Now this seems a bit silly question, but I'm confused about it. The picture shows an inductor which is connected with a battery whose value is 5V. It means we are creating a constant 5 volts current push through the inductor at ANY given time.now suppose in some moment at $t$ (which is ofcourse at the very beginning) the inductor is producing a back-emf of value 3volts, that means, in effect A value of 2 volts (applied voltage-back emf) should appear across the inductor, (at that moment)which contradicts KVL. What is wrong here? I know i'm ignoring the resistance of the inductor, maybe that could be the reason, but can somebody please explain this in a systematic manner?
The back emf from a pure inductor with no series resistance when connected to a 5 volt source is 5 volts. It's as simple as that. If the inductor is producing a back emf of 3 volts then it's not a pure inductor and will have series resistance that is dropping 2 volts.
It means we are creating a constant 5 volts current push through the inductor at ANY given time.
First thing, the volt is a unit of potential difference or emf, not a unit of current. The phrase "5 volts current" makes no more sense than "5 kilograms tall" or "7 meters of weight".
now suppose in some moment at \$t\$ (which is ofcourse at the very beginning) the inductor is producing a back-emf of value 3volts...
There's no reason you can assume a priori that the inductor has a certain back-emf. The back emf will be whatever the rest of the circuit determines it to be, not some arbitrary value.
You must solve the circuit to find out what the back-emf is, not assume a value and apply that to the circuit equations.
hat means, in effect A value of 2 volts (applied voltage-back emf) should appear across the inductor, (at that moment)which contradicts KVL. What is wrong here?
What's wrong is, with no justification whatsoever, you have assumed you knew the inductor's back-emf before solving the circuit. You should have solved the circuit to find out the back-emf instead.
If your instructor has posed this as a problem and asserted that the back-emf is 3 V, then it is just one of those problems (which bears little relation to anything you will ever encounter working as an engineer) where they arbitrarily reveal certain circuit variables and expect you to determine the others. In this case you should probably assume the remaining potential in the KVL loop is taken up by the parasitic resistance of the inductor. But this is inconsistent with your saying that the time $\t$ being analyzed is "at the very beginning", because the inductor current in the instant after the switch is closed will remain at 0 A, and thus the resistive voltage drop will be 0 V.