Back EMF in an inductor with an AC supply
In a real inductor scenario (i.e one with no losses), the sign of the back emf voltage will always be opposite of the voltage source with the magnitude of the the back emf voltage always different from the one of the source voltage. Thus, current flows. My argument actually follows from the popular "electric machinery" textbook from Fizgerald and Kingsley. See the following extract from page 61 of the 6th edition:



The text describes the setup of a no load transformer (i.e. one where the secondary winding is open circuited) which is actually an inductor (with an iron core) attached to a voltage source. Equation 2.3 and especially the last paragraph clearly describes what I previously explained, that is, when the ohmic loses of the inductor are minuscule, the back-emf equals the source voltage. In practice, ohmic losses are always present, and so Usource is not equal to Uemf and thus a current i=(Usource-Uemf)/R flows