Understanding Ampere's Law: A Step-by-Step Derivation

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SUMMARY

This discussion focuses on the derivation of Ampere's Law from the Biot-Savart Law in the context of magnetostatics. The participants analyze the mathematical steps involved, particularly the application of vector calculus identities, such as the curl and divergence operations. The key conclusion is that Ampere's Law in differential form, expressed as \(\nabla \times \vec{B}(\vec{r}) = \mu_0 \vec{J}(\vec{r})\), can be derived by taking the curl of the Biot-Savart Law, with specific attention to the treatment of terms involving the gradient of \(\frac{1}{r}\). The discussion also touches on the relationship between Ampere's Law and the Biot-Savart Law, highlighting the complexity of deriving one from the other.

PREREQUISITES
  • Understanding of vector calculus, particularly curl and divergence operations.
  • Familiarity with Biot-Savart Law and its application in electromagnetism.
  • Knowledge of Ampere's Law in both differential and integral forms.
  • Basic concepts of magnetostatics and magnetic fields.
NEXT STEPS
  • Study the derivation of the Biot-Savart Law from Ampere's Law in detail.
  • Learn about the mathematical identities used in vector calculus, such as the curl of a cross product.
  • Explore the implications of Ampere's Law in various electromagnetic applications.
  • Review the fundamental equations of magnetostatics and their physical interpretations.
USEFUL FOR

Students and professionals in physics, particularly those studying electromagnetism, electrical engineers, and anyone interested in the mathematical foundations of magnetic field theory.

latentcorpse
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This must be a pretty standard proof but I'm having difficulty with part of it.

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

we take the curl of this and show the second term vanishes to leave us with \nabla \times \vec{B}(\vec{r})=\mu_0 \vec{J}(\vec{r}) 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

\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})

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

the dot product doesn't commute when a grad is involved does it?
 
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Try this identity: \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}.
 
ok so when i do that i get everything to work except i have an extra term -(\mathbf{J} \cdot \nabla) \nabla(\frac{1}{r}). why does this vanish?
 
surely here though, the components of \nabla (\frac{1}{r}) are changing and so their gradient is non-zero?
 
Isn't Ampere's law more fundamental than Biot-Savart law? That's how it was presented in my E&M course...
 
i think so. you have the fundamental equations of magnetostatics \nabla \cdot \mathbf{B}=0 , \nabla \times \mathbf{B}=\mu_0 \mathbf{J} 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?
 
Yeah, the prof derived it in class over the course of two lectures!
 

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