# How to find vector magnetic potential given magnetic field?

1. Mar 9, 2011

### yungman

If we are given B, how can we find A? I can fine the magnitude of A by:

$$\int_{s'} \vec B \cdot d\vec{s'} = \int_{s'} (\nabla X \vec A) \cdot d\vec{s'} = \int_{C} \vec A \cdot d\vec{l}$$

So given B and the surface area, you can get the magnitude of A.

But how do you get the direction information? All I know is A is orthogonal to B and $\vec B = \nabla X \vec A$. But still I cannot find a formula to nail down the direction.

Anyone can help?

2. Mar 9, 2011

### clem

It's a bit like long division. There is no general result. For a specific B, you try to find a vector A whose curl is that B. If you can write B in a coordinate system, then you can write partial differential equations whose solution give A.
You can also use divB=-del^2 A and solve that equation for A.

3. Mar 9, 2011

### RedX

How is it true that A is orthogonal to B?

Also, I don't see how:

$$\int_{s'} \vec B \cdot d\vec{s'} = \int_{s'} (\nabla X \vec A) \cdot d\vec{s'} = \int_{C} \vec A \cdot d\vec{l}$$

gets you the magnitude of A.

4. Mar 9, 2011

### RedX

divB=0 is one of the Maxwell equations.

5. Mar 9, 2011

### yungman

The curl of a vector is always orthogonal to the original vector. eg.

$$\hat x \;X\; \hat y = \hat z$$

$$\Phi = \int_{s'} \vec B \cdot d\vec{s'} \;\Rightarrow \;\int_{C} \vec A \cdot d\vec{l}= \Phi$$

So If you are given the B and given the surface that the flux cut through. You get the A within that surface.

6. Mar 9, 2011

### RedX

The vector field (y,0,1) has a curl in the z-direction, but is not perpendicular to the z-direction.

As for your second equation, you can get the line integral of A around the perimeter of the surface, but I don't think you can get A at a particular point on the perimeter, or at the center point of the surface.

addendum: here's a way to visualize why you can't get A at a particular point on the perimeter. This is a tri-force taken from a popular classic video game:

You know the line integral of the middle triangle. But you want the field on the left side of that middle triangle. So how do you isolate just that part? Well you know the line integral of the leftmost triangle, and the rightmost part of the leftmost triangle is what you're looking for. So by considering the leftmost triangle, you do get an extra equation involving the left side of the middle triangle, but unfortunately this equation causes you to have two more sides (the remaining sides of the leftmost triangle), so in general you don't have enough information to get A at all. I hope this is right - someone correct me if I'm wrong!

Last edited: Mar 10, 2011
7. Mar 10, 2011

### Meir Achuz

I think he meant curl B.

8. Mar 10, 2011

### Troels

The simplest way to do it, for a general B, is via a PDE problem:

$$\nabla^2 \vec A =-\mu_0\vec J$$

whilst remembering that

$$\nabla\times \vec B =\mu_0\vec J$$

This, togther with suitable boundary conditions, (eg. specifing B on the boundary) should give you the result you are looking for. Please note, that an analytical solution may not be obtainable.

The "Amperes Law for A" method;

$$\int_S \vec B\cdot d\vec a = \oint_{\partial S} \vec A\cdot d\vec l$$

That you suggested in your OP works only if you have some suitable symmetry in the problem (ie A and dl are parallel at all points along the integration curve). The only case I can think of off the top of my head is the magnetic potential inside an infinitely long solenoid

9. Mar 10, 2011

### yungman

My bad, I keep thinking about A X B is always orthogonal to A or B.

I am still reading the materials in the book.

10. Mar 10, 2011

### yungman

So

$$\int_S \vec B\cdot d\vec a = \oint_{C} \vec A\cdot d\vec l$$

can only be true with certain symetry like a long wire carrying current or a long solenoid?

Yes this is from a problem of a long solenoid.

11. Mar 10, 2011

### yungman

I read Griffiths page 234 in Magnetic Vector Potential in STATIC condition where $\nabla X \vec B = \mu_0 \vec J$.

$$\nabla X \vec B= \nabla X \nabla X \vec A = \nabla(\nabla \cdot \vec A ) - \nabla^2 \vac A = \mu_0 \vec J$$

We invoke gauge where [itex] \nabla \cdot \vec A = 0 [/tex]

$$\Rightarrow\; \nabla^2 \vec A =-\mu_0 \vec J$$

Does this imply in all cases of static condition, A and J are in the same direction in all static case?

On the other note, for time varying condition where:

$$\nabla X \vec B = \mu_0 \vec J + \mu_0 \epsilon_0 \frac {\partial {\vec E}}{ \partial t}$$

A and J are not in the same direction.

Am I correct?

12. Mar 11, 2011

### Meir Achuz

Del^2A need not be in the same direction as A.

13. Mar 11, 2011

### yungman

I skipped a step to show that

$$\nabla^2 \vec A = -\mu_0 \vec J \;\Rightarrow\; \vec A(\vec r) = \frac {\mu_0}{4\pi}\int_{v'} \frac {\vec J(\vec r)}{|\vec r|} dv'$$

And

$$\vec A = \frac {\mu_0}{4\pi}\oint_{C'} \frac {\vec I}{|\vec r|} dl'$$

That shows A is same direction of J. Is this right?

14. Mar 11, 2011

### Troels

Don't get me wrong. The equation - like amperes law - is always true, but - likewise like amperes law - it is only useful in finding the magnetude of A along the integration curve - you have to dream up the direction from a symmetry argument.

To adress your other question, no. In general you cannot assume that A and J are parallel at all points, but this is where symmetry comes in. In case of a a solenoid, you have current running in the $$\hat \phi$$-direction only and thus the PDE becomes:
$$\nabla^2 A_r = 0$$
$$\nabla^2 A_\phi = J_{\phi}$$
$$\nabla^2 A_z = 0$$

A trial solution of Ar=Az=0 satisfies the first and last equation and thus, by the uniqueness theorem, are the solutions, thus A has only a phi-component, in this case.

Last edited: Mar 11, 2011
15. Mar 11, 2011

### Meir Achuz

Still NO.

16. Mar 11, 2011

### yungman

Can you explain? There is only one vector on the left and only one on the right. How can they not be in the same direction. In fact it is common to pull the unit vector out and put them in front of the equation.

$$\vec A = \frac {\mu_0}{4\pi}\oint_{C'} \frac {\vec I}{|\vec r|} dl'$$

$$\Rightarrow\; (\hat A)\; A \;=\; (\hat I)\;[\frac {\mu_0}{4\pi}\oint_{C'} \frac {I}{|\vec r|} dl']$$

$$\Rightarrow\;\hat A =\hat I$$

Last edited: Mar 11, 2011
17. Mar 11, 2011

### kcdodd

18. Mar 11, 2011

### Meir Achuz

The unit vector I CAN NOT be pulled out of the integral.
It can, and commonly does, vary in direction during the integration.

19. Mar 12, 2011