After compared to Before ...A standard quarter is electromagnetically compressed to the size of a dime in, literally, the blink of an eye, a flash of light, and a loud crack.
The forces on the coin are produced by an estimated 100 kA current in a 10-turn copper solenoid winding. The induced current (about 1 MA) in the outer edge of the copper core of the quarter interacts with an estimated 58 T magnetic field produced by the solenoid, compressing the quarter radially. The peak field last for about a millionth of a second because the coil is vaporized by that current.
The energy needed (about 4600 J) is delivered by a 178 uF capacitor charged to 7200 V. The capacitor, which contains about 1.28 coulombs of charge, takes several minutes to reach its final voltage. A display next to the demonstration included a page on the
principles of operation of a quarter shrinker from "Stoneridge Engineering", the
Teslamania web site of Bert Hickman that gave additional details for a system similar to the one we saw. My independent calculations at the bottom of this article roughly confirm what is on that page and the display at the magnet lab.
Side comment. You would get 4600 J of energy if you dropped 47 kg (about 100 pounds) a distance of 10 m (about 33 feet). Now imagine all of that energy concentrated int the small area of a quarter. Splat.
Tools of the trade:An assembled coil is in the background. The coil itself (left foreground) is about 10 turns of 14 gauge wire (wire that would normally carry no more than 15 A in commercial use, and will melt if used with 166 A). The quarter is sandwiched between two cylinders of G10 fiberglass (right background) to center it within the coil. (Wood had been used in the past, but the G10 survives and can be reused from year to year.) Tape holds it all in place.
Centering the quarter and ensuring it is perpendicular to the magnetic field is crucial to making sure the forces compress the coin rather than twist it. Oh, yes, and the vial contains fragments of copper coils used in past experiments. The lighter colored pieces are probably stainless steel chipped off of the box that is used to contain the explosion of the coil.
This last photo shows the interior of the box used to contain the explosion of the magnet coil. You can see the copper from the used coils embedded in it. I also notice that the stainless steel panels on those three sides appear to have been added after the fact to the inside of Lexan panels that appear to have been the original design plan.
The Physics Details:I collected a lot more details this year than I managed to get last year, correcting some of the information I used for back-of-the-envelope example in PHY2049 this year and last. [I had remembered the 1 MA current, but not that it was the induced current in the coin or the number of turns, and did not have enough info to estimate the R or L of the circuit being used here.] I now know he used 14 AWG 200C copper magnet wire with 10 turns in the coil. That info, and my estimate that the coil has an inner diameter of 2.4 cm (to fit around a quarter) and a length of between 1.7 cm (absolute minimum for the wire diameter) and 2 cm, are essential to a qualitative understanding of what is going on.
The length of the coil plus 10 cm for each lead is about 1 m, so we can estimate its resistance at about 0.008 ohm. All other conductors are large bus bars and will contribute little to the resistance of the circuit.
The coil that produces the magnetic field is too short to be correctly modeled as a solenoid. If we do that anyway, however, we require about 92 kA to produce 58 T, while 65 kA (see below) produces 41 T. These numbers assume 10 turns and a 2 cm length. Only 78 kA is needed to make 58 T (or 48 T from 65 kA) if the solenoid is 1.7 cm long. We always want the coil as tightly wound as possible!
We also get an inductance L = 2 uH for a 2 cm coil. An on-line calculator (of unknown reliability) says those dimensions would give 1.8 uH, while 1.7 cm gives 2.0 uH. I will use L = 2 uH as a conservative value. Notice that a shorter length makes the inductance bigger, which is a bad thing.
The RC time constant of 1.4 us (micro seconds), with an initial current of 900 kA, tells us what would happen if there was no coil (inductance) in the circuit. Unfortunately, the coil does not like a rapidly rising current. It would take 250 us (0.25 ms) for the current to reach 570 kA if we had just the coil with an ideal 7200 V battery. However, this is actually an un-driven LCR circuit, with all three elements playing a role. The circuit has an inductive time constant of 500 us and a natural frequency of about 53,000 rad/s. The solution to this problem says a peak current of 65 kA will be reached after 19 us. A smaller inductance makes the current bigger, by the way. The estimated inductance is a critical quantity. Reducing the inductance to 1.8 uH will increase the current to 78 kA. Notice that this is just what we need to produce 58 T in a 1.7 cm coil.
Faraday's Law says the large dB/dt produces a large EMF around the edge of the coin, which acts like a single turn in a 10:1 transformer. This leads to a rough estimate of an induced current of 650 kA (perhaps 780 or 920 kA) around the edge of the coin.
Even with "only" I = 65 kA and B = 41 T, the compressive force due to the 650 kA current induced in the outer edge of the coin would be something like 2 MN. If we really have I = 78 kA and B = 58 T, the force increases to 3.4 MN.
This force only acts for a micro second or so at the peak of the sine function that describes the current (before the coil is torn apart by the equal and opposite repulsive force of the field on the coil). The total impulse is not very large, although the total energy is significant even if a lot of it is wasted.
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