# The solid angle equation for a magnetic field of a loop

1. Jan 6, 2015

### Coffee_

Background: Using Biot-Savart law we proved that for a closed loop with current $I$, the magnetic field at a point P was equal to $\vec{B}=-\frac{\mu_{0} I \nabla{\Omega}}{4 \pi}$ where $\Omega(x,y,z)$ is a function of the position of P that represents the solid angle at which the loop is seen from P. The convention for the sign of $\Omega$ is that using the right hand rule to define the normal vector $\vec{n}$, the solid angle is positive at the side towards which $\vec{n}$ points.

Problem: When trying to derive the circulation/closed loop integral of $\vec{B}$ using this information it's easy to see that it's 0 for any closed path that does not pass through this loop which agrees with Maxwell. However when going through the loop, I get a wrong sign. Let's take the situation where the integration path taken penetrates the loop from the negative side of the solid angle towards the positive. The only contribution for the passing through is from the penetration of the loop. $\nabla \Omega . \vec{ds}=d\Omega=4\pi$ in this case. Using this reasoning I get Circulation($\vec{B}$)=$-\mu_{0}I$ instead of a plus sign. Where have I messed up?

2. Jan 6, 2015

### TSny

Don't forget the sign convention for $I$ in Ampere's circuital law: $\oint \vec{B} \cdot d\vec{s} = \mu_0 I$.

BIG EDIT: Although the sign convention for $I$ is important, I don’t think that’s what’s causing the confusion.

The confusion is dealing with the discontinuous change in the solid angle $\Omega$ when passing through the loop. I'm also confused by it. Although $\Omega$ is discontinuous as you pass through the loop, the gradient of $\Omega$ is essentially continuous as you pass through the loop. It appears that you can get the correct result if you ignore the sudden change in $\Omega$ itself.

For example, consider the picture below which shows a current loop in black and a path of integration of $\vec{\nabla} \Omega$ in blue that passes through the loop from the side where $\Omega$ is negative to the side where it is positive. Let P be a point on the path of integration that is off to the side of the current loop such that the solid angle subtended by the current loop is zero at P. Consider the value of the part of the line integral of $\vec{\nabla} \Omega$ that starts at P and ends at the center of the loop at point O. Then consider the value of the remaining part of the line integral that starts at O (just on the other side of the loop) and returns to point P. Then add the two parts together to get the line integral of $\vec{\nabla} \Omega$ for the entire path of integration. I think you'll see that you get a result in agreement with Ampere's law.

But this is not too satisfactory. The discontinuity of $\Omega$ means that its gradient has a delta-function behavior at the region of passing through the loop. And I have ignored this. It seems to me that if you include the delta function contribution, then you get zero for the line integral of $\vec{\nabla} \Omega$ around the entire path. One argument for ignoring the delta function might be that we know B is continuous as we pass through the loop. Since B is supposed to be proportional to $\vec{\nabla} \Omega$, we need to treat $\vec{\nabla} \Omega$ as continuous (i.e., ignore the delta function behavior). This is a very hand-wavy argument which deserves criticism.

Hopefully, others will chime in. Sorry for the less than satisfactory response.

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• ###### Circuital law.png
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Last edited: Jan 7, 2015
3. Jan 7, 2015

### Coffee_

First of all thank you for the elaborate response.

A small comment on this latter reasoning. I don't see how it gets in agreement with Ampere's law. To me it seems that the way you describe it it would still give zero even though a current penetrates the path.

4. Jan 7, 2015

### TSny

What do you get for the change in $\Omega$ for the half of the path that takes you from P to O? For the other half that goes from O to P?

5. Jan 7, 2015

### Coffee_

Oh! Now I see. Considering it like that does indeed make sense.

6. Jan 7, 2015

### Coffee_

Thanks a lot for the help.

7. Jan 7, 2015

### TSny

I don't know how to do two loops of current as shown below. How would you even define $\Omega$ in this case? The scalar magnetic potential seems to have some limitations in its use.

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