Solving Vector Potential: Calculating Magnetic Field & Difficulties

In summary: So now that I have these, I can solve for ##\mathbf{A}##:In summary, to solve for the vector potential ##A=2Cln\frac{x^2+y^2}{z^2} \hat{z}##, I used the following formula to calculate the magnetic field:I was having problem solving this part, I was thinking about using the same formula in question a) ##\mathbf{B} = \nabla \times \mathbf{A'}## and solve for ##\mathbf{A'}## since I have a known ##\mathbf{B}##, but then I realized it is too
  • #1
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Homework Statement
Consider the vector potential
A = C ln( (x^2+y^2)/z^2 ) \hat{z}

a) Determine the B-field for this potential. Describe a situation that gives a B-field of that shape.
b) Calculate \nabla \cdot A
c) Give a vector potential in coulomb gauge, A’, which gives the same B as in the a-task.
Relevant Equations
B = \nabla x A
My solution for the vector potential ##A=2Cln\frac{x^2+y^2}{z^2} \hat{z}## is:
a) I used the following formula to calculate the magnetic field
$$ \mathbf{B} = \nabla \times \mathbf{A} = \left( \frac{dA_z}{dy} - 0 \right) \hat{x} + \left( 0 - \frac{dA_z}{dx} \right)\hat{y} + 0 \hat{z} = \frac{dA_z}{dy} - \frac{dA_z}{dx} = 2C \frac{x}{x^2+y^2} \hat{x} - 2C \frac{y}{x^2+y^2} \hat{y}. $$
How should I describe a situation with this magnetic field here?

b) I just did the calculation of the formula and got a non-zero value:
$$ \nabla \cdot \mathbf{A} = \left( \frac{d}{dx}, \frac{d}{dy}, \frac{d}{dz} \right) \cdot \mathbf{A} = \frac{d\mathbf{A}}{dx} + \frac{d\mathbf{A}}{dy} + \frac{d\mathbf{A}}{dz} = 2C \left( \frac{y}{x^2+y¨2} + \frac{x}{x^2+y^2} - \frac{1}{z} \right) $$

c) I was having problem solving this part, I was thinking about using the same formula in question a) ##\mathbf{B} = \nabla \times \mathbf{A'}## and solve for ##\mathbf{A'}## since I have a known ##\mathbf{B}##, but then I realized it is too difficult to solve it. I also thought mathematically that this expression ##\mathbf{A'} = Cln(x^2+y^2)\hat{z}## could give us the same value of B as in the a-task. But this is still not solving the problem correctly. What should I do?
 
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  • #2
From the looks of it, you should convert to cylindrical coordinates. Look up the del operator if you don't know it in cylindrical coordinates.
 
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  • #3
kuruman said:
From the looks of it, you should convert to cylindrical coordinates. Look up the del operator if you don't know it in cylindrical coordinates.
How should I convert to cylendrical form? Do you mean re-write ##\mathbf{A}## in cylderincal form or re-do the whole calculation?
 
  • #4
kuruman said:
From the looks of it, you should convert to cylindrical coordinates. Look up the del operator if you don't know it in cylindrical coordinates.
Also, why should it be cylendrical?
 
  • #5
Yes, that's what I mean. Find its form here. It should be cylindrical because that is its symmetry.
 
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  • #6
Also what do you mean by its symmetry? What is its symmetry?
 
  • #7
kuruman said:
Yes, that's what I mean. Find its form here. It should be cylindrical because that is its symmetry.
Okay, so for the part where the curl is I need to calculate in this form:
$$ \nabla \times A = \left[ \frac{1}{s} \frac{\partial A_z}{\partial \phi} - \frac{\partial A_\phi}{\partial z} \right]\hat{s} + \left[ \frac{\partial A_s}{\partial z} - \frac{\partial A_z}{\partial s} \right]\hat{\phi} + \frac{1}{s} \left[ \frac{\partial s A_\phi}{\partial s} - \frac{\partial A_s}{\partial \phi} \right] \hat{z} = \frac{1}{s} \frac{\partial A_z}{\partial \phi}\hat{s} - \frac{\partial A_z}{\partial s} \hat{\phi} $$
How am I going to derivate ##A_z## with respect to ##\phi## or ##s##?
 
  • #8
Gustav said:
How am I going to derivate ##A_z## with respect to ##\phi## or ##s##?
First, you need to write ##A_z## as a function of ##r##, ##\phi## and ##z##? Do that first, then take the partial derivatives.
 
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  • #9
kuruman said:
First, you need to write ##A_z## as a function of ##r##, ##\phi## and ##z##? Do that first, then take the partial derivatives.
Okay, I tried to re-write it and I got something like this:
## x= s \cos(\phi ), y= s \sin(\phi), z=z##
$$ \mathbf{A} = C \ln \left( \frac{x^2+y^2}{z^2} \right) = C \ln \left( \frac{ (s \cos(\phi )^2 + (s \sin(\phi )^2}{z^2} \right) = C \ln \frac{s^2}{z^2} $$
 
  • #10
What is ##s##? Also, please fix the LaTeX.

On edit: Thank you for fixing it.
 
Last edited:
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  • #11
kuruman said:
What is ##s##? Also, please fix the LaTeX.
##s=\rho ##, I am sorry about the LaTeX, I am working on it.
 
  • #12
Now I think I can manage to calculate the curl and the divergence. But I still don't understand why we used cylendrical coordinates?
 
  • #13
Gustav said:
Okay, I tried to re-write it and I got something like this:
## x= s \cos(\phi ), y= s \sin(\phi), z=z##
$$ \mathbf{A} = C \ln \left( \frac{x^2+y^2}{z^2} \right) = C \ln \left( \frac{ (s \cos(\phi )^2 + (s \sin(\phi )^2}{z^2} \right) = C \ln \frac{s^2}{z^2} $$
Also this is in the ##\hat{z}##
 
  • #14
Right. You should put in the ##\hat z## to avoid confusing anyone who might be reading this. Also, conventionally ##s## is ##r## but that's OK if you know what you're doing. Now find the curl and the divergence. Once you have these, I will explain to you why one should use cylindrical coordinates for this problem.
 
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  • #15
kuruman said:
Right. You should put in the ##\hat z## to avoid confusing anyone who might be reading this. Also, conventionally ##s## is##r## but that's OK if you know what you're doing. Now find the curl and the divergence. Once you have these, I will explain to you why one should use cylindrical coordinates.
Okay, so now that I have my ## A_z = C \ln \frac{s^2}{z^2} \hat{z} ## I can calculate the derivatives:
1) Curl:
$$ \nabla \times A = \frac{1}{s}\frac{\partial A_z}{\partial \phi} \hat{s} - \frac{\partial A_z}{\partial s} \hat{\phi} = 0 - \frac{2C}{s}\hat{\phi} = - \frac{2C}{s}\hat{\phi}.$$
2) Divergence:
$$ \nabla \cdot A = 0 + 0 + \frac{\partial A_z}{\partial z} = - \frac{2C}{z} $$
 
  • #16
Very good. Compare these expressions and how easy it was to take derivatives with what you did earlier with Cartesian coordinates. Also, note that from three coordinate ##x##, ##y## and ##z## the problem has been reduced to two ##s## and ##z##.

Look at ##\vec\nabla \times \vec A.## It must be a magnetic field. Can you guess what can possibly generate a magnetic field that looks like this? If so, what is the appropriate value for ##C##?
 
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  • #17
kuruman said:
Very good. Compare these expressions and how easy it was to take derivatives with what you did earlier with Cartesian coordinates. Also, note that from three coordinate ##x##, ##y## and ##z## the problem has been reduced to two ##s## and ##z##.

Look at ##\vec\nabla \times \vec A.## It must be a magnetic field. Can you guess what can possibly generate a magnetic field that looks like this? If so, what is the appropriate value for ##C##?
Nothing is crossing my mind... I don't think I know what could generate this magnetic field. How should I think to get that?
 
  • #18
Have you had a course in introductory physics where you calculated magnetic fields for various current configurations?
 
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  • #19
kuruman said:
Have you had a course in introductory physics where you calculated magnetic fields for various current configurations?
Yes, I have
 
  • #20
So what I think you mean is that the constant ##C=\frac{\mu_0}{2\pi}##. But in that kind of experession we have a current in the dominator.
 
  • #21
So we are talking about a magnetic field in a wire that is circular shaped?
 
  • #22
Gustav said:
So what I think you mean is that the constant ##C=\frac{\mu_0}{2\pi}##. But in that kind of experession we have a current in the dominator.
There is never a current in the denominator of an expression for a magnetic field. The magnetic field is proportional to the current. That's Ampere's law. What current configuration are you thinking of? If you don't remember, look it up. There is no hurry.
 
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  • #23
kuruman said:
There is never a current in the denominator of an expression for a magnetic field. The magnetic field is proportional to the current. That's Ampere's law. What current configuration are you thinking of? If you don't remember, look it up. There is no hurry.
Sorry I meant numerator not denominator! I was thinking about an expression as the following
$$ \mathbf{B} = \frac{\mu_0 I}{2\pi s} \hat{\phi} $$
 
  • #24
kuruman said:
There is never a current in the denominator of an expression for a magnetic field. The magnetic field is proportional to the current. That's Ampere's law. What current configuration are you thinking of? If you don't remember, look it up. There is no hurry.
So I have an example in my book about "finding the magnetic field a distance s from a stright wire", the expression of its magnetic field reminded me of the one I got in the calculation for the curl.
 
  • #25
That's the one. Nice work. Can you finish the problem now?
 
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  • #26
kuruman said:
That's the one. Nice work. Can you finish the problem now?
Now I am stuck when calculating the vector potential in coulomb gauge, A’. How can I find the A'?
 
  • #27
It's not just any A'. It must have a curl that gives the B-field of a straight wire.
 
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  • #28
kuruman said:
It's not just any A'. It must have a curl that gives the B-field of a straight wire.
Yes, but how can I solve for it?
 
  • #29
kuruman said:
It's not just any A'. It must have a curl that gives the B-field of a straight wire.
Does it mean that it is the same as the vector potential A?
 
  • #30
No, it means that the curl of A' gives the same B-field. Furthermore, it must satisfy the Coulomb gauge ##\vec \nabla \cdot \vec A' =0##.
 
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  • #31
kuruman said:
No, it means that the curl of A' gives the same B-field. Furthermore, it must satisfy the Coulomb gauge ##\vec \nabla \cdot \vec A' =0##.
Okay, I am with you so far. But does that mean I need to think of an expression for A' or is there a way to calculate it?
 
  • #32
No way to calculate it. There are many answers that work, just pick one and verify that it does work by taking its curl and divergence.
 
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  • #33
kuruman said:
No way to calculate it. There are many answers that work, just pick one and verify that it does work by taking its curl and divergence.
Oh okay, so does it work if
$$ \vec A' = C \ln \left( \frac{1}{s^2} \right)\hat{z} ? $$
 
  • #34
Because then I will have:
1) Curl:
$$ \vec B = \nabla \times \vec A' = -\frac{2C}{s}\hat{\phi} $$
2) Divergence:
$$ \nabla \cdot \vec A' = 0 $$
 
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  • #35
Gustav said:
Because then I will have:
1) Curl:
$$ \vec B = \nabla \times \vec A' = -\frac{2C}{s}\hat{\phi} $$
2) Divergence:
$$ \nabla \cdot \vec A' = 0 $$
Excellent. You are done.
 
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