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Vector Identity problem

  1. Jun 5, 2013 #1
    1. The problem statement, all variables and given/known data
    I want to compute the electric field knowing the magnetic field using a vector identity

    2. Relevant equations

    E=i [itex]\frac{c}{k}[/itex] (∇[itex]\times[/itex]B)

    B(r,t)=(μ0ωk/4π) ([itex]\hat{r}[/itex]×[itex]\vec{p}[/itex])[1-[itex]\frac{1}{ikr}[/itex]](eikr/r)

    [itex]\vec{p}[/itex]=dipole moment,constant vector

    we have ti use the identity [itex]\nabla[/itex][itex]\times[/itex](A[itex]\times[/itex]B)=(B[itex]\cdot[/itex]∇)A-(A[itex]\cdot[/itex]∇)B +A(∇[itex]\cdot[/itex]B) +B(∇[itex]\cdot[/itex]A)

    the identy simplifies in this situtation because for some reason we take (A[itex]\cdot[/itex]∇)B=0 and A(∇[itex]\cdot[/itex]B)=0
    So applying this we have :

    E(r,t)=ic/k(μ0ωk/4π) [itex]\nabla[/itex][eikr/r2(1-[itex]\frac{1}{ikr}[/itex]]×(r×p)+ic/k(μ0ωk/4π)[eikr/r2(1-[itex]\frac{1}{ikr}[/itex]]∇×(r×p)
    E(r,t)=i(ω/4πε0c)[ik([itex]\frac{1}{r^2}[/itex]-[itex]\frac{1}{ikr^3}[/itex])]eikr r×(r×p) + i(ω/4πε0c)[(eikr/r^2)(1-[itex]\frac{1}{ikr}[/itex])][-∇[itex]\cdot[/itex]r)p+(p[itex]\cdot[/itex]∇)r] the this part says it's equal to -∇[itex]\cdot[/itex]r)p+(p[itex]\cdot[/itex]∇)r=-3p+p=-2p so

    E(r,t)=[itex]\frac{k^2}{4πε0}[/itex](r×p)×r (ei(kr-ωt)/r) + [itex]\frac{1}{4πε0}[/itex][3r(r[itex]\cdot[/itex]p)-p]([itex]\frac{1}{r^3}[/itex]-[itex]\frac{ik}{r^2}[/itex])ei(kr-ωt)

    My problem is i dont know how the vector identy is used here..with this tools we calculate magnetic and electric fields in the approximation zones( near,far-field) when vector potential is given. Can someone give a more simple example than this of what he did in this solution?
     
    Last edited: Jun 5, 2013
  2. jcsd
  3. Jun 5, 2013 #2

    Simon Bridge

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    Lets make sure I follow you first:

    You want to find ##\vec{E}## given:
    $$\vec{E} = i\frac{c}{k}\vec{\nabla}\times\vec{B}\\
    \vec{B}(\vec{r},t)=\frac{\mu_0\omega k}{4\pi}\left (\vec{r}\times\vec{p} \right )
    \left [ 1 - \frac{1}{ik\vec{r}} \right ]\frac{1}{\vec{r}}e^{ikr}\\

    \vec{\nabla}\times(\vec{A}\times\vec{B})=(\vec B\cdot\vec{\nabla})\vec A-(\vec A\cdot\vec \nabla)\vec B + \vec A(\vec \nabla \cdot \vec B)+ \vec B(\vec \nabla \cdot \vec A)

    $$... skipping a bit for now:
    ... if I got the above right, it looks to me that when you do ##\vec \nabla \times \vec B## you will end up with a term involving ##\vec \nabla \times (\vec{r}\times\vec{p})## ... which is where the identity should have come in.

    BTW: the equation editor can be tricky to use.
    It is normally better just to type the LaTeX markup in directly.
     
  4. Jun 5, 2013 #3
    i figured out how the identity works. In this situation i don't know , but as it seems the problems i am into, don't require all of the above but simpler cases.

    Thanks a lot for your time
     
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