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 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 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) 6454 seconds; (b) 26004 seconds; (c) Larger diameter lowers drift velocity.

Step 1: Determine the Cross-sectional Area for 12-gauge Wire

To find the cross-sectional area of a 12-gauge wire, use the formula for the area of a circle: \( A = \pi \left(\frac{d}{2}\right)^2 \), where \(d\) is the diameter. For 12-gauge wire, \(d = 2.05\,\text{mm} = 0.00205\,\text{m}\). So, \( A = \pi \left(\frac{0.00205\,\text{m}}{2}\right)^2 \approx 3.30 \times 10^{-6} \,\text{m}^2 \).

Step 2: Calculate Drift Velocity for 12-gauge Wire

Use the formula for drift velocity: \( v_d = \frac{I}{n \cdot e \cdot A} \), where \(I = 4.85\,\text{A}\), \(n = 8.5 \times 10^{28}\,\text{electrons/m}^3\), and \(e = 1.602 \times 10^{-19}\,\text{C}\). Substitute \(A = 3.30 \times 10^{-6} \,\text{m}^2 \) into the equation to get \( v_d \approx 1.10 \times 10^{-4} \,\text{m/s} \).

Step 3: Calculate Travel Time for 12-gauge Wire

The travel time \( t \) is calculated as \( t = \frac{L}{v_d} \), where \( L = 71.0 \,\text{cm} = 0.71 \, \text{m} \). Substitute \( v_d = 1.10 \times 10^{-4} \, \text{m/s} \) to get \( t \approx 6454 \, \text{seconds} \).

Step 4: Determine the Cross-sectional Area for 6-gauge Wire

For a 6-gauge wire, \(d = 4.12\,\text{mm} = 0.00412\,\text{m}\). Calculate the area: \( A = \pi \left(\frac{0.00412\,\text{m}}{2}\right)^2 \approx 1.33 \times 10^{-5} \,\text{m}^2 \).

Step 5: Calculate Drift Velocity for 6-gauge Wire

Use the same drift velocity formula. For 6-gauge wire, substitute \(A = 1.33 \times 10^{-5} \,\text{m}^2 \) into the equation to get \( v_d \approx 2.73 \times 10^{-5} \,\text{m/s} \).

Step 6: Calculate Travel Time for 6-gauge Wire

Calculate \( t \) for the 6-gauge wire using \(v_d = 2.73 \times 10^{-5} \, \text{m/s} \) and \( L = 0.71 \, \text{m} \), resulting in \( t \approx 26004 \, \text{seconds} \).

Step 7: Effect of Changing the Diameter on Drift Velocity

As the diameter of the wire increases, the cross-sectional area increases, which decreases the drift velocity for the same current. A larger diameter allows more electrons to contribute to the current flow, thus each electron moves slower for the same current.

Key Concepts

Free Electrons in Copper

Copper is an excellent conductor of electricity. It owes this property largely to the presence of its free electrons. These are electrons that can move freely through the metal lattice, facilitating the flow of electric current.
Each cubic meter of copper contains approximately \(8.5 \times 10^{28}\) free electrons. This high density of free electrons allows copper to efficiently transport electrical charges. When a voltage is applied across a copper wire, these free electrons drift toward the positive terminal, creating an electric current.
Given this large number of electrons, the movement or drift of each individual electron is quite slow compared to the speed at which the current propagates through the wire. This slow drift is due to the constant collisions between electrons and atoms in the metal, which impede their progress.

Cross-Sectional Area of Wire

The cross-sectional area of a wire is a crucial factor in determining its electrical properties. It refers to the area of the wire as seen from the cross-section perpendicular to its length, essentially the circle slice of the wire.
The cross-sectional area can be calculated using the formula for the area of a circle: \[ A = \pi \left(\frac{d}{2}\right)^2 \] where \(d\) is the diameter of the wire.

• For a 12-gauge wire with a diameter of 2.05 mm, the area is approximately \(3.30 \times 10^{-6} \text{ m}^2\).
• For a 6-gauge wire with a diameter of 4.12 mm, the area is roughly \(1.33 \times 10^{-5} \text{ m}^2\).

The size of the cross-sectional area influences how easily current can flow through the wire. A larger area allows more electrons to pass through simultaneously, reducing resistance and enabling more current to flow for the same voltage.

Effect of Wire Diameter on Current Flow

Increasing the diameter of a wire increases its cross-sectional area, which significantly affects the current flow in the wire.
For a fixed current, increasing the diameter of the wire provides a greater volume for the free electrons to move through. This results in a decrease in what is known as 'drift velocity'.
Drift velocity (\(v_d\)) is the average speed at which free electrons move through a conductor when subjected to an electric field. It can be calculated using the formula:\[ v_d = \frac{I}{n \cdot e \cdot A} \]where:

• \(I\) is the current.
• \(n\) is the number of free electrons per unit volume.
• \(e\) is the charge of an electron.
• \(A\) is the cross-sectional area of the wire.

With a larger diameter, the same amount of current spreads across more electrons within the increased cross-sectional area, leading to a reduced drift velocity. This reduced speed means that for the same current, electrons take longer to travel along the wire's length. Therefore, doubling the wire's diameter does not necessarily double the current it can carry, but it allows the same current to flow more efficiently with less resistance.