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A copper wire has a square cross \(2.3 \mathrm{~mm}\) on a side. The wire is \(4.0 \mathrm{~m}\) long and carries a current of \(3.6 \mathrm{~A}\). The density of free electrons is \(8.5 \times 10^{28} / \mathrm{m}^{3} .\) Find the magnitudes of (a) the current density in the wire and (b) the electric field in the wire. (c) How much time is required for an electron to travel the length of the wire?

Short Answer

Expert verified
The magnitudes are (a) current density in the wire: \(6.8 \times 10^5 A/m^2\), (b) electric field in the wire: \(5.0 \times 10^{-5} V/m\) and (c) time required for an electron to travel the length of the wire: \(8.0 \times 10^4 s\)

Step by step solution

01

Calculate the cross-sectional area of the wire

The cross-sectional area A of the wire can be calculated using the given cross-sectional side length, \(2.3 mm = 2.3 x 10^{-3} m\), and using the formula \(A = side^2\). Hence, \(A = (2.3 \times 10^{-3})^2 = 5.29 \times 10^{-6} m^2\).
02

Calculate the current density (J)

The current density J can be calculated using the formula \(J = I/A\), where \(I\) is the current and \(A\) is the cross-sectional area. Hence, \(J = 3.6 A/5.29 x 10^{-6} m^2 = 6.8 \times 10^5 A/m^2\).
03

Calculate the magnitude of the electric field (E)

The magnitude of the electric field can be found using the formula \(E = J/ (\sigma)\), where \(\sigma\) is the electrical conductivity and given by \(\sigma = nq\), where \(n\) is the number density of free electrons and q is the charge of an electron. Here, \(q = 1.6 \times 10^{-19} C\) and \(n = 8.5 \times 10^{28}\ m^{-3}\). Therefore, \(\sigma = nq = 8.5 \times 10^{28}\ m^{-3} \times 1.6 \times 10^{-19} C = 1.36x10^{10} S/m \). Thus, \(E = 6.8 \times 10^5 A/m^2 / 1.36x10^{10} S/m = 5.0 \times 10^{-5} V/m \)
04

Calculate the time required for an electron to travel the length of the wire

The time required for an electron to travel the length of the wire can be found using the formula \(t = L/v\), where \(L\) is the length of the wire and \(v\) is the drift velocity. The drift velocity \(v\) can be found using the formula \(v = J/(\sigma)\). Hence, \(v = 6.8 \times 10^5 A/m^2 / 1.36x10^{10} S/m = 5.0 \times 10^{-5} m/s\). Hence, \(t = 4.0 m / 5.0 \times 10^{-5} m/s = 8.0 \times 10^4 s \)

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Key Concepts

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

Electric Field
An electric field is an invisible force field that surrounds all charged particles. It exerts a force on other charges within the field. The electric field \(E\) inside a conductor such as a copper wire can be calculated using the relationship between current density \(J\) and electrical conductivity \(\sigma\), given by the equation \(E = J / \sigma\).
Copper is a great conductor because it allows electrons to move freely within its atomic structure. However, even the best conductors have some resistance, which produces an electric field when current flows. In our exercise, we calculated \(E\) to ensure students understand the forces at play within the wire.
This electric field is crucial to understanding how energy is transmitted through a medium. The field represents how voltage and distance relate, where the units of \(E\) are volts per meter \(\text{V/m}\). A strong electric field indicates a greater force exerted by the electric charges, facilitating the flow of electrons from negative to positive regions in a circuit.
Copper Wire
Copper wire is a common component in electrical and electronic setups due to its excellent conductivity qualities. These characteristics stem from the high density of free electrons available within the material. This quality makes it highly efficient for conducting current.
In the wire used in the exercise, we looked at a specific cross-sectional area to understand how it influences current flow. Given the square cross-section \[2.3 \; \text{mm on a side}\], we ensure our calculations accurately reflect how copper wire behaves under these conditions.
Additionally, the length of the wire, \(4.0 \; \text{m}\), complements the cross-sectional area in understanding electrical factors such as resistance and electric field strength. The relationship between these dimensions and the electrical properties like current density \(J = I / A\) plays an essential role in electrical circuit design.
Electron Drift Velocity
Electron drift velocity pertains to the average speed at which electrons move through a conductor when subjected to an electric field. Despite the staggering number of electrons constantly moving within a conductor like copper, without an external field, these movements are random.
When an electric field is applied, electrons drift in the direction opposite the electric field due to their negative charge. The drift velocity \(v\) is given by the formula \(v = J / (nq)\), where \(n\) is the concentration of conduction electrons and \(q\) is the charge of an electron. This formula helps pinpoint how rapidly electrical signals travel through conductive materials.
Understanding drift velocity is essential for comprehending how quickly an electrical signal can be transmitted through a wire. In practical terms, although an electron's drift speed is relatively slow, the electrical signal itself moves at nearly the speed of light due to the energy transferred through the interactions between electrons.

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Most popular questions from this chapter

An 18 gauge copper wire (diameter \(1.02 \mathrm{~mm}\) ) carries a current with a current density of \(3.20 \times 10^{6} \mathrm{~A} / \mathrm{m}^{2} .\) The density of free electrons for copper is \(8.5 \times 10^{28}\) electrons per cubic meter. Calculate (a) the current in the wire and (b) the magnitude of the drift velocity of electrons in the wire.

A magnetic field of \(0.080 \mathrm{~T}\) is in the \(y\) -direction. The velocity of wire segment \(S\) has a magnitude of \(78 \mathrm{~m} / \mathrm{s}\) and components of \(18 \mathrm{~m} / \mathrm{s}\) in the \(x\) -direction, \(24 \mathrm{~m} / \mathrm{s}\) in the \(y\) -direction, and \(72 \mathrm{~m} / \mathrm{s}\) in the \(z\) -direction. The segment has length \(0.50 \mathrm{~m}\) and is parallel to the \(z\) -axis as it moves. (a) Find the motional emf induced between the ends of the segment. (b) What would the motional emf be if the wire segment was parallel to the \(y\) -axis?

A typical cost for electrical power is 0.120 dollar per kilowatthour. (a) Some people leave their porch light on all the time. What is the yearly cost to keep a \(75 \mathrm{~W}\) bulb burning day and night? (b) Suppose your refrigerator uses \(400 \mathrm{~W}\) of power when it's running, and it runs 8 hours a day. What is the yearly cost of operating your refrigerator?

According to the U.S. National Electrical Code, copper wire used for interior wiring of houses, hotels, office buildings, and industrial plants is permitted to carry no more than a specified maximum amount of current. The table shows values of the maximum current \(I_{\max }\) for several common sizes of wire with varnished cambric insulation. The "wire gauge" is a standard used to describe the diameter of wires. Note that the larger the diameter of the wire, the smaller the wire gauge. $$ \begin{array}{ccc} \text { Wire gauge } & \text { Diameter }(\mathrm{cm}) & I_{\max }(\mathrm{A}) \\\ \hline 14 & 0.163 & 18 \\ 12 & 0.205 & 25 \\ 10 & 0.259 & 30 \\ 8 & 0.326 & 40 \\ 6 & 0.412 & 60 \\ 5 & 0.462 & 65 \\ 4 & 0.519 & 85 \end{array} $$ (a) What considerations determine the maximum current-carrying capacity of household wiring? (b) A total of \(4200 \mathrm{~W}\) of power is to be supplied through the wires of a house to the household electrical appliances. If the potential difference across the group of appliances is \(120 \mathrm{~V},\) determine the gauge of the thinnest permissible wire that can be used. (c) Suppose the wire used in this house is of the gauge found in part (b) and has total length \(42.0 \mathrm{~m}\). At what rate is energy dissipated in the wires? (d) The house is built in a community where the consumer cost of electrical energy is \(\$ 0.11\) per kilowatt-hour. If the house were built with wire of the next larger diameter than that found in part (b), what would be the savings in electricity costs in one year? Assume that the appliances are kept on for an average of 12 hours a day.

The free-electron density in a copper wire is \(8.5 \times 10^{28}\) electrons \(/ \mathrm{m}^{3} .\) The electric field in the wire is \(0.0600 \mathrm{~N} / \mathrm{C}\) and the temperature of the wire is \(20.0^{\circ} \mathrm{C}\). (a) What is the drift speed \(v_{\mathrm{d}}\) of the electrons in the wire? (b) What is the potential difference between two points in the wire that are separated by \(20.0 \mathrm{~cm} ?\)

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