How does current flow if back emf equals to applied voltage Suppose we have a circuit with a voltage source, a switch open and an inductor all in series. If we close the switch, the potential difference of the voltage source is instantaneously applied to the inductor. As the current starts to build up, induced voltage from the inductance opposes it. If the induced voltage (back-emf) is equal and opposite to the applied voltage, and the net voltage is zero, what drives the current then? 


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*Main question would be that if induced emf equals to applied voltage then how does current flow
P.S I have looked everywhere for this question but couldn't find any.
 A: Lenz's law states that the induced current is in such a direction as to oppose the change producing it.
Back emf and a complete "conducting" circuit will result in an induced current.  
The back emf can never exactly equal the applied voltage as then the current would be zero and not changing which would mean that there cannot be an back emf.
So you can think of it as follows.
As soon as a current starts to flow an emf is induced which produces an induced current which tries to oppose that change of current in the circuit produced by the applied voltage.
That induced current slows down the rate at which the current in the circuit increases.
So when deriving equations relating current to time in such a circuit it is convenient to say that at time = 0, when the switch is closed, the current is zero because the applied voltage and the back emf are equal in magnitude. 

What you are usually not concerned with is a time scale equal to that whilst  the switch is being closed.  
Here is an attempt to illustrate the complexity of what happens even as the switch is being closed.
The switch is a capacitor and when open has charges stored on it.
As the switch is being closed the capacitance of the switch increases and so a very small current starts to flow around the circuit as the capacitor charges up.
The change of current is opposed by the induced current produced by the back emf.
As the distance between the contact of the switch continues to get less, the capacitance of the switch continues to increase and a changing current continues to flow.
When the contacts finally close there is already a very small but changing current flowing around the circuit which is being opposed by the induced current produced by the back emf with the back emf slightly less than the applied voltage.  
What actually happens during the switch closing process is complicated by the fact that now you have an inductor, resistor and capacitor in the circuit and also that lumped circuit element analysis might be inappropriate to use over such a small time scale?. 
A: First of all, the condition that you are talking about, induced emf = voltage, is achieved at $t=0$. This is because at that instant no current is flowing in the wire and hence it follows from initial conditions and Kirchoff's law. (No current is flowing at $t=0$ since in $dt$ time from $t=0$ if $di$ has a finite value then $di/dt$ would give an impossibly large emf without anything to support this amount of energy.Update- Therefore di is infinitesimal, current increases infintesimally from 0 and is 0). 
Now, the voltage after this just keeps on decreasing as $di/dt$ decreases. You get this from the equation,
$$
L\ di/dt+iR=V,$$
 which gives $i=I(1-e^{-(R/L)t})$ where $I=V/R$. (Derive this!) The derivative of this function is $di/dt$ and it approaches zero with time. Thus the back emf keeps on decreasing from initial conditions and there is no question of it being equal to voltage. 
