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Chapter 25: Problem 4

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鈥檚 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 heart defibrillator is used to enable the heart to start beating if it has stopped. This is done by passing a large current of 12 A through the body at 25 V for a very short time, usually about 3.0 ms. (a) What power does the defibrillator deliver to the body, and (b) how much energy is transferred?

An external resistor with resistance \(R\) is connected to a battery that has emf \(\varepsilon\) and internal resistance \(r\). Let \(P\) be the electrical power output of the source. By conservation of energy, \(P\) is equal to the power consumed by \(R\). What is the value of \(P\) in the limit that \(R\) is (a) very small; (b) very large? (c) Show that the power output of the battery is a maximum when \(R = r\). What is this maximum \(P\) in terms of \(\varepsilon\) and \(r\)? (d) A battery has \(\varepsilon\) = 64.0 V and \(r =\) 4.00 \(\Omega\). What is the power output of this battery when it is connected to a resistor \(R\), for \(R =\) 2.00 \(\Omega\), \(R =\) 4.00 \(\Omega\), and \(R =\) 6.00 \(\Omega\) ? Are your results consistent with the general result that you derived in part (b)?

Compact fluorescent bulbs are much more efficient at producing light than are ordinary incandescent bulbs. They initially cost much more, but they last far longer and use much less electricity. According to one study of these bulbs, a compact bulb that produces as much light as a 100-W incandescent bulb uses only 23 W of power. The compact bulb lasts 10,000 hours, on the average, and costs \(11.00, whereas the incandescent bulb costs only \)0.75, but lasts just 750 hours. The study assumed that electricity costs $0.080 per kilowatt-hour and that the bulbs are on for 4.0 h per day. (a) What is the total cost (including the price of the bulbs) to run each bulb for 3.0 years? (b) How much do you save over 3.0 years if you use a compact fluorescent bulb instead of an incandescent bulb? (c) What is the resistance of a "100-W" fluorescent bulb? (Remember, it actually uses only 23 W of power and operates across 120 V.)

A 3.00-m length of copper wire at 20\(^\circ\)C has a 1.20-mlong section with diameter 1.60 mm and a 1.80-m-long section with diameter 0.80 mm. There is a current of 2.5 mA in the 1.60- mm-diameter section. (a) What is the current in the 0.80-mmdiameter section? (b) What is the magnitude of \(\vec E\) in the 1.60-mm-diameter section? (c) What is the magnitude of \(\vec E\) in the 0.80-mm-diameter section? (d) What is the potential difference between the ends of the 3.00-m length of wire?

A "540-W" electric heater is designed to operate from 120-V lines. (a) What is its operating resistance? (b) What current does it draw? (c) If the line voltage drops to 110 V, what power does the heater take? (Assume that the resistance is constant. Actually, it will change because of the change in temperature.) (d) The heater coils are metallic, so that the resistance of the heater decreases with decreasing temperature. If the change of resistance with temperature is taken into account, will the electrical power consumed by the heater be larger or smaller than what you calculated in part (c)? Explain.
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