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Magnetostatics: Magnetic Vector Potential

  1. May 3, 2010 #1
    1. The problem statement, all variables and given/known data
    Give an expression for the magnetic field and show that a magnetic vector exists such as [tex]\vec{A}(P) = A(r)\hat{z}[/tex] and [tex]\vec{B}(P) = \vec{\nabla} \times \vec{A}[/tex]
    For the infinite wire shown in figure 1.

    Here is a link to the figure and problem statement. The problem is the second problem on the first page.

    https://docs.google.com/fileview?id...UtOGNkYi00ZGQyLTkxOTktNWVjYzM2MGViNDg3&hl=en"


    2. Relevant equations

    [tex] \vec{\nabla} \times \vec{A} = \left(\frac{1}{r}\frac{\partial A_{z}}{\partial \theta} - \frac{\partial A_{\theta}}{\partial z}\right)\hat{r} + \left(\frac{\partial A_{r}}{\partial z} - \frac{\partial A_{z}}{\partial r}\right)\hat{\theta} + \frac{1}{r}\left(\frac{\partial(rA_{\theta})}{\partial r} - \frac{\partial A_{r}}{\partial \theta}\right)\hat{z}[/tex] curl in polar coordinates

    [tex]\oint \vec{B} \cdot d\vec{l} = \mu_{0}I_{enc}[/tex] Ampere's Law


    3. The attempt at a solution

    First we must solve for the magnetic field at point P. Using Ampere's Law we get,

    [tex] \vec{B}(P) = \frac{\mu_{0}I}{2\pi r}\hat{\theta} [/tex] where the positive theta direction is into the page.

    Now in order to find A we can set the curl equations equal to the magnetic field equation by curl(A) = B.

    Doing so gives,

    [tex]\frac{1}{r}\frac{\partial A_{z}}{\partial \theta} - \frac{\partial A_{\theta}}{\partial z} = 0 [/tex]

    [tex]\frac{\partial A_{r}}{\partial z} - \frac{\partial A_{z}}{\partial r} = \frac{\mu_{0}I}{2\pi r}[/tex]

    [tex]\frac{\partial(rA_{\theta})}{\partial r} - \frac{\partial A_{r}}{\partial \theta} = 0[/tex]

    The answer to this problem (given to us by our instructor) is,

    [tex]\vec{A} = \left(-\frac{\mu_{0}I}{2\pi}ln(r) + K \right)\hat{z}[/tex]

    and the only way I can get there is by saying that because A points most often in the direction of the current (which is the z direction in this case) then,

    [tex]-\frac{dA_{z}}{dr} = \frac{\mu_{0}I}{2\pi r}.[/tex]

    Then solving this differential equation gives the desired result.

    Can someone please point me in the right direction as to how solve this problem in a more rigorous manner?

    Do I need to use the equation [tex]\vec{A} = \frac{\mu_{0}}{4\pi}\int \frac{\vec{J}(\vec{r'})}{r}d\tau '[/tex]?


    Many Thanks in advance,

    KEØM
     
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. May 3, 2010 #2
    You already know that:

    [tex]A_r = 0[/tex]

    and

    [tex]A_{\theta} = 0[/tex]

    and

    [tex]A_z = A(r)[/tex]

    since it was given to you in the problem. This simplifies your 3 diff eq's quite a bit.
     
  4. May 4, 2010 #3
    Thanks for your reply nickjer. So because A is a function of only r then all other derivatives go away and because it is only in the z direction then all other components are equal to zero. This then allows me to let [tex]-\frac{dA_{z}}{dr} = \frac{\mu_{0}I}{2\pi r}.[/tex] Is there anything else I must say for that condition to be satisfied?

    Thanks again,
    KEØM
     
  5. May 4, 2010 #4

    kreil

    User Avatar
    Gold Member

    nope that does it:

    [tex]\vec{B}(P) = \vec{\nabla} \times \vec{A}[/tex]

    [tex]\vec{B}(P) = \frac{\mu_{0}I}{2\pi r}\hat{\theta} [/tex]

    [tex]\vec{\nabla} \times \vec{A}=-\frac{dA_{z}}{dr} [/tex]

    thus [tex]-\frac{dA_{z}}{dr}=\frac{\mu_{0}I}{2\pi r}\hat{\theta}[/tex]
     
  6. May 4, 2010 #5
    Thanks! I really appreciate the help kreil and nickjer.

    KEØM
     
  7. May 4, 2010 #6
    I now have a question concerning the third problem on that page attached.

    Problem Statement:

    We consider now two wires of axis (O,z) and separated by the distance (2a). The currents in the two wires are [tex]+I[/tex] and [tex]-I[/tex] (see figure). Show that the expression of the magnetic potential vector [tex]\vec{A}(P)[/tex] is given by

    [tex]\vec{A}(P) = \frac{\mu_{0}I}{2\pi}ln\left(\frac{r_{1}}{r_{2}}\right)\hat{z}[/tex]

    (the observation point is located at [tex]P(r, \theta, z=0)[/tex]). Note: The constant of integration is integration obtained by taking [tex]\vec{A}(O) = 0[/tex].

    Relevant Equations:

    Same as previous problem.

    Attempt at a solution:

    For this problem can I say the same as before? He did say that our answer should be of the form [tex]\vec{A}(P) = A(r)\hat{z}.[/tex]

    Thanks in advance.
     
    Last edited: May 4, 2010
  8. May 4, 2010 #7
    You solved for the vector potential from a single wire. So for two wires you will just add the vector potentials from each. But be careful with the direction of each of the vectors.
     
  9. May 4, 2010 #8
    I know from just looking at the answer that the -I current must be in the positive z-direction and the +I current must be in the negative z-direction but I don't know why.
     
  10. May 4, 2010 #9
    In the first problem you solved [itex]\vec{A}[/itex] for a current going in the +z direction. So use that to solve this next problem.
     
  11. May 4, 2010 #10
    Thanks for your reply nickjer.

    I think I got it. So because +I flows in the positive direction it must be negative and the opposite must be true for the wire with -I.

    One more question:

    For problem 4 on that document it says,

    Problem Statement
    The wires are now very close to each other. Determine [tex]\vec{A}(r,\theta)[/tex] such as r >> a and by keeping the first and second order terms in (a/r).

    Relevant Equations
    Remember that [tex](1 + \epsilon)^{n} = 1 + n\epsilon[/tex]

    and [tex]ln(1 + \epsilon) = \epsilon, \epsilon << 1[/tex]

    Attempt at a solution

    Now I know he wants us to approximate the answer by expanding[tex]\vec{A}(P) = \frac{\mu_{0}I}{2\pi}ln\left(\frac{r_{1}}{r_{2}}\right)[/tex]

    with a Taylor Expansion in [tex] \epsilon = \frac{a}{r} << 1[/tex] but I don't see how to do this when the distance between the wires [tex]2a[/tex] never comes up in any of our answers.

    I really appreciate all the help nickjer.
     
  12. May 4, 2010 #11
    Now you will need to write out [itex]r_1[/itex] and [itex]r_2[/itex] in terms of r and a. I suggest using the law of cosines.
     
  13. May 8, 2010 #12
    OK. Thanks again for all of your help nickjer.
     
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