Amit & Deepali Sinha Foundation

A Long Walk for a Short Drift

Irodov 3.208DifficultyUniqueness

Problem Irodov 3.208

A copper wire carries a current of density j=1.0j = 1.0 A/mm2^2. Taking one free electron per copper atom (copper: density 8.9×1038.9\times10^{3} kg/m3^3, molar mass 63.563.5 g/mol), estimate the total distance an electron travels while it is displaced by l=10l = 10 mm along the wire.

A length of the copper wire carrying the current density j, and the displacement l along it.

A length of the copper wire carrying the current density j\vect j, and the displacement ll along it.

Given
  • j=1.0j = 1.0 A/mm2^2
  • l=10l = 10 mm
  • one free electron per atom
  • copper: density 8.9×1038.9\times10^{3} kg/m3^3, molar mass 63.563.5 g/mol (table values)
  • room temperature, T=300T = 300 K (assumed)
Find
the distance ss the electron covers

Solution

Step 1How long the drift takes

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

n=d NAM=(8.9×103 kg/m3)(6.02×1023 mol−1)0.0635 kg/mol=8.4×1028 m−3n = \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}

The current density is the charge carried by the drift, so the drift speed and the time to drift through ll follow:

u=jent=lu=e n lj\begin{aligned} u &= \frac{j}{en}\\ t &= \frac{l}{u} = \frac{e\,n\,l}{j} \end{aligned}

Numerically the drift is very slow:

t=(1.60×10−19 C)(8.4×1028 m−3)(0.010 m)1.0×106 A/m2=1.3×102 st = \frac{(1.60\times10^{-19}~\text{C})(8.4\times10^{28}~\text{m}^{-3})(0.010~\text{m})}{1.0\times10^{6}~\text{A/m}^2} = 1.3\times10^{2}~\text{s}
Step 2The distance actually covered
Schematic: between collisions the electron flies in all directions, so its path is far longer than its net displacement l (in reality about 10^9 times longer, not the few times drawn); it starts at the dot and ends at the arrowhead.

Schematic: between collisions the electron flies in all directions, so its path is far longer than its net displacement ll (in reality about 10910^9 times longer, not the few times drawn); it starts at the dot and ends at the arrowhead.

All that time the electron is in its thermal motion, whose mean speed at the temperature TT (room temperature, assumed) is

⟨v⟩=8kTπm=8(1.38×10−23 J/K)(300 K)π (9.11×10−31 kg)=1.1×105 m/s\langle v\rangle = \sqrt{\frac{8kT}{\pi m}} = \sqrt{\frac{8(1.38\times10^{-23}~\text{J/K})(300~\text{K})}{\pi\,(9.11\times10^{-31}~\text{kg})}} = 1.1\times10^{5}~\text{m/s}

The drift adds nothing noticeable to this speed, so the distance covered is the thermal speed times the drift time:

s=⟨v⟩ t=(1.1×105 m/s)(1.3×102 s)s = \langle v\rangle\,t = (1.1\times10^{5}~\text{m/s})(1.3\times10^{2}~\text{s})
s=e n l ⟨v⟩j≈1.4×107 ms = \frac{e\,n\,l\,\langle v\rangle}{j} \approx 1.4\times10^{7}~\text{m}
(1)

That is of order 10710^{7} m, some fourteen thousand kilometres, to advance one centimetre.

Check

Dimensions: e n l/je\,n\,l/j is (C/m3^3)(m)/(A/m2^2) = C/A = s, a time, so ss is a length. The premise that the drift adds nothing to the path: the drift speed u=j/enu = j/en, about 7.4×10−57.4\times10^{-5} m/s, is some 10−910^{-9} of the thermal speed, so it changes the speed between collisions by about one part in a billion.

Remark

The drift is a minute bias on a violent random motion: the electrons race about at 10510^5 m/s while the whole electron gas creeps along at a fraction of a millimetre per second. In a metal the electrons in fact move faster still, at the Fermi speed of about 1.6×1061.6\times10^{6} m/s, so the true path is longer again by an order of magnitude, and the classical estimate above is a lower bound.