If I have that circuit, the equation that describe the circuit is: $\epsilon = L \frac{dI}{dt}$.
Now, if the back e.m.f. is equal to the source e.m.f, how current can pass? and it should pass because I have a variation of current due to the applied voltage across inductor.
. . . . if the back e.m.f. is equal to the source e.m.f, how current can pass?
The current can indeed be zero but the rate of change of current $\dfrac {di}{dt} = (-) \dfrac{\mathcal E_{\rm back}}{L}$ is not zero.
So the current will then change from being zero and if there is still a back emf then the current will continue to change.
In the case of a source of emf $\mathcal E$ and an inductor, $L$, in series with it and there being no resistance in the circuit $\mathcal E = L\dfrac {di}{dt} \Rightarrow \displaystyle \int _0^i di = \int_0^t \dfrac {\mathcal E}{L}\, dt\Rightarrow i= \dfrac {\mathcal E}{L}\, t$, a linear rise in current with time.
With resistance, $R$, in the circuit the rate of change of current falls with time and the current asymptotically reaches a value of $\dfrac {\mathcal E}{R}$.
Current does increase because the induced emf doesn't make the inductor work like a battery with an opposite polarity. Rather, it makes it work exactly like a resistor in the sense that it drops the voltage across it by an amount equals to the negative of the induced emf, i.e., $v_{L}=-emf_{induced}$. However, the voltage drop in the inductor case is dependant upon the instantaneous rate of the current increase, unlike the case with resistors. Therefore, the current can't increase to its maximum value instantaneously.
