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If you change the magnetic flux through a superconducting ring, what happens to induced emf?

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- Thread starter Redbelly98
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- #1

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If you change the magnetic flux through a superconducting ring, what happens to induced emf?

- #2

cabraham

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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

- #3

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Thanks,

Mark

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