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AC magnetic field applied to a superconductor, London Theory

  1. Apr 28, 2012 #1
    Hey all,

    I'm just working through the london theory of superconductors. I've dervied the london penetration depth, the distance for the amplitude to drop by a factor of 1/e... Seems simple...

    Now my book talks about applying a an ac magnetic field to a superconductor. How would the field in the superconductor change as a function of time?

    I think it should just effectively be a waveform of the same form frequency of the ac magnetic field applied to it with a smaller amplitude?

    Any discussion would be greater! Struggling to see any different from this!
  2. jcsd
  3. Apr 28, 2012 #2
    In http://lss.fnal.gov/archive/tm/TM-0991.pdf is an AC measurement on a large dipole magnet with superconducting coils as it was slowly warming up. These magnets were about 22 feet long, and had 114 turns of a superconducting braided wire made of niobium-3 tin strands.
    Look at Fig.5 on page 13. This plot represents the AC inductance of the dipole magnet, measured from 10 Hz to 10,000 Hz. The inductance represents the ability of the magnet volume including the superconducting coil to store magnetic energy. Specifically, the inductance is proportional to the volume integral (1/2μ)∫B·H dV. As the warming magnet temperature reached about 9 kelvin, the inductance rather suddenly increased about 4 milliHenrys at all frequencies. This change in inductance is attributed to the ability of the ac signal to penetrate inside the superconducting coil (Meissner effect) as it became normal. The fact that the ac resistive losses (Fig. 3) did not change implies that the Meissner effect applies down to very low frequencies (probably dc).

    The AC signal for this measurement was about 1 amp. The magnet wire was designed to carry about 4500 amps.

    See http://en.wikipedia.org/wiki/Meissner_effect
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