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: Magnetic Potential

  1. Feb 16, 2009 #1
    hi again, i'm trying to show that [itex]\mathbf{B} = \nabla \times \mathbf{A} [/itex] where [itex]\mathbf{A}[/itex] is the magnetic potential given by:

    [itex]\mathbf{A(r)}=\frac{\mu_0}{4 \pi} \int dV' \frac{\mathbf{J(r')}}{|\mathbf{r-r'}|}[/itex]

    i deduced that since [itex]\nabla=(\frac{\partial}{\partial{r_1}},\frac{\partial}{\partial{r_2}},\frac{\partial}{\partial{r_3}})[/itex] it only acts on r terms and so

    [itex]\nabla \times \mathbf{A}=\frac{\mu_0}{4 \pi} \int dV' \mathbf{J(r')} ( \nabla \times \frac{1}{|\mathbf{r-r'}|} )[/itex]

    i can't figure out how to compute that curl - it should be fairly elementary, no?
  2. jcsd
  3. Feb 16, 2009 #2
    You should choose a specific co-ordinate system and work it.
  4. Feb 17, 2009 #3
    tried it in cartesian and it got messy-

    [itex]\epsilon_{ijk} \partial_j [{(r-r')_k (r-r')_k}]^{-\frac{1}{2}}[/itex]

  5. Feb 17, 2009 #4
    Oops, that had been bad advice. My apologies, I should have paid more attention.

    Here's some better advice : you want to take the curl of this [tex] \frac{\mathbf{J(r ^')}}{|\mathbf{r - r^'}|} [/tex]

    This is a vector multiplied by a scalar. So you gotta use the product rule for curl in such situations.
  6. Feb 17, 2009 #5
    ok so
    [itex]\nabla \times \frac{\mathbf{J(r')}}{\mathbf{|r-r'|}} = \nabla(\frac{1}{\mathbf{|r-r'|}}) \times \mathbf{J(r')} + \frac{1}{\mathbf{|r-r'|}} \nabla \times \mathbf{J(r')}[/itex]

    im confused as to how to take the curl of J(r') because [itex]\nabla=\frac{\partial}{\partial{r_i}}[/itex] not [itex]\nabla=\frac{\partial}{\partial{r'_i}}[/itex] so how can the del operator act on a vector function of r'???
  7. Feb 17, 2009 #6
    No, it can't.
  8. Feb 17, 2009 #7
    so the second term is 0

    how do i write out the first term though

    [itex]\partial_i \left( (r_j-r'_j)(r_j-r'_j) \right)^{-\frac{1}{2}}[/itex] this appears to violate Einstein summation convention?
  9. Feb 17, 2009 #8
    I would suggest forgetting fancy conventions and working it out in some co-ordinate system with all x s and y s, or r s and thetas, whichever you like.
  10. Feb 17, 2009 #9
    yeah wwe're expected to use that notation though.

    i did this:

    [itex]a_i=r_i-r'_i[/itex] then we have

    [itex]\partial_i(a_ja_j)^{-\frac{1}{2}}=-\frac{1}{2}(a_ja_j)^{-\frac{3}{2}}2a_j(\partial_i(r_j-r'_j))=-\frac{1}{(r-r')^3}a_j \partial_i (r_j-r'_j)=-\frac{1}{(r-r')^3}a_j (\delta_{ij}-0)=-\frac{1}{(r-r')^2}[/itex]

    which seems like a suitbale answer???
  11. Feb 17, 2009 #10
    It does indeed.
  12. Feb 17, 2009 #11
    actually, shouldn't this be [itex]-\frac{\mathbf{r-r'}}{(r-r')^3}[/itex] as grad should be a vector?
  13. Feb 17, 2009 #12
    It does not act on [tex] r^' [/tex] . Of course, there should be a unit vector along r .
  14. Feb 17, 2009 #13
    but back in post 9 i got [itex]-\frac{a_i}{(r-r')^3}[/itex] where[itex]a_i=(r-r')_i[/itex] suggesting a vector in both directions???
  15. Feb 17, 2009 #14
    Duh, I am caught napping again. Actually the term in your 11th post is correct. Absolutely and completely perfect.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook