Solving Vector Potential: Calculating Magnetic Field & Difficulties

[Gustav]

Homework Statement

Consider the vector potential
A = C ln( (x^2+y^2)/z^2 ) \hat{z}

a) Determine the B-field for this potential. Describe a situation that gives a B-field of that shape.
b) Calculate \nabla \cdot A
c) Give a vector potential in coulomb gauge, A’, which gives the same B as in the a-task.

Relevant Equations

B = \nabla x A

My solution for the vector potential $A=2Cln\frac{x^2+y^2}{z^2} \hat{z}$ is:
a) I used the following formula to calculate the magnetic field
$$ \mathbf{B} = \nabla \times \mathbf{A} = \left( \frac{dA_z}{dy} - 0 \right) \hat{x} + \left( 0 - \frac{dA_z}{dx} \right)\hat{y} + 0 \hat{z} = \frac{dA_z}{dy} - \frac{dA_z}{dx} = 2C \frac{x}{x^2+y^2} \hat{x} - 2C \frac{y}{x^2+y^2} \hat{y}. $$
How should I describe a situation with this magnetic field here?

b) I just did the calculation of the formula and got a non-zero value:
$$ \nabla \cdot \mathbf{A} = \left( \frac{d}{dx}, \frac{d}{dy}, \frac{d}{dz} \right) \cdot \mathbf{A} = \frac{d\mathbf{A}}{dx} + \frac{d\mathbf{A}}{dy} + \frac{d\mathbf{A}}{dz} = 2C \left( \frac{y}{x^2+y¨2} + \frac{x}{x^2+y^2} - \frac{1}{z} \right) $$

c) I was having problem solving this part, I was thinking about using the same formula in question a) $\mathbf{B} = \nabla \times \mathbf{A'}$ and solve for $\mathbf{A'}$ since I have a known $\mathbf{B}$, but then I realized it is too difficult to solve it. I also thought mathematically that this expression $\mathbf{A'} = Cln(x^2+y^2)\hat{z}$ could give us the same value of B as in the a-task. But this is still not solving the problem correctly. What should I do?

 

[kuruman]

From the looks of it, you should convert to cylindrical coordinates. Look up the del operator if you don't know it in cylindrical coordinates.

 

[Gustav]

From the looks of it, you should convert to cylindrical coordinates. Look up the del operator if you don't know it in cylindrical coordinates.

How should I convert to cylendrical form? Do you mean re-write $\mathbf{A}$ in cylderincal form or re-do the whole calculation?

 

[Gustav]

From the looks of it, you should convert to cylindrical coordinates. Look up the del operator if you don't know it in cylindrical coordinates.

Also, why should it be cylendrical?

 

[kuruman]

Yes, that's what I mean. Find its form here. It should be cylindrical because that is its symmetry.

 

[Gustav]

Also what do you mean by its symmetry? What is its symmetry?

 

[Gustav]

Yes, that's what I mean. Find its form here. It should be cylindrical because that is its symmetry.

Okay, so for the part where the curl is I need to calculate in this form:
$$ \nabla \times A = \left[ \frac{1}{s} \frac{\partial A_z}{\partial \phi} - \frac{\partial A_\phi}{\partial z} \right]\hat{s} + \left[ \frac{\partial A_s}{\partial z} - \frac{\partial A_z}{\partial s} \right]\hat{\phi} + \frac{1}{s} \left[ \frac{\partial s A_\phi}{\partial s} - \frac{\partial A_s}{\partial \phi} \right] \hat{z} = \frac{1}{s} \frac{\partial A_z}{\partial \phi}\hat{s} - \frac{\partial A_z}{\partial s} \hat{\phi} $$
How am I going to derivate $A_z$ with respect to $\phi$ or $s$?

 

[kuruman]

How am I going to derivate $A_z$ with respect to $\phi$ or $s$?

First, you need to write $A_z$ as a function of $r$, $\phi$ and $z$? Do that first, then take the partial derivatives.

 

[Gustav]

First, you need to write $A_z$ as a function of $r$, $\phi$ and $z$? Do that first, then take the partial derivatives.

Okay, I tried to re-write it and I got something like this:
$x= s \cos(\phi ), y= s \sin(\phi), z=z$
$$ \mathbf{A} = C \ln \left( \frac{x^2+y^2}{z^2} \right) = C \ln \left( \frac{ (s \cos(\phi )^2 + (s \sin(\phi )^2}{z^2} \right) = C \ln \frac{s^2}{z^2} $$

 

[kuruman]

What is $s$? Also, please fix the LaTeX.

On edit: Thank you for fixing it.

 

[Gustav]

What is $s$? Also, please fix the LaTeX.

$s=\rho$, I am sorry about the LaTeX, I am working on it.

 

[Gustav]

Now I think I can manage to calculate the curl and the divergence. But I still don't understand why we used cylendrical coordinates?

 

[Gustav]

Okay, I tried to re-write it and I got something like this:
$x= s \cos(\phi ), y= s \sin(\phi), z=z$
$$ \mathbf{A} = C \ln \left( \frac{x^2+y^2}{z^2} \right) = C \ln \left( \frac{ (s \cos(\phi )^2 + (s \sin(\phi )^2}{z^2} \right) = C \ln \frac{s^2}{z^2} $$

Also this is in the $\hat{z}$

 

[kuruman]

Right. You should put in the $\hat z$ to avoid confusing anyone who might be reading this. Also, conventionally $s$ is $r$ but that's OK if you know what you're doing. Now find the curl and the divergence. Once you have these, I will explain to you why one should use cylindrical coordinates for this problem.

 

[Gustav]

Right. You should put in the $\hat z$ to avoid confusing anyone who might be reading this. Also, conventionally $s$ is$r$ but that's OK if you know what you're doing. Now find the curl and the divergence. Once you have these, I will explain to you why one should use cylindrical coordinates.

Okay, so now that I have my $A_z = C \ln \frac{s^2}{z^2} \hat{z}$ I can calculate the derivatives:
1) Curl:
$$ \nabla \times A = \frac{1}{s}\frac{\partial A_z}{\partial \phi} \hat{s} - \frac{\partial A_z}{\partial s} \hat{\phi} = 0 - \frac{2C}{s}\hat{\phi} = - \frac{2C}{s}\hat{\phi}.$$
2) Divergence:
$$ \nabla \cdot A = 0 + 0 + \frac{\partial A_z}{\partial z} = - \frac{2C}{z} $$

 

[kuruman]

Very good. Compare these expressions and how easy it was to take derivatives with what you did earlier with Cartesian coordinates. Also, note that from three coordinate $x$, $y$ and $z$ the problem has been reduced to two $s$ and $z$.

Look at $\vec\nabla \times \vec A.$ It must be a magnetic field. Can you guess what can possibly generate a magnetic field that looks like this? If so, what is the appropriate value for $C$?

 

[Gustav]

Very good. Compare these expressions and how easy it was to take derivatives with what you did earlier with Cartesian coordinates. Also, note that from three coordinate $x$, $y$ and $z$ the problem has been reduced to two $s$ and $z$.

Look at $\vec\nabla \times \vec A.$ It must be a magnetic field. Can you guess what can possibly generate a magnetic field that looks like this? If so, what is the appropriate value for $C$?

Nothing is crossing my mind... I don't think I know what could generate this magnetic field. How should I think to get that?

 

[kuruman]

Have you had a course in introductory physics where you calculated magnetic fields for various current configurations?

 

[Gustav]

Have you had a course in introductory physics where you calculated magnetic fields for various current configurations?

Yes, I have

 

[Gustav]

So what I think you mean is that the constant $C=\frac{\mu_0}{2\pi}$. But in that kind of experession we have a current in the dominator.

 

[Gustav]

So we are talking about a magnetic field in a wire that is circular shaped?

 

[kuruman]

So what I think you mean is that the constant $C=\frac{\mu_0}{2\pi}$. But in that kind of experession we have a current in the dominator.

There is never[/color] a current in the denominator of an expression for a magnetic field. The magnetic field is proportional to the current. That's Ampere's law. What current configuration are you thinking of? If you don't remember, look it up. There is no hurry.

 

[Gustav]

There is never a current in the denominator of an expression for a magnetic field. The magnetic field is proportional to the current. That's Ampere's law. What current configuration are you thinking of? If you don't remember, look it up. There is no hurry.

Sorry I meant numerator not denominator! I was thinking about an expression as the following
$$ \mathbf{B} = \frac{\mu_0 I}{2\pi s} \hat{\phi} $$

 

[Gustav]

There is never a current in the denominator of an expression for a magnetic field. The magnetic field is proportional to the current. That's Ampere's law. What current configuration are you thinking of? If you don't remember, look it up. There is no hurry.

So I have an example in my book about "finding the magnetic field a distance s from a stright wire", the expression of its magnetic field reminded me of the one I got in the calculation for the curl.

 

[kuruman]

That's the one. Nice work. Can you finish the problem now?

 

[Gustav]

That's the one. Nice work. Can you finish the problem now?

Now I am stuck when calculating the vector potential in coulomb gauge, A’. How can I find the A'?

 

[kuruman]

It's not just any A'. It must have a curl that gives the B-field of a straight wire.

 

[Gustav]

It's not just any A'. It must have a curl that gives the B-field of a straight wire.

Yes, but how can I solve for it?

 

[Gustav]

It's not just any A'. It must have a curl that gives the B-field of a straight wire.

Does it mean that it is the same as the vector potential A?

 

[kuruman]

No, it means that the curl of A' gives the same B-field. Furthermore, it must satisfy the Coulomb gauge $\vec \nabla \cdot \vec A' =0$.

 

[Gustav]

No, it means that the curl of A' gives the same B-field. Furthermore, it must satisfy the Coulomb gauge $\vec \nabla \cdot \vec A' =0$.

Okay, I am with you so far. But does that mean I need to think of an expression for A' or is there a way to calculate it?

 

[kuruman]

No way to calculate it. There are many answers that work, just pick one and verify that it does work by taking its curl and divergence.

 

[Gustav]

No way to calculate it. There are many answers that work, just pick one and verify that it does work by taking its curl and divergence.

Oh okay, so does it work if
$$ \vec A' = C \ln \left( \frac{1}{s^2} \right)\hat{z} ? $$

 

[Gustav]

Because then I will have:
1) Curl:
$$ \vec B = \nabla \times \vec A' = -\frac{2C}{s}\hat{\phi} $$
2) Divergence:
$$ \nabla \cdot \vec A' = 0 $$

 

[kuruman]

Because then I will have:
1) Curl:
$$ \vec B = \nabla \times \vec A' = -\frac{2C}{s}\hat{\phi} $$
2) Divergence:
$$ \nabla \cdot \vec A' = 0 $$

Excellent. You are done.

 

[vanhees71]

A much simpler way to see, what's described is to calculate the current density, using
$$\vec{\nabla} \times \vec{B}=\mu \vec{j}.$$
The only subtlety is that you have to be careful how to deal with the singularity along the $z$-axis!

 

[Gustav]

Very good. Compare these expressions and how easy it was to take derivatives with what you did earlier with Cartesian coordinates. Also, note that from three coordinate $x$, $y$ and $z$ the problem has been reduced to two $s$ and $z$.

Look at $\vec\nabla \times \vec A.$ It must be a magnetic field. Can you guess what can possibly generate a magnetic field that looks like this? If so, what is the appropriate value for $C$?

I just want to make sure if I understood this correctly, so we converted from Cartesian coordinates to cylendrical coordinates because it will make the calculations easier? But how is it possible to determine whether one should convert to cylinderical or any other coordinates?

 

[kuruman]

The three main choices of coordinates are Cartesian, cylindrical and spherical. I would say 99% of E&M problems that can be solved analytically use one of these. To find which is more appropriate to a problem, you look at the charge distribution or the potential and imagine what you can do to keep it looking the same or different. Here are some examples.

A volume charge density that depends on the distance from the center only and we write as $\rho(r)$ has spherical symmetry. Say you look at this distribution and then someone rotates it by any amount about any diameter while your back is turned. When you take another look, you wouldn't know that someone rotated it behind your back because everything looks the same as before. We say that this charge distribution has spherical symmetry and use spherical coordinates. If the distribution also depends on angle $\theta$ measured about a special diameter, e.g. north-south pole, it does not have spherical symmetry but has azimuthal symmetry. This means that if someone rotated the sphere about the special axis while you weren't looking, you still wouldn't be able to tell. However if someone rotated it about some axis other than the special axis, you would be able to tell. Nevertheless, one still uses spherical coordinates and writes the volume charge density as $\rho(r,\theta)$.

What about cylindrical symmetry? Well, about what axis can you rotate a cylinder and have it look the same? There is azimuthal symmetry about the long axis which is what the magnetic vector potential has because of its form $\vec A=A(r,z)\hat z$. A lot of situations with cylindrical symmetry have also translational symmetry. This means that if you go along the z-axis only, things look the same, i.e. the physical situation is independent of coordinate $z$. That's what you did in part (d) of the problem to find a magnetic vector potential that is consistent with the Coulomb gauge: you removed the z-dependence so that the potential depends only on $r$ (or $s$).

 

[Gustav]

No, it means that the curl of A' gives the same B-field. Furthermore, it must satisfy the Coulomb gauge $\vec \nabla \cdot \vec A' =0$.

For our vector potential $\vec A$, we got a non-zero value for the divergence and the curl. But now when calculating for the Coulomb gauge $\vec A'$ we need to have a non-zero value equal to the one in $\vec A$ for the curl, but this times the divergence needs to be equal to zero. How can one connect the $\vec A$ with $\vec A'$?

 

[kuruman]

I am not sure I understand what you mean by "connect". They have the same curl, is that what you mean?

 

[Gustav]

I am not sure I understand what you mean by "connect". They have the same curl, is that what you mean?

No I meant how can they be related? Since they have the same curl there must be a relation between them?

 

[kuruman]

$\vec A' = \vec A +\vec \nabla \varphi$ where $\varphi$ is any scalar function.
When you take the curl on both sides, you get $\vec \nabla \times \vec A'=\vec \nabla \times \vec A +0$. Is this what you have in mind?

 

[Gustav]

$\vec A' = \vec A +\vec \nabla \varphi$ where $\varphi$ is any scalar function.
When you take the curl on both sides, you get $\vec \nabla \times \vec A'=\vec \nabla \times \vec A +0$. Is this what you have in mind?

Yes, that part I understand, but why can't A' be equal to A, even though the divergence in A is a non-zero value?

 

[kuruman]

Yes, that part I understand, but why can't A' be equal to A, even though the divergence in A is a non-zero value?

Why should they be equal? It's OK for two vectors to have the same curl but different divergences as this example illustrates. Maybe I don't understand what you are asking.

 

[Gustav]

Why should they be equal? It's OK for two vectors to have the same curl but different divergences as this example illustrates. Maybe I don't understand what you are asking.

Oh okay, I think I understand now.