Drift velocity, mobility, conductivity: Worked examples

1.8.15 · D3 · Physics › Electromagnetism › Drift velocity, mobility, conductivity

Yeh page parent topic ka drill room hai. Parent ne tools banaye; yahan hum har tarah ke questions unpe throw karte hain. Koi bhi numbers se pehle, hum map karte hain ki "har tarah" ka matlab kya hai — taaki koi bhi case skip na ho.

Har symbol jo hum use karte hain ( $v_{d}$ , $n$ , $τ$ , $μ$ , $σ$ , $ρ$ , $J$ , $E$ , $A$ , $e$ , $m$ ) parent mein define kiya gaya tha. Agar koi bhi shaky lag raha hai, pehle parent ke sections 1–3 dobara padho.

The scenario matrix

Har problem is topic par sochna ek cell ko grid mein fill karna hai. Do cheezein hain jo problem-to-problem badlati hain: kaun sa quantity unknown hai, aur kya special situation numbers describe karti hai (normal, zero, extreme, real-world, sign-tricky). Neeche ki table har cell list karti hai. Har worked example us cell ke saath tagged hai jo wo hit karta hai.

Cell	Isme kya special hai	Example
C1 — solve for $v_{d}$	$I = n e A v_{d}$ ka seedha forward use	Ex 1
C2 — solve for $μ$ then $τ$	do definitions chain karo	Ex 2
C3 — solve for $σ, ρ$	micro ko macro se jodo	Ex 3
C4 — geometry twist (area/shape change)	$A$ changes, $I$ fixed	Ex 4 (figure)
C5 — sign & direction	electron $E$ ke against flow karta hai	Ex 5 (figure)
C6 — zero / degenerate input	$E = 0$ , ya $τ \to 0$ , ya $n \to 0$	Ex 6
C7 — limiting / temperature behaviour	$T ↑\Rightarrow τ ↓$	Ex 7
C8 — real-world word problem	full-circuit reasoning	Ex 8
C9 — exam-style twist	do wires, ratio, koi plug-in numbers nahi	Ex 9

Constants jo poore use honge (inhe yaad karo): $e = 1.6 \times 1 0^{- 19} C, m = 9.1 \times 1 0^{- 31} kg .$ Copper ke liye: $n = 8.5 \times 1 0^{28} m^{- 3}$ .

Ex 1 — Forward: drift speed nikalo · Cell C1

Step 1 — Area ko SI mein convert karo. $A = 2 mm^{2} = 2 \times (1 0^{- 3} m)^{2} = 2 \times 1 0^{- 6} m^{2} .$ Yeh step kyun? $I = n e A v_{d}$ tab hi kaam karta hai jab har quantity metres, coulombs, seconds mein ho. Raw " $2$ " in mm² $1 0^{6}$ se off hoga.

Step 2 — Unknown ke liye master relation rearrange karo. $v_{d} = \frac{I}{n e A} .$ Yeh step kyun? $v_{d}$ woh hai jo chahiye, to ise isolate karo. Right side par baaki sab given hai.

Step 3 — Plug in karo. $v_{d} = \frac{3}{( 8.5 \times 1 0 ^{28} ) ( 1.6 \times 1 0 ^{- 19} ) ( 2 \times 1 0 ^{- 6} )} .$ Denominator hai $8.5 \times 1.6 \times 2 \times 1 0^{28 - 19 - 6} = 27.2 \times 1 0^{3} = 2.72 \times 1 0^{4}$ . $v_{d} = \frac{3}{2.72 \times 1 0 ^{4}} \approx 1.1 \times 1 0^{- 4} m/s .$ Yeh step kyun? Sirf arithmetic — lekin ten ke powers dhyan se track karo; wahi par marks jaate hain.

Ex 2 — Sideways: mobility, phir relaxation time · Cell C2

Step 1 — Mobility ki definition use karo. $μ = \frac{v _{d}}{E} = \frac{1.1 \times 1 0 ^{- 4}}{8 \times 1 0 ^{- 3}} .$ Yeh step kyun? Mobility literally drift speed per field hai (parent ki definition). Koi nayi physics nahi — bas divide karo. $μ = \frac{1.1}{8} \times 1 0^{- 4 + 3} = 0.1379 \times 1 0^{- 1} \approx 1.38 \times 1 0^{- 2} m^{2} V^{- 1} s^{- 1} .$

Step 2 — $τ$ nikalo $μ = \frac{e τ}{m}$ se. $τ = \frac{μ m}{e} = \frac{( 1.38 \times 1 0 ^{- 2} ) ( 9.1 \times 1 0 ^{- 31} )}{1.6 \times 1 0 ^{- 19}} .$ Yeh step kyun? Parent ne $μ = e τ / m$ dikhaya; rearranging ek hi tarika hai $τ$ ko measurable quantities se nikalne ka ( $τ$ directly measure nahi ho sakta — it's inferred).

Step 3 — Calculate karo. Numerator $= 1.38 \times 9.1 \times 1 0^{- 2 - 31} = 12.6 \times 1 0^{- 33} = 1.26 \times 1 0^{- 32}$ . $τ = \frac{1.26 \times 1 0 ^{- 32}}{1.6 \times 1 0 ^{- 19}} \approx 7.9 \times 1 0^{- 14} s .$

Ex 3 — Micro → macro: conductivity aur resistivity · Cell C3

Step 1 — $σ = n e μ$ use karo. $σ = (8.5 \times 1 0^{28}) (1.6 \times 1 0^{- 19}) (1.38 \times 1 0^{- 2}) .$ Yeh step kyun? Yeh woh clean link hai jo parent ne derive kiya: $σ = n e μ$ ( $τ$ alag se jaanna zaroori nahi). $σ = 8.5 \times 1.6 \times 1.38 \times 1 0^{28 - 19 - 2} = 18.8 \times 1 0^{7} \approx 1.9 \times 1 0^{8} Ω^{- 1} m^{- 1} .$

Step 2 — Resistivity ke liye invert karo. $ρ = \frac{1}{σ} = \frac{1}{1.9 \times 1 0 ^{8}} \approx 5.3 \times 1 0^{- 9} Ω m .$ Yeh step kyun? Resistivity defined hai $1/ σ$ se; woh same information carry karte hain, bas flipped.

Ex 4 — Geometry twist: ek wire jo patli hoti jaati hai · Cell C4

Figure — Drift velocity, mobility, conductivity

Figure dekho: electrons ka same number per second har cross-section se guzarna chahiye (charge pile up nahi ho sakta). Jahan pipe narrow hai (teal section), har electron ko count maintain karne ke liye tezi se move karna padega — jaise garden hose mein paani jab tum end dabate ho.

Step 1 — Har section ke liye $I = n e A v_{d}$ likho. $I = n e A_{1} v_{d 1} = n e A_{2} v_{d 2} .$ Yeh step kyun? $I$ , $n$ , $e$ dono sections mein identical hain (same wire, same current, same material). Sirf $A$ aur $v_{d}$ differ karte hain.

Step 2 — Common factors cancel karo aur ratio lo. $A_{1} v_{d 1} = A_{2} v_{d 2} \Rightarrow \frac{v _{d 2}}{v _{d 1}} = \frac{A _{1}}{A _{2}} = \frac{4}{1} = 4.$ Yeh step kyun? Ratios se saare messy constants ( $n, e$ ) cancel ho jaate hain — fastest route jab question "compare" pooche.

Step 3 — Patli part mein actual value nikalo. $v_{d 2} = \frac{I}{n e A _{2}} = \frac{2}{( 8.5 \times 1 0 ^{28} ) ( 1.6 \times 1 0 ^{- 19} ) ( 1 0 ^{- 6} )} \approx 1.47 \times 1 0^{- 4} m/s,$ aur $v_{d 1} = v_{d 2} /4 \approx 3.68 \times 1 0^{- 5} m/s$ .

Ex 5 — Sign aur direction: electron kis taraf jaata hai? · Cell C5

Figure study karo — teen arrows, teen colours.

Step 1 — Force direction. Electron ka charge $- e$ hai. Force hai $F = (- e) E .$ Kyunki $E$ $+ x$ point karta hai aur charge negative hai, $F$ $- x$ ki taraf point karta hai (left, plum arrow). Yeh step kyun? Rightward vector ko negative number se multiply karna use flip karta hai. Charge ka sign hi direction rule hai.

Step 2 — Drift direction. $v_{d} = a τ = \frac{- e E}{m} τ$ same direction mein point karta hai jaise $F$ : left ( $- x$ , orange arrow). Yeh step kyun? Drift sirf accumulated acceleration hai; yeh force ki direction inherit karta hai.

Step 3 — Conventional current direction. Conventional current us direction mein flow karta hai jismein positive charge move karta — yani $E$ ke saath: right ( $+ x$ , teal arrow). Yeh step kyun? Definition se current positive-charge motion follow karta hai, jo electrons ke opposite hai.

Step 4 — Magnitudes sign-blind hain. Speed aur current magnitudes use karte hain: $v_{d} = \frac{e E τ}{m}, I = n e A v_{d},$ dono positive. Minus sign sirf direction batata hai, size kabhi nahi.

Ex 6 — Degenerate inputs: kya tootta hai aur kya nahi · Cell C6

Case (a): $E = 0$ . $v_{d} = \frac{e ( 0 ) τ}{m} = 0 \Rightarrow I = n e A \cdot 0 = 0.$ Yeh step kyun? Koi field nahi = koi sideways nudge nahi. Electrons thermally zoom karte rehte hain, lekin average zero hai (parent ki swarm-of-bees picture). Koi net current nahi — bilkul "no battery" baseline.

Case (b): $τ \to \infty$ . $v_{d} = \frac{e E τ}{m} \to \infty.$ Yeh step kyun? Agar electrons kabhi collide na karein, har ek accelerate karta rahega — drift grows without bound, $σ = \frac{n e ^{2} τ}{m} \to \infty$ , resistance $\to 0$ . Yeh idealised superconductor limit hai: collisions hi ek cheez hain jo resistance deti hain, to unhe hatane se resistance hat jaati hai.

Case (c): $n \to 0$ . $I = n e A v_{d} \to 0.$ Yeh step kyun? Koi free carriers nahi matlab kuch move karne ke liye nahi, chahe kitna bhi push karo. Yahi insulator hai. (Real Semiconductors metals aur insulators ke beech exactly isliye hain kyunki $n$ chhota hai lekin zero nahi — aur temperature ke saath badhta hai.)

Ex 7 — Limiting behaviour: wire ko garam karo · Cell C7

Step 1 — Drift velocity. $v_{d} = \frac{e E τ}{m} \propto τ$ , to $v_{d} \to \frac{v _{d}}{3} .$ Yeh step kyun? Sirf $τ$ badla aur $v_{d}$ iske directly proportional hai. Zyada garam ions zyada vibrate karte hain → collisions jaldi aati hain → drift build up karne ke liye kam time milta hai.

Step 2 — Conductivity. $σ = \frac{n e ^{2} τ}{m} \propto τ$ , to $σ \to \frac{σ}{3}, ρ = \frac{1}{σ} \to 3 ρ .$ Yeh step kyun? $τ$ par same direct-proportionality. Resistivity triple ho jaati hai — yahi parent ka " $ρ$ rises with $T$ " hai.

Step 3 — Current. Fixed voltage $V$ par, $I = \frac{V}{R}$ aur $R \propto ρ$ , to $ρ$ triple karne se $I \to \frac{I}{3} .$ Yeh step kyun? Zyada resistivity same push ke liye current ko choke kar deti hai.

Ex 8 — Real-world word problem: light-bulb delay · Cell C8

Step 1 — Ek electron ka travel time. $t = \frac{distance}{speed} = \frac{30}{1 \times 1 0 ^{- 4}} = 3 \times 1 0^{5} s .$ Yeh step kyun? Plain time = distance/speed. $3 \times 1 0^{5} s \approx 83$ hours $\approx 3.5$ days.

Step 2 — Signal time se compare karo. Electric field nearly $c = 3 \times 1 0^{8} m/s$ par spread hoti hai: $t_{signal} = \frac{30}{3 \times 1 0 ^{8}} = 1 \times 1 0^{- 7} s = 0.1 μ s .$ Yeh step kyun? Field — electron nahi — woh cheez hai jo wire ke saare electrons ko drift shuru karne ke liye bolta hai. Yeh light speed ke kareeb travel karta hai.

Step 3 — Paradox resolve karo. Wire already full hai electrons se har jagah (parent ki water-pipe picture). Switch field ko nudge karta hai, jo har electron tak $\sim 0.1 μ s$ mein pahunchti hai; saare ek saath drift karna shuru karte hain, to lamp ke paas already baitha electron turant move karta hai. Kisi ko bhi 3.5-day trip karne ki zaroorat nahi thi.

Ex 9 — Exam twist: do wires, pure ratio · Cell C9

Step 1 — Master relation se drift express karo. $v_{d} = \frac{I}{n e A} \Rightarrow v_{d} \propto \frac{1}{A} (since I, n, e dono ke liye equal hain) .$ Yeh step kyun? Same metal ( $n$ equal), same current ( $I$ equal), same charge $e$ . Sirf area different hai, to baaki sab cancel ho jaata hai — ratio problem ki pehchaan.

Step 2 — Area ko radius se relate karo. Circular cross-section ke liye $A = π r^{2}$ , to $A \propto r^{2}$ . $r_{B} = 2 r_{A}$ ke saath: $\frac{A _{B}}{A _{A}} = (\frac{r _{B}}{r _{A}})^{2} = 2^{2} = 4.$ Yeh step kyun? Radius double karna area quadruple kar deta hai — squaring wahi jagah hai jahan factor 4 (2 nahi) aata hai. Classic trap.

Step 3 — Combine karo. $\frac{v _{d, A}}{v _{d, B}} = \frac{A _{B}}{A _{A}} = 4.$ Yeh step kyun? Kyunki $v_{d} \propto 1/ A$ , chhote area wali wire (A) mein bada drift hoga. A 4× faster hai.

Recall drill

Recall Answers cover karo — kya tum bata sakte ho har ek kaunsa cell hit karta hai?

Q: Same current, wire ¼ area mein narrow hoti hai. Drift speed kaise badlegi? A: ×4 badhti hai (Ex 4, C4) — $A v_{d}$ conserved hai.
Q: Field hata di. Current? A: Zero — $v_{d} = 0$ (Ex 6, C6).
Q: Heating $τ$ ko ek third kar deti hai. Resistivity? A: Triple ho jaati hai (Ex 7, C7) — $ρ \propto 1/ τ$ .
Q: Radius double, same $I$ , same metal. Drift ratio old/new? A: 4 (Ex 9, C9) — $A \propto r^{2}$ , $v_{d} \propto 1/ A$ .
Q: Kaunsa tez hai, field ya electron, aur kitna (Ex 8)? A: Field, ~ $1 0^{12}$ times se.

Connections

← Parent topic par wapis
Current density — Ex 4 aur Ex 9 actually $J = n e v_{d}$ ke statements hain jo wire mein fixed hai.
Ohm's Law — Ex 8 ka $I = V / R$ micro picture ke saath loop close karta hai.
Resistivity and temperature dependence — Ex 7 poora yahan hai.
Semiconductors — Ex 6 ka $n \to 0$ limit yahan rehta hai.
Free electron model of metals — jahan se $n$ aur random motion aate hain.
Electric field inside conductors — Ex 5 aur Ex 8 ka field yahan hai.