# does drift velocity depend on current?

I need conceptual help concerning the equation I = nAvq. The equation states that the current flowing through a conductor is equivalent to the number of charge carriers multiplied by the charge q, cross-sectional area A and drift velocity v. My question is,

if the current is increased would that increase the velocity v, if A and q are kept constant? (as opposed to only n increasing)

I really need an answer to this question thank you

• Yes, drift velocity and current are both proportional to the electric field.
– user137289
Jan 13 '18 at 22:59
• J., what makes you think it might not be so? Jan 13 '18 at 23:00

You can increase the drift current by either increasing the density or velocity of the charge carriers. If the current increases, it could be caused by an increase in density or an increase in velocity of the charge carriers. Usually, in a conducting metal you cannot change the electron density. Therefore, an increase in current is caused by an increase in drift velocity $v=\mu E$ due to an increase of the applied electric field, where $\mu$ is the mobility of the electrons is the eletron mobility of the metal.

You have already narrowed it down to $n$ or $v$ increasing. Now $n$ is number density of electrons (or charge carriers). It could increase, but since like-charged particles repel each other, they are unlikely to bunch up - which is what happens when $n$ increases.

So, barring special circumstances, $v$ increases.

Given the equation

$$I = nqvA$$

the current can depend on the drift velocity, since that is one of the parameters, but it can also depend on other factors as well. In particular, for any given fixed drift velocity $v$, you have three options to obtain a stronger current $I$:

1. choose a material with a higher volumetric number density $n$ of charge carriers, e.g. a metal with a denser crystal structure,

2. use charge carriers with greater charge $q$, e.g. using suitable ions dissolved in some sort of solvent instead of electrons (though you'll probably in real life end up seriously losing on number density because they are much bulkier than electrons so this will in practice result in less current, not to mention the dilution due to all the suspending fluid),

3. increase the cross section $A$ of your conductor, that is, use a fatter piece of wire, so there are more overall charge carriers moving.

Of these, only the third option is really the most feasible in practice. Likewise, if all three of these are fixed , which they will be for a given sample of wire provided the current is not so great as to cause some sort of physical effect or even damage like heating and melting, then yes, increasing $I$ will increase the drift velocity since $v = \frac{I}{nqA}$. If they are not fixed, then the drift velocity can go any way, e.g. if you increase $I$ but you also increase $A$ even more, that is, take a higher current through a much thicker wire, then the drift velocity can actually fall.

This situation is similar to the case of the ideal gas law $pV = nRT$, giving $p = \frac{nRT}{V}$, where there are three ways to increase the pressure: either add more gas (increase $n$) into the container, heat the container up (increase $T$), or crush the container (decrease $V$). It all is fairly easy with a little thought about the basic algebra: increasing a multiplier makes things go up, increasing a divisor makes them go down, and inversely for decreasing either.