- #36

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$$\vec{\nabla} \times \vec{B}=\mu \vec{j}.$$

The only subtlety is that you have to be careful how to deal with the singularity along the ##z##-axis!

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- Thread starter Gustav
- Start date

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 toof

- #36

- 23,737

- 14,406

$$\vec{\nabla} \times \vec{B}=\mu \vec{j}.$$

The only subtlety is that you have to be careful how to deal with the singularity along the ##z##-axis!

- #37

- 57

- 5

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?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##?

- #38

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- 6,846

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

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##).

- #39

- 57

- 5

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'##?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##.

- #40

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- #41

- 57

- 5

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

- #42

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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?

- #43

- 57

- 5

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?

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?

- #44

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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.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?

- #45

- 57

- 5

Oh okay, I think I understand now.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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