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Chapter 25: Problem 5

Copper has \(8.5 \times 10^{28}\) free electrons per cubic meter. A 71.0 - \(\mathrm{cm}\) length of 12 -gauge copper wire that is 2.05 \(\mathrm{mm}\) in diameter carries 4.85 \(\mathrm{A}\) 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?

Short Answer

Expert verified
(a) 6455 seconds for 12-gauge; (b) 26004 seconds for 6-gauge; (c) Larger diameters decrease electron drift velocity.

Step by step solution

01

Understanding Electron Drift Velocity

The drift velocity \( v_d \) can be derived from the relation \( I = nAe v_d \), where \( I \) is the current, \( n \) is the number of free electrons per unit volume, \( A \) is the cross-sectional area of the wire, and \( e \) is the charge of an electron \( (1.6 \times 10^{-19} \text{ C}) \).
02

Calculate Cross-Sectional Area for 12-Gauge Wire

For a 12-gauge copper wire, the diameter is 2.05 mm. The cross-sectional area \( A \) is given by \( A = \pi (r^2) \), where \( r \) is half the diameter. First convert the diameter to meters: \( 2.05 \: \mathrm{mm} = 2.05 \times 10^{-3} \: \mathrm{m} \), then compute the radius \( r = \frac{2.05 \times 10^{-3}}{2} = 1.025 \times 10^{-3} \: \mathrm{m} \). Substitute into the area formula: \( A = \pi (1.025 \times 10^{-3})^2 \approx 3.30 \times 10^{-6} \: \mathrm{m^2} \).
03

Calculate Drift Velocity for 12-Gauge Wire

Using the formula \( v_d = \frac{I}{nAe} \), substitute \( I = 4.85 \: \mathrm{A} \), \( n = 8.5 \times 10^{28} \: \mathrm{m}^{-3} \), \( A = 3.30 \times 10^{-6} \: \mathrm{m^2} \), and \( e = 1.6 \times 10^{-19} \: \mathrm{C} \): \[ v_d = \frac{4.85}{(8.5 \times 10^{28})(3.30 \times 10^{-6})(1.6 \times 10^{-19})} \approx 1.10 \times 10^{-4} \: \mathrm{m/s}. \]
04

Calculate Time for Electron Travel in 12-Gauge Wire

The time \( t \) for an electron to travel the length \( L = 71 \: \mathrm{cm} = 0.71 \: \mathrm{m} \) is given by \( t = \frac{L}{v_d} \). Using the calculated drift velocity: \[ t = \frac{0.71}{1.10 \times 10^{-4}} \approx 6455 \: \mathrm{s}. \]
05

Calculate Cross-Sectional Area for 6-Gauge Wire

For 6-gauge wire, convert its diameter from mm to meters: \( 4.12 \: \mathrm{mm} = 4.12 \times 10^{-3} \: \mathrm{m} \). The radius \( r = \frac{4.12 \times 10^{-3}}{2} = 2.06 \times 10^{-3} \: \mathrm{m} \). So, \( A = \pi (2.06 \times 10^{-3})^2 \approx 1.33 \times 10^{-5} \: \mathrm{m^2} \).
06

Calculate Drift Velocity for 6-Gauge Wire

Using the same formula, \( v_d \) with the new area, \( v_d = \frac{4.85}{(8.5 \times 10^{28})(1.33 \times 10^{-5})(1.6 \times 10^{-19})} \approx 2.73 \times 10^{-5} \: \mathrm{m/s}. \)
07

Calculate Time for Electron Travel in 6-Gauge Wire

Using \( t = \frac{L}{v_d} \) again for the 6-gauge wire with \( v_d = 2.73 \times 10^{-5} \: \mathrm{m/s} \): \[ t = \frac{0.71}{2.73 \times 10^{-5}} \approx 26004 \: \mathrm{s}. \]
08

Discuss Effect of Wire Diameter on Drift Velocity

A larger diameter decreases the drift velocity of electrons for the same current, because the cross-sectional area increases, which lowers the current density \( \frac{I}{A} \). Therefore, electrons travel slower through a wider wire for a given current.

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Key Concepts

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

Electron Mobility
Electron mobility describes how easily electrons can move through a conductor when an electric field is applied. It signifies the velocity at which electrons drift through a material, influenced by material properties and conditions such as temperature.
This mobility is vital for predicting how quickly electrons can travel through a conductor like copper wire, as seen in computations involving drift velocity. It gives insight into the efficiency of charge carriers in a material which helps optimize electrical systems.
Factors affecting electron mobility include
  • Material purity: Higher purity often means fewer obstacles for electron movement.
  • Temperature: Generally, mobility decreases with increasing temperature due to more vibration in the atomic lattice.
  • External forces: Electric fields drive electron movement, affecting their velocity.
Understanding electron mobility can enhance our grasp of systems involving current flow through conductors.
Current Density
Current density is a measure of electric current (I) flowing per unit cross-sectional area (A) through a conductor. It is important because it indicates how concentrated the flow of electrons is within a wire.
Mathematically, current density (J) is expressed as \( J = \frac{I}{A} \), where \( I \) is the current in amperes, and\( A \) is the cross-sectional area in square meters.
High current density can indicate more stress on the conductor as it carries a large amount of current through a relatively small area. This can lead to overheating and potential damage to the wire.
  • Higher current density means rapid electron movement, leading to increased drift velocity.
  • Lower current density allows for more efficient energy distribution and reduced losses.
Understanding current density helps us design systems for safety and efficiency.
Wire Diameter Effect
The wire diameter significantly impacts the drift velocity of electrons in a conductor. Diameter changes affect the cross-sectional area, and this in turn modifies the current density for any given level of electric current.
With a larger diameter:
  • There is a greater cross-sectional area.
  • Current density decreases since the same current is distributed over a larger area.
  • Electrons experience less resistance and drift velocity lowers for the same current.
Conversely, a smaller diameter increases current density and drift velocity, as the same number of electrons must move through a narrower space. This understanding assists in electronic design, ensuring optimal wire sizing for electrical applications.
Cross-sectional Area
The cross-sectional area of a wire is a key factor determining both current density and drift velocity in a conductor. To find this area, it is calculated using the formula for a circle: \( A = \pi r^2 \), where \( r \) is the radius of the wire.
In practice, you would
  • Convert the wire's diameter from millimeters to meters.
  • Divide the diameter by two to find the radius.
  • Use the circle area formula to obtain the cross-sectional area.
For example, if the diameter is 2.05 mm, converting and calculating gives a radius of 1.025 mm or 1.025 \( \times 10^{-3} \) meters, leading to an area calculation useful in determining current density and drift velocity.
Electric Current
Electric current represents the flow of electric charge, commonly realized as electron flow in a conductor like a copper wire. It's measured in amperes (A), with real-world calculations and equations often revolving around this concept.
The relationship between current, electron movement, and wire characteristics such as cross-sectional area and electron mobility is critical in electronics. Drift velocity, for example, is calculated in part using current values.
Other current-related considerations include
  • The ability of the conductor to sustain the current without overheating.
  • Ensuring the charges are transported efficiently through the material.
Understanding and managing electric current is crucial for the safe and efficient operation of electrical circuits.

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Most popular questions from this chapter

A toaster using a Nichrome heating element operates on 120 \(\mathrm{V}\) . When it is switched on at \(20^{\circ} \mathrm{C}\) , the heating element carries an initial current of 1.35 \(\mathrm{A}\) . A few seconds later the current reaches the steady value of 1.23 \(\mathrm{A}\) . (a) What is the final temperature of the element? The average value of the temperature coefficient of resistivity for Nichrome over the temperature range is \(4.5 \times\) \(10^{-4}\left(\mathrm{C}^{\circ}\right)^{-1}\) . (b) What is the power dissipated in the heating element initially and when the current reaches a steady value?

A \(5.00-\mathrm{A}\) 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?

On your first day at work as an electrical technician, you are asked to determine the resistance per meter of a long piece of wire. The company you work for is poorly equipped. You find a battery, a voltmeter, and an ammeter, but no meter for directly measuring resistance (an ohmmeter). You put the leads from the voltmeter across the terminals of the battery, and the meter reads 12.6 \(\mathrm{V}\) . You cut off a 20.0 -m length of wire and connect it to the battery, with an ammeter in series with it to measure the current in the wire. The ammeter reads 7.00 A. You then cut off a 40.0 -m length of wire and connect it to the battery, again with the ammeter in series to measure the current. The ammeter reads 4.20 \(\mathrm{A}\) . Even though the equipment you have available to you is limited, your boss assures you of its high quality: The ammeter has very small resistance, and the voltmeter has very large resistance. What is the resistance of 1 meter of wire?

A silver wire 2.6 \(\mathrm{mm}\) in diameter transfers a charge of 420 \(\mathrm{C}\) in 80 \(\mathrm{mm}\) . Silver contains \(5.8 \times 10^{23}\) free electrons per cubic meter. (a) What is the current in the wire? (b) What is the magnitude of the drift velocity of the electrons in the wire?

A tightly coiled spring having 75 coils, each 3.50 \(\mathrm{cm}\) in diameter, is made of insulated metal wire 3.25 \(\mathrm{mm}\) in diameter. An ohmmeter connected across its opposite ends reads 1.74\(\Omega\) . What is the resistivity of the metal?
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