# Back EMF and opposing current of a super conductor?

A 5 meter long super conductor carries 100 Amps, and it passes through a 1 Tesla magnetic field at 0.010 seconds, Is EMF = - (BL) / (t)?

And since the resistance of the super conducting wire is zero, shouldn't EMF = 0 based on ohms law?

Now, lets take another wire with R = 0.001 ohms, and the the induced EMF was 1V for example, the current is V/R = 1000 Amps?!

This is confusing because the magnetic resistance(Lenz law) is massive!

• What's the question though? – horta Mar 11 '14 at 3:52
• Well there are 2 questions, 1 being if a SC moved around a magnetic field shouldn't -V = 0 since the resistance = 0, thus no current? (Based on ohms law V = IR)? The second one you already answered to it :) – Pupil Mar 11 '14 at 8:43

I asked a physicist this question. Here's what I understand.

First, Ohm's "law" only applies to ideal resistors. A superconductor is a nearly ideal inductor. That means that its voltage and current are related (to a first approximation) by the equation v=L*di/dt, not by Ohm's law.

Faraday's law still applies to superconductors. A change in magnetic field causes a voltage to appear in the superconductor. This voltage causes current to flow. Because voltage is proportional to the derivative of current, a transient voltage (when integrated) results in an enduring current in a superconducting loop. The current will stick around as long as the magnetic field is present. As the magnetic field is being removed, a voltage transient of the opposite polarity appears, which induces a current in the opposite direction, and when the field is gone, the current in the loop is 0 again.

This voltage is in fact what causes the current to flow -- it's not possible for the current in a superconducting loop to change without some kind of voltage present. When the current is constant (dc), as in a perpetual current loop, the voltage is 0; v=L*di/dt is satisfied.

If you have a power supply to drive 1000 Amps then yes, at 1V through a 0.001 ohm resistive wire, you'll end up with 1000 Amps. Normally, power supply limitations prevent that from happening.

The only thing you seem to be confused about is the asymptotic effect generated by the inverse function 1/x (or V/R in this case).