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About the curl of B using Biot-Savart Law

  1. Dec 12, 2009 #1
    I am reading the Griffiths about finding the curl of B using Biot-Savart Law. I do not understanding the step between equation (5.52) and (5.53) which finding the x components of the following:
    [tex](\boldsymbol{J}\cdot\nabla^{\prime})\dfrac{\hat{\xi}}{\xi^{2}}[/tex]
    where
    [tex]\mathbf{\mathbf{\xi}}=\mathbf{r}-\mathbf{r}^{\prime}[/tex]
    I don't know why it said using product rule 5 in his book, can get the following result:
    [tex]\left(\boldsymbol{J}\cdot\nabla^{\prime}\right)\dfrac{x-x^{\prime}}{\xi^{3}}=\nabla^{\prime}\cdot\left[\dfrac{(x-x^{\prime})}{\xi^{3}}\mathbf{J}\right]-\left(\dfrac{x-x^{\prime}}{\xi^{3}}\right)\left(\nabla^{\prime}\cdot\mathbf{J}\right)[/tex]
    Is there anyone know how to deduce this result?
     
  2. jcsd
  3. Dec 12, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi yanyan_leung ! Welcome to PF! :smile:
    That's ∑i (Ji∂/∂xi) (f) = ∑i ∂/∂xi (fJi) - f ∑i ∂/∂xi (Ji) …

    now just use the product rule on the middle term. :wink:
     
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