1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Confusion regarding the scalar potential

  1. Nov 26, 2015 #1
    1. The problem statement, all variables and given/known data
    Consider the following in cylindrical coordinates [itex]\rho,\varphi,z[/itex]. An electric current flows in an infinitely long straight cylindrical wire with the radius [itex]R[/itex]. The magnetic field [itex]\mathbf{B}[/itex] outside of the thread is [tex]\textbf{B}(\textbf{r})=\frac{I\mu_0}{2\pi}\frac{\mathbf{e}_{\varphi}}{\rho}, ρ>R.[/tex] We want to determine wether [itex]\mathbf{B}[/itex] has a scalar potential or not.

    2. Relevant equations

    3. The attempt at a solution
    Since the domain is not simply connected, there is no use showing that [itex]\nabla\times\mathbf{B}=\mathbf{0}[/itex] for [itex]ρ>R[/itex]. Therefore, we must show that for all points [itex]P[/itex] and [itex]Q[/itex] the line integral of [itex]\mathbf{B}[/itex] from [itex]P[/itex] to [itex]Q[/itex] is independent of path, since this implies that [itex]\mathbf{B}[/itex] has a scalar potential.

    Consider a circle, [itex]C[/itex], that is concentric with the cylindrical wire with radius [itex]R_1[/itex] such that [itex]R_1>R[/itex]. Line integration of [itex]\mathbf{B}[/itex] over the closed curve [itex]C[/itex] yields [tex]\oint_C\mathbf{B}⋅d\mathbf{r} = [d\mathbf{r} = R_1d\varphi\mathbf{e}_{\varphi}] = \frac{I\mu_0}{2\pi}\int^{2\pi}_0\mathbf{e}_{\varphi}⋅\mathbf{e}_{\varphi}d\varphi = I\mu_0 \neq 0.[/tex] But this says that [itex]\mathbf{B}[/itex] is dependent of path, and therefore [itex]\mathbf{B}[/itex] does not have a scalar potential.

    However, if we now consider the function [tex]\phi(\varphi) = \frac{I\mu_0}{2\pi}\varphi,[/tex] we observe that [tex]\nabla\phi = \frac{1}{\rho}\frac{I\mu_0}{2\pi}\mathbf{e}_\varphi = \mathbf{B}.[/tex] But this says that [itex]\mathbf{B}[/itex] is independent of path, and therefore [itex]\mathbf{B}[/itex] has a scalar potential.

    And so, we end up with a contradiction. Since I am very confident that the line integral of [itex]\mathbf{B}[/itex] over [itex]C[/itex] is correct, I assume that there is something in the last part that is wrong, but I cannot find the error.
     
    Last edited: Nov 26, 2015
  2. jcsd
  3. Nov 26, 2015 #2

    pasmith

    User Avatar
    Homework Helper

    Your problem here is that [itex]\varphi[/itex] lies between [itex]0[/itex] and [itex]2\pi[/itex], but both [itex]\varphi = 0[/itex] and [itex]\varphi = 2\pi[/itex] represent the same physical half-plane. Thus if [itex]\phi = k\varphi[/itex], [itex]k \neq 0[/itex], then there is a half-plane where [itex]\phi[/itex] is discontinuous: from one side [itex]\phi \to 0[/itex] and from the other [itex]\phi \to 2k\pi[/itex]. This is a problem for the existence of [itex]\nabla \phi[/itex] *everywhere* in [itex]\rho > R[/itex].

    To avoid this problem, physically acceptable scalar or vector fields are always periodic in [itex]\varphi[/itex] with period [itex]2\pi[/itex].

    (The other way to avoid this is to admit multivalued potentials, in which case yes, [itex]\nabla (k\varphi) = (k/\rho)\mathbf{e}_{\varphi}[/itex].)
     
    Last edited: Nov 26, 2015
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Confusion regarding the scalar potential
  1. Scalar potential in 3D (Replies: 1)

Loading...