27.4 Conductivity and Resistivity

The current density $J = n_{e} e v_{d}$ is directly proportional to the electron drift speed $v_{d}$ . We earlier used the microscopic model of conduction to find that the drift speed is $v_{d} = e τ E / m$ , where $τ$ is the mean time between collisions and $m$ is the mass of an electron. Combining these, we find the current density is

J = n_{e} e v_{d} = n_{e} e (\frac{e τ E}{m}) = \frac{n_{e} e^{2} τ}{m} E

(27.15)

A photograph of a woman holding a device with two hands. — This woman is measuring her percentage body fat by gripping a device that sends a small electric current through her body. Because muscle and fat have different resistivities, the amount of current allows the fat-to-muscle ratio to be determined.

The quantity $n_{e} e^{2} τ / m$ depends only on the conducting material. According to Equation 27.15, a given electric field strength will generate a larger current density in a material with a larger electron density $n_{e}$ or longer times $τ$ between collisions than in materials with smaller values. In other words, such a material is a better conductor of current.

It makes sense, then, to define the conductivity $σ$ of a material as

σ = conductivity = \frac{n_{e} e^{2} τ}{m}

(27.16)

Conductivity, like density, characterizes a material as a whole. All pieces of copper (at the same temperature) have the same value of $σ$ , but the conductivity of copper is different from that of aluminum. Notice that the mean time between collisions $τ$ can be inferred from measured values of the conductivity.

With this definition of conductivity, Equation 27.15 becomes

J = σ E

(27.17)

This is a result of fundamental importance. Equation 27.17 tells us three things:

Current is caused by an electric field exerting forces on the charge carriers.
The current density, and hence the current $I = J A$ , depends linearly on the strength of the electric field. To double the current, you must double the strength of the electric field that pushes the charges along.
The current density also depends on the conductivity of the material. Different conducting materials have different conductivities because they have different values of the electron density and, especially, different values of the mean time between electron collisions with the lattice of atoms.

The value of the conductivity is affected by the structure of a metal, by any impurities, and by the temperature. As the temperature increases, so do the thermal vibrations of the lattice atoms. This makes them “bigger targets” and causes collisions to be more frequent, thus lowering $τ$ and decreasing the conductivity. Metals conduct better at low temperatures than at high temperatures.

For many practical applications of current it will be convenient to use the inverse of the conductivity, called the resistivity:

ρ = resistivity = \frac{1}{σ} = \frac{m}{n_{e} e^{2} τ}

(27.18)

The resistivity of a material tells us how reluctantly the electrons move in response to an electric field. TABLE 27.2 gives measured values of the resistivity and conductivity for several metals and for carbon. You can see that they vary quite a bit, with copper and silver being the best two conductors.

TABLE 27.2 Resistivity and conductivity of conducting materials

Material	Resistivity $(Ω m)$	Conductivity $(Ω^{- 1} m^{- 1})$
Aluminum	$2.8 \times 10^{- 8}$	$3.5 \times 10^{7}$
Copper	$1.7 \times 10^{- 8}$	$6.0 \times 10^{7}$
Gold	$2.4 \times 10^{- 8}$	$4.1 \times 10^{7}$
Iron	$9.7 \times 10^{- 8}$	$1.0 \times 10^{7}$
Silver	$1.6 \times 10^{- 8}$	$6.2 \times 10^{7}$
Tungsten	$5.6 \times 10^{- 8}$	$1.8 \times 10^{7}$
Nichrome^*	$1.5 \times 10^{- 6}$	$6.7 \times 10^{5}$
Carbon	$3.5 \times 10^{- 5}$	$2.9 \times 10^{4}$

^*Nickel-chromium alloy used for heating wires.

The units of conductivity, from Equation 27.17, are those of $J / E$ , namely $A C / N m^{2}$ . These are clearly awkward. In the next section we will introduce a new unit called the ohm, symbolized by $Ω$ (uppercase Greek omega). It will then turn out that resistivity has units of $Ω m$ and conductivity has units of $Ω^{- 1} m^{- 1}$ .

EXAMPLE 27.5 The electric field in a wire

A 2.0-mm-diameter aluminum wire carries a current of 800 mA. What is the electric field strength inside the wire?

SOLVE The electric field strength is

E = \frac{J}{σ} = \frac{I}{σ π r^{2}} = \frac{0.80 A}{(3.5 \times 10^{7} Ω^{- 1} m^{- 1}) π (0.0010 {m)}^{2}} = 0.0073 V / m

where the conductivity of aluminum was taken from Table 27.2.

REVIEW This is a very small field in comparison with those we calculated in Chapters 22 and 23. This calculation justifies the claim in Table 23.1 that a typical electric field strength inside a current-carrying wire is $\approx 0.01 V/m$ . It takes very few surface charges on a wire to create the weak electric field necessary to push a considerable current through the wire. The reason, once again, is the enormous value of the charge-carrier density $n_{e}$ . Even though the electric field is very tiny and the drift speed is agonizingly slow, a wire can carry a substantial current due to the vast number of charge carriers able to move.