A resistor R = 6Ω is connected across a battery of emf V = 6V of negligible internal resistance, forming a sin

NCERT Exemplar · Q24

R = 6\ \Omega

(a) How much energy is absorbed by the conduction electrons in going from the initial state of no current (ignore thermal motion) to the steady state in which they move with the drift velocity?

(b) The electrons give up energy at the rate of

R I^2

per second to thermal energy. What time scale would one associate with the energy in part (a)? Take the number density of electrons

n = 10^{29}\ \text{per m}^3

, the length of the circuit

= 10\ \text{cm}

, and the cross-sectional area

A = (1\ \text{mm})^2

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Concept understanding — Drift Velocity

Drift Velocity: The Slow March of Electrons

Electrons in a metal are always moving — but randomly. At room temperature they zip around at roughly $10^6$ m/s, colliding with the lattice ions every few trillionths of a second. Without an electric field this motion cancels out: for every electron heading left another heads right, so the net velocity is zero.

Apply a battery and the field gives every electron a tiny, steady push in one direction. Between collisions the electron accelerates only briefly before smashing into an ion and losing its directed motion. What survives is a very small average velocity along the field — the drift velocity.

Note

The random thermal speed is about $10^5$ m/s, but the drift velocity is only about $10^{-4}$ m/s — about a billion times slower. An electron drifts slower than a snail, yet a lamp lights instantly, because the electric field (not the electrons) propagates at nearly the speed of light and starts every electron drifting almost at once.

The precise definition

Drift velocity ( $v_d$ ) is the average velocity acquired by the charge carriers in a conductor under an applied electric field:

v_d = \frac{eE\tau}{m}

where:

$e$ = electron charge ( $1.6 \times 10^{-19}$ C)
$E$ = electric field inside the conductor (V/m)
$\tau$ = average relaxation time — the mean time between collisions (s)
$m$ = electron mass ( $9.1 \times 10^{-31}$ kg)

v_d = \frac{eE\tau}{m}

Linking to current

Drift velocity connects the microscopic motion of electrons to the current an ammeter reads:

I = n e A v_d

where $n$ is the free-electron number density and $A$ the cross-sectional area. A larger $v_d$ means more current, but $v_d$ stays tiny because $\tau$ is tiny (about $10^{-14}$ s in copper).

Tip

For a copper wire carrying $1$ A with area $1$ mm² and $n \approx 8.5 \times 10^{28}$ m $^{-3}$ :

$v_d = \frac{I}{neA} \approx \frac{1}{(8.5\times10^{28})(1.6\times10^{-19})(10^{-6})} \approx 7 \times 10^{-5}\ \text{m/s}$ …

Why this formula?

Drift Velocity: Why the Formula Holds

Let's build this from first principles — understanding why electrons drift the way they do, not just memorizing the formula.

1. The Core Idea: What is Drift Velocity?

In a conductor, free electrons are constantly moving randomly (thermal motion, speeds ~ $10^5$ m/s). Without an electric field, their net displacement is zero — they're like a swarm of bees buzzing in all directions.

When we apply an electric field $\vec{E}$ , it gently nudges each electron in the opposite direction (since electrons are negatively charged). This small, steady net velocity superimposed on the random motion is drift velocity ( $v_d$ ).

Key insight: Drift velocity is not the speed of individual electrons — it's the average velocity of the entire electron cloud.

2. The Derivation: Step by Step

Step 1: Force on a single electron

An electron of charge $-e$ in an electric field $\vec{E}$ experiences:

\vec{F} = -e \vec{E}

The magnitude of acceleration (opposite to $\vec{E}$ ) is:

a = \frac{F}{m} = \frac{eE}{m}

where $m$ is the electron's mass.

Step 2: What happens between collisions?

Electrons don't accelerate forever — they keep colliding with atoms/ions in the metal lattice. Let the average time between collisions be $\tau$ (relaxation time). Just after a collision an electron's velocity is essentially random (zero average in the field direction); it then accelerates for time $\tau$ before the next collision.

Step 3: Drift velocity

Averaging the field-driven velocity over the relaxation time $\tau$ gives the net drift:

\boxed{v_d = \frac{eE\tau}{m}}

Here $\tau$ is the average time since the last collision, so this expression already averages over electrons at every stage between collisions — it is the standard result used in the NCERT treatment.

3. Connecting to Current: The Big Picture

Drift velocity directly gives us current density $J$ :

J = n e v_d

where $n$ = number of free electrons per unit volume. …

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The steady current is $1\ \text{A}$ , corresponding to a tiny drift speed $\sim 6\times10^{-5}\ \text{m/s}$ . Summing the drift kinetic energy of all $10^{22}$ conduction electrons gives only $\approx 1.8\times10^{-17}\ \text{J}$ . Compared with the $6\ \text{W}$ ohmic dissipation, that energy corresponds to an astonishingly short time scale $\approx 3\times10^{-18}\ \text{s}$ , far shorter than an electron's collision time — showing the drift KE is utterly negligible next to the heat continuously generated.

Given / preliminary

I = \frac{V}{R} = \frac{6}{6} = 1\ \text{A}, \qquad A = (1\ \text{mm})^2 = 10^{-6}\ \text{m}^2, \qquad L = 0.1\ \text{m}.

Drift velocity

From $I = n e A v_d$ ,

v_d = \frac{I}{n e A} = \frac{1}{10^{29}\times(1.6\times10^{-19})\times10^{-6}} = \frac{1}{1.6\times10^{4}} = 6.25\times10^{-5}\ \text{m/s}.

(a) Energy absorbed by the electrons

The number of conduction electrons in the circuit is

N = n\,A\,L = 10^{29}\times10^{-6}\times0.1 = 10^{22}.

Each acquires drift kinetic energy $\tfrac12 m v_d^2$ (with $m = 9.1\times10^{-31}\ \text{kg}$ ), so the total energy absorbed is

KE = \tfrac12 N m v_d^2 = \tfrac12 (10^{22})(9.1\times10^{-31})(6.25\times10^{-5})^2.

(6.25\times10^{-5})^2 = 3.906\times10^{-9}, \quad KE = \tfrac12 (10^{22})(9.1\times10^{-31})(3.906\times10^{-9}) \approx 1.8\times10^{-17}\ \text{J}.

(b) Associated time scale …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 13 on this concept.

KEAM 2026Set eng-2026-04174 marksMCQ
$1 \times 10^{-7}\,m^2$
›Reveal solutionSolution

Drift velocity $v_d=I/(neA)=2\times10^{-3}\ \text{m/s}=2\ \text{mm s}^{-1}$ .

The drift velocity is related to current $I$ , number density $n$ , electron charge $e$ , and cross-sectional area $A$ by $I=neAv_d$ , so:
$v_d=\frac{I}{neA}$
Substituting $I=1.6\ \text{A}$ , $n=5\times10^{28}\ \text{m}^{-3}$ , $e=1.6\times10^{-19}\ \text{C}$ , $A=1\times10^{-7}\ \text{m}^2$ : …
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KEAM 2026Set eng-2026-04194 marksMCQ
$2\\ mm^2$
›Reveal solutionSolution

Drift velocity $v=I/(nAe)\propto I/A$ ; compute the ratio for the two wires.

Drift velocity relation: $I=nAev_d\Rightarrow v_d=\dfrac{I}{nAe}$ , so for the same material $v_d\propto\dfrac{I}{A}$ .

Wire 1: current $I$ , area $2\ \text{mm}^2\Rightarrow v_1\propto\dfrac{I}{2}$ . …
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KEAM 2026Set eng-2026-04204 marksMCQ
$Q. When the electric field applied across a conductor is doubled without changing its temperature, the drift velocity (A) doubles (B) triples (C) quadruples (D) becomes half (E) remains unchanged$
›Reveal solutionSolution

$v_d\propto E$ , so doubling the field doubles the drift velocity.

Reasoning. The drift velocity is $v_d=\dfrac{eE\tau}{m}$ , where the relaxation time $\tau$ depends on temperature (held constant). Since $v_d$ is dir …
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KEAM 2026Set pha-2026-0419F4 marksMCQ
$Q. The drift velocity of electrons in a linear conductor of given length under a potential can be doubled by (A) decreasing the potential 4 times (B) increasing the potential 2 times (C) decreasing the potential 8 times (D) increasing the potential 4 times (E) decreasing the potential 2 times$
›Reveal solutionSolution

$v_d \propto V$ , so doubling the applied potential doubles the drift velocity.

The drift velocity of electrons in a conductor of length $L$ under potential difference $V$ is

$v_d = \frac{eE\tau}{m} = \frac{eV\tau}{mL}.$ …
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KEAM 2025Set eng-2025-04234 marksMCQ
$v_d$
›Reveal solutionSolution
For Ohm's law to hold, the drift velocity is directly proportional to the electric field, $v_d \propto E$ .

Concept and Intuition

Drift velocity is $v_d = \frac{eE\tau}{m}$ , linear in $E$ . This linearity is exactly what makes current proportional to voltage, i.e. Ohm's law.

Step-by-Step Solution

$v_d = \frac{eE\tau}{m}$ .

With $\tau$ constant, $v_d \propto E$ . …
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KEAM 2025Set eng-2025-04254 marksMCQ
$\tau$
›Reveal solutionSolution

Mobility is drift speed per unit field; since $v_d = eE\tau/m$ , dividing by $E$ gives $\mu = e\tau/m$ .

Drift velocity of electrons under field $E$ :
$v_d = \frac{eE\tau}{m}.$
Mobility is defined as drift speed per unit electric field: …
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KEAM 2025Set eng-2025-04284 marksMCQ
$Q. Mobility is the drift velocity per unit (A) charge (B) volume (C) electric field (D) current (E) time$
›Reveal solutionSolution

Mobility is defined as drift velocity per unit electric field: $\mu=v_d/E$ .

The mobility of a charge carrier is the magnitude of drift velocity acquired per unit applied electric field:

$\mu=\frac{|v_d|}{E},$ …
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KEAM 2025Set eng-2025-04284 marksMCQ
$A$
›Reveal solutionSolution

Resistivity $\rho=\dfrac{m}{ne^2\tau}$ , so it is directly proportional to $\dfrac{1}{n}$ .

From the free-electron model,

$\rho=\frac{m}{ne^2\tau}.$ …
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KEAM 2024Set eng-2024-06064 marksMCQ
$Q. n number of electrons flowing in a copper wire for 1 minute constitute a current of 0.5 A. Twice the number of electrons flowing through the same wire for 20 s will constitute a current of (A) 0.25 A (B) 3 A (C) 1 A (D) 1.25 A (E) 2.25 A$
›Reveal solutionSolution

Find the charge of $n$ electrons from the first case, double it, then divide by the new time.

In the first case $n$ electrons flow in 1 min (60 s) giving 0.5 A:

$Q_1=ne=I t=0.5\times60=30\text{ C}.$ …
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KEAM 2024Set eng-2024-06074 marksMCQ
$\mu$
›Reveal solutionSolution

$\mu=\dfrac{e\tau}{m}$ . …

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KEAM 2024Set eng-2024-06084 marksMCQ
$1.25\times10^{7}$
›Reveal solutionSolution

Number per second $=I/e=2/(1.6\times10^{-19})=1.25\times10^{19}$ .

Current is charge per unit time: $I = ne$ , where $n$ is the number of electrons per second and $e = 1.6\times10^{-19}$ C. …
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KEAM 2024Set eng-2024-06094 marksMCQ
$Q. Magnitude of drift velocity per unit electric field is known as (A) displacement current (B) mobility (C) electric resistance (D) electrical conductivity (E) relaxation time$
›Reveal solutionSolution

Mobility is defined as the drift speed acquired per unit applied electric field, $\mu = v_d/E$ .

For a conductor the drift velocity is $v_d = \dfrac{eE\tau}{m}$ , so the magnitude of drift velocity per unit field is

$\mu = \dfrac{|v_d|}{E} = \dfrac{e\tau}{m}.$ …
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Q.A resistor $R = 6\ \Omega$ is connected across a battery of emf $V = 6\ \text{V}$ of negligible internal resistance, forming a single loop in which a steady current $I$ flows.

Concept understanding — Drift Velocity

Drift Velocity: The Slow March of Electrons

The precise definition

Linking to current

Why this formula?

Drift Velocity: Why the Formula Holds

1. The Core Idea: What is Drift Velocity?

2. The Derivation: Step by Step

Step 1: Force on a single electron

Step 2: What happens between collisions?

Step 3: Drift velocity

3. Connecting to Current: The Big Picture

Given / preliminary

Drift velocity

(a) Energy absorbed by the electrons

(b) Associated time scale …

Method: Comparing Microscopic Drift Kinetic Energy to Macroscopic Joule Heating

Steps

Concept and Intuition

Step-by-Step Solution

Q.A resistor R=6 ΩR = 6\ \OmegaR=6 Ω is connected across a battery of emf V=6 VV = 6\ \text{V}V=6 V of negligible internal resistance, forming a single loop in which a steady current III flows.

Given / preliminary

Drift velocity

(a) Energy absorbed by the electrons

(b) Associated time scale …

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Q.A resistor $R = 6\ \Omega$ is connected across a battery of emf $V = 6\ \text{V}$ of negligible internal resistance, forming a single loop in which a steady current $I$ flows.

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