01The question
Q.1 A metal wire of cross-sectional area 0.5 mm2 and length 100 m is connected across a battery of e.m.f. 2 V and internal resistance 1 Ω. The density, atomic mass and electrical conductivity of the metal are 6.35 × 103 kg m−3, 63.5 gm/mole and 2 × 108 mho m−1, respectively. Assuming one conduction electron per atom of the metal, the drift velocity (in mm s−1) of the electrons in the wire is:
[Take Avogadro’s number as 6 × 1023 and charge of the electron as 1.6 × 10−19 C.]
[Take Avogadro’s number as 6 × 1023 and charge of the electron as 1.6 × 10−19 C.]
- (A)0.052
- (B)0.104
- (C)0.208
- (D)0.156
02The experiment
Target:drift velocity vd (mm s−1)
Ready.
Your guess
Free e⁻ per atom
What if…
Race time-lapse
Speed
03Detailed solution
Answer: vd = 0.208 mm s−1
- SI units. A = 0.5 mm2 = 0.5 × 10−6 m2; M = 63.5 g mol−1 = 63.5 × 10−3 kg mol−1; σ = 2 × 108 Ω−1 m−1 (1 mho = 1 S = 1 Ω−1).
- Resistance of the wire. R = L/(σA) = 100 / (2 × 108 × 0.5 × 10−6) = 1 Ω.
- Current. The wire is in series with the battery's internal resistance: I = ε/(R + r) = 2/(1 + 1) = 1 A. Only IR = 1 V of the 2 V e.m.f. appears across the wire; the other 1 V is used up inside the battery.
- Free-electron density. One cubic metre holds ρ/M moles, so n = ρNA/M × (1 electron per atom) = 6.35 × 103 × 6 × 1023 / (63.5 × 10−3) = 6 × 1028 m−3.
- Drift velocity. The charge crossing a section per second is I = neAvd, so vd = I/(neA) = 1 / (6 × 1028 × 1.6 × 10−19 × 0.5 × 10−6) = 1/4800 m s−1 = 2.08 × 10−4 m s−1 = 0.208 mm s−1: option (C).
- Check through the field. E = Vwire/L = 1/100 = 0.01 V m−1; from J = σE = nevd, vd = σE/(ne) = 2 × 108 × 0.01 / (6 × 1028 × 1.6 × 10−19) = 0.208 mm s−1. The same.
- The other options. (A) 0.052, (B) 0.104 and (D) 0.156 are a quarter, a half and three quarters of the true value. Forgetting the internal resistance gives I = 2 A and 0.417 mm s−1, which is not an option at all.
Understand the concept
Switching on a circuit sets up an electric field throughout the wire almost at once, so every free electron starts drifting together and the current appears immediately. The drift itself is extraordinarily slow: about 0.2 mm s−1 here, so one electron would need more than five days to travel the 100 m wire. The current is large because the number of free electrons is huge (6 × 1028 per cubic metre), not because they move fast. Underneath the drift, electrons move randomly at very high speeds, but that random motion averages to zero and carries no current.
Key takeaways
- R = L/(σA): conductivity is the reciprocal of resistivity.
- The current depends on the whole circuit: I = ε/(R + r); the wire feels only the terminal voltage IR.
- Free-electron density n = (ρ/M)NA × (electrons per atom).
- I = neAvd, equivalently vd = σE/(ne): typically about 10−4 m s−1 in metals.
Verification: an independent derivation; two routes that share no code with the page (the circuit route I/(neA) and the field route σE/(ne)), in exact rational arithmetic; a counting simulation of electrons crossing a section; every option audited; and the page's own engine, run by tests/verify_p2phyq01.py. The MathonGo solution was read first only for the approach. The official final answer key (JEE Advanced 2026, Paper 2), read last, gives (C), in agreement. Status: script-verified by the PrayogX simulation factory; no review by a person has been recorded.
How to use this simulation
- Press ▶ START EXPERIMENT. The switch closes, the field and current appear at once, and the magnifier opens into the wire.
- Watch the yellow tracer electron against the 1 mm scale: it shows the real drift speed. The fast zig-zag is random motion that averages out.
- The calculation fills in as each quantity is measured: R, then I, then n, then vd, which is matched to the options.
- Then try What if…: no internal resistance, a thicker wire, or the race to the end of the wire. The sliders change A, r, L and the free electrons per atom.
- Replay runs it again. Reset clears it. Classroom fills the screen for a projector.