Resistivity and conductivity

From the drift to Ohm's law

In the first lesson we found $v_d = eE\tau/m$ and $J = nev_d$. Put them together:

The factor in front of $E$ depends only on the material. It is the conductivity, $\sigma = ne^2\tau/m$, so that

The reciprocal of the conductivity is the resistivity:

The unit of $\rho$ is the ohm metre ($\Omega$ m). The unit of $\sigma$ is S/m, where the siemens (S) is one reciprocal ohm.

Now take a wire of length $L$ and cross section $A$, with a voltage $V$ across it. The field is $E = V/L$ and the current is $I = JA$. So

This is $V = IR$, with $R = \rho L/A$. Ohm's law is not an extra rule. It follows from the drift, as long as $\tau$ does not change with the field. The drift is a tiny change on top of the random motion, so $\tau$ stays the same, and the material is called ohmic.

Resistivity belongs to the material, at one temperature. Resistance belongs to a piece of it, and depends on its length and cross section too.

Which materials conduct well?

The resistivities below are at room temperature, in $\Omega$ m.

• Conductors. Silver $1.6 \times 10^{-8}$, copper $1.7 \times 10^{-8}$, aluminium $2.7 \times 10^{-8}$, tungsten $5.6 \times 10^{-8}$, iron $1.0 \times 10^{-7}$, nichrome $1.0 \times 10^{-6}$.
• Semiconductors. Carbon (graphite) $3.5 \times 10^{-5}$, germanium $0.46$, silicon $2300$.
• Insulators. Glass $10^{10}$ to $10^{14}$, hard rubber $10^{13}$ to $10^{16}$.

The spread is huge. Look at $\sigma = ne^2\tau/m$. A metal has about $10^{29}$ free electrons in each cubic metre. A semiconductor has far fewer, and an insulator has almost none.

How often do the electrons hit the ions?

Turn the formula round: $\tau = \dfrac{m}{ne^2\rho}$. For copper, the bottom is $8.5 \times 10^{28} \times (1.6 \times 10^{-19})^2 \times 1.7 \times 10^{-8} = 3.7 \times 10^{-17}$ in SI units. So

So each electron has about $4 \times 10^{13}$ hits every second. This is the number we used in the first lesson.

There is another way to write the conductivity. Since $\mu = e\tau/m$, we get $\sigma = ne\mu$. For copper, $8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 4.4 \times 10^{-3} \approx 6 \times 10^7$ S/m. This is $1/\rho$ for copper, within rounding.

The resistance of a wire

For a wire of resistivity $\rho$, length $L$ and cross section $A$:

A round wire of diameter $d$ has $A = \pi d^2/4$. Take care: the area uses the square of the diameter.

A worked example. A heater uses a nichrome wire of diameter $0.40$ mm. How long must the wire be to give $24\ \Omega$?

1. The area. $A = \dfrac{\pi d^2}{4} = \dfrac{\pi \times (0.40 \times 10^{-3}\ \text{m})^2}{4} = 1.26 \times 10^{-7}\ \text{m}^2$.
2. The length. We want $L$, so $L = \dfrac{RA}{\rho} = \dfrac{24\ \Omega \times 1.26 \times 10^{-7}\ \text{m}^2}{1.0 \times 10^{-6}\ \Omega\,\text{m}} = 3.0$ m.

What if the wire is drawn thinner? A wire is pulled through a die until its diameter is half. Its volume stays the same, so its length becomes $4$ times larger and its area $4$ times smaller. So $R$ becomes $4 \times 4 = 16$ times larger.

What if a wire is cut in pieces? A wire of $36\ \Omega$ is cut into $3$ equal pieces. Each piece has $12\ \Omega$. Joined side by side, the three pieces give $4\ \Omega$. In general, $n$ pieces side by side give $R/n^2$.

Make the wire thin and the metal a poor conductor. How does the resistance compare with a copper wire of the same size?

Practise with new numbers

Try it in Python

The program finds the resistance of a wire of the same size, made of six different metals. It also gives the conductivity of each.

Common mistakes

• Mixing up resistivity and resistance. Resistivity belongs to the material. Resistance belongs to the wire, and depends on its size.
• Using the diameter as if it were the radius. $A = \pi d^2/4$, or $\pi r^2$ with the radius $r = d/2$.
• Forgetting that $1\ \text{mm}^2 = 10^{-6}\ \text{m}^2$.
• Saying a longer wire has a larger resistivity. The resistivity is the same. The resistance is larger.
• Thinking Ohm's law is a separate rule. It follows from the drift, when $\tau$ does not change with the field.

Remember these

• $J = \sigma E$, with $\sigma = \dfrac{ne^2\tau}{m} = ne\mu$, and $\rho = \dfrac{1}{\sigma}$.
• $R = \rho\dfrac{L}{A}$. A round wire has $A = \dfrac{\pi d^2}{4}$.
• Resistivity: $\rho$ in $\Omega$ m. Conductivity: $\sigma$ in S/m.
• Metals: $\rho$ about $10^{-8}\ \Omega$ m. Semiconductors: a wide middle range. Insulators: $10^{10}\ \Omega$ m or more.
• Cutting a wire into $n$ pieces and joining them side by side gives $R/n^2$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Wire B is made of the same metal as wire A. It is twice as long and has twice the diameter. How does the resistance of B compare with that of A?

2. Why does a thick copper wire resist the current less than a thin one of the same length?

3. A uniform wire of resistivity $\rho$ is cut in half. What happens to the resistivity, and to the resistance of each half?

4. A wire of $18\ \Omega$ is cut into $3$ equal pieces. The pieces are joined side by side. What is the resistance now?