Outside the origin circular loop current density

AI Thread Summary
The discussion focuses on deriving the current density for a circular loop displaced along the y-axis in spherical coordinates. Initially, the user presented an incorrect equation for the current density, prompting feedback about unit balance and the need for clarity regarding the parameters involved. After revisions, the correct expression for the current density was provided, ensuring it satisfies the integral condition for current. The user then clarified their request by suggesting a method to translate the origin-centered solution into Cartesian coordinates before converting it back to spherical coordinates. This approach aims to accurately represent the current path for the displaced circular loop.
Mr. Rho
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Hi, I'm trying to write the current density for such circular loop in spherical coordinates. For a circular loop of radius a that lies in the XY plane at the origin, the current density it's simply:

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

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

Any suggestions?
 
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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?
 
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stedwards said:
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?
Sorry I wrote the equation wrong, just fixed it. I'm using this kind of spherical coordinates:

250px-Spherical_polar.png
 
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.
 
stedwards said:
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.

I don't know what I was thinking, the correct current density is:

\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}​

it satisfies I=\int\mathbf{J}\cdot{d\mathbf{S}}=\int_{0}^{\pi}\int_{0}^{\infty}\mathbf{J}\cdot{\hat{\phi}}rdrd\theta, where \mathbf{S} 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).
 
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.
 
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