What does an induced emf do to a loop of varying current? I know that the magnetic flux through a loop in space is proportional to the current such that $\Phi=LI$. Therefore a change in magnetic flux will produce an emf, so $\mathscr{E}=-L\dfrac{dI}{dt}$. What exactly does this induced emf do to the sole loop of current if there's already something allowing current to flow? Does it make affect the flow of current in anyway?
 A: Yes, in general it affects the current, in particular if that current was going to change, the self inductance $L$ makes it change less quickly than it otherwise would change. Here are some examples:
You could put a resistive loop inside a solenoid as a good example.
The total $\vec{B}$ field is the $\vec{B}$ field due to the solenoid $\vec{B}_1(t)$, and the $\vec{B}$ field due to the loop $\vec{B}_2(t)$, based on the geometry we can set up the mutual inductance and self inductance by looking at the flux through the loop due to $\vec{B}_1$, namely $\Phi_1$ and the flux through the loop due to $\vec{B}_2$, namely $\Phi_2$.  So $\Phi_1=MI_1$ $\Phi_2=MI_2$ where $I_1$ is the current through the solenoid and $I_2$ is the current through the loop.
Now we have $\mathscr{E}=-\frac{d}{dt} \Phi=-\frac{d}{dt} (\Phi_1+\Phi_2)=-M\frac{dI_1}{dt}-L\frac{dI_2}{dt}$.
And based on the resistance $R$ of the loop, $I_2R=\mathscr{E}=-M\frac{dI_1}{dt}-L\frac{dI_2}{dt}$.  This is a differential equation, and it needs $I_1(t)$ as a function to solve it.
You could hook up a resistive loop to a capacitor as another example.
Now there are two sources of EMF, one from the changing current, one from the capacitor (dis)charging. For the capacitor $V=Q/C$, if the gap is small the induced EMF could be approximated by the change in flux, $-\frac{d}{dt} \Phi_2=-L\frac{dI_2}{dt}$. In quasi-statics there is a conservative electrostatic field from the capacitor and a nonconservative field associated with the changing $\vec{B}$ field flux. Now $I_2=-\frac{dQ}{dt}$, and $I_2R=Q/C-L\frac{dI_2}{dt}$, so we get a second order differential equation:
$$-\frac{dQ}{dt}R=Q/C+L\frac{d^2Q}{dt^2}.$$
There are many possibilities.
