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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 strand of wire has resistance 5.60 \(\mu \Omega\). Find the net resistance of 120 such strands if they are (a) placed side by side to form a cable of the same length as a single strand, and (b) connected end to end to form a wire 120 times as long as a single strand.

An electrical conductor designed to carry large currents has a circular cross section 2.50 mm in diameter and is 14.0 m long. The resistance between its ends is 0.104\(\Omega\). (a) What is the resistivity of the material? (b) If the electric-field magnitude in the conductor is 1.28 V/m, what is the total current? (c) If the material has \(8.5 \times 10{^2}{^8}\) free electrons per cubic meter, find the average drift speed under the conditions of part (b).

You apply a potential difference of 4.50 V between the ends of a wire that is 2.50 m in length and 0.654 mm in radius. The resulting current through the wire is 17.6 A. What is the resistivity of the 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?

A 1.50-m cylindrical rod of diameter 0.500 cm is connected to a power supply that maintains a constant potential difference of 15.0 V across its ends, while an ammeter measures the current through it. You observe that at room temperature (20.0\(^\circ\)C) the ammeter reads 18.5 A, while at 92.0\(^\circ\)C it reads 17.2 A. You can ignore any thermal expansion of the rod. Find (a) the resistivity at 20.0\(^\circ\)C and (b) the temperature coefficient of resistivity at 20\(^\circ\)C for the material of the rod.

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