I have the following surface current density $$ \bar{\sigma}_s = \hat{\phi} \sin(kz) |\bar{\sigma}_s| $$ to approximate an infinite number of alternating Helmholtz coils stacked along the z-axis with radii of $R$. I want to check if the magnetic field outside the coils is zero? I am unsure if Ampere's Circuital law applies in this situation, because the current changes sign when displacing along the z-axis. It might be possible that the magnetic field is zero, but I am stuck on how to prove this. I tried calculating the magnetic field using the boundary surface condition $\hat{\rho}\times \bar{B} = \bar{\sigma}_s$, where $\hat{\rho}$ is the surface normal of the cylindrical surface. I have two possible solutions for the magnetic field $$ \bar{B} = \hat{\phi}\times \bar{\nabla}\Phi_1 + \bar{\nabla}\Phi_0 $$ where $$ \Phi_0 = A \cos(kz)K_0(k\rho) \\ \Phi_1 = B \sin(kz)K_1(k\rho) $$ and $K_0(x)$ and $K_1(x)$ are the exponentially decaying modified Bessel functions. Both yield non-zero surface currents $$ \hat{\rho}\times \bar{B} = \hat{\rho}\times (\hat{\phi}\times \bar{\nabla}\Phi_1) + \hat{\rho}\times \bar{\nabla}\Phi_0 \\= \hat{\phi} (\hat{\rho} \cdot\bar{\nabla} \Phi_1) + \hat{\rho}\times \bar{\nabla}\Phi_0 \\= \hat{\phi} \dfrac{\partial \Phi_1}{\partial \rho} - \hat{\phi} \dfrac{\partial \Phi_0}{\partial z} $$ and both satsify $\bar{\nabla}\times \bar{B}=0$ and $\bar{\nabla}\cdot \bar{B}=0$. Should I use Ampere's Circuital law instead?


1 Answer 1


On the one hand, you can't solve for the magnetic field without appropriate boundary conditions (e.g. there could always be an incoming electromagnetic wave that hasn't yet impinged on your cylinder). On the other hand if you have a fixed charge and current distribution you can always use Jefimenko's equations to find a solution to Maxwell's equations, and it will have perfectly fine (physical) radiation boundary conditions.

In your particular case, your currents are steady and your charge density is zero, so Jefimenko's equations reduce to zero for the electric field and you just get the law of Biot-Savart for the magnetic field. Which gives a nonzero magnetic field in the plane $z=\pi/(2k)$ for instance. So your magnetic field is not identically zero everywhere.

As for whether to use Ampere's Law, it can be used, but for instance you pretty much need to know need to know the direction of your magnetic field. But again, in the plane $z=\pi/(2k)$ we know the magnetic field points in the direction -z outside the coils and in the +z direction in the center. And then Ampere's Law tells us that $B_{center}+B_{outside} =\mu_0|\sigma_s|\sin \pi/2.$ And so it comes down to finding the magnetic field at the center of the coils at the plane at $z=\pi/(2k)$. For that part, there is no easy shortcut with Ampere. But it's not a particularly difficult problem for Biot-Savart.

I used the $z=\pi/(2k)$ plane because it was the simplest place to show that the magnetic field is not zero and to show how you can use Jefimenko, Ampere, and Biot-Savart to solve the problem. Using Ampere is optional, and it basically relates the curl to the current enclosed, it isn't really much different than saying $\vec \nabla \times \vec B = \vec 0$ inside and outside and having the jump conditions related to the surface current. It does look like the magnetic field is uniform in those two regions in the plane $z=\pi/(2k)$ so both your proposed solutions look deficient at first glance.

But maybe you neglected that constant functions are also solutions to $\vec \nabla \times \vec B = \vec 0$ and $\vec \nabla \cdot \vec B = 0$?

  • $\begingroup$ Thanks for the answer. I still working on the problem. I am very surprised that the magnetic field is uniform along equally spaced planes with fixed z. I honestly believed that the windings would fight each other, greatly diminishing the outside magnetic field. $\endgroup$ Commented May 11, 2015 at 18:05
  • $\begingroup$ @linuxfreebird How much they fight each other depends on the numerical value of kR. $\endgroup$
    – Timaeus
    Commented May 11, 2015 at 18:08
  • $\begingroup$ I am assuming that the magnetic field solution will be separable with respect to $z$ and $\rho$, this may not be true. $\endgroup$ Commented May 11, 2015 at 18:11
  • $\begingroup$ I think I figured out the problem. I forgot to include the boundary condition $\hat{\rho}\cdot(\bar{B}_1-\bar{B}_2)=0$. My original solution was lined with magnetic monopole charges which gave rise to a non vanishing magnetic field. Oops. $\endgroup$ Commented May 11, 2015 at 21:05
  • $\begingroup$ I apologize. When I said non-vanishing field, I meant in the limit as $k$ goes to zero. I think your solution is valid. $\endgroup$ Commented May 12, 2015 at 13:39

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