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

Creating a current density due to conductivity in a conducting medium causes some energy to be dissipated. Experiments show that this dissipated energy is always observed as heat: this is the Joule phenomenon, discovered by J. P. Joule in 1841. In order to understand it, let us determine the work done to move the charges in a time interval $Δ t$ in a small volume $Δ V$ of the conductor.

The positive charges in $Δ V$ , of total charge $ρ_{+} Δ V$ and velocity $v_{+}$ , are influenced by the Lorentz force

F_{+} = ρ_{+} Δ V (E + v_{+} \times B)

and move a distance $Δ r_{+} = v_{+} Δ t$ . The magnetic part of the force is perpendicular to $v_{+}$ and therefore to $Δ r_{+}$ : it does no work. Hence, in unit time, the work done on them is

Δ W_{+} = \frac{F _{+} \cdot Δ r _{+}}{Δ t} = \frac{ρ _{+} Δ V E \cdot v _{+} Δ t}{Δ t} = (J_{+} \cdot E) Δ V,

where $J_{+} = ρ_{+} v_{+}$ is the current density of the positive carriers. Similarly for the negative charges we find $Δ W_{-} = (J_{-} \cdot E) Δ V$ : both $ρ_{-}$ and the sense of $v_{-}$ are reversed, and $J_{-} = ρ_{-} v_{-}$ points along $E$ just as $J_{+}$ does. For $J_{c} = J_{+} + J_{-}$ the total work per unit time in $Δ V$ becomes $Δ W = (J_{c} \cdot E) Δ V$ , and for the whole conductor

Joule power

P = \int_{V} (J_{c} \cdot E) d V = σ \int_{V} E^{2} d V = \int_{V} \frac{J ^{2}}{σ} d V [W], (4.2.33)

with the power density $p = J_{c} \cdot E = σ E^{2} = J^{2} / σ$ [W/m $^{3}$ ].

In plain words

The field pushes the charges, the charges hand the energy to the lattice in collisions, and the lattice warms up. The power is field times current density, summed over the volume. Only $E$ heats: the $v \times B$ part of the force does no work, because it is always perpendicular to the motion. A magnetic field can bend a current, it can never heat it.

The circuit form of Eq. (4.2.33) is the familiar $P = V I = I^{2} R$ : for the uniform wire of Eq. (4.2.31), $p A l = J E A l = (I / A) (V / l) A l = V I$ . Home exercise 4.2.4 checks this for a copper wire ( $54.9 W$ in $100 m$ carrying $10 A$ ), Exercise 4.2.4 for a heater ( $4.7 kW$ in $2 m$ of nichrome) and Exercise 4.2.5 for a radial current, where the integral over the volume has to be carried out.

The Joule phenomenon, wanted and unwanted

$p = J^{2} / σ$ is large where $σ$ is small and $J$ is high. A heater is where the Joule phenomenon is wanted: the nichrome wire of Exercise 4.2.4 dissipates $12 GW/ m^{3}$ and glows. The copper leads that feed it carry the same current, but their cross-section is chosen to keep $J$ near a few A/mm $^{2}$ , so that their $J^{2} / σ$ is some $1 0^{5}$ times smaller and they stay cool. The $I^{2} R$ of circuit theory is the Joule phenomenon integrated over the conductor.

Quick check

A copper wire and an aluminium wire of the same length and cross-section carry the same current. Which one gets hotter?

The power density is $J^{2} / σ$ with the same $J$ , so the aluminium wire dissipates $5.8/3.5 = 1.66$ times more. This is why aluminium conductors are made thicker than copper ones for the same current.

4.2.7 The Joule phenomenon