Vector potential with current density

Click For Summary
SUMMARY

The discussion focuses on calculating the vector potential \(\vec{A}(\vec{r})\) from a given current density \(\vec{j}=\frac{Q}{4\pi R^2}\delta(r-R)\vec{\omega}\times\vec{r}\). Participants clarify that the vector \(\vec{r}\) in the integral should indeed have a prime, and that the angles \(\theta\) and \(\phi\) refer to the primed position. The symmetry of the current distribution allows for simplifications, indicating that the vector potential \(\vec{A}(z)\) must be perpendicular to the z-axis and rotationally invariant about it.

PREREQUISITES
  • Understanding of vector calculus and integral equations.
  • Familiarity with the concepts of vector potential and current density in electromagnetism.
  • Knowledge of symmetry considerations in physics.
  • Proficiency in using Dirac delta functions in mathematical expressions.
NEXT STEPS
  • Study the derivation of vector potentials in electromagnetic theory.
  • Learn about the application of Dirac delta functions in physics problems.
  • Research the implications of symmetry in vector fields and potentials.
  • Explore the mathematical techniques for evaluating integrals involving singularities.
USEFUL FOR

This discussion is beneficial for physics students, particularly those studying electromagnetism, as well as researchers working on problems involving vector potentials and current distributions.

Faust90
Messages
19
Reaction score
0

Homework Statement


Hey,
I got the current density \vec{j}=\frac{Q}{4\pi R^2}\delta(r-R)\vec{\omega}\times\vec{r} and now I should calculate the vector potential:
\vec{A}(\vec{r})=\frac{1}{4\pi}\int\frac{j(\vec{r})}{|r-r'|}.


Homework Equations


The Attempt at a Solution


here my attempt till now:
http://phymat.de/physics.png
I'm really not sure how to go on now. Is this right what I wrote there?
 
Last edited by a moderator:
Physics news on Phys.org
In the second equation of the notes, should the vector r in the numerator have a prime?

[EDIT: Also, do \theta and \phi refer to the unprimed position or the primed position?]
 
Last edited:
TSny said:
In the second equation of the notes, should the vector r in the numerator have a prime?

[EDIT: Also, do \theta and \phi refer to the unprimed position or the primed position?]

Hi,

thanks for your answer. I think they should all have primes.

Edit: I should only calculate the potential A(r) on the z-axis (for r=z e_z). But I don't know how this can be helpful.
 
Last edited:
Faust90 said:
I should only calculate the potential A(r) on the z-axis (for r=z e_z).

Ahh. If ##\vec{\omega}## is also along the z-axis, then life is looking Wunderbar!
 
TSny said:
Ahh. If ##\vec{\omega}## is also along the z-axis, then life is looking Wunderbar!

Hi, thanks,
Yes Omega is also along the z-Axis, but I don't know how this could help me. Didn't I have to solve the integral first?
 
Faust90 said:
Hi, thanks,
Yes Omega is also along the z-Axis, but I don't know how this could help me. Didn't I have to solve the integral first?

I don't think so. You can get the answer from symmetry considerations. Note that the current ##\vec{j}(r') d^3r'## in a small volume element makes an infinitesitmal contribution to the vector potential at ##r## of amount ##d\vec{A}(r)##, and ##d\vec{A}(r)## has the same direction as ##\vec{j}(r')##.

Do you see that each nonzero value of ##\vec{j}(r') d^3r'## is parallel to the xy plane? So, the total value of ##\vec{A}(r)## must be parallel to the xy plane at any observation point ##r##.

For this problem the observation point is at some z on the z-axis. Note that the current distribution ##\vec{j}(r')## is axially symmetric about the z axis. So ##\vec{A}(z)## would have to point perpendicular to the z-axis and yet be rotationally invariant about the z-axis. There is only one possibility for the value of ##\vec{A}## at z.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
752
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
5
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
11
Views
3K
Replies
8
Views
2K