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Magnetic field due at a point due to a wire outlining an ellipse

  1. Apr 9, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the magnetic field due to a curved wire segment.


    2. Relevant equations
    Biot-Savart Law (differential form)

    dB=[itex]\frac{\mu_{o}i}{4\pi}[/itex] [itex]\frac{d\vec{S}\times \hat{r}}{r^{2}}[/itex]

    3. The attempt at a solution

    In class we found the magnetic field at a point in space (point P) caused by the current running through a wire. Point P is equidistant from every point on the wire call this distance R. dS is a differential element that points along the wire and r hat points toward point P. The angle between dS and r hat is 90 degrees at all points.

    The point P is essentially the center of a circle and the wire outlines the edge of a circle.

    B=[itex]\frac{\mu_{o}i}{4\pi}[/itex]∫[itex]\frac{d\vec{S}\times \hat{r}}{r^{2}}[/itex]

    B=[itex]\frac{\mu_{o}i}{4\pi}[/itex]∫[itex]\frac{|d\vec{S}||\hat{r}|sinθ}{r^{2}}[/itex]

    B=[itex]\frac{\mu_{o}i}{4\pi R^{2}}[/itex]∫[itex]dS(1)sin90^{o}[/itex]

    B=[itex]\frac{\mu_{o}i S}{4\pi R^{2}}[/itex] S=R[itex]θ_{1}[/itex]
    where [itex]θ_{1}[/itex] is the angle swept out between one end of the wire and point P and the other end the wire and point P.

    B=[itex]\frac{\mu_{o}i R θ_{1} }{4\pi R^{2}}[/itex]

    B=[itex]\frac{\mu_{o}i θ_{1} }{4\pi R}[/itex]

    OK punchline.

    I was thinking how could I find the magnetic field at point P due to a piece of wire that outlines a portion of an ellipse. This would mean that the angle between dS and r hat would be a different angle at every point on the wire and the distance between point P and the wire would be different at every point along the wire.

    So the integral would be

    B=[itex]\frac{\mu_{o}i}{4\pi}[/itex]∫[itex]\frac{dS sinθ}{r^{2}}[/itex] where S,θ, and r are all variables. How would I integrate this?

    dS=drdθ where dθ the angle between dS and r hat and dr is the infinitesimal change of the radius as the integral adds from one end of the wire to the other end.

    B=[itex]\frac{\mu_{o}i}{4\pi}[/itex]∫[itex]\frac{drdθ sinθ}{r^{2}}[/itex]

    This is as far as I can get. Any help would be appreciated. Thanks.

    In retrospect maybe I shouldn't say the wire outlines an ellipse. I just want a wire the satisfies the conditions that the angle between dS and r hat changes at every point along the wire and the distance between point P and the wire changes at every point along the wire. Thanks again.
     
    Last edited: Apr 9, 2014
  2. jcsd
  3. Apr 9, 2014 #2
    Are you thinking the wire is a portion of the ellipse or the whole thing? I got the impression you were talking about a piece of wire that was just curved, but if it is a whole ellipse then the field will be zero t the center just like a circle.

    Were I doing it I would use the equation for an ellipse in the integral and go from there.

    An ellipse is

    [itex]r(\theta) = \frac{ab}{\sqrt{(b^2\cos^2\theta)+(a^2\sin^2(\theta)}}[/itex]

    so plugging into B-S law:

    $$B = \frac{\mu_0I}{4\pi} \int \frac{r (b^2\cos^2\theta)+(a^2\sin^2\theta)d\theta}{(ab)^2}= \frac{\mu_0I}{4\pi (ab)}\int \sqrt{(b^2\cos^2\theta)+(a^2\sin^2 \theta)} d\theta$$

    which is a very ugly looking integral.
     
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