PhysicsCBSE Class 12 · Current Electricity

Drift Velocity, Mobility and Resistivity Class 12: Derivations

Updated 15 October 2026 · Fact-checked

In a wire carrying current, free electrons keep their random motion but gain a small average velocity opposite to the field, the drift velocity vd = eEτ/m. This gives I = neAvd, mobility μ = vd/E = eτ/m, and resistivity ρ = m/(ne²τ), which explains Ohm's law and why metals resist more when hot.

Without a field electrons wander and go nowhere; with E each zigzag path slowly shifts opposite to E, giving vd = eEτ/m and I = neAvd.

Illustration drawn for QuestionBuddy.

Understand Drift Velocity, Mobility and Resistivity

When you switch on a lamp, it lights at once, yet the electrons in the wire crawl along at a fraction of a millimetre per second. The idea of drift velocity explains both facts and shows where Ohm's law and resistivity come from.

Free electrons without a field. A metal has a huge number of free electrons, n per unit volume (about $8.5 \times 1 0^{28}$ m⁻³ for copper). They move randomly at high speeds, colliding with the vibrating ions. Their velocities point every which way, so the average velocity is zero and there is no net current.

With a field: drift. Apply a potential difference, so a field $E$ exists in the wire. Each electron gets an acceleration $a = - \frac{e E}{m}$ between collisions. After a collision its velocity is random again, so it only keeps the extra velocity gained since its last collision. Averaging over all electrons, the average time since the last collision is the relaxation time $τ$ (about $1 0^{- 14}$ s). So the electrons acquire an average velocity, the drift velocity: $v_{d} = - \frac{e E}{m} τ$ , opposite to the field. Its size is tiny, around $1 0^{- 4}$ m/s.

Current and drift velocity. In time $Δ t$ , electrons within a length $v_{d} Δ t$ of a cross-section cross it. That is a volume $A v_{d} Δ t$ containing $n A v_{d} Δ t$ electrons, i.e. charge $n e A v_{d} Δ t$ . So $I = n e A v_{d}$ , and the current density (current per unit area) is $j = n e v_{d}$ .

Ohm's law from drift. Substituting $v_{d} = \frac{e E τ}{m}$ : $j = \frac{n e ^{2} τ}{m} E = σ E$ . The conductivity $σ = \frac{n e ^{2} τ}{m}$ and the resistivity $ρ = \frac{1}{σ} = \frac{m}{n e ^{2} τ}$ depend only on the material (n and $τ$ ), not on the shape of the wire. For a wire of length l and area A, $E = \frac{V}{l}$ and $I = j A$ , which gives $V = \frac{ρl}{A} I$ , i.e. Ohm's law with $R = \frac{ρl}{A}$ . Resistivity is measured in $Ω$ m.

Mobility. Mobility is drift velocity per unit field: $μ = \frac{∣ v _{d} ∣}{E} = \frac{e τ}{m}$ . SI unit: m² V⁻¹ s⁻¹. A material whose charge carriers have high mobility conducts well. In terms of mobility, $σ = n e μ$ .

Why resistivity rises with temperature (metals). Heating makes the ions vibrate more, so electrons collide more often; $τ$ decreases. In a metal n hardly changes, so $ρ = \frac{m}{n e ^{2} τ}$ increases. Over a limited range, $ρ_{T} = ρ_{0} [1 + α (T - T_{0})]$ , with $α$ the temperature coefficient of resistivity.

Why the lamp lights at once. The electric field is set up along the whole wire almost at the speed of light, so electrons everywhere in the circuit begin drifting together. The current does not need any one electron to travel from the switch to the lamp.

Drift Velocity, Mobility and Resistivity: formulas and key facts

Drift velocity: $v_{d} = - \frac{e E}{m} τ, v_{d} = \frac{e E τ}{m}$; Opposite to E for electrons; τ is the relaxation time.
Current and drift velocity: $I = n e A v_{d}$; n = free electrons per m³, A = cross-section area.
Current density: $j = \frac{I}{A} = n e v_{d} = σ E$; SI unit A/m²; a vector along E.
Resistivity and conductivity: $ρ = \frac{m}{n e ^{2} τ}, σ = \frac{1}{ρ} = \frac{n e ^{2} τ}{m}$; Depend on material and temperature, not on size or shape.
Mobility: $μ = \frac{v _{d}}{E} = \frac{e τ}{m}, σ = n e μ$; SI unit m² V⁻¹ s⁻¹.
Resistance of a wire: $R = \frac{ρl}{A}$; Stretching a wire to n times its length (same volume) makes R n² times.
Temperature dependence (metals): $ρ_{T} = ρ_{0} [1 + α (T - T_{0})]$; τ falls as T rises, so ρ rises.

How to solve Drift Velocity, Mobility and Resistivity questions

Use this method for drift velocity, current density and resistivity questions.

1Write n, e, A (in m²) and I or E. For a circular wire, $A = \frac{π d ^{2}}{4}$ .
2For drift velocity from current, use $v_{d} = \frac{I}{n e A}$ ; from field and relaxation time, $v_{d} = \frac{e E τ}{m}$ ; from mobility, $v_{d} = μ E$ .
3For ratio questions, decide what stays fixed: in series the current is the same, so $v_{d} \propto \frac{1}{A}$ ; with fixed V across a wire, $E = \frac{V}{l}$ .
4For resistivity or relaxation time, use $ρ = \frac{m}{n e ^{2} τ}$ and rearrange.
5If a wire is stretched, keep volume constant: $A$ falls as $l$ rises.
6Give units: m/s for $v_{d}$ , A/m² for j, $Ω$ m for $ρ$ , m² V⁻¹ s⁻¹ for μ.

Drift Velocity, Mobility and Resistivity in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Drift velocity is tiny

Suppose a phone charging cable has copper conductors of cross-section 0.5 mm² carrying 2 A. Take $n = 8.5 \times 1 0^{28}$ m⁻³ and $e = 1.6 \times 1 0^{- 19}$ C.

(a) Find the drift velocity of the electrons. (b) How long would one electron take to drift along 1 m of cable? (c) Why does charging start the moment you plug in?

Try it first, then show the solutionSolution

$v_{d} = \frac{I}{n e A} = \frac{2}{8.5 \times 1 0 ^{28} \times 1.6 \times 1 0 ^{- 19} \times 0.5 \times 1 0 ^{- 6}} = \frac{2}{6800} \approx 2.94 \times 1 0^{- 4}$ m/s.
Time for 1 m: $\frac{1}{2.94 \times 1 0 ^{- 4}} \approx 3400$ s, nearly an hour.
The field is set up along the entire cable almost at the speed of light, so electrons all along it start drifting at once; current flows immediately.

Answer: (a) about $2.9 \times 1 0^{- 4}$ m/s; (b) about 3400 s; (c) the field, not any one electron, travels fast.

Memory hook: Electrons crawl, the field flies.

Resistivity as a material property

Imagine an engineer replacing a copper wire with an aluminium wire of the same length and the same resistance. Take resistivities $1.7 \times 1 0^{- 8}$ Ω m (copper) and $2.7 \times 1 0^{- 8}$ Ω m (aluminium), and densities 8.9 g/cm³ and 2.7 g/cm³.

(a) By what factor must the aluminium wire's cross-section be larger? (b) Compare the masses of the two wires. (c) Why might aluminium be preferred for long overhead lines?

Try it first, then show the solutionSolution

Same R and l means $\frac{ρ}{A}$ is the same: $\frac{A _{A l}}{A _{C u}} = \frac{2.7}{1.7} \approx 1.59$ .
Mass = density × volume: $\frac{m _{A l}}{m _{C u}} = \frac{2.7 \times 1.59}{8.9} \approx 0.48$ .
The aluminium wire is thicker but weighs less than half as much for the same resistance, which matters for long spans held up by towers.

Answer: (a) about 1.59 times; (b) the aluminium wire has about 0.48 times the mass; (c) it is much lighter for the same resistance.

Memory hook: ρ belongs to the material; R belongs to the wire.

Tricks to remember Drift Velocity, Mobility and Resistivity

Short memory aids. Say each one aloud once, then use it on a question the same day.

I = n e A v ("I never eat a vegetable")
Current = number density × electron charge × area × drift velocity.
Use it when: Writing I = neAvd without hesitation.
Rho is 'm over n e squared tau'
Mass on top, everything that helps conduction (n, e², τ) below.
Use it when: Recalling ρ = m/(ne²τ).
Mobility: speed per field
μ = v/E, just like speed per push.
Use it when: Defining mobility and its unit.
Hot metal, short τ
More shaking ions, more collisions, shorter τ, higher ρ.
Use it when: Explaining temperature dependence.

Drift Velocity, Mobility and Resistivity: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Drift velocity in a copper wire · 2 marks · Very short answerBoard level

A copper wire of cross-section 1.0 mm² carries a current of 1.5 A. The number density of free electrons in copper is $8.5 \times 1 0^{28}$ m⁻³. Find the drift velocity of the electrons.

Try it first, then show the solutionSolution

$v_{d} = \frac{I}{n e A}$ with $A = 1.0 \times 1 0^{- 6}$ m². (1)
$v_{d} = \frac{1.5}{8.5 \times 1 0 ^{28} \times 1.6 \times 1 0 ^{- 19} \times 1 0 ^{- 6}} = \frac{1.5}{13600} \approx 1.1 \times 1 0^{- 4}$ m/s. (1)

Answer: About $1.1 \times 1 0^{- 4}$ m/s.

Example 2: Derivation: drift velocity and I = neAvd · 3 marks · Short answerBoard level

(a) Define relaxation time and drift velocity. Derive $v_{d} = \frac{e E τ}{m}$ . (b) Derive the relation between current and drift velocity.

Try it first, then show the solutionSolution

(a) Relaxation time τ: the average time between successive collisions of a free electron. Drift velocity: the average velocity of the free electrons in a conductor when a field is applied. (½)
Between collisions an electron has acceleration $a = - \frac{e E}{m}$ , so $v_{i} = u_{i} - \frac{e E}{m} t_{i}$ . Averaging over all electrons, $⟨ u_{i} ⟩ = 0$ and $⟨ t_{i} ⟩ = τ$ , so $v_{d} = - \frac{e E τ}{m}$ . (1)
(b) In time Δt, electrons within distance $v_{d} Δ t$ of a cross-section cross it: number $= n A v_{d} Δ t$ , charge $= n e A v_{d} Δ t$ . (1)
$I = \frac{Δ Q}{Δ t} = n e A v_{d}$ . (½)

Answer: $v_{d} = \frac{e E τ}{m}$ (opposite to E); $I = n e A v_{d}$ .

Example 3: Derivation: resistivity in terms of τ · 5 marks · Long answerHarder board level

(a) Starting from $I = n e A v_{d}$ , show that $ρ = \frac{m}{n e ^{2} τ}$ and hence deduce Ohm's law. (3)
(b) Copper has $ρ = 1.7 \times 1 0^{- 8}$ Ω m and $n = 8.5 \times 1 0^{28}$ m⁻³. Estimate the relaxation time. (2)

Try it first, then show the solutionSolution

(a) $j = \frac{I}{A} = n e v_{d} = n e \frac{e E τ}{m} = \frac{n e ^{2} τ}{m} E$ . (1)
So $j = σ E$ with $σ = \frac{n e ^{2} τ}{m}$ , i.e. $ρ = \frac{1}{σ} = \frac{m}{n e ^{2} τ}$ . (1)
For a wire, $E = \frac{V}{l}$ and $j = \frac{I}{A}$ : $\frac{I}{A} = \frac{V}{ρl}$ , so $V = \frac{ρl}{A} I = R I$ , which is Ohm's law (ρ is constant at a fixed temperature). (1)
(b) $τ = \frac{m}{n e ^{2} ρ} = \frac{9.1 \times 1 0 ^{- 31}}{8.5 \times 1 0 ^{28} \times ( 1.6 \times 1 0 ^{- 19} ) ^{2} \times 1.7 \times 1 0 ^{- 8}}$ . (1)
$τ \approx 2.5 \times 1 0^{- 14}$ s. (1)

Answer: (a) $ρ = \frac{m}{n e ^{2} τ}$ , leading to V = IR; (b) about $2.5 \times 1 0^{- 14}$ s.

Example 4: Wire of non-uniform cross-section · 2 marks · Very short answerHarder board level

A steady current flows through a wire whose cross-section at end P is twice that at end Q. Compare at P and Q (a) the current, (b) the current density, (c) the drift velocity and (d) the electric field.

Try it first, then show the solutionSolution

(a) The current is the same everywhere (charge does not pile up). (½)
(b) $j = \frac{I}{A}$ : $j_{Q} = 2 j_{P}$ . (½)
(c) $v_{d} = \frac{I}{n e A}$ : $v_{Q} = 2 v_{P}$ . (½)
(d) $E = ρ j$ : $E_{Q} = 2 E_{P}$ . (½)

Answer: Same current; current density, drift velocity and field are all twice as large at the narrow end Q.

Example 5: Stretching a wire · 3 marks · Short answerHOTS / competency

A wire is connected across a fixed potential difference V. It is then stretched uniformly to twice its length and connected across the same V. Find the factor by which (a) its resistance, (b) the current, and (c) the drift velocity change. Show (c) in two ways.

Try it first, then show the solutionSolution

(a) Volume is constant, so doubling l halves A: $R^{'} = \frac{ρ ( 2 l )}{A /2} = 4 R$ . (1)
(b) $I^{'} = \frac{V}{4 R} = \frac{I}{4}$ . (½)
(c) Method 1: $v_{d} = \frac{I}{n e A}$ , so $v_{d}^{'} = \frac{I /4}{n e ( A /2 )} = \frac{v _{d}}{2}$ . (1)
Method 2: $v_{d} = \frac{e E τ}{m}$ with $E = \frac{V}{l}$ ; doubling l halves E, so $v_{d}$ halves. (½)

Answer: (a) 4 times; (b) one-quarter; (c) half.

Example 6: Mobility and the speed of electrons · 4 marks · Case-basedHarder board level

Read the passage and answer.

In a metal wire 2 m long, the relaxation time of free electrons is $3.0 \times 1 0^{- 14}$ s. A potential difference of 1 V is applied across it. Take $m = 9.1 \times 1 0^{- 31}$ kg.
(a) Find the mobility of the electrons. (1)
(b) Find the drift velocity. (1)
(c) How long would an electron take to drift from one end to the other? (1)
(d) A lamp in this circuit glows the moment the switch is closed. Explain. (1)

Try it first, then show the solutionSolution

(a) $μ = \frac{e τ}{m} = \frac{1.6 \times 1 0 ^{- 19} \times 3.0 \times 1 0 ^{- 14}}{9.1 \times 1 0 ^{- 31}} \approx 5.3 \times 1 0^{- 3}$ m² V⁻¹ s⁻¹. (1)
(b) $E = \frac{1}{2} = 0.5$ V/m; $v_{d} = μ E \approx 2.64 \times 1 0^{- 3}$ m/s. (1)
(c) $t = \frac{2}{2.64 \times 1 0 ^{- 3}} \approx 760$ s, about 13 minutes. (1)
(d) The field is set up throughout the circuit almost instantly, so electrons everywhere, including those in the lamp filament, start drifting at once. (1)

Answer: (a) about $5.3 \times 1 0^{- 3}$ m² V⁻¹ s⁻¹; (b) about $2.6 \times 1 0^{- 3}$ m/s; (c) about 760 s; (d) the field acts everywhere at once.

Where students lose marks in Drift Velocity, Mobility and Resistivity

The slips examiners see most often, and what to write instead.

Mistake: Writing drift velocity along the field for electrons.
Forgetting the negative charge of the electron.
Wrong working
$v_{d} = \frac{e E τ}{m}$ along E
Right working
$v_{d} = - \frac{e E τ}{m}$ , opposite to E
Do this: Keep the minus sign in vector form; give magnitude separately.
Usually costs: ½ mark.
Mistake: Using diameter as radius in A.
Rushing the area calculation.
Wrong working
A = π(1 × 10⁻³)² for a 1 mm diameter wire
Right working
A = π(0.5 × 10⁻³)²
Do this: Write r = d/2 as a separate line.
Usually costs: The final answer mark.
Mistake: Saying resistivity changes when a wire is stretched.
Confusing R and ρ.
Do this: ρ depends on material and temperature only; R changes with l and A.
Usually costs: 1 mark in a conceptual question.
Mistake: Skipping the averaging step in the drift velocity derivation.
Jumping from v = u + at to vd = eEτ/m.
Do this: State that the average of initial random velocities is zero and the average time between collisions is τ.
Usually costs: 1 mark of the derivation.
Mistake: Thinking a smaller drift velocity means a smaller current in a non-uniform wire.
Not separating current (same everywhere) from current density and drift velocity.
Do this: In series or along one wire, I is constant; j and v_d change with area.
Usually costs: 1 mark in ratio questions.

Exam tips

Define relaxation time, drift velocity and mobility in one line each before a derivation; definitions carry marks.
In the ρ = m/(ne²τ) derivation, end by writing Ohm's law V = IR with R = ρl/A.
Ratio questions are easiest when you state what is constant (I, V or material) in the first line.
Board papers have asked drift velocity and its relation with current and mobility as 2- and 3-mark items (for example 2023 55/5/1 Q24 and 2024 55/1/1 Q23(a)).
Remember the order of magnitude: drift speed ~10⁻⁴ m/s, τ ~10⁻¹⁴ s; use it to check answers.

Practice questions on Drift Velocity, Mobility and Resistivity

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The potential difference across a metal wire is doubled at constant temperature. The drift velocity of the electrons
(a) is halved (b) is doubled (c) stays the same (d) becomes four times
Hint
$v_{d} \propto E = V / l$ .
Answer
(b) is doubled.
Q2 · 2 marks · Very short answerBoard level
Define mobility of a charge carrier. Write its SI unit and its relation with relaxation time.
Hint
Drift velocity per unit field.
Answer
Mobility is the drift velocity per unit electric field, $μ = \frac{v _{d}}{E}$ . SI unit m² V⁻¹ s⁻¹. $μ = \frac{e τ}{m}$ .
Q3 · 2 marks · Very short answerHarder board level
Two wires of the same material, with diameters in the ratio 1 : 2, are joined in series and carry a current. Find the ratio of the drift velocities in them.
Hint
Same I; area ∝ d².
Answer
Same I, n and e, so $v_{d} \propto \frac{1}{A} \propto \frac{1}{d ^{2}}$ : $v_{1} : v_{2} = 4 : 1$ .
Q4 · 2 marks · Very short answerHarder board level
A copper wire of diameter 1 mm carries 3 A. Find (a) the current density and (b) the drift velocity ( $n = 8.5 \times 1 0^{28}$ m⁻³).
Hint
$A = π r^{2}$ with r = 0.5 mm.
Answer
(a) $j = \frac{3}{π ( 0.5 \times 1 0 ^{- 3} ) ^{2}} \approx 3.82 \times 1 0^{6}$ A/m². (b) $v_{d} = \frac{j}{n e} = \frac{3.82 \times 1 0 ^{6}}{1.36 \times 1 0 ^{10}} \approx 2.8 \times 1 0^{- 4}$ m/s.
Q5 · 1 mark · Very short answerHarder board level
The mobility of free electrons in a metal is $5 \times 1 0^{- 3}$ m² V⁻¹ s⁻¹. Find their relaxation time.
Hint
τ = μm/e.
Answer
$τ = \frac{μ m}{e} = \frac{5 \times 1 0 ^{- 3} \times 9.1 \times 1 0 ^{- 31}}{1.6 \times 1 0 ^{- 19}} \approx 2.8 \times 1 0^{- 14}$ s.
Q6 · 1 mark · Assertion-reasonHOTS / competency
Assertion (A): Although the drift velocity of electrons is very small, a bulb lights almost as soon as the switch is closed.
Reason (R): The electric field is set up throughout the circuit almost at the speed of light, and electrons everywhere start drifting together.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Which travels fast, the electron or the field?
Answer
(a).
Q7 · 3 marks · Short answerHOTS / competency
Using $ρ = \frac{m}{n e ^{2} τ}$ , explain why the resistivity of a metal increases with temperature. Why does the same argument not apply to a semiconductor?
Hint
Which of n and τ changes in each material?
Answer
In a metal, n is fixed. Heating makes the ions vibrate more, so collisions are more frequent and τ decreases; hence ρ increases. In a semiconductor, n increases sharply with temperature (more electrons freed), which outweighs the fall in τ, so ρ decreases.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Expression for drift velocity?Show answer
$v_{d} = \frac{e E τ}{m}$ .
2.Relation between current and drift velocity?Show answer
$I = n e A v_{d}$ .
3.Resistivity in terms of microscopic quantities?Show answer
$ρ = \frac{m}{n e ^{2} τ}$ .
4.SI unit of mobility?Show answer
m² V⁻¹ s⁻¹.
5.Why does a metal's resistivity rise with temperature?Show answer
τ decreases because collisions with vibrating ions become more frequent.
6.Order of magnitude of drift velocity in a household wire?Show answer
About 10⁻⁴ m/s.

Drift Velocity, Mobility and Resistivity: frequently asked questions

If drift velocity is so small, how does current flow so fast?

The electric field is set up along the whole circuit almost at the speed of light. Electrons everywhere begin drifting at once, so current starts immediately even though each electron moves slowly.

Does drift velocity depend on the length of a wire?

For a fixed potential difference, yes: E = V/l, so a longer wire has a smaller field and a smaller drift velocity. For a fixed current, it depends on the area, not the length.

What is the difference between resistance and resistivity?

Resistance R depends on the material, length and area of a particular wire. Resistivity ρ = RA/l is a property of the material alone (at a given temperature).

What is relaxation time?

The average time between successive collisions of a free electron with the ions of the metal, about 10⁻¹⁴ s in metals at room temperature.

Why is the average velocity of free electrons zero without a field?

Their thermal velocities point in all directions at random, so they cancel when averaged over a large number of electrons.