Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

So far the currents were given. We now ask what makes them flow. Consider a material of volume $V$ that is conducting. Let us suppose we have attached a generator with $(+)$ and $(-)$ terminals to this material and made the charges move (Figure 4.2.13). Then we observe a nonzero electric field in the conducting material, in contrast with the electrostatic conductor of Section 3.4, where $E = 0$ because the charges had come to rest. Also, because of the constant motion of the charges, there is a current density $J$ .

A conducting body connected to a generator. The generator maintains an electric field \vec E in the conductor, and the charges, pushed by \vec E and slowed by collisions, drift with a steady current density \vec J. — **Figure 4.2.13.** A conducting body connected to a generator. The generator maintains an electric field $E$ in the conductor, and the charges, pushed by $E$ and slowed by collisions, drift with a steady current density $J$ .

Naturally, this current density is proportional to the electric field: the charges are pushed by $E$ , and the collisions with the lattice give them a steady drift velocity proportional to the push. For most good conductors we can express this relation as

Ohm's law

J = σ E [A/ m^{2}], (4.2.30)

where $σ$ is the conductivity of the medium, in siemens per metre [S/m]. Its reciprocal $1/ σ$ is the resistivity, in [ $Ω \cdot$ m].

This is the point form of Ohm's law. The circuit form $V = R I$ follows by integrating it over a conductor: for a wire of length $l$ and cross-section $A$ in which $J$ is uniform, $V = E l$ , $I = J A$ and

R = \frac{V}{I} = \frac{l}{σ A} [Ω], (4.2.31)

which Home exercise 4.2.4 evaluates for a copper wire and Exercise 4.2.5 for a radial current in a cylinder, where $J$ is not uniform and the integration must be done properly.

From now on we classify current densities into two kinds:

the current density due to conductivity, $J_{c} = σ E$ , carried by the charges of a conductor drifting in a field, and
the convection current density, $J_{v} = ρ v$ , carried by charges that move freely, such as the electron beam of Application 3.1.4 in Section 3.1 or the charged droplets of an ink-jet printer,

J = J_{v} + J_{c} . (4.2.32)

In plain words

In a conductor the current goes where the field pushes it, and $σ$ says how willingly. Copper ( $σ = 5.8 \times 1 0^{7} S/m$ ) needs only $55 mV$ per metre to carry $3.2 A/m m^{2}$ (Home exercise 4.2.4), whereas a heater wire of nichrome ( $9.1 \times 1 0^{5} S/m$ ) needs $115 V/m$ to carry its current (Exercise 4.2.4).

Table 4.2.2. Conductivities at room temperature.

Material	$σ$ [S/m]
Silver	$6.2 \times 1 0^{7}$	contacts, plating
Copper	$5.8 \times 1 0^{7}$	wires, windings, tracks
Aluminium	$3.5 \times 1 0^{7}$	overhead lines, heat sinks
Iron	$1.0 \times 1 0^{7}$	cores (laminated against eddy currents)
Nichrome	$9.1 \times 1 0^{5}$	heating elements
Sea water	$\approx 4$
Silicon (intrinsic)	$\approx 4 \times 1 0^{- 4}$
Glass, polyethylene	$1 0^{- 12}$ to $1 0^{- 17}$	insulators

Quick check

The same current density $J = 3 A/m m^{2}$ flows in a copper wire and in an aluminium wire. In which is the electric field larger, and by how much?

$E = J / σ$ , so the field is larger in aluminium by the factor $5.8/3.5 = 1.66$ : $52 mV/m$ in copper, $86 mV/m$ in aluminium.

4.2.6 Ohm's law