# Homework Help: Magnetics Questions

1. Apr 10, 2005

### AKG

The magnetic vector potential is given as:

$$\mathbf{A}(\mathbf{r}) = \frac{\mu _0}{4\pi }\int \frac{\mathbf{J}(\mathbf{r'})}{|\mathbf{r} - \mathbf{r'}|}\d\tau '$$

I am asked to show that $\mathbf{\nabla } \cdot \mathbf{A} = 0$ by applying the divergence to the equation above. When I've done that, I've used some product rules to simplify and end up with:

$$\mathbf{\nabla } \cdot \mathbf{A} = -\frac{\mu _0}{4\pi }\int \frac{\mathbf{J}(\mathbf{r'})\cdot(\mathbf{r} - \mathbf{r'})}{|\mathbf{r} - \mathbf{r'}|^3}d\tau '$$

I can't see why this must be zero.

Second question: Consider a surface carrying uniform surface current density. We know that the component of $\mathbf{B}$ parallel to the surface but perpendicular to the direction of flow is discontinuous at the surface, and the "derivative" of $\mathbf{A}$ inherits this discontinuity. This is expressed (supposedly) in the equation:

$$\frac{\partial \mathbf{A}_{above}}{\partial n} - \frac{\partial \mathbf{A}_{below}}{\partial n} = -\mu _0\mathbf{K}$$

I am asked to prove this using the equations:

$$\mathbf{A}_{above} = \mathbf{A} _{below}$$

$$\mathbf{B} _{above} - \mathbf{B} _{below} = \mu _0 (\mathbf{K} \times \hat{\mathbf{n}})$$

$$\mathbf{\nabla } \cdot \mathbf{A} = 0$$

Now, I'm not entirely sure what the equation I'm trying to prove even says. Earlier in my book, we have the equation:

$$\frac{\partial V}{\partial n} \equiv \mathbf{\nabla }V\cdot \hat{\mathbf{n}}$$

This gives a definition for the normal derivative for a scalar field, but not for a vector field. Would it just be:

$$\frac{\partial \mathbf{A}_{above}}{\partial n} = \frac{\partial \mathbf({A_x\hat{\mathbf{x}} + A_y\hat{\mathbf{y}} + A_z\hat{\mathbf{z}}})_{above}}{\partial n} = \frac{\partial A_x}{\partial n}\hat{\mathbf{x}} + \frac{\partial A_y}{\partial n}\hat{\mathbf{y}} + \frac{\partial A_z}{\partial n}\hat{\mathbf{z}}\ ?$$

Also, doesn't the given equation:

$$\mathbf{A}_{above} = \mathbf{A} _{below}$$

Imply that the right side of the equation I have to prove should be 0, and not $-\mu _0\mathbf{K}$?

Last edited: Apr 10, 2005
2. Apr 10, 2005

### StatusX

For the first question, use:

$$\nabla (4 \pi \delta^3(\mathbf{r}) ) = \frac{\matbf{\hat r}}{r^2}$$

Then you have:

$$\nabla \cdot \mathbf{A} = \int \nabla (4 \pi \delta^3(\mathbf{r}) ) \cdot \mathbf{J}$$

Which can be integrated by parts using the product rule:

$$\nabla \cdot(f \mathbf{A}) = f(\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot (\nabla f)$$

It should be straightforward from there.

For the second question, your derivative formula is right, and you should be able to simplfy it to get a defintion very similar to that for a scalar. Also, just because the potential is equal at a surface doesn't mean its derivatives are. Look at x and x^2 at x=0.

Last edited: Apr 10, 2005
3. Apr 10, 2005

### AKG

Where does that come from? My book gives:

$$\mathbf{\nabla }\cdot \left (\frac{\hat{\mathbf{r}}}{r^2}\right ) = 4\pi \delta ^3(\mathbf{r})$$
Okay, thanks. I'll try that.

4. Apr 10, 2005

### StatusX

Sorry about that, let me try again. First, you can add a prime to the del operator at the expense of an extra term:

$$\nabla \cdot (\mathbf{J}\frac{1}{|r-r'|}) = \frac{1}{|r-r'|}\nabla \cdot \mathbf{J} + \mathbf{J} \cdot \nabla (\frac{1}{|r-r'|}) = \mathbf{J} \cdot \nabla (\frac{1}{|r-r'|})$$

$$\nabla ' \cdot (\mathbf{J}\frac{1}{|r-r'|}) = \frac{1}{|r-r'|}\nabla ' \cdot \mathbf{J} + \mathbf{J} \cdot \nabla '(\frac{1}{|r-r'|})$$

Now in this second term, the del can be changed to unprimed at the cost of a minus sign, and solving for this will lead you to:

$$\nabla \cdot (\mathbf{J}\frac{1}{|r-r'|}) = \frac{1}{|r-r'|}\nabla ' \cdot \mathbf{J} - \nabla ' \cdot (\mathbf{J}\frac{1}{|r-r'|})$$

But for steady currents, the divergence of J is 0, so this leaves:

$$\nabla \cdot \mathbf{A} = - \int d\tau ' \nabla ' \cdot (\mathbf{J}\frac{1}{|r-r'|})$$

This can be turned into a surface integral, which you can easily show will vanish, since the integral must extend to enclose all current.

Last edited: Apr 10, 2005
5. Apr 11, 2005

### AKG

Okay, I managed to get the $\mathbf{\nabla }\cdot \mathbf{A}$ question, thanks very much for the help. I'm still not getting the other one. I set up a Cartesian co-ordinate system where $\hat{\mathbf{z}} = \hat{\mathbf{n}}$, and $\mathbf{K} = K\hat{\mathbf{x}}$.

Then I can re-write the given equations as:

$$\mathbf{A} _{above} = \mathbf{A} _{below}$$

$$\mathbf{B} _{above} - \mathbf{B} _{below} = -\mu _0 K\hat{\mathbf{y}}$$

$$\mathbf{\nabla }\cdot \mathbf{A} = 0$$

and I need to prove:

$$\frac{\partial \mathbf{A}_{above}}{\partial n} - \frac{\partial \mathbf{A}_{below}}{\partial n} = -\mu _0 K\hat{\mathbf{x}}$$

where

$$\frac{\partial \mathbf{A}}{\partial n} = \frac{\partial A_x}{\partial z}\hat{\mathbf{x}} + \frac{\partial A_y}{\partial z}\hat{\mathbf{y}} + \frac{\partial A_z}{\partial z}\hat{\mathbf{z}}$$

This amounts to proving:

$$\frac{\partial A_{x-above}}{\partial z} - \frac{\partial A_{x-below}}{\partial z} = -\mu _0 K$$

$$\frac{\partial A_{y-above}}{\partial z} - \frac{\partial A_{y-below}}{\partial z} = 0$$

$$\frac{\partial A_{z-above}}{\partial z} - \frac{\partial A_{z-below}}{\partial z} = 0$$

Given that $\mathbf{B} = \mathbf{\nabla } \times \mathbf{A}$, I can state, using the second given equation:

$$\left (\frac{\partial A_{x-above}}{\partial z}\right ) - \left (\frac{\partial A_{z-above}}{\partial x}\right ) - \left (\frac{\partial A_{x-below}}{\partial z}\right ) + \left (\frac{\partial A_{z-below}}{\partial x}\right ) = \mu _0 K$$

$$\left (\frac{\partial A_{x-above}}{\partial y}\right ) - \left (\frac{\partial A_{y-above}}{\partial x}\right ) - \left (\frac{\partial A_{x-below}}{\partial y}\right ) + \left (\frac{\partial A_{y-below}}{\partial x}\right ) = 0$$

$$\left (\frac{\partial A_{y-above}}{\partial z}\right ) - \left (\frac{\partial A_{z-above}}{\partial y}\right ) - \left (\frac{\partial A_{y-below}}{\partial z}\right ) + \left (\frac{\partial A_{z-below}}{\partial y}\right ) = 0$$

And just to clarify, things like $\partial A_{y-above} / \partial z$ should be $(\partial A_y / \partial z) _{above}$, i.e. the derivative of $A_y$ on its own evaluated above, not the derivative of some function called $A_{y-above}$.

The other two equations give me:

$$A_{x-above} - A_{x-below} = 0$$

$$A_{y-above} - A_{y-below} = 0$$

$$A_{z-above} - A_{z-below} = 0$$

and

$$\left (\frac{\partial A_x}{\partial x}\right )_{above} + \left (\frac{\partial A_y}{\partial y}\right )_{above} + \left (\frac{\partial A_z}{\partial z}\right )_{above}= 0$$

$$\left (\frac{\partial A_x}{\partial x}\right )_{below} + \left (\frac{\partial A_y}{\partial y}\right )_{below} + \left (\frac{\partial A_z}{\partial z}\right )_{below}= 0$$

So I've got 5 equations with derivatives in them, with 18 unknown derivatives 3 (for each component) x 3 (for each $\partial x,\ \partial y,\ \partial z$) x 2 (for above and below), and I need to derive from this 3 equations in 6 of those unknowns. I'm pretty lost here...

Last edited: Apr 11, 2005
6. Apr 11, 2005

### StatusX

Like I said, the directional derivative of a vector valued function can be expressed simply using vector derivatives. I'll let you derive it, but I came up with:

$$\frac{\partial \mathbf{A}}{\partial n} = (\mathbf{\hat n} \cdot \nabla) \mathbf{A}$$

You can use vector product rules and the fact that:

$$\mathbf{\hat n} \times (\mathbf{K} \times \mathbf{\hat n}}) = \mathbf{K}$$

(since K and n are perpendicular) to get the desired result.

Last edited: Apr 11, 2005
7. Apr 11, 2005

### AKG

$$(\hat{\mathbf{n}} \cdot \mathbf{\nabla })\mathbf{A} = \left (\frac{\partial }{\partial z}\right )\mathbf{A} = \frac{\partial A_x}{\partial z}\hat{\mathbf{x}} + \frac{\partial A_y}{\partial z}\hat{\mathbf{y}} + \frac{\partial A_z}{\partial z}\hat{\mathbf{z}} = \frac{\partial A_x}{\partial n}\hat{\mathbf{x}} + \frac{\partial A_y}{\partial n}\hat{\mathbf{y}} + \frac{\partial A_z}{\partial n}\hat{\mathbf{z}} = \frac{\partial \mathbf{A}}{\partial n}$$

$$\left (\frac{\partial \mathbf{A}}{\partial n}\right )_a - \left (\frac{\partial \mathbf{A}}{\partial n}\right )_b$$

$$= \left [\mathbf{\nabla }(\hat{\mathbf{n}} \cdot \mathbf{A}) - \hat{\mathbf{n}} \times (\mathbf{\nabla }\times \mathbf{A}) - \mathbf{A} \times (\mathbf{\nabla } \times \hat{\mathbf{n}}) - (\mathbf{A} \cdot \mathbf{\nabla })\hat{\mathbf{n}}\right ]_a - \left [\mathbf{\nabla }(\hat{\mathbf{n}} \cdot \mathbf{A}) - \hat{\mathbf{n}} \times (\mathbf{\nabla }\times \mathbf{A}) - \mathbf{A} \times (\mathbf{\nabla } \times \hat{\mathbf{n}}) - (\mathbf{A} \cdot \mathbf{\nabla })\hat{\mathbf{n}}\right ]_b$$

$$= \left [\mathbf{\nabla }(\hat{\mathbf{n}} \cdot \mathbf{A}) - \hat{\mathbf{n}} \times (\mathbf{\nabla }\times \mathbf{A})\right ]_a - \left [\mathbf{\nabla }(\hat{\mathbf{n}} \cdot \mathbf{A}) - \hat{\mathbf{n}} \times (\mathbf{\nabla }\times \mathbf{A})\right ]_b$$

Since $\hat{\mathbf{n}}$ is constant, so any derivatives of it will be zero.

$$= (\mathbf{\nabla}A_z)_a - (\mathbf{\nabla}A_z)_b - \hat{\mathbf{n}}\times (\mathbf{B}_a - \mathbf{B}_b)$$

$$= (\mathbf{\nabla}A_z)_a - (\mathbf{\nabla}A_z)_b + \mu _0K\hat{\mathbf{z}}\times \hat{\mathbf{y}}$$

$$= (\mathbf{\nabla}A_z)_a - (\mathbf{\nabla}A_z)_b - \mu _0K\hat{\mathbf{x}}$$

$$= (\mathbf{\nabla}A_z)_a - (\mathbf{\nabla}A_z)_b - \mu _0\mathbf{K}$$

At this point, I haven't used the facts that $\mathbf{A}_{above} = \mathbf{A}_{below}$ (in fact, I can't see how to use that one since we're dealing with the derivatives here, not the actual values) or that $\mathbf{\nabla }\cdot \mathbf{A} = 0$. It remains to show that:

$$(\mathbf{\nabla}A_z)_a - (\mathbf{\nabla}A_z)_b = \mathbf{0}$$

I'm not really sure what to do here. I think that if we consider only a very small portion of the surface, we can consider it to be flat, and, by symmetry, we can say that the magnetic vector potential will not vary with x and y, and so the above equation becomes:

$$\left (\frac{\partial A_z}{\partial z}\hat{\mathbf{z}}\right )_a - \left (\frac{\partial A_z}{\partial z}\hat{\mathbf{z}}\right )_b = \mathbf{0}$$

$$\left (\frac{\partial A_z}{\partial z}\right )_a - \left (\frac{\partial A_z}{\partial z}\right )_b = 0$$

$$\left (\frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y}\right )_a - \left (\frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y}\right )_b = 0$$

Using $\mathbf{\nabla }\cdot \mathbf{A} = 0$. This equation above holds if I'm right in saying that the potential will not vary with x or y, and if this holds, the proof is complete. I still haven't used the fact that the potential is the same above as it is below. Well, actually, my book says that the because the divergence of A is zero, the normal components of the potential are continuous at the surface, and because the curl of A is B, the tangential components are continuous, and it is because of these two facts that the potential is the same above and below. So it could be that using the fact that the potential is the same above and below implies using the two facts that make it continuous, one of those facts being that the divergence is zero (which I used), and the other being that the curl of A is B, which I also used. $\mathbf{\nabla }\times \mathbf{A} = \mathbf{B}$ was known without knowing that $\mathbf{A}$ was continuous, in fact it is true by definition, so you might say that I didn't really use the fact that $\mathbf{A}_{above} - \mathbf{A}_{below} = 0$.

So is that fact really useless to this proof, or did I do something wrong in not using this fact? Aside from that, is there anything else wrong?

Again, thank you very much.

8. Apr 11, 2005

### StatusX

You can't say the potential won't vary with x or y, but you can say its derivatives are continuous as you move in the x or y directions, since B is, which gives the same result. I think the fact that A is continuous is sort of a seperate boundary condition, and I don't think you need it to prove this.

9. Apr 11, 2005

### AKG

Why does the continuity of the derivatives give the desired result? Also, I was specifically asked to use the three equations to solve the problem, one of which was that $\mathbf{A}_{above} = \mathbf{A}_{below}$. It seems strange that I didn't have to use it at all.

10. Apr 11, 2005

### StatusX

I'm starting to think this isn't the simplest way to prove this, so if anyone else knows a better way, please let us know. Anyway, I think I got it. Since a gradient field has zero curl, you can prove the parellel components have to be continuous the same way you do for the E field. Now, to prove the perpendicular component is also continuous, let's look at the divergence of this function:

$$\nabla \cdot \nabla(\mathbf{\hat n \cdot A}) = \nabla^2 (\mathbf{\hat n \cdot A}) = (\nabla^2 \mathbf{A})_n$$

$$\nabla^2 \mathbf{A} = \nabla(\nabla \cdot \mathbf{A}) - \nabla \times (\nabla \times \mathbf{A}) = -\nabla \times \mathbf{B} = -\mu_0 \mathbf{J}$$

so:

$$\nabla \cdot \nabla(\mathbf{\hat n \cdot A}) =-\mu_0 \mathbf{J}_n$$

Now, integrating over a box straddling the surface of vanishing thickness, you get:

$$\int d\tau \nabla \cdot \nabla(\mathbf{\hat n \cdot A}) = \mbox{area} \times((\nabla(\mathbf{\hat n \cdot A})_{above})_n - (\nabla(\mathbf{\hat n \cdot A})_{below})_n ) = \int d\tau( -\mu_0 \mathbf{J}_n )$$

But the component of the current perpendicular to the plane will be finite, so this will vanish as the box gets infinitely thin. So we have:

$$(\nabla(\mathbf{\hat n \cdot A})_{above})_n - (\nabla(\mathbf{\hat n \cdot A})_{below})_n = 0$$

and since the parellel component is also continuous:

$$\nabla(\mathbf{\hat n \cdot A})_{above} - \nabla(\mathbf{\hat n \cdot A})_{below} = 0$$

Last edited: Apr 11, 2005