Problem 3 - University Physics with Modern Physics

Question

A 5.00 -A current runs through a 12 -gauge copper wire (diameter 2.05 \(\mathrm{mm}\) ) and through a light bulb. Copper has \(8.5 \times\) \(10^{28}\) free electrons per cubic meter, (a) How many electrons pass through the light bulb each second? (b) What is the current density in the wire? (c) At what speed does a typical electron pass by any given point in the wire? (d) If you were to use wire of twice the diameter, which of the above answers would change? Would they increase or decrease?

Step-by-Step Solution

Answer

(a) \(3.13 \times 10^{19}\) electrons/s; (b) \(1.52 \times 10^6\) A/m²; (c) \(1.10 \times 10^{-4}\) m/s; (d) Current density and drift velocity decrease.

Step 1: Calculate the Number of Electrons Passing per Second

The current (I) of 5.00 A represents the flow of charge per second. One ampere corresponds to 1 coulomb per second (C/s). Each electron has a charge of approximately \(1.6 \times 10^{-19} \text{C}\). The number of electrons \(n\) flowing per second can be calculated by dividing the current by the charge of one electron: \[ n = \frac{I}{e} = \frac{5.00 \, \text{C/s}}{1.6 \times 10^{-19} \, \text{C/electron}} \].Perform the calculation to find \( n \):\[ n \approx 3.13 \times 10^{19} \text{ electrons/second} \].

Step 2: Determine the Cross-Sectional Area of the Wire

The current density (J) is the current per unit area (\(A\)). First, find the cross-sectional area of the wire, which has a circular shape. The diameter given is 2.05 mm, so the radius \(r\) is \(1.025 \, \text{mm}\) or \(1.025 \times 10^{-3} \, \text{m}\). The area \(A\) of a circle is \(\pi r^2\), so:\[ A = \pi (1.025 \times 10^{-3} \, \text{m})^2 \approx 3.30 \times 10^{-6} \, \text{m}^2 \].

Step 3: Compute Current Density

Current density \(J\) is given by the formula:\[ J = \frac{I}{A} \].Substitute the known values of \(I = 5.00 \, \text{A}\) and \(A \approx 3.30 \times 10^{-6} \, \text{m}^2\):\[ J \approx \frac{5.00}{3.30 \times 10^{-6}} \approx 1.52 \times 10^6 \, \text{A/m}^2 \].

Step 4: Find the Drift Velocity

The drift velocity (\(v_d\)) is defined by the equation \(I = n e A v_d\), where \(n\) is the number density of free electrons and \(e\) is the electron charge. Rearranging the formula gives:\[ v_d = \frac{I}{n e A} \].Substitute the values \(I = 5.00 \, \text{A}\), \(n = 8.5 \times 10^{28} \, \text{electrons/m}^3\), \(e = 1.6 \times 10^{-19} \, \text{C/electron}\), and \(A \approx 3.30 \times 10^{-6} \, \text{m}^2\):\[ v_d \approx \frac{5.00}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 3.30 \times 10^{-6}} \approx 1.10 \times 10^{-4} \, \text{m/s} \].

Step 5: Effect of Doubling the Diameter

Doubling the diameter of the wire results in quadrupling the area (since area is proportional to the square of the diameter). The current density \(J\) would decrease because \(J = \frac{I}{A}\), and the area \(A\) increases. The drift velocity \(v_d\) would also decrease since \(v_d = \frac{I}{n e A}\) and \(A\) is larger. The number of electrons passing per second remains unchanged because it only depends on the current, which is constant.

Key Concepts

Drift Velocity

Drift velocity is a key concept in understanding how electric current flows through a conductor like a copper wire. While electrons themselves move at incredibly high speeds, their overall drift velocity is much slower. This drift refers to the net velocity component in one direction due to the electric field applied across the wire.

To calculate drift velocity, use the formula: \[ v_d = \frac{I}{n e A} \] where

• \(v_d\) is the drift velocity,
• \(I\) is the current flowing through the conductor,
• \(n\) is the number density of electrons (electrons per unit volume),
• \(e\) is the charge of an electron,
• and \(A\) is the cross-sectional area of the wire.

The drift velocity is usually very small. In our example, it is approximately \(1.10 \times 10^{-4} \text{ m/s}\). This slow net speed is paradoxical considering the rapid random motion of electrons within the wire, emphasizing that electric signals propagate at nearly the speed of light, not the individual electrons.

Number of Electrons

The number of electrons passing through a section of the wire every second is directly linked to the electric current flowing through it. Current can be thought of as the number of charge carriers (in this case, electrons) moving through the wire per second. Since current \(I\) in amperes corresponds to charge per unit time and given the elementary charge \(e\) of an electron, the number of electrons \(n\) is derived from \[ n = \frac{I}{e} \] Using our example, where the current is 5.00 A, we can find the number of electrons passing through any point in the circuit each second by substituting the given charge of an electron \(1.6 \times 10^{-19} \text{ C}\).

Performing this calculation: \[ n \approx \frac{5 \text{ C/s}}{1.6 \times 10^{-19} \text{ C/electron}} \approx 3.13 \times 10^{19} \text{ electrons/second} \] shows the immense number of electrons involved in electric current, highlighting the collaborative action of countless electrons needed to produce even a modest electrical flow.

Cross-Sectional Area of Wire

The cross-sectional area of a wire plays a significant role in determining its electrical characteristics, particularly current density and resistance. When a wire's diameter is specified, the area is calculated assuming a circular cross section, as wires are typically cylindrical.

Given a diameter of 2.05 mm, the radius \(r\) of the wire is half that, or 1.025 mm. You can convert this to meters to 1.025 × 10^-3 meters for calculation purposes. The formula for the cross-sectional area \(A\) for circular wires is: \[ A = \pi r^2 \] In our example, substituting the known radius into this formula gives: \[ A \approx \pi \times (1.025 \times 10^{-3} \text{ m})^2 \approx 3.30 \times 10^{-6} \text{ m}^2 \]

This calculation is crucial for understanding properties such as current density \(J\), which describes how much current flows through a unit area of the wire: \[ J = \frac{I}{A} \] The area also impacts the wire's resistance and the drift velocity of electrons: larger diameters lower resistance and drift velocity, given constant current.