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

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Homework Help Overview

The discussion revolves around the derivation of Ampere's Law from the Biot-Savart Law, focusing on the mathematical identities involved in the process. Participants are exploring the relationships between these laws in the context of magnetostatics.

Discussion Character

  • Mathematical reasoning, Conceptual clarification

Approaches and Questions Raised

  • Participants are attempting to apply vector calculus identities to derive Ampere's Law and are questioning specific steps in the derivation, particularly regarding the treatment of terms involving gradients and the implications of the dot product in this context.

Discussion Status

The discussion is active, with participants sharing their attempts to reconcile different terms in the derivation. Some have provided alternative identities for consideration, while others are questioning the assumptions made in the derivation process. There is no explicit consensus yet, but multiple lines of reasoning are being explored.

Contextual Notes

Participants are noting the complexity of the relationships between Ampere's Law and the Biot-Savart Law, with references to their presentations in coursework and the foundational nature of these laws in magnetostatics.

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 [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?
 
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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].
 
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?
 
surely here though, the components of [itex]\nabla (\frac{1}{r})[/itex] 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 [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?
 
Yeah, the prof derived it in class over the course of two lectures!
 

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