# Hypothetical situation: circuit with 0 resistance

1. Jan 9, 2015

### Coffee_

1. Consider a closed loop which already has a current $I$ running through it while the resistance of this loop is $R=0$. I now suddenly start tuning on a homogenous magnetic field through the loop which gives raise to an emf in the loop $\epsilon=-\frac{d\Phi}{dt}$. What will happen?

My personal reasoning is that this emf can be interpreted the same as turning on a sudden voltage in this loop. In this case since the resistance is zero the current wil go to infinity or minus infinity depending on the induced field.

This does not happen it seems according to what I remember from class and has something to do with self-induction. I don't remember the full argument though so I would appreciate it if someone could help me out to reason this through correctly.

Last edited by a moderator: Jan 9, 2015
2. Jan 9, 2015

### Staff: Mentor

You mean you have a current circulating in a superconducting coil? That's about the only way I can think of having a current continuously circulating in a zero resistance coil.

Anyway, for your question, a zero resistance coil acts as a "shorted turn" for external flux changes piercing the coil. The shorted turn supports a changing reverse current that is the correct magnitude to cancel out the changing flux that is piercing the coil (it is not an infinite current).

3. Jan 9, 2015

### willem2

If you have an emf of ε in the loop the current I will continuously rise according to ε = L (dI/dt), (with L the self-inductance of the loop) but the current will also produce an emf.

4. Jan 9, 2015

### Coffee_

Well not really a superconductor because we assume the B-field can be non zero here. It's just a hypothetical exercise.

5. Jan 9, 2015

### Staff: Mentor

But that's the point. If the resistance of this shorted turn is zero, you get zero changing flux through it. If there is a finite resistance, you get a reduced changing flux through it.

6. Jan 9, 2015

### Coffee_

So, tell me if I get it correctly. The change in external $B$ field will want to make $I$ go to infinity as there is no resistance but since as $I$ starts rising quickly it will induce a counter acting emf to keep this for happening. What will the final situation be?

7. Jan 9, 2015

### Coffee_

You get zero changing flux but you can still have a constant flux all the time. From what I understood about superconductors they can't have a flux at all in them.

8. Jan 9, 2015

### Staff: Mentor

Think of the external flux changing sinusoidally (since you can't ramp a flux up forever). That sinusoidal flux will induce a reverse current in the 0-Ohm coil that opposes the external flux, resulting in zero net flux through the coil That's what a shorted turn does, and it does not take "infinite current".

9. Jan 9, 2015

### Staff: Mentor

Bingo! Because the zero resistance acts as an ideal shorted turn! :-)

10. Jan 10, 2015

### rude man

You would not be able to change the net flux thru the coil. The slightest external dφ/dt would produce an arbitrarily high counter-flux to keep the existing φ from changing. There would be no net change in the net flux, which would remain at φ = Li where L is the coil's self-inductance.

Superconducting levitated rail works that way. If the train tries to drop towards the superconducting rail coils, the reult is an immediate "infinite" coil current which works against the train's magnets so no dφ/dt can take place, hence no vertical train motion.

11. Jan 10, 2015

### Coffee_

Thanks for answering. I understand now that this happens but how do you know this. What is the formal reasoning to go from $\epsilon=-\frac{d\Phi}{dt}$ towards the fact that the net flux can't change? I know all the seperate phenomena like self induction and $L\frac{dI}{dt}$ and so on but seem to not be able to put them together in a strong argument in this case.

12. Jan 10, 2015

### rude man

If R=0 then emf must = 0 too since emf = Ri. So $\epsilon=-\frac{d\Phi}{dt}$ = 0. So φ can't change.
OK let's 'do the math':
Assume R is finite for the moment. The NET dφ/dt comprises the rate of change of the external flux φe plus that of the self-induced flux φi, the latter being of opposite sign by Lenz's law:
dφ/dt = dφe/dt + dφi/dt = emf
but dφi/dt = -L di/dt so
dφ/dt = dφe/dt - L di/dt = Ri.
Let's assume a step external input ramp in time: φe(t) = φedot t, i.e. ramp rate = φedot = constant.
We then have a first-order, linear ODE, easily solved for i(t). We now impose R → 0; I will leave the details to you:
i(t) = φedot t/L + i0
where i0 = your original current.
So di/dt =φedot/L
and L di/dt =φedot
so if we add the two separate flux rates we get
dφ/dt = L di/dt - L di/dt = 0.
So the net flux thru the coil never changes. i builds up linearly over time to just balance out the applied external flux change.

You can use other forcing functions for the external rate, for example
φe = φ0sin(wt) etc. to see the same result.

Last edited: Jan 10, 2015