Objectives:

• To state what is meant by the term mean drift velocity of charge carriers;
• To be able to select and use the equation $I = Anev$;
• To describe the difference between conductors, semiconductors and insulators in terms of the number density n.

Mean Drift Velocity

In order for a current to flow in a wire the charge carriers, i.e. electrons, must be moving. How fast they are moving depends on several factors, some of which include the potential difference across the wire, the resistance of the wire and the number of electrons present.

When there is no potential difference across a particular wire, each individual charge carrier (in most cases electrons) will be moving in its own direction with a speed dependent on the amount of energy it has previously gained; this energy could depend on the thermal energy it has gained from its surroundings. So a circuit like the following; with a wire, as located in the circuit, will have electrons in it which are moving with their own internal energy;

When a potential difference (voltage) is supplied across the end of a wire, each electron is forced to move in a certain direction. The potential difference forces the electrons along – work is done on the charged particles;

Here we will be discussing how fast the electrons need to be travelling in order to produce an ‘adequate’ current. Since $I = \frac{\Delta Q}{\Delta t}$; a current can only flow if the charges move some distance in a time t seconds.

Deriving the current equation

For this derivation we will be making the assumption that all the electrons will be travelling at the same speed, this speed will actually be the average speed, or mean drift velocity, of the electrons.

If a detector is placed at position and is capable of timing a single charge flowing past this point;

Suppose over a time $t$ second, the highlighted electron will travel a distance $l$.

Assuming all the electrons are travelling at the same speed, all the electrons in the cylinder of length $l$ will have in fact flowed past the detector;

In order to work out how much current has flowed, we first need to work out how much charge has passed the detector. The total amount of charge that will have passed the detector would fill the volume, and this volume can be calculated;

Volume of the cylinder, $V = l \times A$, where $A$ is the cross sectional area of the wire.

This cylinder will not be entirely made up of electrons, there will be ions for example as well, however for a specific material there will be a certain concentration of electrons per metre cubed $n$.

The number of electrons in this cylinder can then be written as;

$number\ of\ electrons\ in\ cylinder = n \times A \times l$

If each electron carries a charge $q$ then;

The charge charge by the cylinder, $Q = n \times A \times l \times q$

Since this cylinder is dependent on how long the detector is timing for, an element of time should be in the equation for the $Q$. If all the electrons are travelling at a mean drift velocity $v$ for a time $t$ seconds, and  $v = \frac{l}{t}$  we can write $l = v \times t$

therefore;                                        $Q = n \times A \times v \times t \times q$

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

we can write;                                  $I = \frac{Q}{t} = \frac{n \times A \times v \times t \times q}{t}$

The $t$‘s on the right hand side of this equation cancel to give:

$I = n \times A \times v \times q$

Since the charge of an electron, in this case  $q$,  is  $1.6 \times 10^{-19}$  or also denoted with an $e$, we can write;

$I = n \times A \times v \times e$

$I = nAve$

where;
$I$ is the current, measured in $A$
$n$ is the number of electrons per metre cubed, therefore measured in $m^{-3}$
$A$ is the cross sectional area, measured in  $m^{2}$
$v$ is the mean drift velocity of the electrons, measured in  $ms^{-1}$
$e$ is the charge of an electron,  $1.6 \times 10^{-19}$, measure in $C$

One method of checking whether this equation is valid is to check the units are conserved on each side;
$I = nAve$

$A = m^{-3} \times m^{2} \times ms^{-1} \times C$
$A = s^{-1} \times C$  Since,  $I = \frac{Q}{t}$  and therefore  $A = \frac {C}{s}$,  the equation holds true.

So, to address the original question how fast do electrons need to be travelling in order to produce an ‘adequate’ current?

Lets plug in some realistic numbers to determine the mean drift velocity:

If a current of  $0.50 \ A$ flows through a copper wire measured to be  $0.2019 \ mm$  in thickness, using a micrometer, and copper has $8.5 \times 10^{28}$  electrons per metre cubed:
$I = 0.50 \ A$
$n = 8.5 \times 10^{28} \ m^{-3}$
$d = 0.2019 \ mm = 0.2019 \times 10^{-3} \ m$

A ‘non-deformed’ wire can be assumed to be cylindrical and therefore the cross sectional area can be said to be circular. So $A = \pi r^{2}$

$A = \pi (\frac{d}{2})^{2} = \pi (\frac{0.2019 \times 10^{-3}}{2})^{2} = 3.202 \times 10^{-8} m^{2}$

Rearranging  $I = nAve$  to give $v = \frac{I}{nAe}$ and substituting everything in gives:

$v = \frac{0.50}{8.5 \times 10^{28} \times 3.202 \times 10^{-8} \times 1.6 \times 10^{-19}}$

$v = \frac{0.50}{435.47}$

$v = 1.148 \times 10^{-3}$

$v = 1.1 \times 10^{-3} ms^{-1} \ ({2s.f})$

This shows that the mean drift velocity of electrons flowing down a standard gauge wire with a current of $0.50 \ A$ is very slow, approximately $1 mms^{-1}$. Many people think that electrons travel down wires extremely quickly, which would explain why a light bulb would turn on almost immediately after a switch has been flicked). However, they actually travel very slowly and the reason a bulb still turns on as if instantaneously is because all electrons repel one another, as soon as one electrons is propelled along by a power source it repels its neighbouring electrons, which does the same to its neighbouring electrons, and so on and so forth until the bulb (this repulsive force os what causes a light bulb to light up so quickly.

Electrons as incredibly small, they have a mass of  $9.11 \times 10^{-31} kg$,  they are not visible to any human instrument (they can only be detected so far), so the fact that they can travel even  $1 mm$  in just one second is actually quite impressive.