Outside the origin circular loop current density

Mr. Rho
Messages
14
Reaction score
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:
on Phys.org
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:
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:

[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).
 
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:

Similar threads

  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
16K
Replies
8
Views
2K