I’m on sabbatical working on an text (Introductory Computational Physics using Python) and, to take a break from writing, I am also writing interesting questions to better prepare our physics students for the physics GRE, and well, because it’s fun to think about physics.

So, here’s a question I wrote (thinking that the answer wouldn’t be too hard to derive—ha!) that turns out to be more interesting than I thought. First, the question as stated, with the geometry shown in the figure at the right.


Geometry for the rotating ring. The ring rotates about a the z-axis and the magnetic field is into the page along the -x direction.

Geometry for the rotating ring. The ring rotates about a the z-axis
and the magnetic field is into the page along the -x direction.

Suppose that a thin copper wire of diameter a is bent into a circular ring of radius r >> a. The ring is rotating without friction about a diameter and that this diameter is perpendicular to a magnetic field

    \[\vec{B} = -B_0 \hat{x} = - 0.01\, \hat{x}\;\mathrm{Tesla}\]

If the intial angular velocity of the copper ring is \omega_0, calculate the time it will take for it to drop to \frac{1}{e} of this initial value, using the assumption that the dissipated energy goes into Joule heat. You may use the fact that the resistivity of copper is 18\times 10^{-9}\,\Omega\cdotm and its density is 8900 kg/m^3.

How does your answer depend on the radius of the ring, r and it’s cylindrical wire radius a?


This problem has a lot of depth and subtlety. Sit down with a pen and paper and derive the initially non-linear differential equation for the angular velocity (by considering kinetic energy dissipation), make an appropriate approximation and presto, your equation is simple to solve.

The above hint is the only analytical way I can see to solve this problem…approximately, anyway. And, if you’re paying attention, you might wonder why I asked you to consider kinetic energy dissipation and not total energy dissipation.  Why not include the magnetic dipole potential energy term -\vec{\mu}\cdot\vec{B} ?  I think that the answer to this is going to be non-trivial and might depend on fully understanding the January 1999 paper in the American Journal of Physics entitled: Magnetic dipole orientation energetics by G. H. Goedecke and Roy C. Wood.

In any case, one can work out the “exact” dynamics by equating the magnetic torque on the induced dipole moment to the rate of change of angular momentum and avoid energy considerations entirely, obtain a non-linear differential equation and find the resulting dynamics. I turned to Python and SciPy to write a quick snippet to solve this in the iPython Notebook. More when I post a solution.

Tagged with:

Data about our Helmholtz Coil pair:

  1. Diameter = 1.016 \pm 0.008 m
  2. R_{\mathrm{coils}} = 0.939 \pm 0.001\;\Omega
  3. R_{\mathrm{coils\, in\, parallel\, with \,leads}} = 0.516 \pm 0.001\;\Omega
  4. R_{\mathrm{coils\, in\, series\, with\, leads}} = 2.067 \pm 0.001\;\Omega
  5. diameter of coil wire = 0.075 \pm 0.001 in = 1.905\pm 0.025 mm
  6. Predicted number of turns = 49.9\pm 1.4 turns

These (home made) coils which I’ve inherited are clearly  marked as having 43 turns, so I’ve got quite a discrepancy here. I’m inclined to believe my measurements, but will have to check by putting a known current and measuring the magnetic field. A good project for a student measurement.

I’m adding a sensing resistor (1.420 \pm 0.001\; \Omega ) to the above in order to measure the current using the LabJack A/D. This resistance value includes the lead wires to the LabJack.

Addendum: (12 April 2012): I think that I have overestimated the resistance; will have to re-do measurements! Think this is not supposed to happen in a laboratory notebook? Unfortunately, it happens all the time, but hopefully we catch our mistakes.


Tagged with:

A mock-up of our pendulum hanging in our enclosure. The magnets are gold plated Neodymium magnets (1" diameter), and they sandwich the fiber between to identical aluminum disks. The final pendulum will have 4 magnets and the outer magnets will have 50mm diameter front surfaced mirrors super-glued in place; the aluminum disks are designed so that the pendulum will have equal moments of inertia about each principle axis.

Our machinist (Steve Innes) has finished the main enclosure. Now what’s left is for me to give him plans for the optical system. I’ll be designing two different optical systems. The first will mount on the rotating plate visible in the photograph and will be used for calibrating the pendulum (finding out the torsion constant and the moment of inertia — simultaneously; but that’s a story for a future post). This internal system will be removed after the calibration is complete, and replaced with an optical system on the exterior plate (removed so that you can see the interior).
Now, I’m awaiting a new quadrant photodiode, as I fried the old one by back-biasing the diode accidentally. If there’s good news in this, it is that I found a source that sells the same detector for a cheaper price — Pacific Silicon Sensor.

Tagged with:

Aluminum oxide layer

On 2011-Jul-16, in experimental, Materials, physics, by paul

According to this source: http://www.finishesltd.com/anodizing.html , the naturally occurring oxide layer on aluminum is about 500 nm thick. Much more than I would have guessed.

Tagged with:


Our ace machinist, Steve Innes, sent me this picture of the main apparatus enclosure in the center of which will hang our torsion pendulum. Steve has yet to mount the 1 meter long “chimney” onto the cube, and has to make the interior optical lever platform which will allow us to calibrate and monitor the fluctuations of our pendulum.

Here’s a question. Since the pendulum will be surrounded by aluminum (save a small opening at the bottom through which the wiring will emerge, one would expect pretty good shielding. However, aluminum oxidizes readily, and I guess then we have a layer (how thick? 1 atom? 10’s of atoms? more?) of aluminum oxide with dielectric constant of between 9-12.
(I gave a range of values, because the proper way to express the dielectric constant with a crystalline structure is with a susceptibility tensor, and aluminum oxide has diagonal elements which are not all equal.)

Now, the question: Should I be concerned about the aluminum oxide layer? Should I paint the interior of the cube with something conductive—like graphite in alcohol?

I’ll have to think about this. I don’t know the answer, but if the pendulum itself was a perfect conductor (which is likely a good approximation) and was electrically connected to the body of the aluminum enclosure, I think I know the answer, or at least how to get there.

Tagged with: