Drift of Electrons and the Origin of Resistivity

Current Electricity · CBSE

Theory

Derivation of the drift velocity — write it exactly like this

Consider $N$ free electrons in the conductor. With no field their velocities are random, so

$\dfrac{1}{N}\sum_{i=1}^{N}\mathbf{v}_i = 0$

Switch on a field $\mathbf{E}$ . Each electron of charge $-e$ and mass $m$ is accelerated:

$\mathbf{a} = \dfrac{-e\mathbf{E}}{m}$

Look at the $i$ -th electron at time $t$ . Let its last collision have been a time $t_i$ earlier and let $\mathbf{v}_i$ be its velocity immediately after that collision. Then its velocity now is

$\mathbf{V}_i = \mathbf{v}_i + \left(\dfrac{-e\mathbf{E}}{m}\right) t_i$

Average over all $N$ electrons. The average of the $\mathbf{v}_i$ is zero, because the direction just after any collision is completely random. The average of the $t_i$ is the relaxation time $\tau$ , the average time between two successive collisions. Hence the average velocity — the drift velocity — is

$\mathbf{v}_d = \left\langle \mathbf{V}_i \right\rangle = 0 - \dfrac{e\mathbf{E}}{m}\tau = \dfrac{-e\mathbf{E}\tau}{m}$

The result is remarkable: although each electron is being accelerated, the average velocity is constant in time. The minus sign says the drift is opposite to $\mathbf{E}$ — electrons drift towards the higher potential.
In magnitude, and using $E = V/l$ for a conductor of length $l$ across which $V$ is applied,

$\lvert v_d \rvert = \dfrac{eE\tau}{m} = \dfrac{eV\tau}{ml}$

Definitions CBSE wants word-perfect

Drift velocity: the average velocity acquired by the free electrons of a conductor, superposed on their random thermal motion, under the influence of an applied electric field; it is directed opposite to the field. Its physical significance is that it is the drift, not the random motion, that produces the net current.
Relaxation time: the average time that elapses between two successive collisions of a free electron with the lattice ions. Its significance is that it controls the resistivity, and its variation with temperature explains how resistivity varies with temperature.

Current in terms of drift velocity

Method 1 — the slab of length $v_d\Delta t$ (NCERT). Take a plane area $A$ inside the conductor with its normal along $\mathbf{E}$ . In a time $\Delta t$ , every electron within a distance $\lvert v_d\rvert \Delta t$ on one side crosses it. If $n$ is the free-electron density, that is $n A \lvert v_d\rvert \Delta t$ electrons, carrying charge $neA\lvert v_d\rvert \Delta t$ . Since this charge is $I\Delta t$ by definition,

$I = neAv_d$

Method 2 — the whole wire (the version CBSE prints). For a conductor of length $l$ , area $A$ and free-electron density $n$ , the total number of free electrons is $N = nAl$ , carrying charge $q = Ne = nAle$ . An electron drifting with speed $v_d$ crosses the length in $t = l/v_d$ . Hence

$I = \dfrac{q}{t} = \dfrac{nAle}{l/v_d} = neAv_d$

Dividing by $A$ gives the current density:

$j = \dfrac{I}{A} = nev_d$

Origin of resistivity, and Ohm's law deduced

Substituting $v_d = eE\tau/m$ into $j = nev_d$ ,

$j = \dfrac{ne^{2}\tau}{m}E \qquad \text{i.e.} \qquad \mathbf{j} = \dfrac{ne^{2}\tau}{m}\mathbf{E}$

Comparing with $\mathbf{j} = \sigma\mathbf{E}$ identifies the conductivity, and hence the resistivity:

$\sigma = \dfrac{ne^{2}\tau}{m}, \qquad \rho = \dfrac{1}{\sigma} = \dfrac{m}{ne^{2}\tau}$

For the whole conductor, $I = neAv_d = \dfrac{ne^2A\tau}{m}E$ , and with $E = V/l$ ,

$I = \dfrac{ne^{2}A\tau}{ml}V \qquad \Rightarrow \qquad \dfrac{V}{I} = \dfrac{ml}{ne^{2}\tau A} = \dfrac{\rho l}{A} = R$

So $V = IR$ : Ohm's law, with $R = ml/ne^2\tau A$ a constant for a given conductor at a given temperature. The derivation assumes $n$ and $\tau$ do not depend on $E$ — which is exactly why Ohm's law has limits.

Mobility (NCERT 3.5.1)

Conductivity arises from mobile charge carriers: electrons in a metal, electrons and positive ions in an ionised gas, positive and negative ions in an electrolyte.
Mobility $\mu$ is defined as the magnitude of the drift velocity per unit electric field:

$\mu = \dfrac{\lvert v_d \rvert}{E} = \dfrac{e\tau}{m}$

SI unit $\text{m}^2\text{V}^{-1}\text{s}^{-1}$ ; this is $10^4$ times the practical unit $\text{cm}^2\text{V}^{-1}\text{s}^{-1}$ . Mobility is always positive.
Conductivity in terms of mobility: $\sigma = ne\mu$ .
Because $\mu = e\tau/m$ contains only $e$ , $m$ and $\tau$ , mobility does not change when the applied potential difference is doubled at fixed temperature, and does not change along a wire of varying thickness.

Numbers and comparisons CBSE quotes

Copper wire of area $1.0\times10^{-7}\ \text{m}^2$ carrying $1.5\ \text{A}$ , with $n = 8.5\times10^{28}\ \text{m}^{-3}$ :

$v_d = \dfrac{I}{neA} = \dfrac{1.5}{8.5\times10^{28}\times1.6\times10^{-19}\times1.0\times10^{-7}} = 1.1\times10^{-3}\ \text{m s}^{-1}$

With $n = 9\times10^{28}\ \text{m}^{-3}$ instead, the same calculation gives $1.04\times10^{-3}\ \text{m s}^{-1} \approx 1\ \text{mm s}^{-1}$ .
Thermal speed of a copper atom at 300 K is about $2\times10^{2}\ \text{m s}^{-1}$ , so the drift speed is about $10^{-5}$ times the thermal speed.
The field travels along the conductor at the speed of an electromagnetic wave, $3.0\times10^{8}\ \text{m s}^{-1}$ ; the drift speed is smaller by a factor of about $10^{-11}$ .

The five conceptual answers (NCERT Example 3.2)

Current is established almost instantly because the field is set up throughout the circuit at nearly the speed of light and starts a local drift everywhere at once; nobody waits for an electron to travel from one end to the other. It does take a little while for the current to settle to its steady value.
Electrons do accelerate, but each one is stopped and randomised by a collision with an ion; on the average they therefore acquire only a steady drift speed, not an ever-growing one.
Large currents are possible despite the tiny drift speed and the tiny charge because the free-electron density is enormous, $\sim 10^{29}\ \text{m}^{-3}$ .
The drift velocity is superposed on the large random velocities, so it is not true that all the free electrons move in one direction.
Between collisions the paths are straight lines when there is no field, and in general curved when there is one.

Standard results

$v_d = eE\tau/m = eV\tau/ml$ ; $v_d \propto V$ , $v_d \propto \tau$ , $v_d \propto 1/l$ .
$I = neAv_d$ ; $j = nev_d$ ; $\sigma = ne^2\tau/m$ ; $\rho = m/ne^2\tau$ ; $R = ml/ne^2\tau A$ ; $\mu = e\tau/m$ ; $\sigma = ne\mu$ .
The slope of a $v_d$ against $V$ graph is $e\tau/ml$ , which is inversely proportional to length: the line with the smaller slope belongs to the longer wire.
A conductor of length $L$ replaced by one of length $5L$ at the same voltage: $v_d$ becomes $v_d/5$ . Stretched to $3L$ at the same voltage: $v_d$ becomes $v_d/3$ .
Doubling $V$ at constant temperature doubles $v_d$ but leaves $\mu$ and $\tau$ unchanged.
Raising the temperature reduces $\tau$ , hence reduces $v_d$ at a fixed field.
Two wires of the same diameter in series with $n_A = 1.5\,n_B$ : equal currents give $n_AeAv_{dA} = n_BeAv_{dB}$ , so $v_{dA}/v_{dB} = n_B/n_A = 2/3$ .
Two conductors of the same material and length but areas $A_1 \ne A_2$ in parallel across a cell: $v_d = V/(ne\rho L)$ for both, so their drift velocities are equal.
Along a wire whose thickness decreases from A to B, both $E$ and $v_d$ increase from A to B (and $j$ with them), while $\mu$ stays the same.

Theory Questions Asked on This Subtopic

Real board questions. Answers are at the end of the page.

1
How does the mobility of electrons in a conductor change, if the potential difference applied across the conductor is doubled, keeping the length and temperature of the conductor constant?
1 markCBSE Class 12 Physics — Very Short Answer · 2020 · Section A, 1 markAnswer 1 ↓
2
How is the drift velocity in a conductor affected with the rise in temperature?
1 markCBSE Class 12 Physics — Very Short Answer · Delhi 2019 · Section A, 1 markAnswer 2 ↓
3
Define the term drift velocity of charge carriers in a conductor and write its relationship with the current flowing through it.
1 markCBSE Class 12 Physics — Very Short Answer · Delhi 2014 · Section A, 1 markAnswer 3 ↓
4
Write the expression for the drift velocity of charge carriers in a conductor of length ' $l$ ' across which a potential difference ' $V$ ' is applied.
1 markCBSE Class 12 Physics — Very Short Answer · AI 2014C, An · Section A, 1 markAnswer 4 ↓
5
Define the term 'drift velocity' of electrons in a current carrying conductor. Obtain the relationship between the current density and the drift velocity of electrons.
2 marksCBSE Class 12 Physics — SA · 2020 · Section B, 2 marksAnswer 5 ↓
6
Define the term 'mobility' of charge carriers in a current carrying conductor. Obtain the relation for mobility in terms of relaxation time.
2 marksCBSE Class 12 Physics — SA · 2020, U · Section B, 2 marksAnswer 6 ↓
7
Using the concept of drift velocity of charge carriers in a conductor, deduce the relationship between current density and resistivity of the conductor.
2 marksCBSE Class 12 Physics — SA · Delhi 2015C, An · Section B, 2 marksAnswer 7 ↓
8
Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area $1.0 \times 10^{-7}\,\text{m}^2$ carrying a current of 1.5 A. Assume the density of conduction electrons to be $9 \times 10^{28}\,\text{m}^{-3}$ .
2 marksCBSE Class 12 Physics — SA · AI 2014 · Section B, 2 marksAnswer 8 ↓
9
Define drift velocity of electrons in a conductor connected across a battery. The figure shows the variation of the drift velocity $(v_d)$ of electrons in two copper wires $A$ and $B$ of different lengths, versus the potential difference $(V)$ applied across their ends. Line $A$ is the steeper of the two. (i) What does the slope of the line represent? (ii) Which one of the two wires is longer?
3 marksCBSE Class 12 Physics — SA · 2021C · Section C, 3 marksAnswer 9 ↓
10
The thickness of a conductor continuously decreases from its one end ( $A$ ) to another end ( $B$ ). It is connected across the terminals of a battery. What will be the effect on the value of (a) electric field, (b) current density, and (c) mobility of the electron, at a point on the conductor as one moves from end $A$ to end $B$ ?
3 marksCBSE Class 12 Physics — SA · 2020C · Section C, 3 marksAnswer 10 ↓
11
(a) Define the terms 'drift velocity' and 'relaxation time' giving their physical significance. (b) A conductor of length $L$ is connected across a dc source of emf $E$ . If the conductor is replaced by another of the same material and area of cross-section but of length $5L$ , by what factor will the drift velocity change?
3 marksCBSE Class 12 Physics — SA · 2019C · Section C, 3 marksAnswer 11 ↓
12
(a) Define the term 'conductivity' of a metallic wire. Write its SI unit. (b) Using the concept of free electrons in a conductor, derive the expression for the conductivity of a wire in terms of number density and relaxation time. Hence obtain the relation between current density and the applied electric field $E$ .
3 marksCBSE Class 12 Physics — SA · 2018 · Section C, 3 marksAnswer 12 ↓
13
(a) Find the relation between the drift velocity and relaxation time of charge carriers in a conductor. (b) A conductor of length $L$ is connected to a dc source of emf $V$ . If the length of the conductor is tripled by stretching it, keeping $V$ constant, explain how the drift velocity would be affected.
3 marksCBSE Class 12 Physics — SA · Section C, 3 marksAnswer 13 ↓
14
A steady current flows in a metallic conductor of non-uniform cross-section. Which of these quantities is constant along the conductor: current, current density, electric field, drift speed?
5 marksCBSE Class 12 Physics — Long Answer · 1/3, Delhi 2015C · Section E, 5 marksAnswer 14 ↓
15
(a) Deduce the relation between the current $I$ flowing through a conductor and the drift velocity $v_d$ of the electrons. (b) The figure shows a plot of current ' $I$ ' flowing through the cross-section of a wire versus time ' $t$ '. Use the plot to find the charge flowing in 10 s through the wire. [Graph: $I$ rises linearly from 0 to 5 A over $t=0$ to $5$ s, then stays constant at 5 A from $t=5$ s to $10$ s.]
5 marksCBSE Class 12 Physics — Long Answer · AI 2015C · Section E, 5 marksAnswer 15 ↓
16
(i) Define mobility of electrons. Give its SI unit. (ii) A steady current flows through a wire $AB$ of non-uniform cross-section (thickness decreasing from $A$ to $B$ ), as shown in the figure. What happens to the electric field and the drift velocity along the wire? Justify your answer. (iii) Consider the circuit shown in the figure. Find the effective resistance of the circuit and the current drawn from the battery.
5 marksCBSE Class 12 Physics — Long Answer · 2023 · Section E, 5 marksAnswer 16 ↓
17
(i) Explain how free electrons in a metal at constant temperature attain an average velocity under the action of an electric field. Hence obtain an expression for it. (ii) Consider two conducting wires $A$ and $B$ of the same diameter but made of different materials, joined in series across a battery. The number density of electrons in $A$ is 1.5 times that in $B$ . Find the ratio of drift velocity of electrons in wire $A$ to that in wire $B$ .
5 marksCBSE Class 12 Physics — Long Answer · Delhi 2017, Ev · Section E, 5 marksAnswer 17 ↓
18
(i) Derive an expression for the drift velocity of electrons in a conductor. Hence deduce Ohm's law. (ii) A wire whose cross-sectional area is increasing linearly from one end to the other is connected across a battery of $V$ volts. Which of the following quantities remain constant in the wire: drift speed, current density, electric current, electric field? Justify your answer.
5 marksCBSE Class 12 Physics — Long Answer · Delhi 2017/2023, Ev · Section E, 5 marksAnswer 18 ↓
19
Define the term 'drift velocity' of charge carriers in a conductor. Obtain the expression for the current density in terms of relaxation time.
2 marksCBSE Class 12 Physics — SA · 2/5, Foreign 2014 · Section B, 2 marksAnswer 19 ↓

Numericals

Every numerical set on this subtopic. Full working is at the end of the page.

20
If $n$ , $e$ , $\tau$ and $m$ have their usual meanings, then the resistance of a wire of length $l$ and cross-sectional area $A$ is given by
- (a)
  $\dfrac{ne^2A}{2m\tau l}$
- (b)
  $\dfrac{ml}{ne^2\tau A}$
- (c)
  $\dfrac{m\tau A}{ne^2l}$
- (d)
  $\dfrac{ne^2\tau A}{2ml}$
1 markCBSE Class 12 Physics — Multiple Choice · Term I 2021-22, U · Section A, 1 markAnswer 20 ↓
21
The potential difference applied across a given conductor is doubled. The mobility of the electrons in the conductor
- (a)
  is doubled
- (b)
  is halved
- (c)
  remains unchanged
- (d)
  becomes four times
1 markCBSE Class 12 Physics — Multiple Choice · AI 2019 · Section A, 1 markAnswer 21 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1.↑ back to the question
Mobility $\mu = e\tau/m$ depends only on the relaxation time $\tau$ (fixed by temperature) and constants $e, m$ ; it does not depend on the applied potential difference. Hence, keeping the temperature constant, the mobility of the electrons remains unchanged even when the potential difference is doubled.
2.↑ back to the question
Drift velocity $v_d = \dfrac{eE\tau}{m}$ . As temperature rises, the ions of the conductor vibrate more and collide with the free electrons more frequently, so the relaxation time $\tau$ decreases. Since $v_d \propto \tau$ , the drift velocity decreases with the rise in temperature (for a fixed applied field).
3.↑ back to the question
Drift velocity is the average velocity acquired by the free electrons of a conductor, superposed on their random thermal motion, under the influence of an externally applied electric field; it is directed opposite to the field. Relationship with current: $I = neAv_d$ , where $n$ is the free-electron density, $A$ the area of cross-section and $e$ the electronic charge.
4.↑ back to the question
$v_d = \dfrac{eE\tau}{m}$ , and since $E = V/l$ , $v_d = \dfrac{eV\tau}{ml}$
5.↑ back to the question
Drift velocity: the average velocity with which free electrons drift towards the positive end of a conductor under an applied electric field (superposed on their random thermal motion). Since $I = neAv_d$ , current density $J = \dfrac{I}{A} = nev_d$
6.↑ back to the question
Mobility ( $\mu$ ) is defined as the magnitude of the drift velocity acquired by a charge carrier per unit electric field: $\mu = |v_d|/E$ . Since $v_d = \dfrac{eE\tau}{m}$ , $\mu = \dfrac{e\tau}{m}$
7.↑ back to the question
Drift velocity: $v_d = \dfrac{eE\tau}{m}$ Current density: $J = nev_d = \dfrac{ne^2\tau}{m}E = \sigma E$ , where $\sigma = \dfrac{ne^2\tau}{m}$ Since resistivity $\rho = 1/\sigma = \dfrac{m}{ne^2\tau}$ , $J = \dfrac{E}{\rho}$
8.↑ back to the question
$v_d = \dfrac{I}{neA} = \dfrac{1.5}{9 \times 10^{28} \times 1.6 \times 10^{-19} \times 1.0 \times 10^{-7}}$ $= 1.042 \times 10^{-3}\,\text{m s}^{-1} \approx 1\,\text{mm s}^{-1}$
9.↑ back to the question
Drift velocity: the average velocity with which free electrons drift towards the positive end of a conductor under an applied field. Since $v_d = \dfrac{eV\tau}{ml}$ , (i) Slope $= \dfrac{v_d}{V} = \dfrac{e\tau}{ml}$ , i.e. the slope of the $v_d$ – $V$ line represents $e\tau/ml$ , which is inversely proportional to the length $l$ of the wire. (ii) Since slope $\propto 1/l$ , the wire with the smaller slope (line B) has the greater length. Hence wire $B$ is longer.
[Note on this record: the original figure was lost in import. Which wire is longer depends only on which line is the steeper, so that one fact has been stated in the question. Nothing else about the question or its answer has changed.]
10.↑ back to the question
Since the current $I$ is the same throughout in the steady state, and $I = JA = \dfrac{EA}{\rho}$ , i.e. $E = \dfrac{I\rho}{A}$ and $J = \dfrac{I}{A}$ : (a) As the area $A$ decreases from $A$ to $B$ , the electric field $E \left(= I\rho/A\right)$ increases from end $A$ to end $B$ . (b) Current density $J = I/A$ also increases from end $A$ to end $B$ , since $A$ decreases. (c) Mobility $\mu = e\tau/m$ depends only on the material and temperature, not on the area of cross-section. Hence the mobility of the electron remains unchanged from end $A$ to end $B$ .
11.↑ back to the question
(a) Drift velocity: the average velocity acquired by free electrons in a conductor under an applied electric field, superposed on their random thermal motion; it determines the net current in the conductor. Relaxation time: the average time elapsed between two successive collisions of a free electron with the lattice ions; it is a measure of how a conductor's resistivity depends on temperature. (b) $v_d = \dfrac{eE\tau}{m} = \dfrac{eV\tau}{mL}$ , where $V (=E)$ is the applied emf and $L$ is the conductor's length. For the new conductor of length $5L$ (with the same $V$ ): $v_d' = \dfrac{eV\tau}{m(5L)} = \dfrac{v_d}{5}$ So the drift velocity becomes one-fifth of its original value.
12.↑ back to the question
(a) Conductivity ( $\sigma$ ) is the reciprocal of resistivity, $\sigma = 1/\rho$ . SI unit: $\Omega^{-1}\text{m}^{-1}$ . (b) $v_d = \dfrac{eE\tau}{m}$ Current density $J = nev_d = \dfrac{ne^2\tau}{m}E$ Comparing with $J = \sigma E$ : $\sigma = \dfrac{ne^2\tau}{m}$ Hence the relation between current density and applied field: $J = \sigma E$
13.↑ back to the question
(a) Force on an electron, $F = -eE$ ; acceleration $a = -eE/m$ . Starting from rest after each collision (on average) and accelerating for the relaxation time $\tau$ : $v_d = a\tau = \dfrac{eE\tau}{m}$ (b) $v_d = \dfrac{eV\tau}{mL}$ . When the conductor is stretched to length $3L$ keeping $V$ constant, $v_d' = \dfrac{eV\tau}{m(3L)} = \dfrac{v_d}{3}$ So the drift velocity reduces to one-third of its original value.
14.↑ back to the question
In the steady state, charge cannot accumulate anywhere in the conductor, so the same current $I$ must flow through every cross-section. Since $I = JA = neAv_d = \dfrac{EA}{\rho}$ , and the area $A$ varies along a non-uniform conductor, the current density $J$ , drift speed $v_d$ and electric field $E$ must all vary (each is inversely proportional to $A$ ) so that their product with $A$ stays equal to the constant current $I$ . Hence only the current $I$ remains constant along the conductor.
15.↑ back to the question
(a) Consider a conductor of length $l$ , area of cross-section $A$ , with $n$ free electrons per unit volume. Total number of free electrons, $N = nAl$ . Time taken for an electron drifting with speed $v_d$ to cross the length $l$ : $t = l/v_d$ . Charge crossing a cross-section in this time: $q = Ne = nAle$ $\therefore I = \dfrac{q}{t} = \dfrac{nAle}{l/v_d} = neAv_d$ (b) Charge = area under the $I$ – $t$ graph: $Q = \dfrac{1}{2}\times 5 \times 5 + (10-5)\times 5 = 12.5 + 25 = 37.5$ C
16.↑ back to the question
(i) Mobility $\mu = |v_d|/E = e\tau/m$ . SI unit: $\text{m}^2\text{V}^{-1}\text{s}^{-1}$ . (ii) Since the current $I = neAv_d$ is the same everywhere, and the area decreases from $A$ to $B$ , the drift velocity $v_d\, (\propto 1/A)$ is smaller at $A$ (larger area) and greater at $B$ (smaller area) — i.e. drift velocity increases from $A$ to $B$ . Also $v_d \propto E$ , so the electric field is correspondingly smaller at $A$ and greater at $B$ — i.e. the electric field also increases from $A$ to $B$ . (iii) The $10\,\Omega$ (right) and $14\,\Omega$ resistors are in series: $R_{s1} = 10+14 = 24\,\Omega$ . The $10\,\Omega$ (upper) and $10\,\Omega$ (lower-right) resistors are in series: $R_{s2} = 10+10 = 20\,\Omega$ . $R_{s1}$ , $R_{s2}$ and $30\,\Omega$ are all in parallel between the battery terminals: $\dfrac{1}{R_p} = \dfrac{1}{30}+\dfrac{1}{24}+\dfrac{1}{20} = \dfrac{4+5+6}{120} = \dfrac{15}{120}$ $R_p = \dfrac{120}{15} = 8\,\Omega$ Current from battery: $I = \dfrac{V}{R_p} = \dfrac{6}{8} = \dfrac{3}{4}$ A
17.↑ back to the question
(i) The field $E$ exerts a force $F=-eE$ on each electron, giving it an acceleration $a = -eE/m$ between collisions. Since successive collisions randomize the electron's velocity, the average velocity gained (the drift velocity) is $v_d = a\tau = \dfrac{-eE\tau}{m}$ , where $\tau$ is the average time between collisions (relaxation time); the negative sign shows $v_d$ is opposite to $E$ . (ii) $n_A = 1.5\,n_B$ . Since $A$ and $B$ are in series, the current is the same: $I_A = I_B$ . $n_A eA_A v_{dA} = n_B eA_B v_{dB}$ Since the wires have the same diameter, $A_A = A_B$ , so: $\dfrac{v_{dA}}{v_{dB}} = \dfrac{n_B}{n_A} = \dfrac{n_B}{1.5\,n_B} = \dfrac{2}{3}$
18.↑ back to the question
(i) As derived above, $v_d = \dfrac{eE\tau}{m}$ . Now $I = neAv_d = \dfrac{ne^2A\tau}{m}E$ , and since $E=V/l$ : $I = \dfrac{ne^2A\tau}{ml}V \ \Rightarrow\ \dfrac{V}{I} = \dfrac{ml}{ne^2A\tau} = \dfrac{\rho l}{A} = R$ So $V = IR$ , which is Ohm's law, with $R = \dfrac{ml}{ne^2A\tau}$ constant for a given conductor at constant temperature. (ii) The rate of flow of charge through a conductor of non-uniform cross-section is the same at every section (steady state), so the electric current remains constant. Since the area of cross-section varies along the wire, the current density $J(=I/A)$ , drift speed $v_d(=J/ne)$ and electric field $E(=J\rho)$ all vary — decreasing as the area increases.
19.↑ back to the question
Drift velocity: the average velocity with which free electrons drift towards the positive end of a conductor under an applied electric field. $v_d = \dfrac{eE\tau}{m} = \dfrac{eV\tau}{ml}$ Current density: $J = \dfrac{I}{A} = nev_d = \dfrac{ne^2E\tau}{m}$
20.Answer: (B)↑ back to the question
$R = \dfrac{\rho l}{A}$ and $\rho = \dfrac{m}{ne^2\tau}$ $\therefore R = \dfrac{ml}{ne^2\tau A}$
21.Answer: (C)↑ back to the question
Answer: (c) remains unchanged Mobility $\mu = eE\tau/m \div E = e\tau/m$ depends only on the relaxation time $\tau$ and the mass and charge of the electron — not on the applied potential difference (or field).