I would like to design a long, uniform magnetic field using Hemholtz coils but am not sure how far I can separate them before the field lines diverge and become two loops that go around each coil rather than one which goes through both.

I have formulae for finding the magnetic flux at different points but I am not sure how to use them to determine this distance I am trying to calculate.

Does anyone have experience in this area who could offer some advice?


1 Answer 1


Short answer: Not very far.

I had this question one or two years ago. At that time I did a simple calculation of $B$ along the symmetry line for Helmholtz coils with non-standard distance, i.e. the position of the two coils being different from $+R/2$ and $-R/2$ ($R$ being the radius of the two coils). The homogeneity of $B$ decreases quite quickly when deviating from the standard.

Wikipedia gives you the equation. For a single conductor loop, according to Biot-Savart, $$ B(x)=\frac{\mu_0 I}{2}\cdot \frac{R^2}{(R^2+x^2)^{\frac 32}} $$ with $B$ being the magnetic flux density (more precisely its vector component along the symmetry axis of the loop), $\mu_0$ the vacuum permeability, $I$ the current along the conductor loop, $R$ the radius of the conductor loop and $x$ the distance to the center of the loop along the symmetry axis.

The following plot shows $B(x-d)+B(x+d)$ for some $d \geq R/2$ for $\frac{\mu_0 I}{2} = R = 1$.

enter image description here

As you can see, $B$ drops significantly in the center between the two coils once deviating more than 20% from $d=R/2$. You can calculate for yourself how much deviation you wish to accept.

Be reminded that this is just an evaluation of the magnetic flux density along the central symmetry line of a coil pair. If you are interested in the full 2D field, things become more complicated. But this simple evaluation gives you some idea what happens when you move the two coils apart.

For a longer homogeneous field you need more coils.

  • $\begingroup$ That's awesome, thank you for that explanation. $\endgroup$
    – user88720
    Commented Dec 26, 2015 at 11:03

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