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Homework Help: Magnetic vector potential in the case of specific current distribution

  1. Nov 14, 2012 #1
    1. Use equation for the magnetic vector potential in the case of specific current distribution and show by direct differentiation that ∇[itex]\bullet[/itex]A=0
    A(r)= µ[itex]_{0}[/itex]/4[itex]\pi[/itex] [itex]\int J(r')/|r-r'|[/itex] dv'

    2. Relevant equations
    ∇[itex]\times[/itex] B(r)= µ0J(r)

    3. The attempt at a solution

    We know that: curl of B(r) = µ0J(r)
    and that the divergence of a curl is equal to zero

    This seems so simple can it really be the solution?
  2. jcsd
  3. Nov 14, 2012 #2


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    Hi, Zamot40 and welcome to Physics Forums!

    The solution is not really that simple. :frown:

    When you take the divergence of A(r), the derivatives are with respect to the argument of A; i.e., the unprimed coordinates. But in the integral, the argument of the current density J is r' (primed coordinates) which denote the integration variables. The only place that the unprimed coordinates appear in the integral is in the denominator |r-r'|. Thus, you should justify that

    [itex]\cdot[/itex](J(r')/|r-r'|) = J(r')[itex]\cdot[/itex](1/|r-r'|)

    Then comes a sleight of hand to convert the divergence derivatives over to the primed coordinates using (1/|r-r'|) = -'(1/|r-r'|). Proof left for you.

    See if you can proceed from there. There's still a trick or two you'll need to invoke. :cool:
    Last edited: Nov 14, 2012
  4. Nov 15, 2012 #3
    First of, thank you for the help.

    Taking the gradient with respect to r' gives


    Hence the integral becomes [itex]\int J(r')\bullet\frac{r-r'}{|r-r'|^{3}} dv'[/itex]

    I discussed your reply with a friend and I would like to know how one of the two in the equation below is zero
    ∇[itex]\bullet(f*\vec{F}) [/itex]= ∇f[itex]\bullet \vec{F}[/itex] + [itex]f∇\bullet \vec{F}[/itex]

    I still have trouble seeing how we get to zero?
    The integral is just the specific volume fx a box, J(r') can be seen as a cross section, but I don't see how it's possible to say that r-r' is perpendicular to J(r')
  5. Nov 15, 2012 #4


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    Good. And, of course, in the same way you can show taking the gradient with respect to the unprimed r gives [itex]\frac{r'-r}{|r-r'|^{3}}[/itex] which is just the negative of the gradient with respect to r'.

    So, you have shown

    (1/|r-r'|) = -'(1/|r-r'|)

    Therefore, you have ##\cdot## A(r) = ∫##\cdot##[J(r')/|r-r'|] dV' = ∫J(r')##\cdot##∇(1/|r-r'|) dV' = -∫J(r')##\cdot##∇'(1/|r-r'|) dV'

    Now see if you can use your identity [itex]∇'\bullet(f*\vec{F}) [/itex]= [itex](∇'f)\bullet \vec{F}[/itex] + [itex]f∇'\bullet \vec{F}[/itex] (where I have added primes to the ∇ operator) to rewrite the last integral as two terms and then show each term is zero. (Note that the last integral above has an integrand of the form of the first term on the right hand side of the identity.)
    Last edited: Nov 15, 2012
  6. Nov 17, 2012 #5
    The divergence of J(r')=0 making the second term zero


    [itex]\int \frac{r−r′}{|r−r′|^{3}} \bullet J(r') dv'[/itex]
    where the gradient of [itex]1/|r-r'|[/itex]

    I thought of several solutions but I can't really pin down the right one.
    1: Making it into a closed surface integral and J(r') somehow equals to zero
    [itex]\oint 1/|r-r'| \bullet J(r') \bullet n ds'[/itex]
    2: Integrating with respect to r' in a closed line integral would somehow end in a zero.
    Last edited: Nov 17, 2012
  7. Nov 17, 2012 #6


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    We've gotten to ##\cdot## A(r) = -∫J(r')##\cdot##∇'(1/|r-r'|) dV', which you want to show is zero.

    Rearrange the identity [itex]∇\:'\cdot(f\vec{F}) = (∇\:'f)\cdot \vec{F}+ f\:∇\:'\cdot \vec{F}[/itex] as

    [itex]\vec{F}\cdot(∇\:'f) = ∇\:'\cdot(f\vec{F}) -f\:∇\:'\cdot \vec{F}[/itex]

    Use this to rewrite the integral ∫J(r')##\cdot##∇'(1/|r-r'|)dV' as two integrals. (Do not write out ∇'(1/|r-r'|) as (r-r')/|r-r'|3).

    You have hit on the right ideas to show that these two integrals will be zero: '##\cdot## J(r') = 0 and converting one of the integrals into a surface integral.
  8. Nov 18, 2012 #7
    I'm trying to solve the same problem and got to the same conclusion. But how do you justify that the surface integral is zero?
  9. Nov 18, 2012 #8


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    You can think of the volume integrals as extending over all of infinite 3D space. So, the surface integral will be over a surface at infinity where you can assume there is no current.
  10. Nov 19, 2012 #9
    Thank you very much for this TSny, my whole class had problems with this excercise.
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