Electric current through wire If you have a wire of area A and length x with a constant current flowing through it, is it reasonable to say that:
$I=\dfrac{dQ}{dt}\Rightarrow dQ=Idt\Rightarrow Q=It \Rightarrow Q=\dfrac{Ix}{v_d}$? 
 A: I believe the answer is no. Current is not defined as the velocity of charge in a conductor. "Electric current through a surface is defined as the rate of charge transport through that surface". (Re NCEE reference handbook for the PE FE exam in Electrical and Computer Engineering). Picture yourself looking edgewise at the cross section of a conductor and watching the electrons go by. The number of coulombs of electrons that pass through the surface per second is the current in amperes.
A: Based on comments on other answers, let's be a bit more careful. 
Let's say we have a uniform current density $J=\frac IA\space$  as well as a uniform charge carrier density $n=\frac NV\space$  Where 
$N$ is the number of charge carriers in the volume $V$ of the wire. 
Now let's think about how much charge $dQ$ passes through a cross section of the wire with area $A$. Well the amount of charge in some volume $dV$ is ${dQ}=qn{dV}=nqA{dx}$, where $dx$ is a small length along the wire, and $q$ is the amount of charge in one charge carrier. Now if the charge carriers are moving with some average drift velocity $v_d$, then the charges in that volume $dV$ will move a distance ${dx}=v_d{dt}$. Therefore,
$$dQ=nqv_dA{dt}$$
Since the current is $I=\frac{dQ}{dt}$ we have $$I=nqv_dA$$
Now let's move towards what you have. If the wire is uniform, then $A=\frac VL$ where $L$ is the length of the wire. Therefore, $$I=\frac NV qv_d\frac VL =\frac{Nqv_d}{L}$$
Rearranging this we get $$Nq=\frac{IL}{v_d}$$
If we wanted to define a total charge carried in the wire as $Nq=Q$ then we get what you were asking about. You just have to keep in mind that this $Q$ is different than the amount of charge that is passing through the area we were looking at earlier. You also should notice that if we have a different carrier density (for example, our wire is made of a different material), then we get a different current. This is lost in the final formula we arrive at.
As you can see, you need to assume things about the system such as uniform current and carrier densities. You also need to be aware of what each variables physically means. But once you do it all works out.
A: Yes. First of all you can write anything in mathematical exp. until and unless it's is validates. Ampere's law and biot-savart law is holding up that equation. I think it is valid but some external factors must be there like +C (for constant).
