11.1 Electric Current

Electric current represents the flow of electric charge through a conductor, driven by an electric potential difference. When charges move through a material, they create a current that can be measured and analyzed. This fundamental concept forms the basis for understanding circuits, electrical systems, and many applications in modern technology.

Movement of Electric Charges

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Current Through Cross-Sectional Area

Current occurs when charge carriers travel through a conductor with an average drift velocity. Electric potential difference (also called electromotive force or emf, $\mathcal{E}$) provides the driving force that pushes these charges through the circuit.

The current $I$ passing through a cross-sectional area $A$ of a wire can be calculated using:

$I=n q v_{d} A$

Where:

• $n$ is the charge carrier density (number of carriers per unit volume)
• $q$ is the charge of each carrier
• $v_d$ is the drift velocity of the carriers
• $A$ is the cross-sectional area of the conductor

When charge carriers have no net motion in a section of wire, the current in that section equals zero, even though individual carriers may still be moving randomly at the microscopic level. This is similar to how water molecules in a still pond move randomly but create no current flow.

Current Density

Current density describes the flow of charge per unit area within a conductor. This concept helps us understand how current is distributed throughout the cross-section of a conductor.

The current density vector $\vec{J}$ is calculated as:

$\vec{J}=n q \vec{v}_{d}$

An electric field $\vec{E}$ within a conductor creates the force that moves the charge carriers. This field is related to both the resistivity $\rho$ of the conductor and the current density $\vec{J}$ by:

$\vec{E}=\rho \vec{J}$

Current density is a vector quantity, possessing both magnitude and direction. This means it points in the direction of charge flow and its magnitude indicates the amount of charge flowing per unit area per unit time. In uniform conductors, current density is typically constant across the cross-section, but in non-uniform conductors, it may vary with position.

Total Current from Density

When current density varies across a conductor's cross-section, we can find the total current by integrating the current density over the entire cross-sectional area:

$I_{\text{tot}}=\int \vec{J}(r) \cdot d \vec{A}$

This integration accounts for variations in current density throughout the conductor, allowing us to calculate the total current even when the flow isn't uniform. This approach is particularly useful for analyzing conductors with complex geometries or non-uniform properties.

Direction of Current

Current has a defined direction, though it's not a vector in the traditional sense. By convention, current direction is defined as the direction in which positive charges would move.

In most circuits, electrons (which carry negative charge) are the actual charge carriers that move. This creates an interesting situation where:

• Conventional current flows from positive to negative terminals
• Electron flow is in the opposite direction (from negative to positive)

Because of this unique directional property, current doesn't follow the laws of vector addition and doesn't have vector components. It's better understood as a scalar quantity with an associated direction rather than a true vector.

Practice Problem 1: Current Calculation

Solution

To find the current, we can use the equation: $I = nqv_dA$

Where:

• $n = 8.5 \times 10^{28} \text{ electrons/m}^3$
• $q = 1.6 \times 10^{-19} \text{ C}$ (charge of an electron)
• $v_d = 3.0 \times 10^{-4} \text{ m/s}$
• $A = 2.0 \times 10^{-6} \text{ m}^2$

Substituting these values: $I = (8.5 \times 10^{28} \text{ electrons/m}^3)(1.6 \times 10^{-19} \text{ C})(3.0 \times 10^{-4} \text{ m/s})(2.0 \times 10^{-6} \text{ m}^2)$

$I = 8.16 \text{ A}$

Therefore, the current in the wire is approximately 8.16 amperes.

Practice Problem 2: Current Density

Solution

To find the current density, we need to divide the current by the cross-sectional area: $J = \frac{I}{A}$

First, let's calculate the cross-sectional area of the wire: $A = \pi r^2 = \pi (1.2 \times 10^{-3} \text{ m})^2 = \pi (1.44 \times 10^{-6} \text{ m}^2) = 4.52 \times 10^{-6} \text{ m}^2$

Now we can find the current density: $J = \frac{5.0 \text{ A}}{4.52 \times 10^{-6} \text{ m}^2} = 1.11 \times 10^6 \text{ A/m}^2$

Therefore, the current density in the wire is $1.11 \times 10^6 \text{ A/m}^2$.