXB = ABsinθ.
    Here sinθ = x/r.
     
  11. Nov 4, 2009 #10

    fluidistic

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    I understand all until you wrote "Here sinθ = x/r". So you've considered the angle θ as [tex]OPA[/tex] ? If so then everything's fine.
    I still don't see why I needed [tex]\sin \theta[/tex] instead of [tex]\cos \theta[/tex].
     
  12. Nov 4, 2009 #11

    rl.bhat

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    If A' is the point on the wire vertically above O, to find the field at P, I want the angle A'AP and sine of that angle. To find the field we need the angle between current element I*Δl and r.
     
  13. Nov 4, 2009 #12

    fluidistic

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    Ah ok, I think I see a bit better. At least I understand why we search this angle [tex]\theta[/tex], but still not its sine.
    However in this case I have that the distance between A' and P is [tex]\sqrt {\left ( \frac{a^2}{2} \right ) +x^2}[/tex] and not x. Thus [tex]\sin \theta = \sqrt {\left ( \frac{a^2}{2} \right ) +x^2} \cdot \frac{1}{r}= \sqrt {\left ( \frac{a^2}{2} \right ) +x^2} \cdot \sqrt{l^2+x^2}[/tex].
     
  14. Nov 4, 2009 #13

    rl.bhat

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    Yes. Your sinθ is correct. But what is l? dl is a small segment on the wire. So put OA = sqrt(OA'^2 + A'A^2). And write 1/r
     
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