UY1: Current, Drift Velocity And Current Density

This is meant to be a refresher course on current for undergrads.

An electric current consists of charges in motion from one region to another. The current I through a cross-sectional area A is the net charge flowing through the area per unit time.

$$I = \frac{dQ}{dt}$$


Current In Metal

In a metal, each atom in the metallic crystal gives up one or more of its outer electrons. These free electrons move through the crystal, colliding at intervals with the massive, nearly stationary positive ions. If there is no electric field, the electrons move in straight lines between collisions, the directions of their velocities are random, and on average they never get anywhere. There is no net flow of charge in any direction and hence no current – electrostatic conditions.

If a constant, steady electric field $\vec{E}$ is present, the paths curve slightly because of the acceleration caused by electric field forces.

The average time between collisions is called the mean free time, denoted by $\tau$.

Suppose at time $t \leq 0$, an electron has velocity $\vec{v}_{0}$ and the average velocity of all electrons, $(\vec{v}_{0})_{av}) = 0$. At time t = 0, a constant, steady electric field $\vec{E}$ is turned on. The electric field $\vec{E}$, established inside the metal, exerts a steady force on each free electron inside the metal, which causes a steady acceleration in the direction of the force.

$$\begin{aligned} \vec{F} &= q \vec{E} \\ \vec{a} &= \frac{q \vec{E}}{m} \end{aligned}$$

where m is the electron mass.

At time $t = \tau$, $\vec{v} = \vec{v}_{0} + \vec{a} \tau$.

The velocity of an “average electron” at time $t = \tau$ is $\vec{v}_{av} = (\vec{v}_{0})_{av} + \vec{a} \tau$, Since $(\vec{v}_{0})_{av}) = 0$,

$$\vec{v}_{av} = \vec{a} \tau$$

$\vec{v}_{av}$ is more commonly known as the drift velocity:

$$\vec{v}_{d} = \frac{q \tau}{m} \vec{E}$$

After time $t = \tau$, the tendency of the electric field to increase $\vec{v}_{av}$ just balances the tendency of the collisions to decrease $\vec{v}_{av}$.

The very slow drift of the moving electrons as a group in the direction of $\vec{F}$ results in a net current in the conductor.


Current Density

Consider a conductor with cross-sectional area A and an electric field $\vec{E}$ directed from left to right. Suppose there are n charged particles per unit volume and assume that all the particles move with the same drift velocity with magnitude $v_{d}$

$$dQ =A v_{d} \, dt \times n \times |q|$$

Hence, the current is:

$$\begin{aligned} I &= \frac{dQ}{dt} \\ &= n |q| v_{d} A \end{aligned}$$

The current density (current per unit cross-sectional area):

$$\begin{aligned} J &= \frac{I}{A} \\ &= n |q| v_{d} \end{aligned}$$

Vector current density:

$$\vec{J} = nq \vec{v}_{d}$$

Since $\vec{v}_{d} = \frac{q \tau}{m} \vec{E}$,

$$\vec{J} = \frac{nq^{2} \tau}{m} \vec{E}$$


Next: Resistance And Resistivity

Previous: Dielectrics In Capacitors

Back To Electromagnetism (UY1)

Back To University Year 1 Physics Notes

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