Is it possible to have both a conduction current and a displacement current at the same time? According to Ampere's law, is it possible to have both a conduction current and a displacement current at the same time? 
In the classical derivation of the displacement current using a parallel plate capacitor, the conduction current equals the displacement current. But only one of the latter are present at a given point in space (between the plates or in the wires to the capacitor). In a general case, can the two currents exist at the same time? And if so, is it possible for the effects of the two kinds of current to cancel each other exactly so that no magnetic field is produced?
 A: Writing Ampere's law as
$$\nabla \times \vec{H} = \vec{J} + \epsilon \frac{\partial \vec{E}}{\partial t},$$
the first term on the RHS is the conduction current density and the second term is the displacement current density.
In a conductor, the induced current density $\vec{J} = \sigma \vec{E}$. So now if we induce a current in a conductor using an oscillating electric field there is both a conduction current density and a displacement current density.
However, because displacement current density features the first time derivative of the electric field, they are always out of phase and cannot cancel. i.e. if $E =E_0 \sin \omega t$, then the conduction current will be proportional to $\sin \omega t$, whilst the displacement current density will be proportional to $\cos \omega t$. 
The usual capacitor-based "derivation" of the displacement current term has to make the assumption that the current in the wire is changing relatively slowly in order to ignore its displacement current term. I note that many if not most descriptions of this setup omit to mention this...
In fact the idea of a conduction current without a displacement current is somewhat curious. For a steady current, the net charge within a volume must stay the same, but a current moves charge and the charge has to come from somewhere and from the continuity condition
$$ \oint \left( \vec{J} + \epsilon \frac{\partial \vec{E}}{\partial t} \right) \cdot d\vec{A}=0$$
there must be a displacement current somewhere.
A: Certainly it is possible to have both a conduction current and a displacement current at the same time. When your parallel plate capacitor is charging up, a different wire that runs through the volume between the plates can carry any current you wish, and then you have both sources for a magnetic field.
It is also possible to arrange for cancellation. Consider a fully charged parallel plate capacitor, no wires connected. Now discharge it by connecting a wire from the center of one plate through its volume to the center of the other. [OK, put a resistor in series to slow down this event and make it safer!]
Now you can see that the two terms contribute equal and opposite amounts to the B-field just outside the capacitor. Let's see how.
Since $Q = CV$ we have $\dot V = (1/C)dQ/dt = I/C = V/RC$.
In turn, the displacement current is the area integral of $\epsilon_0 \dot E$, which is $-\epsilon_0 \dot V A / d$, where $A$ and $d$ are the area and plate separation of the capacitor, as usual, and we used $E V / d$ and the fact that $V$ is ${\it decreasing}$. Using further $C = \epsilon_0 A / d$, we see that the displacement current is $-C\dot V = -\dot Q = -I$.
