Irodov 3.209: A Month in Transit, a Meganewton Push – Amit & Deepali Sinha Foundation

Problem Irodov 3.209

A straight copper wire of length $l = 1000$ m and cross-section $S = 1.0$ mm $^2$ carries a current $I = 4.5$ A. Taking one free electron per copper atom, and for copper the density $8.9\times10^{3}$ kg/m $^3$ , the molar mass $63.5$ g/mol and the resistivity $\rho = 1.7\times10^{-8}$ $\Omega\cdot$ m, find:

the time an electron takes to drift from one end of the wire to the other
the total electric force on all the free electrons in the wire

The straight copper wire of length l carrying the current I. — The straight copper wire of length $l$ carrying the current $I$ .

Given

$l = 1000$ m
$S = 1.0$ mm $^2$
$I = 4.5$ A
one free electron per atom
copper: density $8.9\times10^{3}$ kg/m $^3$ , molar mass $63.5$ g/mol, resistivity $\rho = 1.7\times10^{-8}~\Omega\cdot$ m (table values)

Find

(a) the drift time

t

through the wire, (b) the total electric force

F

on its free electrons

Solution

Step 1The electron concentration

With one free electron per atom, the concentration of free electrons is the number of atoms per unit volume, from copper’s density $d$ and molar mass $M$ (table values):

n = \frac{d\,N_A}{M} = \frac{(8.9\times10^{3}~\text{kg/m}^3)(6.02\times10^{23}~\text{mol}^{-1})}{0.0635~\text{kg/mol}} = 8.4\times10^{28}~\text{m}^{-3} \tag{1}

Step 2(a) The drift time

The current is the charge carried through a cross-section per second by the drift, so the drift speed $u$ and the time to cover the length $l$ are

\begin{aligned} I &= e\,n\,u\,S\\ t &= \frac{l}{u} = \frac{e\,n\,l\,S}{I} \end{aligned}

Numerically, with (1):

t = \frac{(1.60\times10^{-19}~\text{C})(8.4\times10^{28}~\text{m}^{-3})(1000~\text{m})(1.0\times10^{-6}~\text{m}^2)}{4.5~\text{A}}

t = \frac{e\,n\,l\,S}{I} \approx 3.0~\text{Ms}

(2)

That is about $35$ days, at a drift speed of a third of a millimetre per second, although a lamp at the far end lights at once: the field, not the electrons, travels fast.

Step 3(b) The total electric force

By Ohm’s law in its local form, the field inside the wire is the resistivity (a table value) times the current density:

E = \rho\,j = \frac{\rho I}{S} \tag{3}

Each free electron feels the force $eE$ , and the wire holds $nSl$ of them; with (3) the total is

F = nSl\cdot eE = nSl\cdot e\,\frac{\rho I}{S}

Numerically:

F = (1.60\times10^{-19}~\text{C})(8.4\times10^{28}~\text{m}^{-3})(1000~\text{m})(1.7\times10^{-8}~\Omega\cdot\text{m})(4.5~\text{A})

F = e\,n\,l\,\rho I \approx 1.0~\text{MN}

(4)

It points against the current, the way the electrons drift.

Check

Dimensions: $\rho I$ is $\Omega\cdot$ m $\cdot$ A = V $\cdot$ m, and $enl$ is C/m $^2$ , so $F$ is C $\cdot$ V/m = J/m = N. Power balance: the force on the electrons times their drift speed is the work the field does on them each second, which must reappear as Joule heat. With $u = l/t$ , about $3.3\times10^{-4}$ m/s, and the resistance $\rho l/S$ , that is $17~\Omega$ :

\begin{aligned} Fu &= (1.0\times10^{6}~\text{N})(3.3\times10^{-4}~\text{m/s}) = 3.4\times10^{2}~\text{W}\\ I^2R &= (4.5~\text{A})^2(17~\Omega) = 3.4\times10^{2}~\text{W} \end{aligned}

Remark

A meganewton acts on the electron gas, yet the electrons do not accelerate and the wire does not move. The electrons hand the force on to the lattice in collisions, which is exactly what resistance is; and the positive ions, carrying the opposite charge in the same field, feel an equal and opposite electric force, so the wire as a whole feels none. The huge numbers are a reminder of how much charge a metal holds: over $10^{7}$ C of free electrons in one kilometre of thin wire.