Magnetic Potential: Computing the Curl of $\mathbf{A}$

In summary, you attempted to show that the magnetic potential given by: \mathbf{A(r)}=\frac{\mu_0}{4 \pi} \int dV' \frac{\mathbf{J(r')}}{|\mathbf{r-r'}|}acts only on r terms. You found that the curl of this vector is: \nabla \times \frac{\mathbf{J(r')}}{\mathbf{|r-r'|}}=\nabla(\frac{1}{\mathbf{|r-r'|}}) \times \mathbf{J(r')}+
  • #1
latentcorpse
1,444
0
hi again, I'm trying to show that [itex]\mathbf{B} = \nabla \times \mathbf{A} [/itex] where [itex]\mathbf{A}[/itex] is the magnetic potential given by:

[itex]\mathbf{A(r)}=\frac{\mu_0}{4 \pi} \int dV' \frac{\mathbf{J(r')}}{|\mathbf{r-r'}|}[/itex]

i deduced that since [itex]\nabla=(\frac{\partial}{\partial{r_1}},\frac{\partial}{\partial{r_2}},\frac{\partial}{\partial{r_3}})[/itex] it only acts on r terms and so

[itex]\nabla \times \mathbf{A}=\frac{\mu_0}{4 \pi} \int dV' \mathbf{J(r')} ( \nabla \times \frac{1}{|\mathbf{r-r'}|} )[/itex]

i can't figure out how to compute that curl - it should be fairly elementary, no?
 
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  • #2
You should choose a specific co-ordinate system and work it.
 
  • #3
tried it in cartesian and it got messy-

[itex]\epsilon_{ijk} \partial_j [{(r-r')_k (r-r')_k}]^{-\frac{1}{2}}[/itex]

?
 
  • #4
Oops, that had been bad advice. My apologies, I should have paid more attention.

Here's some better advice : you want to take the curl of this [tex] \frac{\mathbf{J(r ^')}}{|\mathbf{r - r^'}|} [/tex]

This is a vector multiplied by a scalar. So you got to use the product rule for curl in such situations.
 
  • #5
ok so
[itex]\nabla \times \frac{\mathbf{J(r')}}{\mathbf{|r-r'|}} = \nabla(\frac{1}{\mathbf{|r-r'|}}) \times \mathbf{J(r')} + \frac{1}{\mathbf{|r-r'|}} \nabla \times \mathbf{J(r')}[/itex]

im confused as to how to take the curl of J(r') because [itex]\nabla=\frac{\partial}{\partial{r_i}}[/itex] not [itex]\nabla=\frac{\partial}{\partial{r'_i}}[/itex] so how can the del operator act on a vector function of r'?
 
  • #6
No, it can't.
 
  • #7
so the second term is 0

how do i write out the first term though

[itex]\partial_i \left( (r_j-r'_j)(r_j-r'_j) \right)^{-\frac{1}{2}}[/itex] this appears to violate Einstein summation convention?
 
  • #8
I would suggest forgetting fancy conventions and working it out in some co-ordinate system with all x s and y s, or r s and thetas, whichever you like.
 
  • #9
yeah wwe're expected to use that notation though.

i did this:

[itex]a_i=r_i-r'_i[/itex] then we have

[itex]\partial_i(a_ja_j)^{-\frac{1}{2}}=-\frac{1}{2}(a_ja_j)^{-\frac{3}{2}}2a_j(\partial_i(r_j-r'_j))=-\frac{1}{(r-r')^3}a_j \partial_i (r_j-r'_j)=-\frac{1}{(r-r')^3}a_j (\delta_{ij}-0)=-\frac{1}{(r-r')^2}[/itex]

which seems like a suitbale answer?
 
  • #10
It does indeed.
 
  • #11
actually, shouldn't this be [itex]-\frac{\mathbf{r-r'}}{(r-r')^3}[/itex] as grad should be a vector?
 
  • #12
It does not act on [tex] r^' [/tex] . Of course, there should be a unit vector along r .
 
  • #13
but back in post 9 i got [itex]-\frac{a_i}{(r-r')^3}[/itex] where[itex]a_i=(r-r')_i[/itex] suggesting a vector in both directions?
 
  • #14
Duh, I am caught napping again. Actually the term in your 11th post is correct. Absolutely and completely perfect.
 

1. What is magnetic potential?

Magnetic potential is a concept used in electromagnetism to describe the magnetic field in terms of a scalar potential. It is analogous to the electric potential in electrostatics.

2. How is the magnetic potential related to the magnetic field?

The magnetic potential is related to the magnetic field by the curl operator. The curl of the magnetic potential, denoted by $\mathbf{B}$, gives the magnetic field vector $\mathbf{B}$. This relationship is described by the Maxwell's equations for electromagnetism.

3. What is the importance of computing the curl of $\mathbf{A}$?

Computing the curl of $\mathbf{A}$ is important in understanding the behavior of magnetic fields. It helps in predicting the direction and strength of magnetic fields, as well as in solving problems related to electromagnetic induction.

4. How is the curl of $\mathbf{A}$ calculated?

The curl of $\mathbf{A}$ is calculated using the vector calculus operator called the curl operator, denoted by $\nabla \times$. It is applied to the vector $\mathbf{A}$ to give the magnetic field vector $\mathbf{B}$. In mathematical notation, this can be written as $\mathbf{B} = \nabla \times \mathbf{A}$.

5. Are there any practical applications of computing the curl of $\mathbf{A}$?

Yes, there are many practical applications of computing the curl of $\mathbf{A}$. It is used in various fields such as electromagnetics, electrical engineering, and physics. Some examples include designing magnetic sensors, predicting behavior of magnetic materials, and understanding the dynamics of charged particles in magnetic fields.

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