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Superconductors and emf

  1. May 10, 2008 #1

    Redbelly98

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    After following the thread "ideal conductor vs superconductor" a while, I am wondering about something that must have a simple answer.

    If you change the magnetic flux through a superconducting ring, what happens to induced emf?
     
  2. jcsd
  3. May 10, 2008 #2
    The net emf is the difference of the emf associated with the external ac magnetic field, and that of the internal ac magnetic field. Whenever induction takes place, the external magnetic field (time-changing or "ac" will be understood) is not the only magnetic field, herein referred as "H". The induced current in the superconductiong ring has a magnetic field as well. By Lenz' law, the induced mmf, emf, H, and E (electric field) are directed in opposite orientation to the external quantities. Thus an external H also includes an external E as well. In the ac domain, E and H are mutually inclusive, i.e. one cannot exist without the other.

    When a superconducting ring is placed in the ac H/E field (they always co-exist), a current is induced. This current must have an ac H field associated with it. But the ac H field must co-exist with an ac E field. The H due to the induced current opposes the external H. Likewise for the two E fields. Cancellation occurs so that Ohms law is always upheld. SInce a current in a zero ohm resistance mandates zero voltage, the E fields must cancel. The emf, or V, is the line integral around the closed path of the net E field. Since the E field consists of equal and opposite values (extrenal plus self induced), the net value is zero.

    When the conducting ring has a small resistance, the E fields do not perfectly cancel, so that a small emf exists. Ohms law is upheld. For low resistance rings, the induced current is approx. constant, and the induced emf varies with the resistance. Thus I is approx. fixed, and V varies with R per Ohms law, i.e. V = I*R.

    For a high resistance ring, the induced current is small, as well as the associated E/H fields. Thus the net E/H values are approx. those of the external E/H fields. Then, the induced emf is approx. constant, and the induced current is I = V/R.

    Does that clear things up? If you still have trouble, rest assured that brilliant people struggle with this. It isn't trivial. BR.

    Claude
     
  4. May 10, 2008 #3

    Redbelly98

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    Yes, thank you. If I understand you, the induced current generates a flux that will cancel the flux change due to the external field change. Zero flux change --> zero emf.

    Thanks,

    Mark
     
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