3.5 Drift of Electrons and the Origin of Resistivity

Drift velocity is defined as the velocity with which the free electrons get drifted towards the positive terminal under the effect of the applied external electric field. In addition to its thermal velocity, due to acceleration given by applied electric field, the electron acquires a velocity component in a direction opposite to the direction of the electric field. The gain in velocity due to the applied field is very small and is lost in the next collision.

Let us denote by \(\tau\), the average time between successive collisions. Then at a given time, some of the electrons would have spent time more than \(\tau\) and some less than \(\tau\). Therefore the average value of time then is \(\tau\) (known as relaxation time). Therefore at any given time, an electron has a velocity \(\vec{v}_1=\vec{u}_1+\vec{a} \tau_1\), where \(\vec{u}_1=\) the thermal velocity and \(\vec{a} \tau_1=\) the velocity acquired by the electron under the influence of the applied electric field. \(\tau_1=\) the time that has elapsed since the last collision.
Similarly, the velocities of the other electrons are
\(
\vec{v}_2=\vec{u}_2+\vec{a} \tau_2, \vec{v}_3=\vec{u}_3+\vec{a} \tau_3, \ldots \vec{v}_N=\vec{u}_N+\vec{a} \tau_N .
\)
The average velocity of all the free electrons (assume there are \(N\) electrons) in the conductor is equal to the drift velocity \(\vec{v}_d\) of the free electrons
\(
\vec{v}_d=\frac{\vec{v}_1+\vec{v}_2+\vec{v}_3+\ldots \vec{v}_N}{N}=\frac{\left( u _1+\overrightarrow{ a } \tau_1\right)+\left(\overrightarrow{ u }_2+\overrightarrow{ a } \tau_2\right)+\ldots+\left(\overrightarrow{ u }_{ N }+\overrightarrow{ a } \tau_{ N }\right)}{ N }=\frac{\left(\overrightarrow{ u }_1+\overrightarrow{ u }_2+\ldots+\overrightarrow{ u }_{ N }\right)}{ N }+\overrightarrow{ a }\left(\frac{\tau_1+\tau_2+\ldots+\tau_{ N }}{ N }\right)
\)
\(
\because \frac{\overrightarrow{ u }_1+\overrightarrow{ u }_2+\ldots+\overrightarrow{ u }_{ N }}{ N }=0
\)
\(
\therefore \vec{v}_d=\vec{a}\left(\frac{\tau_1+\tau_2+\ldots+\tau_N}{N}\right)
\)
\(
\Rightarrow \vec{v}_d=\vec{a} \tau=-\frac{e \vec{E}}{m} \tau \dots(3.17)
\)
This last result is surprising. It tells us that the electrons move with an average velocity which is independent of time, although electrons are accelerated. This is the phenomenon of drift and the velocity \(v _{ d }\) in Eq. (3.17) is called the drift velocity.

Note: Order of drift velocity is \(10^{-4} m / s\).

Relation between current and drift velocity

Let \(n =\) number density of free electrons and \(A =\) area of the cross-section of the conductor.
Number of free electrons in conductor of length \(L=n A L\), Total charge on these free electrons \(\Delta q=n e A L\).
Time taken by drifting electrons to cross conductor \(\Delta t=\frac{L}{v_d}\)
\(\therefore\) current \(I =\frac{\Delta q }{\Delta t }=\) \(neAL\) \(\left(\frac{ v _{ d }}{ L }\right)= neA v _{ d }\)

Example 17: Find free electrons per unit volume in a metallic wire of density \(10^4 kg / m ^3\), atomic mass number 100, and number of free electrons per atom is one.

Solution: Number of free charge particle per unit volume \(( n )=\frac{\text { total free charge particle }}{\text { total volume }}\)
\(\therefore\) No. of free electron per atom means total free electrons \(=\) total number of atoms \(=\frac{N_A}{M_w} \times M\)
\(
\text { So } n =\frac{\frac{ N _{ A }}{ M _{ w }} \times M }{ V }=\frac{ N _{ A }}{ M _{ w }} \times d =\frac{6.023 \times 10^{23} \times 10^4}{100 \times 10^{-3}}=6.023 \times 10^{28}
\)

Example 18: (a) Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area \(1.0 \times 10^{-7} m^2\) carrying a current of 1.5 A. Assume that each copper atom contributes roughly one conduction electron. The density of copper is \(9.0 \times 10^3 kg / m ^3\), and its atomic mass is 63.5 u. (b) Compare the drift speed obtained above with, (i) the thermal speeds of copper atoms at ordinary temperatures, (ii) speed of propagation of electric field along the conductor which causes the drift motion.

Solution: (a) The direction of the drift velocity of conduction electrons is opposite to the electric field direction, i.e., electrons drift in the direction of increasing potential. The drift speed \(v_d\) is given by \(v_d=(I / n e A)\)
Now, \(e=1.6 \times 10^{-19} C , A=1.0 \times 10^{-7} m^2, I=1.5 A\). The density of conduction electrons, \(n\) is equal to the number of atoms per cubic metre (assuming one conduction electron per Cu atom as is reasonable from its valence electron count of one). A cubic metre of copper has a mass of \(9.0 \times 10^3 kg\). Since \(6.0 \times 10^{23}\) copper atoms have a mass of 63.5 g.
\(
\begin{aligned}
n & =\frac{6.0 \times 10^{23}}{63.5} \times 9.0 \times 10^6 \\
& =8.5 \times 10^{28} m^{-3}
\end{aligned}
\)
which gives,
\(
\begin{aligned}
v_d & =\frac{1.5}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 1.0 \times 10^{-7}} \\
& =1.1 \times 10^{-3} m s ^{-1}=1.1 mm s ^{-1}
\end{aligned}
\)
(b) (i) At a temperature \(T\), the thermal speed* of a copper atom of mass \(M\) is obtained from \(\left[<(1 / 2) M v^2>=(3 / 2) k_{ B } T\right]\) and is thus typically of the order of \(\sqrt{k_B T / M}\), where \(k_B\) is the Boltzmann constant. For copper at 300 K, this is about \(2 \times 10^2 m / s\). This figure indicates the random vibrational speeds of copper atoms in a conductor. Note that the drift speed of electrons is much smaller, about \(10^{-5}\) times the typical thermal speed at ordinary temperatures.
(ii) An electric field travelling along the conductor has a speed of an electromagnetic wave, namely equal to \(3.0 \times 10^8 m s ^{-1}\). The drift speed is, in comparison, extremely small; smaller by a factor of \(10^{-11}\).

Example 19: (a) The electron drift speed is estimated to be only a few \(mm s ^{-1}\) for currents in the range of a few amperes? How then is current established almost the instant a circuit is closed?
(b) The electron drift arises due to the force experienced by electrons in the electric field inside the conductor. But force should cause acceleration. Why then do the electrons acquire a steady average drift speed?
(c) If the electron drift speed is so small, and the electron’s charge is small, how can we still obtain large amounts of current in a conductor?
(d) When electrons drift in a metal from lower to higher potential, does it mean that all the ‘free’ electrons of the metal are moving in the same direction?
(e) Are the paths of electrons straight lines between successive collisions (with the positive ions of the metal) in the (i) absence of electric field, (ii) presence of electric field?

Solution: (a) Electric field is established throughout the circuit, almost instantly (with the speed of light) causing at every point a local electron drift. Establishment of a current does not have to wait for electrons from one end of the conductor travelling to the other end. However, it does take a little while for the current to reach its steady value.
(b) Each ‘free’ electron does accelerate, increasing its drift speed until it collides with a positive ion of the metal. It loses its drift speed after collision but starts to accelerate and increases its drift speed again only to suffer a collision again and so on. On the average, therefore, electrons acquire only a drift speed.
(c) Simple, because the electron number density is enormous, \(\sim 10^{29} m^{-3}\)
(d) By no means. The drift velocity is superposed over the large random velocities of electrons.
(e) In the absence of electric field, the paths are straight lines; in the presence of electric field, the paths are, in general, curved.

Mobility

As we have seen, conductivity arises from mobile charge carriers. In metals, these mobile charge carriers are electrons; in an ionised gas, they are electrons and positive charged ions; in an electrolyte, these can be both positive and negative ions.
An important quantity is the mobility \(\mu\) defined as the magnitude of the drift velocity per unit electric field:
\(
\mu=\frac{\left| \vec {v _d}\right|}{E} \dots(3.18)
\)
The SI unit of mobility is \(m ^2 / Vs\) and is \(10^4\) of the mobility in practical units ( \(cm ^2 / Vs\) ). Mobility is positive. We know that
\(
v_d=\frac{e \tau E}{m}
\)
Hence,
\(
\mu=\frac{v_d}{E}=\frac{e \tau}{m} \dots(3.19)
\)
where \(\tau\) is the average collision time for electrons.

Example 20: A constant voltage is applied across a wire of constant length. How does the drift velocity of electrons depends on the area of the cross-section of the wire?

Solution: \(v_d=\mu E , \quad \mu=\) mobility of free electrons \(=\frac{e \tau}{m} \quad\) and \(\quad E=\) electric field \(=\frac{V}{L}\)
so \(\quad v_d=\mu \frac{ V }{ L } \quad\); here \(\mu\) is constant.
If \(V\) and \(L\) are constant then \(v_{ d }\) does not depend on area.