Electric field created from time varying current in solenoid I know how to calculate the magnetic field due to current inside the solenoid.
Obviously, when the current is varying, then the magnetic field will be time dependent.  By the 3rd Maxwell's equation, an electric field will be created.
I'm wondering that, why can I still use Ampere's law to calculate the magnetic field within the solenoid? (in fact, can i?)  I thought that the time varying electric field will also produce a magnetic field, meaning that the total magnetic field is not just that calculated from Ampere's law, but the sum of that calculated from Ampere's law and that produced by time varying electric field.
 A: When you say Ampere's law, do you mean this:
$$ \nabla \times \vec B = \mu_0 \vec J $$
Because that is not the general version of Ampere's law. If you are using Ampere's law without the E-field term (see below), then you are assuming the fields vary slowly in time -- called the 'quasi-static' approximation. If you don't assume slowly varying fields, then the general answer can only be obtained by solving both Ampere's Law and Faraday's Law simultaneously as a set of partial differential equations:
$$ \nabla \times \vec E = - \frac{\partial \vec B}{\partial t} $$ 
$$ \nabla \times \vec B = \mu_0 \vec J + \mu_0 \epsilon_0 \frac{\partial \vec E}{\partial t}$$
In that case, both the E-field and B-field feed off each other, producing a wave-like action (i.e. electromagnetic radiation), which can be quite complicated when close to the original source $\vec J$ ( but eventually settle down into plane waves when further from the source).
A: Isn't it the case that the induced emf (electric field created) produces an induced current if there is a complete electrical circuit and that induced current produces a changing magnetic field which is opposition to the changing magnetic field producing the induced emf. - Faraday + Lenz.
The derivation of the approximate value of the self inductance of a coil uses the magnetic field formula (which can be derived from Ampere's law) for an infinite solenoid and that changing magnetic field depends on the current flowing in the coil.
A: You have the right idea – which is that of self-induction. 
To be more precise, though, it isn't the induced electric field that itself produces its own magnetic field. The induced electric field, integrated round the N turns of the solenoid, produces an induced emf, given (if we ignore end-effects) by$$\mathscr{E}_{ind}=-N\frac{d \Phi}{dt}$$
 that opposes the externally applied emf, $\mathscr{E}$. So the magnetic field is less than it would be without this effect, because the current is$$I=\frac{1}{R}(\mathscr{E}+\mathscr{E}_{ind}).$$ This is the current that we use in Ampère's Law, so, inside a solenoid much longer than its diameter, $B=\mu_0 \frac{N}{L}I$, so the flux through the solenoid is $$\Phi=\mu_0 \frac{N}{L}AI$$ in which $A$ is the cross-sectional area of the solenoid (assumed single-layered).
You can combine these equations into one…$$\mathscr{E}=RI+L\frac{dI}{dt}$$ in which $L$, called the self-inductance of the coil, is given by $$L=\mu_0 \frac{N^2}{L}A.$$ Note that the equation $\mathscr{E}=RI+L\frac{dI}{dt}$ is correct for any size or shape of coil, but $L$ is given by $L=\mu_0 \frac{N^2}{L}A\ $ only for a very long, thin, single-layered coil with air (or vacuum) inside and out.
