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To clarify my question, I made the illustration below, assuming I understood the Faraday-Lenz law:

enter image description here

Above a copper loop theoretically stays in parallel between the plates A and B.

The magnetic field B hence the flux between the plates A and B is continuously increased.

The copper loop circulates a current such that to oppose the increasing flux.

Above I showed the directions of increasing flux and self generated opposing flux.

My question is, will the total changing flux always be zero no matter the conductor is perfect or not?

If the applied flux was constant(not varying by time) the total flux would be non zero.

But in case the flux is increasing, will the total changing flux be zero?

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No - there is an easy counterexample. The EMF generated by the flux is independent of the material used in the wire. At a constant EMF, the current in the wire will be proportional to the resistance of the wire (i.e. the conductivity $\sigma$). Therefore, if at some conductivity $\sigma_o$ the "reaction" flux perfect cancels the changing flux, at some lower conductivity it will not.

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  • $\begingroup$ I dont understand what you mean. Lets say the applied flux through the loop was constant for 10 hours and lets say the applied flux started to increase at a constant rate so lets say the increase in flux per second is M. But now the Lenz law says the current will create an opposing magnetic field to stop the change. And to stop the change the opposing flux should be -M per second isnt it? Can you explain in this manner? $\endgroup$ – atmnt Jul 6 '17 at 18:26
  • $\begingroup$ Or how can we formulate the emf by including the opposing magnetic field? In d(phi)/dt is d(phi) the total changing flux or change in the applied flux? $\endgroup$ – atmnt Jul 6 '17 at 18:28
  • $\begingroup$ @user134429 Lenz law says what the direction of the opposing magnetic field should be but not the magnitude. $\endgroup$ – Señor O Jul 6 '17 at 19:09
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Lenz tells you that the induced current will try and oppose the change producing it.
If Lenz wins and stops the change in magnetic flux there will be no induced emf and hence no induced current.
If there is no induced current there is no opposition to the change of magnetic flux.

Hence Lenz cannot stop the change; Lenz can only reduce the change.

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The EMF ${\cal E}$ depends on the rate of change of the flux ${\cal E}=-\frac{d}{dt}\Phi(t)$ but the current induced in the loop will depend on the load of the loop. In your example, you could insert a $10\Omega$ resistor to get some current $I={\cal E}/10$ or a $1\Omega$ resistor to get $10I$. Hence the $\vec B$ field produced by loop can be changed without affecting $\frac{d}{dt}\Phi(t)$ and "the total changing flux" need NOT "always be zero no matter the conductor is perfect or not". x

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  • $\begingroup$ What if the loop resistance goes to zero i.e. just the curling Electric filed exists around an imaginary loop. Would the curling electric field generate an opposing magnetic field as well? $\endgroup$ – atmnt Jul 7 '17 at 8:32
  • $\begingroup$ physics.stackexchange.com/questions/343447/… $\endgroup$ – atmnt Jul 7 '17 at 8:43

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