I am just surprised why EMF and Current are not induced in an unclosed wire. I know Maxwell's equation only defines closed integral. But it doesn't mean that it is not generated in a unclosed wire. Unclosed wire is stationary and magnetic flux is changing with time.
Maybe this diagram will help:
The charge carriers inside the wire will experience a force when they move relative to a magnetic field. This will cause them to be displaced in the wire - as they move, an electric field (emf) is induced in the wire that makes them want to "move the other way". When the two forces are balanced, the charges stop moving.
In an AC situation, current will keep flowing (see this question about antennas); in a DC situation, you just end up with a potential difference between the ends of the (open) wire.
The relationship between the amount of charge that flows, and the potential difference that is set up, is the capacitance of the wire. In the DC case, there will just be a transient charge flow (when the wire first starts moving) to set up the e.m.f. - once that has been established, no further current can (or needs to) flow as the forces are balanced.
The diagram below is a practical realisation of a metal bar passing through a magnetic field if the galvanometer (sensitive ammeter) is omitted.
As was explained in the answer from @Floris charges move along the bar until the electric field inside the wire is strong enough to stop further movement of charges.
Connecting a galvanometer as in the diagram gives a conducting path, an induced current flows and the galvanometer gives a reading.
Now how might this all be explained using Faraday’s law?
You would say that there is a closed loop: the bar, the galvanometer and the connecting wires with a changing magnetic flux through the closed loop and so an emf is induced.
There is no mention by Faraday where that emf resides other than in the loop.
Faraday also does not specify what the loop should be made of so it could be made of an insulator.
This then resembles the situation in the answer given by Floris – a metal rod with an insulator (the air) completing the circuit.
Now that loop made up of the rod and air can be of any size and the size of the loop will determine the rate of change of flux through it and hence the induced emf.
So you cannot say that the emf is set up across the rod or can you?