Exercises — Drift velocity, mobility, conductivity

WHAT: Hum chahte hain ki ek second mein cross hone wale charges ki sankhya, times jo charge har ek carry karta hai. WHY this formula: Figure s01 dekho. Time $\Delta t$ mein har woh electron jo line se $v_d\Delta t$ distance ke andar hai, use cross kar leta hai. Yeh $A\cdot v_d\Delta t$ volume ka ek cylinder hai. $n$ (electrons per m³) se multiply karo to count milega, $e$ se multiply karo to charge milega, $\Delta t$ se divide karo to charge per second = current milega. $I=neA{v}_d$

• $n$ = carriers per m³, $e$ = charge each, $A$ = "darwaaze" ki area, $v_d$ = average electron kitni dhimi raftaar se creep karta hai.

WHAT: $v_d$ ki units ko $E$ ki units se divide karo. $v_d$ $\text{m/s}$ mein hai. Field $E$ $\text{V/m}$ mein hai. $[\mu ]=\frac{\text{m/s}}{\text{V/m}}=\frac{\text{m}}{\text{s}}\cdot \frac{\text{m}}{\text{V}}={\text{m}}^2{\text{V}}^{-1}{\text{s}}^{-1}$

WHAT: Master relation $I=neA{v}_d$ ko invert karo taaki $v_d$ isolate ho. WHY: Hum sab kuch jaante hain sivaay $v_d$ ke, toh uske liye solve karo. Pehle area convert karo: $2{\text{mm}}^2=2\times 10^{-6}{\text{m}}^2$ (kyunki $1\text{mm}=10^{-3}\text{m}$, aur area mein yeh square ho jaata hai). $v_d=\frac{I}{neA}=\frac{3}{(8.5\times 10^{28})(1.6\times 10^{-19})(2\times 10^{-6})}$ Denominator $=8.5\times 1.6\times 2\times 10^{28-19-6}=27.2\times 10^3=2.72\times 10^4$. $v_d=\frac{3}{2.72\times 10^4}\approx 1.10\times 10^{-4}\text{m/s}$ Lagbhag $0.11\text{mm/s}$ — snail speed confirm hai.

WHAT: Mobility literally drift speed per unit field hai. WHY divide: $\mu =v_d/E$ by definition — koi aur tool nahi chahiye. $\mu =\frac{v_d}{E}=\frac{1.10\times 10^{-4}}{0.05}=2.21\times 10^{-3}{\text{m}}^2{\text{V}}^{-1}{\text{s}}^{-1}$

WHAT: $\mu =e\tau /m$ ko rearrange karo taaki $\tau$ mile. WHY: $\mu$ mein pehle se $e$, $\tau$, $m$ packed hain; sirf $\tau$ unknown hai. $\tau =\frac{\mu m}{e}=\frac{(2.21\times 10^{-3})(9.1\times 10^{-31})}{1.6\times 10^{-19}}$ Numerator $=2.21\times 9.1\times 10^{-34}=20.1\times 10^{-34}=2.01\times 10^{-33}$. $\tau =\frac{2.01\times 10^{-33}}{1.6\times 10^{-19}}=1.26\times 10^{-14}\text{s}$ Lagbhag $12$ femtoseconds collisions ke beech.

WHAT: $\sigma =ne\mu$ use karo, phir $\rho =1/\sigma$. WHY $\sigma =ne\mu$ ($\tau$ wala form nahi): Hamare paas pehle se $\mu$ hai, toh yeh ek multiplication hai $\tau$ dobara derive karne ki jagah. $\sigma =ne\mu =(8.5\times 10^{28})(1.6\times 10^{-19})(2.21\times 10^{-3})$ $=8.5\times 1.6\times 2.21\times 10^{28-19-3}=30.1\times 10^6=3.01\times 10^7{\Omega }^{-1}{\text{m}}^{-1}$. $\rho =\frac{1}{\sigma }=\frac{1}{3.01\times 10^7}=3.32\times 10^{-8}\Omega \text{m}$ Real copper ka $\rho \approx 1.7\times 10^{-8}\Omega \text{m}$ hai; hamara illustrative $E$ hume factor ~2 se off karta hai, lekin method bilkul exact hai.

WHAT: $I=neA{v}_d$ se, fixed $I$, $n$, $e$ par: $v_d\propto 1/A$. WHY: Same current ka matlab hai ki same charge-per-second squeeze hona chahiye. Bada darwaza har electron ko slow chalane deta hai. $\frac{v_{d,B}}{v_{d,A}}=\frac{A_A}{A_B}=\frac{A_A}{2A_A}=\frac{1}{2}$ Wire B ke electrons aadhi raftaar se drift karte hain.

WHAT: Dono $v_d=eE\tau /m$ aur $\sigma =n{e}^2\tau /m$ directly $\tau$ ke proportional hain. WHY: $E$, $n$, $e$, $m$ constant hain, toh sirf $\tau$ answer ko move karta hai. $\frac{v_{d,\text{new}}}{v_{d,\text{old}}}=\frac{{\tau }_{\text{new}}}{{\tau }_{\text{old}}}=0.60,\frac{{\sigma }_{\text{new}}}{{\sigma }_{\text{old}}}=0.60$ Drift aur conductivity dono $60\%$ tak girte hain; resistivity $1/0.60\approx 1.67\times$ tak badhti hai. Yehi reason hai ki garam metals zyada resist karte hain — dekho Resistivity and temperature dependence.

WHAT & WHY: Har part agle ko feed karta hai — yeh topic ki poori logical chain hai.

(a) $v_d=\frac{I}{neA}$. Area $=1{\text{mm}}^2=10^{-6}{\text{m}}^2$. Denominator $=(5.86\times 10^{28})(1.6\times 10^{-19})(10^{-6})=9.376\times 10^3$. $v_d=\frac{2}{9.376\times 10^3}=2.13\times 10^{-4}\text{m/s}$

(b) $\mu =v_d/E=\frac{2.13\times 10^{-4}}{0.04}=5.33\times 10^{-3}{\text{m}}^2{\text{V}}^{-1}{\text{s}}^{-1}$.

(c) $\sigma =ne\mu =(5.86\times 10^{28})(1.6\times 10^{-19})(5.33\times 10^{-3})=5.00\times 10^7{\Omega }^{-1}{\text{m}}^{-1}$.

(d) $\tau =\frac{\mu m}{e}=\frac{(5.33\times 10^{-3})(9.1\times 10^{-31})}{1.6\times 10^{-19}}=3.03\times 10^{-14}\text{s}$.

WHAT (symbolic): $J=ne{v}_d$ se shuru karo aur drift expression swap in karo. $J=ne{v}_d=ne\cdot \frac{eE\tau }{m}=\frac{n{e}^2\tau }{m}E=\sigma E$ toh dono forms construction se identical hain — dekho Ohm's Law aur Current density.

WHAT (numeric check): L4.1 ke liye, $J=I/A=2/10^{-6}=2\times 10^6{\text{A/m}}^2$. Via $\sigma E$: $J=(5.00\times 10^7)(0.04)=2\times 10^6{\text{A/m}}^2$. ✓ Bilkul match karta hai.

(a) WHAT: creeping electron ke liye time = distance / speed. $t_{\text{electron}}=\frac{L}{v_d}=\frac{3}{1\times 10^{-4}}=3\times 10^4\text{s}\approx 8.3\text{hours}$

(b) WHAT: field signal ke liye time at speed $c$. $t_{\text{field}}=\frac{L}{c}=\frac{3}{3\times 10^8}=1\times 10^{-8}\text{s}=10\text{ns}$

(c) WHY bulb turant jalta hai: Ratio $t_{\text{electron}}/t_{\text{field}}=3\times 10^4/10^{-8}=3\times 10^{12}$ — twelve orders of magnitude. Wire pehle se electrons se bhari hui hai (jaise marbles se bhari poori pipe). Field nanoseconds mein pahunchti hai aur har electron ko ek saath nudge karti hai, toh bulb ke paas jo electron pehle se baitha hai woh turant drift karne lagta hai. Kisi ek electron ko switch se bulb tak travel nahi karna. Dekho Free electron model of metals aur Electric field inside conductors.

WHAT: $I=neA{v}_d$ ko $A$ ke liye solve karo. WHY: $I$, $n$, $e$, $v_d$ sab fixed hain; $A$ free design variable hai, aur $v_d$ ghataane se bada darwaza force hota hai. $A=\frac{I}{ne{v}_d}=\frac{10}{(8.5\times 10^{28})(1.6\times 10^{-19})(5\times 10^{-5})}$ Denominator $=8.5\times 1.6\times 5\times 10^{28-19-5}=68\times 10^4=6.8\times 10^5$. $A=\frac{10}{6.8\times 10^5}=1.47\times 10^{-5}{\text{m}}^2=14.7{\text{mm}}^2$ Kaafi moti wire — allowed drift ko aadha karne par required area double ho jaayegi (kyunki $A\propto 1/v_d$).

WHAT: $J=ne{v}_d=ne\mu E$ use karo. WHY: Same $J$ aur same $E$ ka matlab hai ki $ne\mu$ dono materials ke liye equal hona chahiye (kyunki $J=\sigma E=ne\mu E$). Bahut kam carriers ke saath, product $n\mu$ tabhi equal reh sakta hai jab $\mu$ (hence $v_d=\mu E$) semiconductor mein bahut zyada ho. $n_{\text{semi}}{\mu }_{\text{semi}}=n_{\text{Cu}}{\mu }_{\text{Cu}}\Rightarrow {\mu }_{\text{semi}}={\mu }_{\text{Cu}}\cdot \frac{n_{\text{Cu}}}{n_{\text{semi}}}\gg {\mu }_{\text{Cu}}$ Real semiconductors ki mobilities metals ki mobilities se 10–100 guna hoti hain; unki conduction chhoti carrier count se limit hoti hai, sluggish carriers se nahi. Dekho Semiconductors.

$v_d$ isolate karo current se ::: $v_d=I/(neA)$ Mobility drift aur field se ::: $\mu =v_d/E$ $\tau$ mobility se ::: $\tau =\mu m/e$ Conductivity $n,e,\mu$ se ::: $\sigma =ne\mu$ Same $I$, double area → drift ::: aadhi ho jaati hai ($v_d\propto 1/A$) Heat wire, $\tau$ gire → $\sigma$ ::: same proportion mein giregi Bulb turant kyun jalta hai ::: field $\sim c$ pe travel karti hai, saare electrons ko ek saath nudge karti hai