Ohm's law — microscopic origin, resistivity: Exercises

The three tools you need are all from the parent: $ρ = \frac{m}{n e ^{2} τ}, R = ρ \frac{L}{A}, v_{d} = \frac{I}{n e A} = \frac{e τ}{m} E, J = n e v_{d} = σ E .$

Constants used throughout: electron charge magnitude $e = 1.6 \times 1 0^{- 19} C$ , electron mass $m = 9.1 \times 1 0^{- 31} kg$ .

L1 — Recognition

Problem 1.1

A copper wire has resistivity $ρ = 1.7 \times 1 0^{- 8} Ω m$ , length $L = 3 m$ , cross-section area $A = 2 \times 1 0^{- 6} m^{2}$ . Find its resistance $R$ .

Recall Solution 1.1

WHAT: we want $R$ from geometry. WHY this formula: $R = ρ L / A$ is the only one linking a material's $ρ$ to a specific shape. $R = ρ \frac{L}{A} = \frac{( 1.7 \times 1 0 ^{- 8} ) ( 3 )}{2 \times 1 0 ^{- 6}} = \frac{5.1 \times 1 0 ^{- 8}}{2 \times 1 0 ^{- 6}} = 0.0255 Ω.$ Answer: $R = 2.55 \times 1 0^{- 2} Ω$ .

Problem 1.2

Write the microscopic form of Ohm's law and name each symbol.

Recall Solution 1.2

$J = σ E .$

$J$ = current density (A/m²), current per unit area.
$σ$ = conductivity (S/m), $σ = n e^{2} τ / m$ .
$E$ = electric field inside the conductor (V/m). The law says the response $J$ is proportional to the cause $E$ , because $σ$ does not depend on $E$ .

L2 — Application

Problem 2.1

A nichrome heating element ( $ρ = 1.1 \times 1 0^{- 6} Ω m$ ) must have resistance $R = 22 Ω$ . If the wire's area is $A = 0.4 mm^{2}$ , what length $L$ do you need?

Recall Solution 2.1

WHAT: solve $R = ρ L / A$ for $L$ . WHY: all three of $R, ρ, A$ are known. First convert area: $A = 0.4 mm^{2} = 0.4 \times 1 0^{- 6} m^{2}$ . $L = \frac{R A}{ρ} = \frac{( 22 ) ( 0.4 \times 1 0 ^{- 6} )}{1.1 \times 1 0 ^{- 6}} = \frac{8.8 \times 1 0 ^{- 6}}{1.1 \times 1 0 ^{- 6}} = 8 m .$ Answer: $L = 8 m$ .

Problem 2.2

A current $I = 5 A$ flows in an aluminium wire with $n = 1.8 \times 1 0^{29} m^{- 3}$ free electrons per unit volume and $A = 2 \times 1 0^{- 6} m^{2}$ . Find the drift speed $v_{d}$ .

Recall Solution 2.2

WHAT: solve $I = n e A v_{d}$ for $v_{d}$ . WHY: current is the charge swept per second by drifting carriers. $v_{d} = \frac{I}{n e A} = \frac{5}{( 1.8 \times 1 0 ^{29} ) ( 1.6 \times 1 0 ^{- 19} ) ( 2 \times 1 0 ^{- 6} )} .$ Denominator $= 1.8 \times 1 0^{29} \times 1.6 \times 1 0^{- 19} \times 2 \times 1 0^{- 6} = 5.76 \times 1 0^{4}$ . $v_{d} = \frac{5}{5.76 \times 1 0 ^{4}} \approx 8.68 \times 1 0^{- 5} m/s .$ Answer: $v_{d} \approx 8.7 \times 1 0^{- 5} m/s$ — slower than a snail, as always.

L3 — Analysis

Problem 3.1

A copper wire is stretched (keeping its volume constant) until its length doubles. By what factor does its resistance change?

Figure — Ohm's law — microscopic origin, resistivity

Recall Solution 3.1

WHAT: find new $R$ in terms of old $R$ . WHY volume is the key constraint: stretching conserves the amount of metal, so $V = L A$ is constant — if $L$ doubles, $A$ must halve. Let old wire: length $L$ , area $A$ , so $R = ρ L / A$ . New wire: length $L^{'} = 2 L$ . Volume constant $\Rightarrow L^{'} A^{'} = L A \Rightarrow A^{'} = \frac{L A}{2 L} = \frac{A}{2}$ . $R^{'} = ρ \frac{L ^{'}}{A ^{'}} = ρ \frac{2 L}{A /2} = ρ \frac{4 L}{A} = 4 R .$ Answer: resistance becomes $4 \times$ the original. Look at the red new-wire in the figure: it is twice as long AND half as thick, and both changes push resistance up — that's why the factor is 4, not 2.

Problem 3.2

Two rods of the same material carry the same current. Rod 2 has twice the diameter of Rod 1. Compare their drift speeds.

Recall Solution 3.2

WHAT: compare $v_{d}$ using $v_{d} = I / (n e A)$ . WHY: same material means $n$ is identical; same $I$ too — so $v_{d} \propto 1/ A$ . Area $\propto$ diameter². Double the diameter $\Rightarrow$ area $\times 4$ . $\frac{v _{d, 2}}{v _{d, 1}} = \frac{A _{1}}{A _{2}} = \frac{1}{4} .$ Answer: Rod 2's electrons drift one-quarter as fast. Fatter wire = more parallel lanes, so each electron crawls slower for the same total current.

L4 — Synthesis

Problem 4.1

A wire dissipates power via Joule heating: $P = I^{2} R$ . A nichrome wire ( $ρ = 1.1 \times 1 0^{- 6} Ω m$ ), $L = 5 m$ , $A = 0.5 \times 1 0^{- 6} m^{2}$ , carries $I = 3 A$ . Find (a) $R$ , (b) the power dissipated $P$ .

Recall Solution 4.1

WHAT & WHY: get $R$ from geometry, then feed it into the Joule heating formula. (a) $R = ρ \frac{L}{A} = \frac{( 1.1 \times 1 0 ^{- 6} ) ( 5 )}{0.5 \times 1 0 ^{- 6}} = \frac{5.5 \times 1 0 ^{- 6}}{0.5 \times 1 0 ^{- 6}} = 11 Ω.$ (b) $P = I^{2} R = (3)^{2} (11) = 9 \times 11 = 99 W .$ Answers: $R = 11 Ω$ , $P = 99 W$ .

Problem 4.2

Estimate the relaxation time $τ$ for a metal with $ρ = 2.4 \times 1 0^{- 8} Ω m$ and $n = 6.0 \times 1 0^{28} m^{- 3}$ .

Recall Solution 4.2

WHAT: invert $ρ = m / (n e^{2} τ)$ . WHY: everything except $τ$ is known, and $τ$ is the microscopic quantity hiding inside a measurable $ρ$ . $τ = \frac{m}{n e ^{2} ρ} = \frac{9.1 \times 1 0 ^{- 31}}{( 6.0 \times 1 0 ^{28} ) ( 1.6 \times 1 0 ^{- 19} ) ^{2} ( 2.4 \times 1 0 ^{- 8} )} .$ Denominator: $(1.6 \times 1 0^{- 19})^{2} = 2.56 \times 1 0^{- 38}$ ; times $6.0 \times 1 0^{28} = 1.536 \times 1 0^{- 9}$ ; times $2.4 \times 1 0^{- 8} = 3.686 \times 1 0^{- 17}$ . $τ = \frac{9.1 \times 1 0 ^{- 31}}{3.686 \times 1 0 ^{- 17}} \approx 2.47 \times 1 0^{- 14} s .$ Answer: $τ \approx 2.5 \times 1 0^{- 14} s$ (~25 femtoseconds), consistent with the parent note's copper value.

L5 — Mastery

Problem 5.1

A composite conductor is made of two segments in series: segment 1 is copper ( $ρ_{1} = 1.7 \times 1 0^{- 8} Ω m$ , $L_{1} = 2 m$ , $A_{1} = 1 \times 1 0^{- 6} m^{2}$ ); segment 2 is aluminium ( $ρ_{2} = 2.8 \times 1 0^{- 8} Ω m$ , $L_{2} = 3 m$ , $A_{2} = 2 \times 1 0^{- 6} m^{2}$ ). Find the total resistance, then the ratio of the electric fields $E_{1} / E_{2}$ inside the two segments when the same current $I$ flows.

Recall Solution 5.1

Part A — total resistance. In series resistances add because the same current passes through each and the voltages stack. $R_{1} = \frac{( 1.7 \times 1 0 ^{- 8} ) ( 2 )}{1 \times 1 0 ^{- 6}} = 3.4 \times 1 0^{- 2} Ω, R_{2} = \frac{( 2.8 \times 1 0 ^{- 8} ) ( 3 )}{2 \times 1 0 ^{- 6}} = 4.2 \times 1 0^{- 2} Ω.$ $R_{tot} = R_{1} + R_{2} = 3.4 \times 1 0^{- 2} + 4.2 \times 1 0^{- 2} = 7.6 \times 1 0^{- 2} Ω.$ Part B — field ratio. The field links to current via $J = σ E = E / ρ$ , so $E = ρ J = ρ I / A$ . Since $I$ is shared: $\frac{E _{1}}{E _{2}} = \frac{ρ _{1} / A _{1}}{ρ _{2} / A _{2}} = \frac{ρ _{1} A _{2}}{ρ _{2} A _{1}} = \frac{( 1.7 \times 1 0 ^{- 8} ) ( 2 \times 1 0 ^{- 6} )}{( 2.8 \times 1 0 ^{- 8} ) ( 1 \times 1 0 ^{- 6} )} = \frac{3.4 \times 1 0 ^{- 14}}{2.8 \times 1 0 ^{- 14}} \approx 1.214.$ Answers: $R_{tot} \approx 7.6 \times 1 0^{- 2} Ω$ ; $E_{1} / E_{2} \approx 1.21$ . Note the field is not uniform along a series conductor — it jumps at the junction (red boundary in the figure) because $ρ$ and $A$ change there, even though $I$ does not.

Problem 5.2

A tungsten filament has resistivity that rises with temperature. At $2 0^{\circ} C$ its resistance is $R_{0} = 10 Ω$ ; the temperature coefficient is $α = 4.5 \times 1 0^{- 3} K^{- 1}$ , so $R = R_{0} [1 + α (T - T_{0})]$ . Find $R$ at operating temperature $T = 202 0^{\circ} C$ , and explain microscopically why it rose.

Recall Solution 5.2

WHAT: plug into the linear temperature law. WHY it's linear here: over this range $τ$ falls roughly linearly with $T$ , and $ρ \propto 1/ τ$ . $Δ T = 2020 - 20 = 2000 K .$ $R = 10 [1 + (4.5 \times 1 0^{- 3}) (2000)] = 10 [1 + 9] = 10 \times 10 = 100 Ω.$ Answer: $R = 100 Ω$ — a tenfold rise. Microscopic reason: hotter lattice ions vibrate with larger amplitude, so an electron collides more often → relaxation time $τ$ drops → $ρ = m / (n e^{2} τ)$ rises → $R = ρ L / A$ rises. This is exactly why a lamp's cold resistance is small (big inrush current at switch-on) but its hot running resistance is large.

Recall Quick self-check ledger (reveal to confirm your numbers)

1.1 ::: $R = 0.0255 Ω$ 2.1 ::: $L = 8 m$ 2.2 ::: $v_{d} \approx 8.7 \times 1 0^{- 5} m/s$ 3.1 ::: $R^{'} = 4 R$ 3.2 ::: $v_{d, 2} / v_{d, 1} = 1/4$ 4.1 ::: $R = 11 Ω$ , $P = 99 W$ 4.2 ::: $τ \approx 2.5 \times 1 0^{- 14} s$ 5.1 ::: $R_{tot} \approx 0.076 Ω$ , $E_{1} / E_{2} \approx 1.21$ 5.2 ::: $R = 100 Ω$

Connections

Ohm's law — microscopic origin, resistivity (Hinglish)
Resistors in series and parallel
Joule heating — power dissipation
Temperature dependence of resistance
Electric field inside a conductor
Drude model of conduction
Electric current and current density