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Outside the origin circular loop current density

  1. Aug 13, 2015 #1
    Hi, I'm trying to write the current density for such circular loop in spherical coordinates. For a circular loop of radius [itex]a[/itex] that lies in the [itex]XY[/itex] plane at the origin, the current density it's simply:

    [itex]\mathbf{J}= \frac{I}{2\pi\sin\theta}\delta(\theta-\frac{\pi}{2})\frac{\delta(r-a)}{a}\hat{\phi}[/itex]​

    I want the current density of the circular loop of radius [itex]a[/itex] displaced a distance [itex]c[/itex] towards the [itex]y[/itex] axis.

    Any suggestions?
    Last edited: Aug 13, 2015
  2. jcsd
  3. Aug 13, 2015 #2
    I can't make sense of this. Your units don't balance. What is the relationship between a, c, and r? Could you try again?
    Last edited: Aug 13, 2015
  4. Aug 13, 2015 #3
    Sorry I wrote the equation wrong, just fixed it. I'm using this kind of spherical coordinates:

  5. Aug 13, 2015 #4
    No, really. Think about it a bit and restate the entire question. 'cuse, now current and current density have the same units, and nobody knows what ##c## is. I'd sleep on it.
  6. Aug 13, 2015 #5
    I don't know what I was thinking, the correct current density is:

    [itex]\mathbf{J}=I\delta(\theta-\frac{\pi}{2})\frac{\delta(r-a)}{a}\hat{\phi} = I\sin\theta\delta(\cos\theta)\frac{\delta(r-a)}{a}\hat{\phi}[/itex]​

    it satisfies [itex]I=\int\mathbf{J}\cdot{d\mathbf{S}}=\int_{0}^{\pi}\int_{0}^{\infty}\mathbf{J}\cdot{\hat{\phi}}rdrd\theta[/itex], where [itex]\mathbf{S}[/itex] is a surface perpendicular to the current direction.

    Sorry for not making myself clear for what I'm asking. I hope this image makes things clear:

    The current density I present is case (i) and the current density I need is case (ii).
  7. Aug 14, 2015 #6
    To begin with, take the origin-centered solution for a circle of radius ##a##, change to Cartesian coordinates, translate to the right (##x \leftarrow x' = x + c##), then back to spherical coordinates.

    It will give the equation for the current path you want.
    Last edited: Aug 14, 2015
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