A lightbulb filament has diameter 0.050 mm and carries 0.833 A. Find the current density (a) in the filament and (b) in the 12-gauge wire (diameter \(2.1 \mathrm{mm}\) ) supplying current to the lightbulb.

Electrical conductivity is a fundamental property that describes how well a material can conduct electric current. It is usually denoted by the symbol \(\sigma\) and is defined as the reciprocal of electrical resistivity. Materials with high electrical conductivity, such as metals, allow electric charges to flow through them easily.

When an electric field is applied across a material, charged particles respond by moving, creating a flow of electric current. The ease with which these particles can move defines the material's conductivity. In the context of the exercise, a filament in a lightbulb typically has high electrical conductivity to allow sufficient current to flow and produce light. Understanding the concept of electrical conductivity is crucial, as it directly impacts the current density, which is the amount of electric current flowing per unit cross-sectional area of a material.

The area of a filament relates to its cross-sectional size, which is vital for calculating its current density. In the example provided, the filament is cylindrical, and hence, its area is calculated using the area formula for a circle, \(A = \pi r^2\), where \(r\) is the radius of the filament.

To find this area precisely, it is important to first convert the diameter into radius and then into the correct units (meters for SI units). As we see in the solution, the diameter is halved to find the radius, and this value is squared and multiplied by \(\pi\) to yield the area. The area of the filament is a critical factor in determining how much current the filament can carry, as a larger area allows for more current to flow without overheating, a fact that's especially relevant in the design of lightbulbs and other electrical devices.

Current in physics refers to the flow of electric charge, and it is commonly measured in amperes (A). Mathematically, current \(I\) is defined as the rate at which charge \(Q\) flows through a surface or point, quantified by the formula: \(I = dQ/dt\), where \(dQ\) is the charge and \(dt\) is the time interval.

In practical terms, current can be visualized as the movement of electrons through a conductor, like the flow of water through a pipe. The current in the lightbulb filament from the exercise is relatively small (0.833 A), representing a controlled flow of electrons that generates light without damaging the filament due to excessive heating. Understanding the flow of current and its relationship with other electrical parameters is essential for working with electrical circuits and ensuring their safe and efficient operation.