Physics Bootcamp

Subsection 35.3.1 Conductivity and Resistivity

Instead of the conductivity, \(\sigma\text{,}\) we often use the inverse of conductivity, called the resistivity, which is denoted by \(\rho\text{.}\)

Temeprature-dependence of Resistivity

The resistivity \(\rho\) (or equivalently the conductivity \(\sigma\)) of a material depends on the temperature. For most metals the dependence is nearly linear over narrow ranges of temperature. Using the value of resistivity \(\rho_0\) at some reference temperature \(T_0\) in a range we express the resistivity \(\rho\) at some other temperature \(T\) near the reference temperature as follows.

The proportionality constant \(\alpha\) is called the temperature coefficient. The temperature coefficient is small, but can be important in some applications. For instance, for copper it is \(\alpha_\text{Cu} = .00386\text{ per }^\circ\text{C}\text{.}\) If copper wire heats us by \(100^\circ\text{C}\text{,}\) you can see the resitivity change by \(38.6\%\text{.}\)

Subsection 35.3.2 Conductance and Resistance

We now work out an important consequence of Ohm's law that is very useful in the study of electric circuits.

Consider a uniform metallic wire of cross-section area \(A\) and length \(L\) that carries current \(I\text{.}\) Suppose we multiply Ohm's law in Eq. (35.3.1) by \(A\text{.}\) This will give Ohm's law in another form.

The quantity \(\sigma A /L \) is called conductance and is denoted by \(G\text{.}\) Therefore, current through the wire is proportional to the potential difference across the wire.

with conductance \(G\) of the cylindrical wire to be

Conductance depends on geometric properties of the wire, thickness given by area of cross-section \(A\) and length \(L\text{,}\) and an intrinsic property, the conductivity, \(\sigma\text{,}\) which varies between materials and also with temperature.

The inverse of conductance is called resistance, which is denoted by \(R\text{.}\) Therefore, resitance of the wire is

where I have replaced \(1/\sigma\) by \(\rho\text{,}\) the resistivity. This relation says that a thicker wire with larger area of cross-section will have lower resistance and a thinner wire of the same material and same length will have larger resistance. Additionally, a longer wire will have larger resistance than a shorter wire of the same material and the same thickness. One can understand this by noting that in a thicker wire there are more charge carriers in the cross-section, and therefore, there are more paths for carrying the current, and hence less resistance.

In terms \(R\text{,}\) Eq. (35.3.4) becomes

which is often written as

where \(V\) refers to voltage, which is same as the potential difference.

Equations (35.3.4) and (35.3.7) are also called Ohm's law. Although, we have derived these expressions for a cylindrical conductor, they are applicable fo any shape.

In SI system of units, the unit of resistance is \(\text{ohm}\) (denoted by the Greek letter \(\Omega\text{.}\) According to Ohm's law in the form given in Eq. (35.3.7), a current of one Ampere will flow through a wire of resistance one Ohm if a potential drop of one Volt is maintained between the ends.

Subsection 35.3.3 Current Through a Junction of Two Materials

Suppose you weld together a copper and a platinum wire with the same cross-section and pass a steady current through the composite conductor as shown in Figure 35.3.3. Since the same current \(I\) would flow through the two conductors of the same cross-sectional area, their volume current densities \(J\) would be equal.

That means

But since, \(\rho_{Pt} \approx 6\times \rho_{Cu}\text{,}\) i.e., \(\sigma_{Cu} \approx 6\times \sigma_{Pt}\text{.}\) That will make electric field to be lower in Cu than in Pt.

How can the electric field be different in the two metals that are welded together? You can trace the difference in the electric field to an accumulation of charges at the interface layer. As shown in Figure 35.3.3, if current goes from copper into platinum a layer of positive charges will build up at the interface. The excess charges at the interface create electric field that adds to the external electric field inside platinum and opposes inside copper.

We can use Gauss's law to deduce the charge density that builds up at the interface. Let \(\sigma_{12}\) be the charge density at the interface, and \(E_1\) and \(E_2\) be the magnitudes of the electric fields in the two wires. Now imagine a Gaussian surface in the shape of a pill box about the interface as shown in Figure 35.3.4.

Note the flux is non-zero through the ends while zero through the side. Let \(A\) be the area of cross-section of the pill box at the interface. Then from Gauss's law we obtain the following.

Hence, the charge density at the interface is proportional to the difference in the magnitudes of the electric field in the two wires.

Question: What would you expect the electric fields if you weld platinum to nichrome? See if you can figure it out.