I understand that the magnetic field produced by a single Helmholtz coil is approximately :


Where $x$ is the distance from the centre of the coil to the point of interest along the central axis of the coil, R is the radius of the coil, N is the number of turns in the coil, I, is the current in the coil and $\mu_0$ is the permeability of free space (in vacuum).

The question I have is, how do you calculate the separation distance between two Helmholtz coils, such that the magnetic field between them is uniform.

On my attempt, I used the principle of super-position, with the aid of the following diagram:

enter image description here

I said the following must be true:

$$ B_{constant} = \frac{\mu_0NIR^2}{2}[\frac{1}{(R^2+x^2)^{3/2}}+\frac{1}{(R^2 +(d-x)^2)^{3/2}}] $$

I am wondering now if this approach will work, and also if there is an easier method as there seems to be a lot of algebra involved and a potential polynomial in d of degree 12.


1 Answer 1


You don't require the magnetic field to be uniform - you require that the correction to the magnetic field amplitude for small $\delta x$ shift from the centre is at most $\mathcal{O}(\delta x^3)$.

The formula you use is correct. Try calculating the maximum of the field intensity along the axis of symmetry for two equal coils (in your configuration) with current $I$ separated by distance $d$. Then require $\frac{\partial^2 B}{\partial x^2} = 0$.

Added a few plots of the magnetic field $B$ versus the distance $x$ on the axis for various configurations. Note the 'flatness' around the maximum for $\frac{d}{R} = 1$enter image description here

  • $\begingroup$ ok, but should that be the second (partial) derivative of B wrt x or the first derivative of B wrt x ? $\endgroup$
    – Think
    Nov 8, 2017 at 16:54
  • $\begingroup$ First derivative vanishes anyway because you are at a field extremum. $\endgroup$
    – DrLRX
    Nov 8, 2017 at 16:59
  • $\begingroup$ But surely the requirement would be that $\frac{\partial{B}}{\partial{x}} = 0$ such that B doesn't change with x? $\endgroup$
    – Think
    Nov 8, 2017 at 17:08
  • 1
    $\begingroup$ Yes, but this is the requirement that the field has a maximum. Which is at $d/2$ whether or not $d = R$ as in Helmholtz configuration. The key difference is that in generic $d/R$ ratio the $B_{xx}$ doesn't necessarily vanish - only if $d/R = 1$ it does - try writing it out! $\endgroup$
    – DrLRX
    Nov 8, 2017 at 18:01
  • $\begingroup$ In fact, by symmetry, the deviations from the central field are proportional to $\delta x^4$. (Since the magnetic field is obviously an even function of $\delta x$, its power series can only contain even terms.) $\endgroup$ Nov 8, 2017 at 18:24

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