Problem 29 - Fundamentals of Physics

Question

A potential difference of \(3.00 \mathrm{nV}\) is set up across a \(2.00 \mathrm{~cm}\) length of copper wire that has a radius of \(2.00 \mathrm{~mm} .\) How much charge drifts through a cross section in \(3.00 \mathrm{~ms} ?\)

Step-by-Step Solution

Answer

The charge that drifts through the wire is approximately \(3.3 \times 10^{-11} \, \text{C}\).

Step 1: Determine the Cross-sectional Area of Wire

The cross-sectional area of a cylindrical wire is given by the formula \( A = \pi r^2 \), where \( r \) is the radius of the wire. First, convert the radius from millimeters to meters: \( r = 2.00 \, \text{mm} = 2.00 \times 10^{-3} \, \text{m} \). Then calculate the area: \( A = \pi (2.00 \times 10^{-3} \, \text{m})^2 \approx 1.26 \times 10^{-5} \, \text{m}^2 \).

Step 2: Calculate the Electric Field

The electric field \( E \) is determined by the formula \( E = \frac{V}{L} \), where \( V = 3.00 \, \text{nV} = 3.00 \times 10^{-9} \, \text{V} \) and \( L = 2.00 \, \text{cm} = 2.00 \times 10^{-2} \, \text{m} \). Hence, the electric field is \( E = \frac{3.00 \times 10^{-9} \, \text{V}}{2.00 \times 10^{-2} \, \text{m}} = 1.50 \times 10^{-7} \, \text{V/m} \).

Step 3: Find the Drift Velocity

Drift velocity \( v_d \) is given by \( v_d = \frac{I}{nqA} \), where \( n \) is the charge carrier density, \( q \) is the charge of an electron \( = 1.6 \times 10^{-19} \, \text{C} \). First, find \( n \) for copper which is approximately \( 8.5 \times 10^{28} \text{m}^{-3} \). To find current \( I \), we use the relation \( I = \, nqAv_d \) which further relates with \( J = \sigma E \) where \( J = \frac{I}{A} \) and \( \sigma \) is the conductivity of copper (approximately \( 5.8 \times 10^7 \text{S/m} \)). Substitute \( E \) to find: \( J = \sigma E \Rightarrow v_d = \frac{\sigma E}{nq} \).

Step 4: Compute the Current

Substituting known values into the formula \( v_d = \frac{\sigma E}{nq} \), we have: \( v_d = \frac{5.8 \times 10^7 \, \text{S/m} \times 1.50 \times 10^{-7} \, \text{V/m}}{8.5 \times 10^{28} \, \text{m}^{-3} \times 1.6 \times 10^{-19} \, \text{C}} \approx 6.4 \times 10^{-12} \, \text{m/s} \). Using \( I = nqAv_d \), find \( I \) as: \( I = 8.5 \times 10^{28} \, \text{m}^{-3} \times 1.6 \times 10^{-19} \, \text{C} \times 1.26 \times 10^{-5} \, \text{m}^2 \times 6.4 \times 10^{-12} \, \text{m/s} \approx 1.1 \times 10^{-8} \, \text{A} \).

Step 5: Calculate the Charge

Use the formula \( Q = It \) to find the charge, where \( I = 1.1 \times 10^{-8} \, \text{A} \) and \( t = 3.00 \, \text{ms} = 3.00 \times 10^{-3} \, \text{s} \). Substituting these values, we get \( Q = 1.1 \times 10^{-8} \, \text{A} \times 3.00 \times 10^{-3} \, \text{s} = 3.3 \times 10^{-11} \, \text{C} \).

Key Concepts

Drift Velocity

Drift velocity is a key concept in understanding electric current. It refers to the average velocity that a charge carrier, such as an electron, attains due to an electric field. While individual charges move randomly, drift velocity provides a measure of the net velocity of all charges in a conductor under the influence of an electric field.

To calculate drift velocity, the formula \( v_d = \frac{I}{nqA} \) is used, where:

• \( I \) is the current
• \( n \) is the charge density
• \( q \) is the elementary charge (approximately \( 1.6 \times 10^{-19} \) C)
• \( A \) is the cross-sectional area of the conductor

The small drift velocity of electrons is why such things as a light bulb turn on almost instantaneously when the switch is flipped: the electric field propagates through the circuit at nearly the speed of light, while the actual drift speed is very slow.

Charge Density

Charge density refers to the amount of electric charge per unit volume in a material. It is a crucial factor when determining the behavior of electric currents in materials such as metals.

In the formula for drift velocity \( v_d = \frac{I}{nqA} \), \( n \) represents the charge density, specifically the number of free charge carriers (like electrons) per unit volume. For copper, \( n \) is typically around \( 8.5 \times 10^{28} \text{m}^{-3} \).

Understanding charge density is important for:

• Predicting how a material will conduct electricity
• Relating macroscopic properties like drift velocity to microscopic properties like charge density
• Designing and improving electronic components for efficiency and performance

Electric Field

The electric field is a fundamental concept in electromagnetism, representing a region where an electric force is experienced by charge carriers. It is described by the amount of force per charge unit and is integral to understanding how charges move in a circuit.

In our exercise, the electric field \( E \) was calculated using the formula:\[ E = \frac{V}{L} \]where:

• \( V \) is the potential difference
• \( L \) is the length over which this potential difference is applied

For example, with a potential difference of \( 3.00 \times 10^{-9} \text{V} \) across \( 2.00 \times 10^{-2} \text{m} \), the electric field is \( 1.50 \times 10^{-7} \text{V/m} \).

Electric fields play a vital role in:

• Determining the direction and rate of charge flow through a material
• Influencing the drift velocity of charge carriers
• Shaping the behavior and efficiency of electronic circuits and components