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

#### 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?

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#### TSny

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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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#### Coffee_

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.
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.

#### TSny

Homework Helper
Gold Member
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?

#### Coffee_

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

#### Coffee_

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?
Thanks a lot for the help.

#### TSny

Homework Helper
Gold Member
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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#### Oxvillian

I stumbled on this thread when struggling with and then googling the same issue - sorry to be 4 years late!

I think we can resolve things by treating the solid angle subtended as a variable that is only defined modulo $4\pi$. The decision to "reset" it from $2\pi$ to $-2\pi$ as we cross through the loop is purely arbitrary. We might just as well keep on increasing it from $2\pi$ all the way up to $4\pi$ (the "inside" of the loop becomes the "outside") and then "reset" it back to zero. Or even better, not bother with the resets and pick up an extra $4\pi$ every time we complete a circuit.

One might make a similar complaint when integrating around a pole, twice, in the complex plane. There is no delta function to integrate over when $\theta$ "resets" from $2\pi$ to zero.

I guess the implication of all this is that we are forced into the conclusion that the magnetic scalar potential is a sort of "multi-valued function". But that's ok because only its gradient matters.

#### TSny

Homework Helper
Gold Member
Hi @Oxvillian . Interesting comments. I don't feel comfortable with letting $\Omega$ be multi-valued. Most textbooks, that I've seen, state that you should restrict your domain of definition of $\Omega$ so that $\Omega$ is single valued at all points of the domain.

For example, in the picture of post #2, we can imagine that the yellow-shaded area is a surface such that all points of the surface are excluded from the domain of definition of $\Omega$. So, we are not allowed to pick a path of integration of $\vec \nabla \Omega$ that punctures that surface. In addition, we exclude from our domain any points of the current $I$ itself.

It occurred to me that this restriction of domain of $\Omega$ does not prevent us from using $\Omega$ to evaluate $\oint \vec B \cdot \vec {ds}$ around a closed path that does pierce the yellow surface.

To see this, consider the blue path shown in the figure of post #2 and imagine that this is not quite a closed path. The path starts at a point, $A$, say, just on the far side of the yellow surface and infinitesimally close to the surface at point $O$. The path then comes around through point $P$ and ends at a point $B$ that is on the near side of the surface and infinitesimally close to $O$. This path is not a closed path. We are allowed to set up the integral $I_1 = -\frac{\mu_0 I}{4 \pi} \int_A^B \vec \nabla \Omega \cdot \vec {ds} = \int_A^B \vec B \cdot \vec{ds}$ for this path since every point of the path lies within our domain of definition of $\Omega$.

But we can also see that $\int_A^B \vec B \cdot \vec {ds} = \oint \vec B \cdot \vec {ds}$ where the path of integration on the right is a closed path that includes piercing through the surface to connect $B$ to $A$. These two integrals are equal since $\vec B$ is continuous along the infinitesimal part from $B$ to $A$ (through the surface). So the contribution to the closed path integral from this infinitesimal part of the path from B to A is negligible.

So, even though we are not allowed to integrate $\vec \nabla \Omega$ around the complete closed path, we can nevertheless conclude that

$\oint \vec B \cdot \vec {ds} = -\frac{\mu_0 I}{4 \pi} \int_A^B \vec \nabla \Omega \cdot \vec {ds}$, where the path of integration on the right side starts at A, passes through P, and ends a B.

Moreover, we can evaluate the integral on the right as $-\frac{\mu_0 I}{4 \pi} (\Omega_B - \Omega_A ) = -\frac{\mu_0 I}{4 \pi} \left[(-2 \pi) – 2 \pi\right] = \mu_0 I$. So, we get a result in agreement with Ampere’s law.

"The solid angle equation for the magnetic field of a loop"

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