UCSB Science Line If a magnet is moved in and out of a copper coil, a current is generated in the coil. This is due to Faraday's Law. The faster the rate of change of magnetic field the higher the current. If I replace the copper coil with a superconductor Type-2 coil, would the current be HIGHER if the rate of change of magnetic field was faster? Does the rate of changing magnetic field changes the current, or the current remains the same in a superconductor-Type 2? Question Date: 2005-05-18 Answer 1:What is happening is that a changing magnetic field produces an electric field - i.e. a voltage. Increasing the rate of change in the magnetic field increases this voltage. A conductor, located across an electric field, will pass a current equal to the strength of the voltage divided by the resistance of the conductor. Thus, with a given voltage, a better conductor (like a superconductor) will pass more current than a less good conductor, just as increasing the voltage will also increase the current. Answer 2:Your question is very interesting.If we mathematically analyze it, we have to make clear that the answer is different when the voltage arising from a changing magnetic field is held constant, than when it is a varying one (as when you move the magnet back and forth in a simple harmonic motion). Here you also have to consider that the resistance R in a superconductor is zero, R=0 (a characteristic of conducting metals when they behave as a superconductors), then all the results will depend on the voltage, the inductance of the wires and time.In the first case, when the voltage is constant (there is not a change of magnetic field), the current in the superconductor increases without a boundary (which is unrealistic, since the magnetic field would also have to increase without boundary (Faradays law)). This situation does not happen with a copper conductor, which has a current limited by the voltage and the value of the resistance. In the case when the voltage is a varying one, the current in the superconductor will increase more than the current in the copper coil, but it will be directly limited by the magnitude of the voltage, and inversely limited by the inductive effect and the frequency at which the voltage signal moves. The magnitude of the current will be given by:Voltage/(inductance x 2pi x frequency of voltage signal) Then, replacing the copper coil with a superconductor Type-2 coil, will make a higher current, but will still be limited by the rate of changing magnetic field.Click Here to return to the search form.    Copyright © 2017 The Regents of the University of California, All Rights Reserved. UCSB Terms of Use