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

Before we can say what field a current produces we need a precise description of the current itself. A current is charge in motion. Consider a region in which charge of volume density $ρ$ moves with velocity $v$ , and take a small flat surface element $Δ S$ with unit normal $\overset{n}{^}$ at a point P (Figure 4.1.3). In the time $Δ t$ every charge that was within the slanted prism of base $Δ S$ and edge $v Δ t$ passes through the element. The height of that prism is $Δ n = v \cdot \overset{n}{^} Δ t$ , its volume $Δ S v \cdot \overset{n}{^} Δ t$ , and the charge it contains is $ρ$ times that volume. The current through $Δ S$ is the charge per unit time:

I_{S} = \frac{Δ Q}{Δ t} = \frac{ρ Δ S v \cdot n ^ Δ t}{Δ t} = ρ v \cdot Δ S \overset{n}{^} . (4.1.6)

The combination $ρ v$ is all that the current through any surface depends on, and it deserves a name of its own.

Definition · Current density

The current density of charge of density $ρ$ moving with velocity $v$ is

J = ρ v [A/ m^{2}] . (4.1.7)

The current through a surface $S$ is the flux of $J$ through it,

⟹ I_{S} = J \cdot Δ S \overset{n}{^} I_{S} = \int_{S} J \cdot d S [A] . (4.1.8)

The current element. The charge that crosses the surface element \Delta S (normal \hat n) during \Delta t is the charge inside the slanted prism of base \Delta S and edge \vec v\,\Delta t, whose height is \Delta n=\vec v\cdot\hat n\,\Delta t. Dividing by \Delta t gives I_S=\rho\vec v\cdot\hat n\,\Delta S=\vec J\cdot\hat n\,\Delta S: only the component of \vec J along the normal carries charge through the element. — **Figure 4.1.3.** The current element. The charge that crosses the surface element $Δ S$ (normal $\overset{n}{^}$ ) during $Δ t$ is the charge inside the slanted prism of base $Δ S$ and edge $v Δ t$ , whose height is $Δ n = v \cdot \overset{n}{^} Δ t$ . Dividing by $Δ t$ gives $I_{S} = ρ v \cdot \overset{n}{^} Δ S = J \cdot \overset{n}{^} Δ S$ : only the component of $J$ along the normal carries charge through the element.

Equation (4.1.8) should look familiar: it is the flux integral of Section 3.2 with $D$ replaced by $J$ . The current through a surface counts how much $J$ passes through it, with the same $cos θ$ rule: a surface edge-on to the flow carries no current.

In plain words

$J$ is "amperes per square metre of cross-section", pointing in the direction in which positive charge moves. A current is a flux of $J$ , exactly as electric flux is a flux of $D$ .

Three forms of the current density are met in practice, and the problems of this section use all of them:

a volume current $J$ $[A/ m^{2}]$ , as in a thick wire or busbar (Problems 4.1.5 and 4.1.6),
a surface current $J_{S}$ $[A/m]$ flowing in a thin sheet: current per unit width across the flow, as on a rotating charged disk (Problem 4.1.3), on the winding of a solenoid (Home exercise 4.1.7) or in a flat coil (Problem 4.1.8), and
a line current $I$ $[A]$ in a thin wire (Problems 4.1.1, 4.1.2 and 4.1.4).

They are related exactly as the three charge densities of Section 3.1: a volume element $J d v^{'}$ becomes $J_{S} d S^{'}$ for a sheet and $I d l^{'}$ for a wire, where $d l^{'}$ points along the current. With the Dirac delta of Section 3.2 a thin wire on the $z$ -axis is written $J = I δ (x^{'}) δ (y^{'}) a_{z}$ , which integrates over any cross-section to $I$ .

Whenever charge moves, there is a current, even if nothing "flows" in the everyday sense. A charged disk spinning about its axis carries the convection current $J_{S} = ρ_{S} v$ , with $v = ω ρ^{'} a_{ϕ}$ (Problem 4.1.3), and a charged ring turning at $ω$ is a current loop with $I = Q ω /2 π$ (Exercise 4.1.17). Rowland showed in 1876 that such convection currents produce exactly the same magnetic field as conduction currents in a wire.

Quick check

A copper wire of cross-section $2.5$ mm $^{2}$ carries $10$ A. What is $∣ J ∣$ ? Copper has $n = 8.5 \times 1 0^{28}$ electrons per m $^{3}$ : how fast do they drift?

$J = I / A = 10/ (2.5 \times 1 0^{- 6}) = 4 \times 1 0^{6}$ A/m $^{2}$ . With $J = n q v$ , $v = J / (n e) = 4 \times 1 0^{6} / (8.5 \times 1 0^{28} \times 1.602 \times 1 0^{- 19}) = 2.9 \times 1 0^{- 4}$ m/s, less than a third of a millimetre per second (Worked example 4.1.2).

4.1.2 Current and current density