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How do you know the vector potential inside a solenoid only depends on r?

  1. Sep 27, 2009 #1
    How can you argue, by symmetry, that the vector potential inside a solenoid depends only on [itex]\rho[/itex], the perpendicular distance from the axis of the solenoid? And how can you argue that there is only a [itex]\hat{\phi}[/itex] (azimuthal) component of the vector potential, such that [itex]\vec{A}[/itex] takes the form

    [tex]\vec{A}(\vec{r}) = A_\phi(\rho) \hat{\phi}[/tex]

    My E&M teacher is fond of talking about how we can "use symmetry arguments" to reach these conclusions without actually providing the arguments for us. It's a little frustrating...what symmetry arguments is he talking about?
     
    Last edited: Sep 27, 2009
  2. jcsd
  3. Sep 27, 2009 #2
    Ok, I think I understand why [itex]A[/itex] has only an azimuthal component. The equation for the vector potential reads
    [tex]
    \vec{A}(\vec{r}) = \frac{\mu_0}{4 \pi} \int\int \frac{\vec{K}(\vec{r}') dA}{|\vec{R}|},
    [/tex]
    and since [itex]\vec{K}(\vec{r}) = \frac{N}{L} I \hat{\phi}[/itex] (where N is the # of turns and L is the length), and we can't obtain other components from performing the surface integral, we're stuck with only a [itex]\phi[/itex] component for the vector potential.

    Is that right?!
     
  4. Sep 27, 2009 #3

    Born2bwire

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    Gold Member

    Basically, the geometry of the problem is invariant in the z hat and phi hat dimensions. There is no change in the problem with respect to these variables and dimensions. If the entire system is invariant in a dimension, then any other properties of the system must also be invariant.
     
  5. Sep 28, 2009 #4

    diazona

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    Homework Helper

    To put it another way: the vector potential has to be determined by the physical charge and current configuration, right? And in a cylindrical solenoid, that configuration exhibits radial symmetry - that is, you can rotate it around its axis, and the charge/current configuration looks just the same as it did before.

    Now suppose the vector potential did depend on [itex]\phi[/itex]. Whatever kind of angular dependence you can imagine, it provides some way to distinguish one direction from another... there has to be some "feature" of the vector potential that points in some particular direction, for it to have angular dependence. Let's say that you start by orienting the solenoid so that this feature points upwards (i.e. in a chosen direction which we'll designate as "up"). Then rotate the solenoid around by 90 degrees. Now the feature of the vector potential points to the left. But the charge and current distribution generating the vector potential still looks the same! How can the same distribution create two different vector potentials, one with this "feature" pointing up and one with it pointing to the left?

    (The answer, of course, is that it can't, which is why we know that the vector potential must be radially symmetric)
     
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