1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Vector potential in cylindrical coordinates

  1. May 25, 2007 #1
    1. The problem statement, all variables and given/known data
    What current density would produce the vector potential, [itex] A = k \hat{\phi} [/itex] where k is a constant, in cylindrical coordinates?

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

    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 [itex] k\hat{\phi} [/tex]

    how do you take the derivative of a unit vector?

    Thanks for the help!
  2. jcsd
  3. May 25, 2007 #2
    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.
  4. May 27, 2007 #3
    yea that was the way suggested by a friend of mine too

    i was wondering if using my formula was applicable as well
  5. May 27, 2007 #4
    Since you vector potential has only [itex]\phi[/itex] dependency, you can take the second derivative with respect to [itex]\phi[/itex] only.
  6. May 28, 2007 #5
    wouldnt that yield zero???
  7. May 29, 2007 #6
    Yes, this is the case of a solenoid. The current density vector [itex]\vec J[/itex] is axial i.e. directed along the 'r'. So there wouldn't be a current in the [itex]\hat \phi[/itex] direction. Hence it is necessary to know the magnetic field in this case.
  8. May 30, 2007 #7
    First of all, its the Laplacian of a vector. So the full operator [itex]\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}[/itex] 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:

    [tex]\vec{\nabla} \times \vec{A} = B[/tex]

    [tex]\oint \vec{B} \bullet \vec{dl} = \oint \vec{\nabla} \times \vec{A} \bullet d\vec{l} = \int \vec{A}\bullet d\vec{s}[/tex]

    This is not for your problem, but for physical insight.
  9. May 30, 2007 #8


    User Avatar

    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.
  10. May 30, 2007 #9
    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.
  11. May 30, 2007 #10
    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook