Problem 5 - University Physics with Modern Physics

Question

Copper has \(8.5 \times 10^{28}\) free electrons per cubic meter. A 71.0 -cm length of 12 -gauge copper wire that is 2.05 \(\mathrm{mm}\) in diameter carries 4.85 A of current. (a) How much time does it take for an electron to travel the length of the wire? (b) Repeat part (a) for 6-gauge copper wire (diameter 4.12 \(\mathrm{mm}\) ) of the same length that carries the same current.(c) Generally speaking, how does changing the diameter of a wire that carries a given amount of current affect the drift velocity of the electrons in the wire?

Step-by-Step Solution

Answer

(a) 6514 seconds; (b) 26300 seconds; (c) Larger diameter reduces drift velocity.

Step 1: Understand Electron Drift Velocity

The drift velocity (\(v_d\)) is the average velocity of free electrons moving in a conductor when a current is applied. It's given by \[ v_d = \frac{I}{nAe} \] where \(I\) is the current, \(n\) is the electron density, \(A\) is the cross-sectional area, and \(e\) is the electron charge \( (1.6 \times 10^{-19}\mathrm{C}) \).

Step 2: Calculate the Cross-Sectional Area for 12-Gauge Wire

For a 12-gauge wire with diameter 2.05 mm, convert the diameter to meters: \(2.05\, \mathrm{mm} = 2.05 \times 10^{-3}\, \mathrm{m}\). The cross-sectional area \(A\) is \(\pi r^2\), where \(r\) is the radius. Thus, \(r = \frac{2.05 \times 10^{-3}}{2}\) m. Calculate: \[ A = \pi \left(\frac{2.05 \times 10^{-3}}{2}\right)^2 \approx 3.30 \times 10^{-6}\, \mathrm{m^2} \].

Step 3: Calculate Drift Velocity for 12-Gauge Wire

Using the formula for drift velocity: \[ v_d = \frac{4.85}{(8.5 \times 10^{28})(3.30 \times 10^{-6})(1.6 \times 10^{-19})} \approx 1.09 \times 10^{-4} \mathrm{m/s} \].

Step 4: Calculate Time for Electron to Travel Wire in 12-Gauge

Given the wire length \(L = 0.71\; \mathrm{m}\), time \(t\) is calculated using: \[ t = \frac{L}{v_d} = \frac{0.71}{1.09 \times 10^{-4}} \approx 6514 \; \mathrm{seconds} \].

Step 5: Calculate Cross-Sectional Area for 6-Gauge Wire

For a 6-gauge wire with diameter 4.12 mm, convert the diameter to meters: \(4.12\, \mathrm{mm} = 4.12 \times 10^{-3}\, \mathrm{m}\). The cross-sectional area \(A\) is \(\pi r^2\), where \( r = \frac{4.12 \times 10^{-3}}{2} \) m. Calculate: \[ A = \pi \left(\frac{4.12 \times 10^{-3}}{2}\right)^2 \approx 1.333 \times 10^{-5} \mathrm{m^2} \].

Step 6: Calculate Drift Velocity for 6-Gauge Wire

Using the formula for drift velocity: \[ v_d = \frac{4.85}{(8.5 \times 10^{28})(1.333 \times 10^{-5})(1.6 \times 10^{-19})} \approx 2.70 \times 10^{-5}\, \mathrm{m/s} \].

Step 7: Calculate Time for Electron to Travel Wire in 6-Gauge

With the same length \(L = 0.71\; \mathrm{m}\), time \(t\) is: \[ t = \frac{0.71}{2.70 \times 10^{-5}} \approx 2.63 \times 10^4\, \mathrm{seconds} \].

Step 8: Discuss Influence of Wire Diameter on Drift Velocity

A larger diameter increases the wire's cross-sectional area, reducing drift velocity for a given current because more electrons can flow simultaneously, acting to distribute the current over a larger number of electrons.

Key Concepts

Electron Density

In a conductor like copper, electrons are responsible for carrying charge through the wire. The number of these free-moving electrons per unit volume is called the electron density. For copper, this is approximately \(8.5 \times 10^{28}\) electrons per cubic meter. This immense number implies that even a small cross-section of copper, there are tremendously many electrons available to carry current. Understanding electron density is crucial because it determines how well a material can conduct electricity. When current flows, it's these mobile electrons that move, though very slowly, compared to how fast the current itself is perceived to travel. Electron density provides the backbone for calculating the drift velocity, which is the average speed at which electrons shift throughout the material under the influence of an electric field.

Current Flow

Current is the flow of electric charge in a conductor. It's measured in amperes (A), where one ampere equals one coulomb of charge passing through a cross-section per second. The process of current flow in metals such as copper is dominated by the movement of free electrons. When a voltage is applied, it causes these electrons to drift from the negative side towards the positive side, thus implying a flow of current in the opposite direction. Current flow in copper wire is efficient due to its high electron density and low resistivity, which allows electrons to move more freely. For example, a current flow of 4.85 A suggests a continuous movement and supply of charge through the wire, sustained by the metal's free electrons. This established flow results in the generation of drift velocity, indicating the actual movement speed of these electrons.

Cross-Sectional Area

The cross-sectional area of a wire is a critical factor determining how much current it can carry. It refers to the area of the wire's cut surface perpendicular to the current flow direction and is often expressed in square meters (m²). This area is computed using the formula \(A = \pi r^2\), where \(r\) is the radius of the wire. A larger cross-sectional area means more space for the electrons to move and hence can carry more current at a lower drift velocity. When the cross-section is increased, electrons face less resistance, allowing more electrons to pass through simultaneously. This ability to accommodate more flowing electrons is why thicker wires are used for high-current applications, as they provide less resistance to the electron flow. Consequently, a larger cross-sectional area generally contributes to a reduction in resistive heat, making it more energy-efficient.

Wire Diameter

Wire diameter, often expressed in millimeters (mm), is directly related to the cross-sectional area of a wire. A larger diameter reflects a greater radius, resulting in a larger area. When the diameter increases, as seen from the transition from a 12-gauge to a 6-gauge wire, the cross-sectional area increases significantly (from approximately \(3.30 \times 10^{-6} \text{ m}^2\) to \(1.333 \times 10^{-5} \text{ m}^2\)). This impacts the drift velocity, as a larger cross-sectional area allows a given current to be spread over a wider path, enabling more electrons to pass with less resistance. As a result, for the same current, drift velocity decreases since electrons have more space to move through. Additionally, less resistance to current flow reduces energy loss via heat, making thicker wires more suitable for carrying higher currents efficiently.