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Ampere's Law Derivation

  1. Apr 15, 2009 #1
    This must be a pretty standard proof but I'm having difficulty with part of it.

    So we have from Biot Savart law that [itex]\vec{B}(\vec{r})=\frac{\mu_0}{4 \pi} \int_V dV' \vec{J}(\vec{r'}) \times \nabla(\frac{1}{r})[/itex]

    we take the curl of this and show the second term vanishes to leave us with [itex]\nabla \times \vec{B}(\vec{r})=\mu_0 \vec{J}(\vec{r})[/itex] which is Ampere's law in differential form.

    however we take the curl of this using the identity a x (b x c) = b(a.c)-c(a.b). here this gives

    [itex]\nabla \times \left(\vec{J}(\vec{r'}) \times \nabla(\frac{1}{r})\right) = \nabla^2(\frac{1}{r}) \vec{J}(\vec{r'}) - (\nabla \cdot \vec{J}(\vec{r'})) \nabla(\frac{1}{r})[/itex]

    but in the book they have the scond term as [itex](\vec{J}(\vec{r'}) \cdot \nabla) \nabla(\frac{1}{r})[/itex]. why is this allowed?

    the dot product doesn't commute when a grad is involved does it?
  2. jcsd
  3. Apr 15, 2009 #2


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    Try this identity: [tex] \nabla \times (\mathbf{A} \times \mathbf{B}) = \mathbf{A} (\nabla \cdot \mathbf{B}) - \mathbf{B} (\nabla \cdot \mathbf{A}) + (\mathbf{B} \cdot \nabla) \mathbf{A} - (\mathbf{A} \cdot \nabla) \mathbf{B} [/tex].
  4. Apr 15, 2009 #3
    ok so when i do that i get everything to work except i have an extra term [itex]-(\mathbf{J} \cdot \nabla) \nabla(\frac{1}{r})[/itex]. why does this vanish?
  5. Apr 16, 2009 #4
    surely here though, the components of [itex]\nabla (\frac{1}{r})[/itex] are changing and so their gradient is non-zero?
  6. Apr 16, 2009 #5
    Isn't Ampere's law more fundamental than Biot-Savart law? That's how it was presented in my E&M course...
  7. Apr 16, 2009 #6
    i think so. you have the fundamental equations of magnetostatics [itex]\nabla \cdot \mathbf{B}=0 , \nabla \times \mathbf{B}=\mu_0 \mathbf{J}[/itex] from which you can establish that the magnetic flux through a closed surface is zero and also ampere's law in integral form.

    however in my notes, Biot Savart law is constructed from the example of force between two wires and from that we establish div and curl of B.

    isn't it pretty copmlicated to establish Biot savart formula for B using ampere's law though?
  8. Apr 16, 2009 #7
    Yeah, the prof derived it in class over the course of two lectures!
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