Published by Patrick Mutisya · 14 days ago
Electric current is the rate at which charge flows through a conductor. It is defined as the amount of charge \$Q\$ passing a given cross‑section per unit time \$t\$:
\$I = \frac{dQ}{dt}\$
For a uniform conductor carrying a steady current, the current can be expressed in terms of the properties of the charge carriers:
\$I = A\,n\,v\,q\$
Consider a segment of conductor of length \$L\$ and cross‑section \$A\$. In a time interval \$t\$, charge carriers travel a distance \$v t\$. The volume of material swept out is \$A v t\$, which contains \$n A v t\$ carriers. Each carrier carries charge \$q\$, so the total charge that passes through the cross‑section in time \$t\$ is
\$\Delta Q = n A v t \, q\$
Dividing by \$t\$ gives the current:
\$I = \frac{\Delta Q}{t} = A n v q\$
| Parameter | Symbol | Typical Units | Explanation |
|---|---|---|---|
| Cross‑sectional area | \$A\$ | m² | Size of the conductor perpendicular to the direction of flow. |
| Number density | \$n\$ | m⁻³ | Number of charge carriers per unit volume of the material. |
| Drift speed | \$v\$ | m s⁻¹ | Average speed of carriers due to the electric field (very small, ≈10⁻⁴ m s⁻¹ in copper). |
| Charge per carrier | \$q\$ | C | Elementary charge \$e = 1.60\times10^{-19}\$ C for electrons; \$+e\$ for positive ions. |
Re‑arranging the expression gives the drift speed:
\$v = \frac{I}{A n q}\$
Example
Solution:
\$A = \pi r^{2} = \pi (1.0\times10^{-3}\ \text{m})^{2} = 3.14\times10^{-6}\ \text{m}^{2}\$
\$v = \frac{5.0\ \text{A}}{(3.14\times10^{-6}\ \text{m}^{2})(8.5\times10^{28}\ \text{m}^{-3})(1.60\times10^{-19}\ \text{C})}\$
\$v \approx 1.2\times10^{-4}\ \text{m s}^{-1}\$
The drift speed is only a few hundred micrometres per second, illustrating why the electric signal propagates much faster than the individual carriers.