Solving Vector Potential: Calculating Magnetic Field & Difficulties

In summary: So now that I have these, I can solve for ##\mathbf{A}##:In summary, to solve for the vector potential ##A=2Cln\frac{x^2+y^2}{z^2} \hat{z}##, I used the following formula to calculate the magnetic field: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
  • #36
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!
 
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  • #37
kuruman said:
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?
 
  • #38
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##).
 
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  • #39
kuruman said:
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'##?
 
  • #40
I am not sure I understand what you mean by "connect". They have the same curl, is that what you mean?
 
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  • #41
kuruman said:
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?
 
  • #42
##\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?
 
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  • #43
kuruman said:
##\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?
 
  • #44
Gustav said:
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.
 
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  • #45
kuruman said:
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.
 
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