The standard definition of current

Question

The book says current is the rate of flow of charge per unit time, but I don't understand whether it is rate of flow of charge through a single cross-sectional area per unit time or the entire amount of charge which makes it through both terminals in unit time. I have seen the answers, but how can you explain this definition $i=nqAv$, where $n$ is charge per unit volume, here a single cross-section is not taken but volume.

Answer 1

Current is defined as the rate at which charge flows through a given surface:

$$I=\frac{\mathrm{d}Q}{\mathrm{d}t}$$

and in a circuit this surface is any cross-sectional area (perpendicular to the flow).

You can often simplify that to charge through a cross-section per unit time and write $I=\frac{Q}{t}$

Answer to the comment:

Another but equivalent way to define current is: $$I=nqAv_d$$

where $n$ is charge carrier concentration (no. of charges per volume), $q$ is charge of each charge carrier (in a typical circuit these might be electrons), $A$ is cross-sectional area, and $v_d$ is drift speed (the average speed in the length direction).

Let's break it up:

• $Av_d$ is area-times-distance-per-second. This is volume-per-second. This whole volume moves through a cross-section every second.
• In better words, all electrons within that volume will move through one cross-section within the next second. To find out how many that is, we multiply with the no. of electrons per unit volume $n$.
• So, $nAv_d$ is the number of electrons moving through a cross-section every second. Thus we multiply with the charge of each electron, and we finally have the total charge moving through a cross-section every second.

These are two equal expressions of current, and:

$$I=\frac{Q}{t}=nqAv_d$$

Remember that $Q$ is total charge, while $q$ is charge-per-charge-carrier.

Answer 2

Consider, lets say, a wire of cross sectional area $A$, with charges (each of the same magnitude, $q$) flowing through it. If you consider a section of the wire of some length $x$, the volume of this region would be $Ax$. Because $n$ is the no of charges per unit volume, the charge in this region would simply be $q(nAx)$ ($nAx$ is the number of charged particles in your volume $Ax$.)

Since we are concerned with finding the current through the region (which is the rate of change of charge with time), and not the charge enclosed in it, we can consider this volume to be an infinitesimal one. ( Primarily because current is defined as the charge flowing through a surface, not a volume, we can express this mathematically by letting $x$ tend to zero.)

This gives:

$$dQ=nqAdx$$

and then, $$I=\frac{dQ}{dt}=nqA(\frac{dx}{dt})=nqAv$$ (where $v$ is defined as the drift velocity of the charged particles.)

Answer 3

It's the amount of charge flowing through a surface per unit time.