Inductors in AC circuits Although I checked similar questions I still can't understand why the voltage across an inductor connected to an AC source is equal to $L\, dI/dt$ without the negative sign of Lenz's law, why isn't the emf on the opposite direction as to oppose the change in current? Or does this emf ($\mathscr E$) indicate the $V$ source? If so why it is not in phase with the current?
 A: First, you need to understand that emf is not voltage.
Emf is a measure of effect of induced electric field (that part of electric field that is due to accelerated charges and that loops around similarly to the coil) on the motion of the charge carriers in some segment of a wire. It's value is determined from Faraday's law and conveniently expressed as
$$
emf = -L\frac{dI}{dt}.
$$
The minus sign is due to convention that emf sign should indicate whether it supports the current in the positive direction of rotation in the circuit (emf is positive) or opposes it (emf is negative).
Now, voltage drop on a coil is a completely different concept: it means difference of electric potential on the terminals of the coil. The sign of voltage drop too obeys the convention that it is positive if the voltage drop supports the current in the positive direction of rotation in the circuit.
Both emf and voltage have the same units, however, they can have different values and often do so. In case of ideal coil with no ohmic resistance, total electric field inside must be zero. This is possible because the electrostatic field due to charges external and those on the wire surface opposes and exactly cancels the induced field. This opposition means that when induced field opposes the positive current, the electrostatic field must support it. Thus, negative emf means equally large but positive voltage drop. Hence voltage drop on ideal coil is
$$
v.d. = L\frac{dI}{dt}
$$
(without the minus sign). This voltage drop can then be used as a term in the Kirchhoff's Voltage Law.
In case of a real coil with non-zero ohmic resistance $R$, voltage drop across its terminals has no simple relation to emf like above, but one can write generalized Ohm's law equation for the coil:
$$
v.d. + emf = RI
$$
Using the Faraday law, this can be written as
$$
v.d. - L\frac{dI}{dt} = RI
$$
so the voltage drop is related both to emf and current, but is not given simply by emf nor $RI$.
