Helmholtz coil approximations

My friend Joss Ives got this blog post rolling with this tweet:

What I decided to do to check this out was to integrate the field for a true coiled wire. The first thing I did was figure out the parametric equation for a coil. Here’s what I came up with:

f(u)=\left[x(u),y(u),z(u)\right]=\left[\left(1+\frac{u}{200}\right)\cos(u),\left(1+\frac{u}{200}\right)\sin(u),\frac{1}{20}\sin\left(\frac{u}{4}\right)\right]

and here’s what that curve looks like:

plot of the coil parametric function

For that picture I had u go from 0 to 16 pi or 8 round trips.

Now that I had a mathematical expression for the coil, I could go about determining the exact magnetic field for it, rather than assuming it was 8 perfect circular loops. The field is given by:

\vec{B}(\vec{r})=\frac{\mu_0}{4\pi}\int\frac{\vec{dl}\times\left(\vec{r}-\vec{r'}\right)}{\left|\vec{r}-\vec{r'}\right|^{3/2}}

The trick to getting this to work was to figure out \vec{dl}. I got that by calculating the directional derivative of the parametric curve, f’(u). I wasn’t, at first, convinced that f’(u) would be right, so I thought about an easier integral first, namely, the length of the parametric curve. For that, you need \left|\vec{dl}\right| which is given by \sqrt{x'(u)^2+y'(u)^2+z'(u)^2}, also known as the length of f’(u). For the vector form of dl, I needed a normalized unit vector in the direction of dl times its length, but that’s just the same as the original vector, f’(u).

Ok, then I was off and running. I could figure out r and r’ from the coordinates, and I just need to ask Mathematica to numerically integrate all three components of the field. Here’s the code that did that:

Mathematica syntax for integrating the magnetic field

You give bfield the location where you want to calculate the field.

To answer Joss’ question, I decided to look at the field strength on the axis, and compare it to the field for an 8-turn coil with an adjustable radius. That latter field is given by:

B=\frac{\mu_0}{4\pi}\frac{(2\pi R)R}{\left(R^2+z^2\right)^{3/2}}

I’ve purposely factored the numerator to show where the length of the coil comes into play. What I did was to calculate the coil field at several points on the axis and then fit that data to the equation above. The result was:

fitting the coil field to a simple circular wire field

I also show the value of R for the best fit along with the error on that. What’s interesting is how close the best value of R is to the midpoint of the coil:

R_\text{mid}=1+\frac{1}{2}\frac{16\pi}{200}=1.12566

However, it’s important to realize that the length of the two coils I’m comparing are not the same. The best fit length is:

\text{length}_\text{best fit}=16\pi R_\text{best}=56.47

while the length of the spiraled coil is given by integrating the dl mentioned above to get 56.61.

Thanks, Joss, that was fun!

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About Andy "SuperFly" Rundquist

Associate professor of physics at Hamline.
This entry was posted in mathematica, physics. Bookmark the permalink.

6 Responses to Helmholtz coil approximations

  1. ProfeJMarie (Janet) says:

    But what do these coils DO? Are they changing the world? Are they helping cure cancer? Huh? Huh?

    • Andy "SuperFly" Rundquist says:

      For those who haven’t guessed, that’s my better half joking about questions I always get from family about research I do. Basically, if it’s not medical research, it isn’t worth anything, or so it feels like sometimes.

  2. Joss Ives says:

    Clever work Andy. It looks like I know into whose ear I should place a bug!

    I’m still trying to decide if I find it surprising that there was such good agreement of the radius with the midpoint of the coil given that you only used 8 coils.

    • Andy "SuperFly" Rundquist says:

      I might go do a longer coil to check, though the integration takes a while so I can’t just whip it out. It would be interesting to see effects from non-integer coils, too.

  3. Pingback: Numerical Calculation for Magnetic Field « Intro Physics II

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