What happens if I have a square conducting wire being permeated by a magnetic field and the field suddenly disappears? Suppose there is a square conducting wire in a magnetic field. The two vertical branches will have the same emf, and the two horizontal branches will have 0 emf, resulting in a net emf of zero and zero current. Now, suppose the magnetic field disappears. Faraday's law says that there should be an induced emf. My question is, how does that work? Wouldn't the two opposite branches still have the same emf, leading to a net of zero and no current?
I really hope the answer is something that'll make me smack my head, because it's bugging me no end.
Thank you!
 A: The situation you posed is an ideal situation or somewhat fantasy in my view.Even though consider a loop with a particular magnetic flux and of course it is not possible for the magnetic field to not be there at the the time it is there so some change in time will be there but infinitely small.This would be the situation when rate of change of magnetic flux would be $$\frac{-d¢}{dt}$$ where ¢ is magnetic flux.This is a large value so you may think emf to be large that the impulse on electrons would be so large possible for the electrons to get thrown tangential to induced electric field lines where possible.A huge amount of heat may be produced to burn the wire.It may not be so easy to create such a situation since by Following lens law you will also have to encounter such a huge force when you are changing flux.To my imagination it may work like seatbelt of a car.
A: The misunderstanding here is in thinking that the two branches would have equal emf. The trick here is to note that a rate of change in a magnetic field does not induce an electric field, they induce a curled electric field. I personally visualise this as a circle of an electric field wrapping around the rate of change of magnetism line. This means that, far from both wires having an emf in the same direction, they would have emfs in the opposite direction, and thus a current would flow - this is an application of something known as the Kelvin-Stokes theorem (right at the bottom of this page). It's this which gives rise to the equation given my Sikander: That the emf is equal to the negative of the rate of change of the flux through the loop. And so, rather than emf being something which a single wire can experience, and something for which you need to add the effects on all the wires to build up a net response, the emf is actually a function of the entire loop and the integral of the rate of change of flux in its enclosed area.
A: Generally, any change in a magnetic field, will result in a current flow that will try to maintain that field. However, with the field you show and the available conductors, that will not be possible. Any current flowing in the conductors cannot create magnetic fields as you have shown. And the shape of the conductor is not relevant.
A: I'm not sure what you say about the voltages in the wire before the B-field is turned off is true. If the B-field is constant at the start then the induced EMF through the wire is 0 because there is no change in flux. Then since the wire is a conductor and the situation is static you must conclude the potential is the same everywhere on the wire.
As for the time-varying case.
Let's regularize the procedure of switching off the B-field by defining a function that can be used to describe the behavior in an appropriate limit:
$$B(t) = B_0 \frac{(-t + T)}{T}$$
This is a linear ramp that starts at value $B_0$ and goes to 0 by time $T$. We can calculate the induced voltage:
$$\mathcal{E} = -\frac{d \Phi}{dt} = - A B_0/T.$$
So as we take the time of the ramp protocol to zero, the induced EMF during the ramp goes to infinity. I.e. your wire loop would have massive currents induced and get very hot.
