# Vector potential

1. May 25, 2007

### stunner5000pt

1. The problem statement, all variables and given/known data
What current density would produce the vector potential, $A = k \hat{\phi}$ where k is a constant, in cylindrical coordinates?

2. Relevant equations
$$\nabla^2 A = -\mu_{0} J$$
In cylindrical coordinates for radial and z symmetry
$$\nabla^2 t = \frac{1}{s^2} \frac{\partial^2 t}{\partial \phi^2}$$

3. The attempt at a solution
Now i m wondering how to take the Laplacian of A
I need to take the second derivative wrt phi of $k\hat{\phi} [/tex] how do you take the derivative of a unit vector? Thanks for the help! 2. May 25, 2007 ### neelakash Given A,find B by B=curl A Possibly,you have to refer to the formula...but do not worry...almost all the terms will cancel.One will survive. Then use curl B=(mu)J I think it is a problem from Griffiths. 3. May 27, 2007 ### stunner5000pt yea that was the way suggested by a friend of mine too i was wondering if using my formula was applicable as well 4. May 27, 2007 ### Reshma Since you vector potential has only [itex]\phi$ dependency, you can take the second derivative with respect to $\phi$ only.

5. May 28, 2007

### stunner5000pt

wouldnt that yield zero???

6. May 29, 2007

### Reshma

Yes, this is the case of a solenoid. The current density vector $\vec J$ is axial i.e. directed along the 'r'. So there wouldn't be a current in the $\hat \phi$ direction. Hence it is necessary to know the magnetic field in this case.

7. May 30, 2007

### maverick280857

First of all, its the Laplacian of a vector. So the full operator $\nabla^{2}[/tex] operates on all components of the vector potential. Secondly, as stated, the magnetic field will be axial (along the axis, not along [itex]\hat{s}$ which would be radial).

This is the magnetic vector potential of a long solenoid. Only if it is "long" (infinite in length) does the magnetic field outside go to zero.

Consider this:

$$\vec{\nabla} \times \vec{A} = B$$

$$\oint \vec{B} \bullet \vec{dl} = \oint \vec{\nabla} \times \vec{A} \bullet d\vec{l} = \int \vec{A}\bullet d\vec{s}$$

This is not for your problem, but for physical insight.

8. May 30, 2007

### Dox

The Laplacian operator acting on a vector is a mess because as you note, the basis in general change. The easier way to compute this is in rectangular coordinates, 'cause the basis are constant.

So, I suggest, write your vector potential in rectangular coordinates, get the answer and then return back to cylindric ones.

9. May 30, 2007

### maverick280857

Yes, thats correct..it was merely an observation from me. It'll be a good exercise to do it...but for this problem, you don't need to stunner, just follow the physical arguments given in my previous post if you haven't figured it out by now.

10. May 30, 2007

### stunner5000pt

well someone hinted in class to find B and then J foir this queston rather than use the Laplaciani was just wondering why my way wouldnt work

now i know why

thanks everyone, for the help