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

In summary, the vector potential inside a solenoid depends only on \rho, the perpendicular distance from the axis of the solenoid due to the invariant geometry of the problem. Additionally, there is only a \hat{\phi} (azimuthal) component of the vector potential because any other component would violate the symmetry of the problem. This is determined by the physical charge and current configuration, which exhibits radial symmetry in a cylindrical solenoid. Therefore, the vector potential must also exhibit radial symmetry.
  • #1
AxiomOfChoice
533
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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?
 
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  • #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?!
 
  • #3
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.
 
  • #4
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)
 

1. How do you determine the vector potential inside a solenoid?

The vector potential inside a solenoid can be determined using the Biot-Savart law, which states that the magnetic field created by a current-carrying wire is directly proportional to the magnitude of the current and inversely proportional to the distance from the wire.

2. Why does the vector potential inside a solenoid only depend on r?

This is because a solenoid is a long, tightly wound coil of wire that creates a uniform magnetic field inside. The magnetic field lines are parallel and evenly spaced, which means the magnitude of the field does not change with distance from the center of the solenoid. Therefore, the vector potential only depends on the distance from the center, or the radius (r).

3. Can the vector potential inside a solenoid change with time?

Yes, the vector potential inside a solenoid can change with time if there is a changing current or if the solenoid is moving. This is because a changing magnetic field induces an electric field, which in turn affects the vector potential.

4. How does the number of turns in a solenoid affect the vector potential inside?

The number of turns in a solenoid has no direct effect on the vector potential inside. As long as the solenoid has a uniform current and is tightly wound, the magnetic field and vector potential will remain constant within the solenoid.

5. Is the vector potential inside a solenoid affected by the material of the core?

No, the vector potential inside a solenoid is not affected by the material of the core. The core only serves to concentrate and strengthen the magnetic field, but the vector potential is still determined solely by the current and distance from the center of the solenoid.

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