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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.