Formula
Drift velocity
v_d = -(eEτ)/m, where τ is relaxation time, e is electron charge. I = n e A v_d, where n is electron density, A is cross-sectional area.
-- NCERT Class 12 Physics, Ch. 3, p. 86Drift Velocity Mobility
8 MCQs1 revision card9-step worked example
Source: NCERT Current ElectricityPYQ coverage: NEET 2020, 2021Official key: NTA-verifiedLast updated: 21 Sep 2026
Drift Velocity Mobility, explained for NEET
The common confusion: "Drift speed is tiny, so a bulb should take hours to light." Both halves of that thought mix up two different speeds. For a copper wire carrying 1.5 A, NCERT estimates the drift speed at only about 1.1 × 10⁻³ m/s (about 1 mm/s), while the electric field propagates at about 3 × 10⁸ m/s (NCERT Class 12 Physics Part I, Chapter 3, Example 3.1, page 87). The current is set up almost at once because the field is established throughout the circuit and electrons everywhere start drifting together (Example 3.2, page 88).
Drift velocity. Free electrons move randomly; an applied field E adds a small average velocity v_d = −(eE/m)τ, where τ is the average time between collisions (Equation 3.17, page 86). The minus sign means electrons drift opposite to E.
Current from drift velocity. I = neA|v_d| (Equation 3.18, page 86), with n the free-electron density and A the cross-section. Rearranged, v_d = I/(neA): at a fixed current, a narrower section means a faster drift. Watch the area units — mm² is 10⁻⁶ m², not 10⁻³ m².
Mobility. μ = |v_d|/E (Equation 3.24, page 88), and for electrons μ = eτ/m (Equation 3.25, page 89). NCERT states that mobility is positive. SI unit: m²/(V s); 1 m²/(V s) = 10⁴ cm²/(V s).
Bridge to NEET. Questions in this area usually ask you to (a) compute v_d from I, n, A; (b) compute μ from v_d and E, or from τ; (c) compare drift speeds in parts of a conductor carrying the same current; or (d) explain why a bulb lights instantly. Do not confuse μ = eτ/m with conductivity σ = ne²τ/m (Equation 3.23, page 87).
Watch-out: μ is drift speed per unit field, not field per unit drift speed, and not drift speed per unit current. If an answer's unit comes out as V s m⁻² or m/s, the ratio was built the wrong way.
Can you answer these Drift Velocity Mobility MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
What is the SI unit of the mobility of a charge carrier?
Show answer and why every option is right or wrong
Answer: B. B is correct because μ = |v_d|/E has unit (m/s)/(V/m) = m²/(V s) (NCERT Class 12 Physics Part I, Chapter 3, page 88).
Why A is wrong: A is wrong because m/s is the unit of drift velocity itself; mobility divides that by the electric field.
Why C is wrong: C is wrong because V s m⁻² is the unit of E/v_d, the inverted ratio; mobility is v_d/E.
Why D is wrong: D is wrong because V/s results from multiplying v_d by E, (m/s)(V/m); mobility is a quotient, not a product.
MCQ 2Easy RecallPractice
For an electron of charge e and mass m with average collision time τ, the mobility is given by:
Show answer and why every option is right or wrong
Answer: D. D is correct: NCERT gives μ = eτ/m for electrons (NCERT Class 12 Physics Part I, Chapter 3, Equation 3.25, page 89).
Why A is wrong: A is wrong because m/(eτ) is the reciprocal of the mobility expression — the ratio has been inverted.
Why B is wrong: B is wrong because ne²τ/m is the conductivity σ (Equation 3.23), not the mobility; it carries the extra factor ne.
Why C is wrong: C is wrong because eEτ/m is the magnitude of the drift velocity (Equation 3.17); mobility is that divided by E.
MCQ 3Easy RecallPractice
Mobility of a charge carrier is defined as:
Show answer and why every option is right or wrong
Answer: D. D is correct: μ = |v_d|/E, the magnitude of drift velocity per unit electric field, and it is a positive quantity (NCERT Class 12 Physics Part I, Chapter 3, Equation 3.24, page 88).
Why A is wrong: A is wrong because current density per unit field, j/E, is the conductivity σ (from j = σE), not the mobility.
Why B is wrong: B is wrong because mobility is defined per unit electric field, not per unit current; drift speed per unit current would depend on n and A.
Why C is wrong: C is wrong because E/|v_d| is the inverted ratio; mobility puts drift speed in the numerator.
MCQ 4Direct ApplicationPractice
A copper wire of cross-sectional area 2.0 mm² carries a steady current of 3.2 A. The free-electron density is 8.0 × 10²⁸ m⁻³ and e = 1.6 × 10⁻¹⁹ C. The drift speed of the electrons is:
Show answer and why every option is right or wrong
Answer: C. C is correct: A = 2.0 mm² = 2.0 × 10⁻⁶ m², so v_d = I/(neA) = 3.2/(8.0 × 10²⁸ × 1.6 × 10⁻¹⁹ × 2.0 × 10⁻⁶) = 3.2/(2.56 × 10⁴) = 1.25 × 10⁻⁴ m/s (NCERT Class 12 Physics Part I, Chapter 3, Equation 3.18, page 86).
Why A is wrong: A is wrong because it treats 2.0 mm² as if it were 2.0 cm² (2.0 × 10⁻⁴ m²), making the area 100 times too large.
Why B is wrong: B is wrong because it converts 2.0 mm² as 2.0 × 10⁻³ m² (using the length factor 10⁻³ without squaring), making the area 1000 times too large.
Why D is wrong: D is wrong because it substitutes A = 2.0 with no unit conversion at all, as if the area were 2.0 m².
MCQ 5Direct ApplicationPractice
Electrons in a conductor acquire a drift speed of 2.0 × 10⁻⁴ m/s in a uniform electric field of 0.50 V/m. Their mobility is:
Show answer and why every option is right or wrong
Answer: A. A is correct: μ = |v_d|/E = (2.0 × 10⁻⁴)/(0.50) = 4.0 × 10⁻⁴ m²/(V s) (NCERT Class 12 Physics Part I, Chapter 3, Equation 3.24, page 88).
Why B is wrong: B is wrong because 2.5 × 10³ is E/v_d, the inverted ratio.
Why C is wrong: C is wrong because 1.0 × 10⁻⁴ is v_d × E; the drift speed must be divided by the field, not multiplied.
Why D is wrong: D is wrong because 4.0 is the mobility expressed in cm²/(V s) (4.0 × 10⁻⁴ m²/(V s) × 10⁴ cm²/m²) but mislabelled with the SI unit m²/(V s).
MCQ 6Direct ApplicationPractice
In a metal, the average collision time of free electrons is τ = 2.5 × 10⁻¹⁴ s. Taking e = 1.6 × 10⁻¹⁹ C and m = 9.1 × 10⁻³¹ kg, the electron mobility is approximately:
Show answer and why every option is right or wrong
Answer: C. C is correct: μ = eτ/m = (1.6 × 10⁻¹⁹ × 2.5 × 10⁻¹⁴)/(9.1 × 10⁻³¹) = (4.0 × 10⁻³³)/(9.1 × 10⁻³¹) ≈ 4.4 × 10⁻³ m²/(V s) (NCERT Class 12 Physics Part I, Chapter 3, Equation 3.25, page 89).
Why A is wrong: A is wrong because 2.3 × 10² is m/(eτ), the inverted expression.
Why B is wrong: B is wrong because 2.7 × 10¹⁶ is τ/m — the charge e has been dropped from the numerator.
Why D is wrong: D is wrong because 1.8 × 10¹¹ is e/m — the collision time τ has been dropped from the numerator.
MCQ 7CalculationPractice
A metal wire of non-uniform cross-section carries a steady current. At point P the radius of the wire is twice its radius at point Q. The ratio of electron drift speed at P to that at Q, v_P : v_Q, is:
Show answer and why every option is right or wrong
Answer: B. B is correct. Step 1: the same current I passes every section, and n and e are the same, so v_d = I/(neA) gives v_d ∝ 1/A. Step 2: A = πr², so r_P = 2r_Q gives A_P = 4A_Q. Hence v_P/v_Q = A_Q/A_P = 1/4 (NCERT Class 12 Physics Part I, Chapter 3, Equation 3.18, page 86).
Why A is wrong: A is wrong because it uses the radius ratio directly as the area ratio; area scales as r², so doubling the radius quadruples the area.
Why C is wrong: C is wrong because it treats drift speed as proportional to area; v_d = I/(neA) makes it inversely proportional.
Why D is wrong: D is wrong because it assumes equal current means equal drift speed; with I fixed, v_d depends inversely on the cross-sectional area.
MCQ 8Concept TrapPractice
The drift speed of electrons in a copper wire carrying an ordinary current is of the order of a millimetre per second, yet a bulb connected by such a wire lights almost as soon as the switch is closed. The reason is:
Show answer and why every option is right or wrong
Answer: A. A is correct: the field is established throughout the circuit almost instantly (it propagates at about 3 × 10⁸ m/s), so electrons already present in every part of the wire, including the bulb filament, acquire their average drift speed together (NCERT Class 12 Physics Part I, Chapter 3, Examples 3.1 and 3.2, pages 87–88).
Why B is wrong: B is wrong because it assigns the field's propagation speed to the electrons themselves; the electrons only drift at about a millimetre per second.
Why C is wrong: C is wrong because the drift speed stays small (about 1 mm/s for NCERT's copper-wire example); it is the field, not the drift, that spreads at about 3 × 10⁸ m/s.
Why D is wrong: D is wrong because current in a metal is carried by the drift of free electrons, I = neA|v_d|; with no drift there would be no current.
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Drift Velocity Mobility: quick recall before you leave
How do you solve a Drift Velocity Mobility question? A worked example
- 1
Given
A copper wire of cross-sectional area A = 1.0 × 10⁻⁷ m² carries a steady current I = 1.5 A. The free-electron density of copper is n = 8.5 × 10²⁸ m⁻³. Electron charge e = 1.6 × 10⁻¹⁹ C. (Data as in NCERT Class 12 Physics Part I, Chapter 3, Example 3.1, page 87.)
- 2
Required
The average drift speed v_d of the conduction electrons, and a comparison with the speed at which the electric field propagates.
- 3
Concept
The current through any cross-section equals the charge carried per second by electrons drifting through it: n electrons per unit volume, each of charge e, sweeping through area A at speed v_d.
- 4
Formula
I = neA|v_d| (Equation 3.18, page 86), rearranged to v_d = I/(neA).
- 5
Substitution
v_d = 1.5 / (8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1.0 × 10⁻⁷)
- 6
Calculation
Denominator: 8.5 × 1.6 = 13.6; powers of ten: 10²⁸ × 10⁻¹⁹ × 10⁻⁷ = 10². So neA = 13.6 × 10² = 1.36 × 10³ A s/m.
v_d = 1.5 / (1.36 × 10³) ≈ 1.10 × 10⁻³ m/s.
No exact counting numbers or mathematical constants enter this calculation; every input (I, A, n, e) is a measured or rounded quantity given to two significant figures, so the result is quoted to two significant figures. - 7
Final answer
v_d ≈ 1.1 × 10⁻³ m/s (about 1.1 mm/s). This is about 3 × 10¹¹ times smaller than the propagation speed of the electric field, about 3 × 10⁸ m/s, which NCERT gives in the same example.
- 8
Common trap
Two slips are common here. First, reporting v_d as the speed at which "electricity travels" — the bulb lights at once because the field, not the individual electrons, spreads through the circuit at about 3 × 10⁸ m/s. Second, area-unit errors: if the area were given in mm², it must be multiplied by 10⁻⁶, not 10⁻³, to get m².
- 9
Similar NEET-style question
"A wire of cross-sectional area 4.0 × 10⁻⁶ m² carries a current of 1.6 A. If the free-electron density is 5.0 × 10²⁸ m⁻³, find the drift speed (e = 1.6 × 10⁻¹⁹ C)."
Strategy: neA = 5.0 × 10²⁸ × 1.6 × 10⁻¹⁹ × 4.0 × 10⁻⁶ = 3.2 × 10⁴ A s/m, so v_d = 1.6/(3.2 × 10⁴) = 5.0 × 10⁻⁵ m/s.
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What to remember before solving Drift Velocity Mobility questions
Which Drift Velocity Mobility formulas do you need for NEET?
1 formula — click to collapse
Current from drift velocity
Macroscopic current expressed in terms of microscopic drift velocity. n = electron density.
| Symbol | Quantity | SI Unit |
|---|---|---|
| I | current | A |
| n | electron density | 1/m^3 |
| e | electron charge | C |
| A | cross-section | m^2 |
| v_d | drift velocity | m/s |
Valid when
- Steady current
- Single-carrier conductor
More in Current Electricity: 3 exam traps and mistakes · 6 formulas · 6 question patterns from its other lessons.
Drift Velocity Mobility questions from past NEET papers
2 questions from NEET 2020, 2021. Answers verified against NTA official keys. — click to collapse
NEET 2021
1(A) - (R), (B) - (Q), (C) - (P), (D) - (S)
2(A) - (R), (B) - (P), (C) - (S), (D) - (Q)
3(A) - (Q), (B) - (R), (C) - (S), (D) - (P)
4(A) - (Q), (B) - (P), (C) - (S), (D) - (R)
NTA Answer: Option 4(final)
NEET 2020
12.25 × 10⁻¹⁵
22.25 × 10¹⁵
32.5 × 10⁶
42.5 × 10⁻⁶
NTA Answer: Option 3(final)
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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