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Homework Help: 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


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    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
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