Is there no net voltage over an ideal inductor/coil? No voltage drop? If you have a coil with self inductance:
$$ \varepsilon= - L \frac{dI}{dt} $$
Then the current is lagging behind the voltage.
If you attach a AC source on the coil/inductor and have an AC power source, then at the highest current there is no internal resisting voltage produced bij the self inductor.
$$ V_{\text{produced by battery}} - \varepsilon =0 (?)$$
My question. Is there no net voltage over the inductor at any moment? I.e. the inductor will always perfectly cancel the voltage produced by the AC source? How do you calculate this?
 A: In AC circuits "voltage on inductor" means difference of electric potential between the inductor's two terminals. In your example where ideal voltage source is driving the circuit with ideal inductor, drop of potential on inductor is determined by the voltage source. For the simplest AC source,  it is sinusoidal function of time:
$$
V_0\sin \omega t,
$$
where $V_0$ is amplitude and $\omega$ frequency of the source.
There is no cancellation of voltage. Cancellation can happen in the sense that in ideal inductor, induced EMF is cancelled by the potential drop. Why? In any inductor free of external forces, the induced EMF is
$$
-L\frac{dI}{dt}.
$$
Provided the inductor is ideal (made of zero resistance conductor, zero capacitance), potential drop is
$$
L\frac{dI}{dt}.
$$
These two forces act in opposite directions (hence the opposite sign) and have the same magnitude, so they cancel each other in the sense that there is no remaining electric field inside the conductor making up the ideal inductor.
In real inductor, this is no longer true, because there is some residual electric field inside the conductor and voltage isn't given by $LdI/dt$.
Voltage on inductor is only due to electrostatic component of electric field. On ideal inductor, this is zero only in special time instants when $dI/dt =0$.
A: There is a net voltage over the inductor most of the time.
Let's say that the AC source has a sinusoidal voltage:
$$U=U_0\sin\omega t$$
Using Kirchhoff's law, the sum of the voltages is equal to zero:
$$U_0\sin\omega t-L\frac{dI}{dt}=0$$
$$U_0\sin\omega t=L\frac{dI}{dt}$$
Separating the variables,
$$U_0\sin\omega t\, dt=L\, dI$$
And integrating,
$$\int U_0\sin\omega t\, dt=\int L\, dI$$
$$I+C=-\frac{U_0}{\omega L}\cos\omega t$$
Because of symmetry, the current amount needs to be symmetric over the $t$ axis, and $C=0$:
$$I=-\frac{U_0}{\omega L}\cos\omega t=-\frac{1}{\omega L}U_0\sin\left(\omega t+\frac{\pi}{2}\right)$$
Therefore, the current really lags behind the voltage, and voltage across the inductor is nonzero, except at the moments of time when $\omega t=(2N+1)\frac \pi 2$.
A: The standard Kirchoffs law does not hold. This is because there are changing magnetic  fields.
Imagine I have a battery and an inductor, such that the inductance L is the inductance of the entire circuit.
Using faradays law:
$$\int \vec{E} \cdot \vec{dl} = -\frac{d \phi_{B}}{dt}$$
$$-V_{batt} + \epsilon_{wire} = -\frac{d \phi_{B}}{dt}$$
$$-V_{batt} + \epsilon_{wire} = - L \frac{dI}{dt}$$
$$ \epsilon_{wire} = V_{batt} - L \frac{dI}{dt}$$
Now, in general, $\epsilon_{wire}$ does not need to be zero.
Using the steady state approximation:
$$\epsilon_{wire} = IR$$
Our equation reduces to:
$$ IR = V_{batt} - L \frac{dI}{dt}$$
When $R=0, \epsilon_{wire} = 0$, and thus the potential across the  inductor is zero.
You find that in order for this to be true.
$$V_{batt} = L \frac{dI}{dt}$$
Aka the contribution to the pd of the battery, cancels out the induced emf.
I would like to point out, that the validity of using ohms law here is not particularly rigourous
