# Outside the origin circular loop current density

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1. Aug 13, 2015

### Mr. Rho

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?

Last edited: Aug 13, 2015
2. Aug 13, 2015

### stedwards

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
3. Aug 13, 2015

### Mr. Rho

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

4. Aug 13, 2015

### stedwards

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.

5. Aug 13, 2015

### Mr. Rho

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:

The current density I present is case (i) and the current density I need is case (ii).

6. Aug 14, 2015

### stedwards

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