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Chapter 26: Problem 29

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 \(3.00 \mathrm{~ms} ?\)

Short Answer

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

Step by step solution

01

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: \( A = \pi (2.00 \times 10^{-3} \, \text{m})^2 \approx 1.26 \times 10^{-5} \, \text{m}^2 \).
02

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 \( E = \frac{3.00 \times 10^{-9} \, \text{V}}{2.00 \times 10^{-2} \, \text{m}} = 1.50 \times 10^{-7} \, \text{V/m} \).
03

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: \( J = \sigma E \Rightarrow v_d = \frac{\sigma E}{nq} \).
04

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: \( I = 8.5 \times 10^{28} \, \text{m}^{-3} \times 1.6 \times 10^{-19} \, \text{C} \times 1.26 \times 10^{-5} \, \text{m}^2 \times 6.4 \times 10^{-12} \, \text{m/s} \approx 1.1 \times 10^{-8} \, \text{A} \).
05

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 \( Q = 1.1 \times 10^{-8} \, \text{A} \times 3.00 \times 10^{-3} \, \text{s} = 3.3 \times 10^{-11} \, \text{C} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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

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Most popular questions from this chapter

A beam contains \(2.0 \times 10^{8}\) doubly charged positive ions per cubic centimeter, all of which are moving north with a speed of \(1.0 \times 10^{5} \mathrm{~m} / \mathrm{s}\). What are the (a) magnitude and (b) direction of the current density \(\vec{J} ?\) (c) What additional quantity do you need to calculate the total current \(i\) in this ion beam?

When \(115 \mathrm{~V}\) is applied across a wire that is \(10 \mathrm{~m}\) long and has a \(0.30 \mathrm{~mm}\) radius, the magnitude of the current density is \(1.4 \times 10^{8} \mathrm{~A} / \mathrm{m}^{2} .\) Find the resistivity of the wire.

The current density in a wire is uniform and has magnitude \(2.0 \times 10^{6} \mathrm{~A} / \mathrm{m}^{2},\) the wire's length is \(5.0 \mathrm{~m},\) and the density of conduction electrons is \(8.49 \times 10^{28} \mathrm{~m}^{-3}\). How long does an electron take (on the average) to travel the length of the wire?

An electric immersion heater normally takes 100 min to bring cold water in a well-insulated container to a certain temperature, after which a thermostat switches the heater off. One day the line voltage is reduced by \(6.00 \%\) because of a laboratory overload. How long does heating the water now take? Assume that the resistance of the heating element does not change.

An 18.0 W device has \(9.00 \mathrm{~V}\) across it. How much charge goes through the device in \(4.00 \mathrm{~h} ?\)
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