11.1 Electric Current

Current is how much charge passes a point each second: $I = \dfrac{dq}{dt}$. Its unit is the ampere (A), and 1 A = 1 C/s. If the current is steady, this is just $I = \dfrac{\Delta q}{\Delta t}$.

Going the other way, the charge that flows past in a time interval is the area under a current-versus-time graph: $\Delta q = \displaystyle\int I\,dt$. Expect to read this area off a graph or integrate a given I(t).

Current needs two things: a closed conducting path and a potential difference to drive it. A battery or generator supplies that push, called the emf, short for electromotive force. Despite the name, emf isn't a force. It's the energy the source gives each coulomb of charge that passes through it, measured in volts and written $\mathcal{E}$.

In a metal wire, the charges that move are electrons. But circuit diagrams use conventional current, which points the way positive charge would move: out of the positive terminal of a battery, through the circuit, and into the negative terminal. Electrons actually drift the opposite way. Every rule in this unit uses conventional current, so stick with it unless a question asks about the electrons themselves.

Current has a direction, but it isn't a vector. The direction just says which way charge flows along the wire, so it follows the wire around every bend. You never split a current into x- and y-components or add currents with vector rules. At a junction, currents simply add as numbers (11.7). Current density, below, is the vector version.

Inside a conductor that carries current there is an electric field along the wire. (That's different from Unit 10, where a conductor sits in equilibrium with zero field inside.) The field pushes the free electrons, they collide constantly with the metal's atoms, and the result is a slow average drift along the wire. That average speed is the drift velocity, $v_d$.

Picture a wire with cross-sectional area A. It contains n free charge carriers per cubic meter (the charge carrier density), each with charge q, drifting at $v_d$. In a time $\Delta t$, all the carriers within a length $v_d \Delta t$ pass a point, so the current is $I = nqv_dA$.

Drift velocities in ordinary wires are tiny, often less than a millimeter per second. A light still turns on almost instantly because the electric field spreads through the circuit near the speed of light and starts every electron in the wire moving at once. The electrons near the bulb don't have to travel from the switch.

For the same current, a thinner wire has a faster drift velocity, because fewer carriers have to share the job. A material with more free carriers per cubic meter needs a slower drift for the same current.

Zero current doesn't mean the electrons are sitting still. In a wire with no current, the free electrons still zip around randomly at high speeds, but they move equally in every direction, so their net motion along the wire is zero. A current is that slow net drift added on top of the random motion.

Current density $\vec{J}$ is current per unit cross-sectional area, a vector that points along the conventional current. When the current is spread evenly, $J = \dfrac{I}{A}$, and from the drift model $\vec{J} = nq\vec{v}_d$. Its unit is A/m².

When J changes across the wire, you can't just multiply by the area. Add up the current through thin pieces instead: $I = \displaystyle\int \vec{J} \cdot d\vec{A}$. For a round wire where J depends only on the distance r from the center, the pieces are thin rings of area $dA = 2\pi r\,dr$, so $I = \displaystyle\int_0^R J(r)\,2\pi r\,dr$.

Current density is tied to the field inside the conductor: $\vec{E} = \rho\vec{J}$, where $\rho$ is the resistivity you'll meet in 11.3. A bigger field drives a denser current.