Confusion regarding the scalar potential

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1. Nov 26, 2015

lampCable

1. The problem statement, all variables and given/known data
Consider the following in cylindrical coordinates $\rho,\varphi,z$. An electric current flows in an infinitely long straight cylindrical wire with the radius $R$. The magnetic field $\mathbf{B}$ outside of the thread is $$\textbf{B}(\textbf{r})=\frac{I\mu_0}{2\pi}\frac{\mathbf{e}_{\varphi}}{\rho}, ρ>R.$$ We want to determine wether $\mathbf{B}$ 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 $\nabla\times\mathbf{B}=\mathbf{0}$ for $ρ>R$. Therefore, we must show that for all points $P$ and $Q$ the line integral of $\mathbf{B}$ from $P$ to $Q$ is independent of path, since this implies that $\mathbf{B}$ has a scalar potential.

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

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

And so, we end up with a contradiction. Since I am very confident that the line integral of $\mathbf{B}$ over $C$ 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. Nov 26, 2015

pasmith

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

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

(The other way to avoid this is to admit multivalued potentials, in which case yes, $\nabla (k\varphi) = (k/\rho)\mathbf{e}_{\varphi}$.)

Last edited: Nov 26, 2015