Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:

    250px-Spherical_polar.png
     
  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:

    Untitled.png
    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook