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Homework Help: Vector Potential of Long Solenoid

  1. Apr 7, 2010 #1
    Let me preface by saying that I am a freshman in an introductory level Electricity and Magnetism course. My professor has assigned this problem, as he briefly introduced the idea of vector potentials, along with curl and divergence operators. I am VERY much lacking in knowledge of any of these concepts and would appreciate guidance. I understand that this site is not supposed to "do the work" for you, but in this case, an explanation would prove very useful.

    Problem Statement and Relevant Equations

    Recall Ampere's Circuital Law
    [tex]\oint{B\bullet \partial l}=\mu_0 I_{enc}=\mu_0 \int{J\bullet dA}[/tex]

    But Stokes Theorem states:

    [tex]\oint{B\bullet \partial l}=\int_{surface}{(\nabla \times B)\bullet dA}[/tex]

    Therefore:

    [tex]\nabla \times B=\mu_0 J[/tex]

    We can apply similar reasoning to:

    [tex]\oint{A\bullet \partial l}=\int{(\nabla \times B)\bullet dA}=\int_{surface}{B\bullet dA}=\phi_{enc}[/tex]

    Apply this to a long solenoid.

    [tex]\oint{A\bullet \partial l}=\phi_{enc}[/tex] is similar to [tex]\oint{B\bullet\partial l}=\mu_0 I_{enc}[/tex]

    Pursue this analogy by drawin the appropriate Amperian loop (radius r) to show:

    [tex]A=\frac{\mu_0 NI}{2}r\hat{\phi}[/tex] inside the solenoid, and:

    [tex]A=\frac{\mu_0 NI}{2}\cdot\frac{R^2}{r}\hat{\phi}[/tex] outisde the solenoid.

    Note that A follows the current direction. Also note that the 1/r dependence gives B=0 outside the coil. Explain and sketch the lines of A in both cases.

    Questions Regarding Problem
    1. Is the A in the problem referring to the vector potential?
    2. What is all of this saying?

    As I stated previously, I am very new to this and any help is greatly appreciated. Thanks
     
  2. jcsd
  3. Apr 8, 2010 #2

    vela

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    Yes.
    The upshot is there's an Ampere's law for the vector potential. If you calculate the line integral of the vector potential A around a closed path, it will equal the magnetic flux through the loop.
     
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