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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 ductile metal wire has resistance \(R\). What will be the resistance of this wire in terms of \(R\) if it is stretched to three times its original length, assuming that the density and resistivity of the material do not change when the wire is stretched? (\(Hint:\) The amount of metal does not change, so stretching out the wire will affect its cross-sectional area.)

The battery for a certain cell phone is rated at 3.70 V. According to the manufacturer it can produce \(3.15 \times 10^4\) J of electrical energy, enough for 5.25 h of operation, before needing to be recharged. Find the average current that this cell phone draws when turned on.

When a resistor with resistance \(R\) is connected to a 1.50-V flashlight battery, the resistor consumes 0.0625 W of electrical power. (Throughout, assume that each battery has negligible internal resistance.) (a) What power does the resistor consume if it is connected to a 12.6-V car battery? Assume that \(R\) remains constant when the power consumption changes. (b) The resistor is connected to a battery and consumes 5.00 W. What is the voltage of this battery?

A 1.50-m cylinder of radius 1.10 cm is made of a complicated mixture of materials. Its resistivity depends on the distance \(x\) from the left end and obeys the formula \(\rho (x) = a + bx^2\), where a and b are constants. At the left end, the resistivity is 2.25 \(\times\) 10\(^{-8} \Omega\) \(\cdot\) m, while at the right end it is 8.50 \(\times\) 10\(^{-8}\Omega\) \(\cdot \) m. (a) What is the resistance of this rod? (b) What is the electric field at its midpoint if it carries a 1.75-A current? (c) If we cut the rod into two 75.0-cm halves, what is the resistance of each half?

A person with body resistance between his hands of 10 k\(\Omega\) accidentally grasps the terminals of a 14-kV power supply. (a) If the internal resistance of the power supply is 2000 \(\Omega\), what is the current through the person's body? (b) What is the power dissipated in his body? (c) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the above situation to be 1.00 mA or less?

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