# Magnetic field/simple current loop

1. Oct 7, 2008

### granpa

given a current loop, centered at the origin, and tilted 90 degrees so that it enters at x=1 and exits at x=-1, carrying a unit amount of current, and completely disregarding the z axis.

is the strength of the magnetic field at any point in the xy plane proportional to 1/(distance from 1,0)^2 - 1/(distance from -1,0)^2

in other words does it have an inverse square relation to the current passing through those two points.

I know there are better ways to calculate the net field but I'm looking to understand what is happening here at an intuitive level.

I need the whole field. not just the far field or some sort of approximation.

Last edited: Oct 7, 2008
2. Oct 7, 2008

### granpa

what I mean is:

can I find the contribution of each of the 2 currents separately by simply finding the vector which is at a right angle to the vector from that point to the current and making its magnitude equal to 1/(distance to the current)^2

and is adding those 2 vectors all that i need to do to calculate the total field

3. Oct 7, 2008

### clem

If it is a closed loop, B does not go like 1/r^2.

4. Oct 7, 2008

5. Oct 7, 2008

### granpa

ok. so evidently thet wont work.

thinking 3 dimensionally, if we look at only one of the 2 currents (due to current along line element dL) then we can see from the symmetry that its field is confined to a wedge extending from the origin through the endpoints of that line element.

I have no idea how to go about calculating the resulting field. I've never seen or done anything like it.

6. Oct 7, 2008

### Ben Niehoff

To find the magnetic field of a finite loop, you will need to use elliptic functions. It's complicated.

However, it can be proven in general that

$$\vec B(\vec r) \propto \nabla \Omega(\vec r)$$

where $\Omega(\vec r)$ is the (oriented) solid angle subtended by the current loop at the observation point.

7. Oct 7, 2008

### granpa

any connection to the scalar potential of a magnetic field?

8. Oct 8, 2008

### clem

Yes, Omega is the magnetic scalar potential, which happens to equal the solid angle subtended by a current loop of any shape.
For a circular current loop, the MSP can be found be either an elliptic integral, or by a Legendre polynomial, partial wave expansion.

9. Oct 8, 2008

### granpa

so the msp isnt so much discontinuous as multi-valued (like an angle)?

Last edited: Oct 8, 2008
10. Oct 8, 2008

### Ben Niehoff

It's multivalued, or discontinuous if you make a branch cut.

However, in free space (in the absence of any currents), it is single-valued, as in this case (there is no J at the observation point).