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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.

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. . . . 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 $\dfrac {\mathcal E}{R}$.

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