The data taken last week showed a linear dependence between the voltage measured in the pick-up coil when the superconductor is levitated and the frequency of the current driving the levitating electromagnet.
I'd like to see if the linear increase in the voltage required to attain levitation is just a result of the increasing impedance of the electromagnet. My first task was to determine a relationshiop between the pick-up coil voltage and the voltage driving the electromagnet. To do that, I attached one channel of the oscilloscope to the supply leads of the electromagnet and another channel to the pick-up coil leads. With the oscilloscope in x-y mode, the first channel is used as the x sweep voltage and the second channel is used as the y sweep voltage. This resulted in the waveform shown below:
Taking the slope of the loop, gives about 26 mV on the pick-up coil for every 5V on the electromagnet supply. Using this slope, the data taken last week can be related back to the electromagnet's supply voltage:
The data point near 180 V on the calculated graph seemed somewhat unrealistic at first because the amplifier docuemntation specifies a maximum voltage swign of 93 V into 8 ohms. The electromagnet, however, has an impedance of about 28 ohms at these frequencies. Just to perform a sanity check, I'm plugging an estimated rms output power of 1000 watts into the following equation:
Using a power estimate of 1000 Watts and a driving frequency of 230 Hz, I wind up with a peak voltage of roughly 229 volts, so my data fits within the estimated maximum output voltage from the amplifier.
If the voltage-frequency line in the levitation data was just due to impedance effects I'd expect to see a flat line when graphing the current through the electromagnet vs. frequency. In other words, the voltage just had to be increased in order to keep the current constant. I'm modelling the electromagnet as the following circuit based on the reading of an impedance meter:
While reading an article on a susceptometer for superconductors, I came across the graph shown below that shows the decrease in the magnetic field of a solenoid driven at 5 V rms as frequency is increase. A solenoid is an inductor with an impedance that is linearly dependent on the frequency of the current flowing through it. The drop in the magnetic field is a result of of the impedance of the solenoid increasing with increasing frequency and reducing the current trough the coil.
I'd like to see if the linear increase in the voltage required to attain levitation is just a result of the increasing impedance of the electromagnet. My first task was to determine a relationshiop between the pick-up coil voltage and the voltage driving the electromagnet. To do that, I attached one channel of the oscilloscope to the supply leads of the electromagnet and another channel to the pick-up coil leads. With the oscilloscope in x-y mode, the first channel is used as the x sweep voltage and the second channel is used as the y sweep voltage. This resulted in the waveform shown below:
The data point near 180 V on the calculated graph seemed somewhat unrealistic at first because the amplifier docuemntation specifies a maximum voltage swign of 93 V into 8 ohms. The electromagnet, however, has an impedance of about 28 ohms at these frequencies. Just to perform a sanity check, I'm plugging an estimated rms output power of 1000 watts into the following equation:
which can be rearranged to give an rms voltage of:
If the voltage-frequency line in the levitation data was just due to impedance effects I'd expect to see a flat line when graphing the current through the electromagnet vs. frequency. In other words, the voltage just had to be increased in order to keep the current constant. I'm modelling the electromagnet as the following circuit based on the reading of an impedance meter:
This gives me a calculated current vs. frequency that looks like:
The current line isn't flat, so there are other things going on. Bean's model of type II superconductors in AC magnetic fields predicts a power loss that is linear with increasing frequency. I'll take a look at that tomorrow.
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