NEET Physics · Current Electricity · Revision Sheet
Built around one formula wearing four faces. Worked problems, magnitude anchors, proportionality rules, and the conceptual traps that separate a metal from a semiconductor.
Every question in this topic is one of these four faces. Choosing the right one is most of the work.
Pick by whichever quantity the problem hands you
| Form | Fastest when |
|---|---|
| vd = I/(nAe) | Current and geometry are given |
| vd = j/(ne) | Current density is given |
| vd = (eτ/m)E | Relaxation time is given |
| vd = μE = μV/L | Mobility or a potential difference is given |
Bridging relations to keep at hand: j = n e vd = σE, σ = ne²τ/m, μ = eτ/m, R = ρL/A.
Drift speed is governed by the electric field, not by thickness and not by current directly. Area only ever enters through the field or through I = nAevd — never on its own.
A logarithmic ladder spanning fourteen decades. Drift speed is the slowest thing on it — slower than a snail.
Speed / m s⁻¹ · logarithmic
The ratio vdrift / vthermal is about 10⁻⁹ to 10⁻¹⁰ — a standard MCQ in its own right.
When a potential difference is applied across a cylindrical wire, the average drift speed of free electrons is V. The wire is now stretched to twice its original length and the same potential difference is applied across it. The new drift speed will be —
V / 2 V/4 was the wrong pick on the key
L′ = 2L, and volume is conserved: A·L = A′·(2L) → A′ = A/2
R′ = ρ(2L)/(A/2) = 4·(ρL/A) = 4R General rule: stretch by factor k → R grows by k²
I′ = ΔV/R′ = ΔV/4R = I/4
v′ = I′/(n A′ e) = (I/4) / (n·(A/2)·e) = (1/4)·(2)·I/(nAe) = v/2 → V/2
Use vd = (eτ/m)(ΔV/L). For a given material, drift speed depends only on the field, so only on ΔV and L. Cross-section is irrelevant. Here ΔV is unchanged and L doubles, so E halves — vd′ = V/2.
It is tempting to stop at "current becomes I/4, so drift speed becomes V/4". But the area also shrinks to A/2, which pushes drift speed back up by a factor of 2. Net effect one-half, not one-quarter.
| Quantity | Scales as |
|---|---|
| Length | k |
| Area | 1/k |
| Resistance | k² |
| Current (same ΔV) | 1/k² |
| Drift speed (same ΔV) | 1/k |
| Drift speed (same current) | k |
Attempt each before opening the solution. Between them they cover almost every drift-velocity question NEET sets.
A copper wire of cross-sectional area 2 mm² carries a current of 3.2 A. If the free electron density is 8 × 10²⁸ m⁻³, the drift speed of electrons is —
(a) 1.25 × 10⁻⁴ m/s
Formula: I = nAevd → vd = I/(nAe)
2 mm² = 2 × 10⁻⁶ m²
This is where most marks are lost: 1 mm² = 10⁻⁶ m², not 10⁻³.
v = 3.2 / (8×10²⁸ × 2×10⁻⁶ × 1.6×10⁻¹⁹)
8 × 2 × 1.6 = 25.6 exponents: 28 − 6 − 19 = 3 → 2.56 × 10⁴
v = 3.2 / 2.56×10⁴ = 1.25 × 10⁻⁴ m/s
For copper at ordinary currents, drift speed always lands around 10⁻⁴–10⁻³ m/s. If your answer comes out as 10⁻¹ or 10⁻⁸, you made a power-of-ten slip — eliminate and recheck.
A conductor of non-uniform cross-section has radius r at end A and 2r at end B. The ratio of drift speeds vA : vB is —
(d) 4 : 1
Key concept: current is the same through every cross-section (charge conservation), and n, e are fixed for one material.
v_d ∝ 1/A ∝ 1/r² v_A / v_B = A_B / A_A = (2r)² / r² = 4 : 1
Narrow section ⇒ fast electrons — the same idea as water speeding up in a pinched pipe. And since A ∝ r², the ratio is a square. The trap option is always the un-squared one, 2 : 1.
Two copper wires of the same length but radii r and 2r are connected in parallel across a battery. The ratio of drift speeds of electrons in them is —
(a) 1 : 1
Parallel → same V across both; same length L R = ρL/A, so R ∝ 1/A → I = V/R ∝ A v_d = I/(nAe) ∝ A/A = constant → 1 : 1
Use vd = (eτ/m)E with E = V/L. Both wires have the same V and the same L, hence the same field, hence the same drift speed. Area never enters.
Series (same current, different A) → vd ∝ 1/A. Parallel (same V, same L) → vd is equal. Identify which situation you are in before anything else.
In a conductor the relaxation time of free electrons is 2.5 × 10⁻¹⁴ s. An electric field of 0.1 V/m is applied. Find (i) the mobility and (ii) the drift speed. Take m = 9.1 × 10⁻³¹ kg.
μ ≈ 4.4 × 10⁻³ m²V⁻¹s⁻¹ · vd ≈ 4.4 × 10⁻⁴ m/s
Formulas: acceleration a = eE/m, so vd = aτ = (eE/m)τ, and mobility μ = vd/E = eτ/m.
μ = (1.6×10⁻¹⁹ × 2.5×10⁻¹⁴) / 9.1×10⁻³¹ numerator = 4 × 10⁻³³ μ = 4×10⁻³³ / 9.1×10⁻³¹ = 4.4 × 10⁻³ m²V⁻¹s⁻¹
v_d = μE = 4.4×10⁻³ × 0.1 = 4.4 × 10⁻⁴ m/s
Compute μ = eτ/m once, then vd = μE. Note that μ does not depend on E or on the wire's geometry — only on material and temperature. This is exactly what Q133 of the paper tested.
A bulb is 10 m from a switch. When the switch is closed the drift speed of electrons in the connecting wire is 2 × 10⁻⁴ m/s. The time taken by a given electron to travel from switch to bulb is —
(a) 5 × 10⁴ s ≈ 13.9 hours
t = distance / drift speed = 10 / (2×10⁻⁴) = 5 × 10⁴ s ≈ 13.9 hours
Because the bulb does not wait for electrons from the switch to arrive. Closing the switch sets up an electric field along the entire wire at nearly the speed of light — about 3 × 10⁻⁸ s for 10 m. Every electron everywhere in the circuit, including those already inside the filament, starts drifting essentially at once.
Signal speed ≈ c. Drift speed ≈ 10⁻⁴ m/s. Ratio ≈ 10¹². Never confuse the two.
| Pattern | Governing idea |
|---|---|
| Current density j = nevd = I/A | vd ∝ j for a given material |
| Stretching a wire | Same V ⇒ vd ∝ 1/L; same I ⇒ vd ∝ 1/A |
| Doubling temperature | τ decreases ⇒ μ and vd decrease ⇒ R increases |
| n from density and molar mass | n = (ρmass/M) × NA × (free electrons per atom) |
| Random speed vs drift speed | Thermal ~10⁵ m/s ≫ drift ~10⁻⁴ m/s |
| Assertion–Reason | "vd is tiny yet current flows immediately" — A true, R true, R is the correct explanation |
Sanity checks, memorised constants and traps, grouped by how they help you in the hall.
| Quantity | Order of magnitude | Use it to… |
|---|---|---|
| Drift speed, vd | 10⁻⁴ – 10⁻³ m/s | Reject any answer near 10⁻¹ or 10⁻⁸ |
| Thermal (random) speed | ~10⁵ m/s | Contrast question: thermal ≫ drift |
| vdrift / vthermal | ~10⁻⁹ – 10⁻¹⁰ | Classic ratio MCQ |
| Signal / field propagation | ~3 × 10⁸ m/s | Explains instant bulb glow |
| Relaxation time, τ | ~10⁻¹⁴ s | If you compute 10⁻⁶ s, you slipped |
| Mean free path (vth × τ) | 10⁻⁹ – 10⁻⁸ m | Nanometre scale, near atomic spacing |
| Mobility in copper, μ | ~10⁻³ m²V⁻¹s⁻¹ | Check your μ = eτ/m result |
| n for copper | 8.5 × 10²⁸ m⁻³ | Often not given — memorise it |
| Current density, household wire | 10⁶ – 10⁷ A/m² | Sanity check on j = I/A |
| Electric field inside a wire | 10⁻² – 10⁻¹ V/m | Very small — it is a good conductor |
| Fact | Value | Why it saves time |
|---|---|---|
| 1 A of current | 6.25 × 10¹⁸ electrons/s | Instantly answers "electrons per second" (1/1.6 = 6.25) |
| e | 1.6 × 10⁻¹⁹ C | — |
| me | 9.1 × 10⁻³¹ kg | — |
| e/m | 1.76 × 10¹¹ C/kg | Speeds up vd = (e/m)Eτ |
| 1 mm² | 10⁻⁶ m² | The single most common conversion error |
| 1 cm² | 10⁻⁴ m² | — |
| 1 mm/s | 10⁻³ m/s | Answers are often asked in mm/s |
| Trap | What to do |
|---|---|
| Diameter given, not radius | A = πd²/4, not πd² |
| Area in mm² or cm² | Convert to m² first, always |
| Answer asked in mm/s | Multiply your SI answer by 10³ |
| n not given; density and molar mass are | n = (ρ/M) × NA × (e⁻ per atom) |
| "Free electrons per atom = 2" | Multiply n by 2 — easy to skip |
| Sign of electron charge | Use magnitude in vd = I/nAe; sign only matters for direction |
Which quantity is held fixed decides everything.
| Held constant | Rule | Note |
|---|---|---|
| Same current I | vd ∝ 1/A | Narrow section ⇒ faster electrons |
| Same current I | independent of L | Length never appears in I = nAevd |
| Same potential difference V | vd ∝ 1/L, no A | Because vd = μV/L |
| Same material, same wire | vd ∝ I | Double the current, double the drift |
| Different materials, same I and A | vd ∝ 1/n | Semiconductors: small n ⇒ much larger vd |
| Wires in series | vd ∝ 1/A | Same current through both |
| Wires in parallel, same length | vd equal | The classic trap |
| Stretching, same V | vd ∝ 1/L | Halves if stretched to 2L |
| Stretching, same I | vd ∝ 1/A ∝ L | Doubles if stretched to 2L |
| Statement | Truth |
|---|---|
| "Electrons move from switch to bulb to light it" | False — the field is established at ~c; local electrons start moving at once |
| "Drift velocity increases indefinitely under E" | False — collisions reset it every τ, giving a steady vd = eEτ/m |
| "Average electron velocity is zero with the field on" | False — it is zero without a field; drift is superposed on random motion |
| "Drift speed depends on wire thickness at fixed voltage" | False — vd = μV/L, no area term |
| "Mobility depends on the applied field" | False — μ = eτ/m depends only on material and temperature |
| "Raising T raises drift speed" | False for metals — τ falls, so μ and vd fall at the same field |
| "vd is the speed of the current" | False — three different speeds: random 10⁵, drift 10⁻⁴, signal 10⁸ |
| Direction of vd | Opposite to E and opposite to conventional current |
| Sign of mobility | Always taken as positive, for electrons and holes alike |
"Raising T raises drift speed" is false for metals — but the statement needs splitting, because two different things happen for two different reasons.
Separate the two factors. Drift speed is vd = μE with μ = eτ/m. Current is I = nAevd. Temperature attacks τ and n independently. In metals only τ moves. In semiconductors both move, in opposite directions — and n wins for current, while μ still governs drift speed.
| Material | n with rising T | τ and μ with rising T | Net on ρ | Net on vd at fixed E |
|---|---|---|---|---|
| Metals Cu, Ag, Al | Constant | Falls (phonon scattering) | ρ increases, α > 0 | Decreases |
| Alloys nichrome, manganin, constantan | Constant | Nearly constant — disorder scattering dominates | ρ almost flat, α ≈ 0 | Almost unchanged |
| Semiconductors / metalloids Si, Ge | Rises exponentially | Falls, μ ∝ T−3/2 | ρ drops sharply, α < 0 | Still decreases |
| Non-metals carbon, graphite | Rises | Falls | ρ decreases, α < 0 | Decreases |
| Insulators | Rises (large Eg, needs high T) | Falls | ρ decreases | Decreases |
| Electrolytes ionic conduction | Roughly constant | Rises — viscosity drops, ions move freely | ρ decreases | Increases |
For a semiconductor at fixed applied field, drift speed still goes down with temperature. Silicon's mobility falls roughly as T−3/2, exactly like a metal, because the same lattice vibrations scatter the carriers. What makes its resistance drop is not faster electrons but vastly more of them — carrier density roughly doubles every 8–10 °C near room temperature, an exponential effect that swamps the modest mobility decline.
Heating never makes charge carriers intrinsically faster in a solid — it makes them more numerous. The only common case where heating genuinely raises drift speed is ionic / electrolytic conduction, where the medium gets less viscous.
| Material | vd at fixed current I |
|---|---|
| Metal | Essentially unchanged — n is fixed, and vd = I/nAe contains no τ |
| Semiconductor | Falls steeply — n rises exponentially, so each carrier need drift far more slowly to deliver the same I |
This is a favourite trick: at fixed current, drift speed in a metal does not care about temperature at all, because τ has vanished from the formula.
In a doped semiconductor at low temperature, scattering is dominated by ionised impurities rather than phonons, and there μ ∝ T+3/2 — mobility and drift speed genuinely rise with temperature. Mobility peaks at intermediate T, then falls once lattice scattering takes over. NEET will not test this, but it is why "μ always falls with T" is not universally true.
| Quantity | Copper | Silicon (intrinsic, 300 K) |
|---|---|---|
| Carrier density n | ~8.5 × 10²⁸ m⁻³ | ~1.5 × 10¹⁶ m⁻³ |
| Mobility μ | ~4 × 10⁻³ m²V⁻¹s⁻¹ | ~0.14 m²V⁻¹s⁻¹ (≈35× larger) |
| Drift speed at equal current density | ~10⁻⁴ m/s | enormously larger |
Semiconductor carriers are individually far more mobile than metal electrons — there are just astronomically fewer of them.