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An 18-gauge copper wire (diameter 1.02 mm) carries a current with a current density of \(3.20 \times 10{^6} A/m{^2}\). 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 drift velocity of electrons in the wire.

Short Answer

Expert verified
The current is approximately 1.31 A, and the drift velocity is about 9.5 \times 10^{-4} m/s.

Step by step solution

01

Determine the cross-sectional area of the wire

To find the current, we first need the cross-sectional area of the wire. Since the wire is round, we use the formula for the area of a circle \(A = \pi r^2\). The radius \(r\) is half of the diameter, so \(r = \frac{1.02 \text{ mm}}{2} = 0.51 \text{ mm} = 0.51 \times 10^{-3} \text{ m}\). Thus, \(A = \pi (0.51 \times 10^{-3})^2\).
02

Calculate the current using current density

The formula for current \(I\) is given by \(I = J \cdot A\), where \(J\) is the current density. We have \(J = 3.20 \times 10^6 \text{ A/m}^2\) and from Step 1, we found \(A\). Thus, \(I = 3.20 \times 10^6 \times \pi (0.51 \times 10^{-3})^2\).
03

Calculate the drift velocity of electrons

The drift velocity \(v_d\) is given by \(v_d = \frac{I}{n \cdot A \cdot e}\), where \(n = 8.5 \times 10^{28}\) m^{-3} is the electron density, and \(e = 1.6 \times 10^{-19}\) C is the charge of an electron. Substitute the value of \(I\) from Step 2, and \(A\) from Step 1 into this formula to find \(v_d\).

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Key Concepts

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

Current Density
Current density ( J ) is a measure of how much current flows through a unit area of a conductor. It's like counting how many electrons pass through a tiny window in the wire per second. Current density is measured in amperes per square meter ( A/m^2 ).
To calculate the current in a wire, knowing the current density is quite essential.
If you think of the wire as a highway, the current density indicates how crowded the highway is with electrons.

Knowing Current Density

  • It tells us how much electric current flows per unit area.
  • Higher current density can lead to increased heat in the wire.
  • This is crucial for safety and efficiency when designing electrical circuits.
Copper Wire
Copper wire is popular in electrical applications due to its excellent electrical conductivity. This makes it very effective in allowing the easy flow of electricity.
In this exercise, the focus is on an 18-gauge copper wire. This refers to the specific size of the wire, and the smaller number in gauge means a larger diameter.

Why Copper?

  • Copper has low resistance, making it ideal for conducting electricity.
  • It is durable and can be bent without breaking easily.
  • Copper's high electron density makes it an efficient conductor.
These properties mean the copper wire allows for efficient current flow with minimal energy loss, crucial in electrical wiring and electronic devices.
Cross-Sectional Area
The cross-sectional area of a wire affects how easily current can flow through it. Basically, it’s the space available for electrons to move.
To find this area in a cylindrical wire, like the one in the exercise, use the formula for the area of a circle: \(A = \pi r^2\).
This helps in calculating the current when you know the current density.

Importance of Cross-Sectional Area

  • Larger areas allow more current to flow, much like a wide river.
  • A smaller area can restrict current flow, similar to a narrow path.
  • Influences resistance; smaller areas mean higher resistance.
Understanding this concept helps in creating efficient electrical designs, as the cross-sectional area needs to match the required current flow.
Electron Drift Velocity
Electron drift velocity (\(v_d\)) is the average velocity of electrons moving through a conductor when subjected to an electric field. It's surprisingly slow compared to the speed of current, which is almost instantaneous.

Why Drift Velocity Matters

  • Helps in understanding the actual movement of electrons in a conductor.
  • Important for calculating conductive properties in materials.
  • Low drift velocity means electrons are "bumping" forward in a crowded space.
The drift velocity can be calculated using the formula: \(v_d = \frac{I}{n \cdot A \cdot e}\), where \(I\) is the current, \(n\) is the electron density, \(A\) is the cross-sectional area, and \(e\) is the charge of an electron.
Appreciating this helps in understanding how electricity behaves in different materials and why the material choice is critical in electrical engineering.

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Most popular questions from this chapter

A lightning bolt strikes one end of a steel lightning rod, producing a 15,000-A current burst that lasts for 65 \(\mu\)s. The rod is 2.0 m long and 1.8 cm in diameter, and its other end is connected to the ground by 35 m of 8.0-mm-diameter copper wire. (a) Find the potential difference between the top of the steel rod and the lower end of the copper wire during the current burst. (b) Find the total energy deposited in the rod and wire by the current burst.

The capacity of a storage battery, such as those used in automobile electrical systems, is rated in ampere-hours (A dot h). A 50-A dot h battery can supply a current of 50 A for 1.0 h, or 25 A for 2.0 h, and so on. (a) What total energy can be supplied by a 12-V, 60-A dot h battery if its internal resistance is negligible? (b) What volume (in liters) of gasoline has a total heat of combustion equal to the energy obtained in part (a)? (See Section 17.6; the density of gasoline is 900 kg/m\(^3\).) (c) If a generator with an average electrical power output of 0.45 kW is connected to the battery, how much time will be required for it to charge the battery fully?

A 2.0-m length of wire is made by welding the end of a 120-cm-long silver wire to the end of an 80-cm-long copper wire. Each piece of wire is 0.60 mm in diameter. The wire is at room temperature, so the resistivities are as given in Table 25.1. A potential difference of 9.0 V is maintained between the ends of the 2.0-m composite wire. What is (a) the current in the copper section; (b) the current in the silver section; (c) the magnitude of \(\vec E\) in the copper; (d) the magnitude of \(\vec E\) in the silver; (e) the potential difference between the ends of the silver section of wire?

Electric eels generate electric pulses along their skin that can be used to stun an enemy when they come into contact with it. Tests have shown that these pulses can be up to 500 V and produce currents of 80 mA (or even larger). A typical pulse lasts for 10 ms. What power and how much energy are delivered to the unfortunate enemy with a single pulse, assuming a steady current?

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

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