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An electrical conductor designed to carry large currents has a circular cross section 2.50 \(\mathrm{mm}\) in diameter and is 14.0 \(\mathrm{m}\) 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 \(1.28 \mathrm{V} / \mathrm{m},\) what is the total current? (c) If the material has \(8.5 \times 10^{28}\) free electrons per cubic meter, find the average drift speed under the conditions of part (b).

Short Answer

Expert verified
(a) \(3.65 \times 10^{-8} \, \Omega \cdot \text{m}\), (b) \(12.31 \, \text{A}\), (c) \(1.87 \times 10^{-4} \, \text{m/s}\)

Step by step solution

01

Find the Cross-sectional Area

The cross-sectional area \( A \) of the conductor can be calculated using the formula for the area of a circle: \[ A = \pi \left( \frac{d}{2} \right)^2 \]Given that the diameter \( d = 2.50 \, \text{mm} = 0.00250 \, \text{m}, \) \[ A = \pi \left( \frac{0.00250}{2} \right)^2 \approx 4.91 \times 10^{-6} \, \text{m}^2 \]
02

Calculate the Resistivity

We use the formula for resistance: \[ R = \frac{\rho L}{A} \]where \( R = 0.104 \, \Omega, \) \( L = 14.0 \, \text{m}, \) and \( A \approx 4.91 \times 10^{-6} \, \text{m}^2. \)Solving for resistivity \( \rho, \)\[ \rho = \frac{R \cdot A}{L} = \frac{0.104 \cdot 4.91 \times 10^{-6}}{14.0} \approx 3.65 \times 10^{-8} \, \Omega \cdot \text{m} \]
03

Calculate Total Current

To find the current, we use Ohm's law in the form \( E = I \cdot R, \) which we reconfigure to solve for \( I \):\[ I = \frac{E}{R} \]where \( E = 1.28 \, \text{V/m}, \) and \( R = 0.104 \, \Omega. \)\[ I = \frac{1.28}{0.104} \approx 12.31 \, \text{A} \]
04

Calculate Drift Speed

The drift speed \( v_d \) is given by the formula:\[ I = n \cdot A \cdot e \cdot v_d \]where \( I = 12.31 \, \text{A}, \) \( n = 8.5 \times 10^{28} \, \text{electrons/m}^3, \) \( e = 1.6 \times 10^{-19} \, \text{C}, \) and \( A \approx 4.91 \times 10^{-6} \, \text{m}^2. \)Solving for \( v_d, \)\[ v_d = \frac{I}{n \cdot A \cdot e} = \frac{12.31}{8.5 \times 10^{28} \cdot 4.91 \times 10^{-6} \cdot 1.6 \times 10^{-19}} \approx 1.87 \times 10^{-4} \, \text{m/s} \]

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Key Concepts

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

Resistivity
Resistivity is a fundamental property of materials that measures how strongly a material opposes the flow of electric current. It is denoted by the symbol \(\rho\). The resistivity of a material is constant under constant conditions, and it depends on the material's nature and temperature.

The unit of resistivity is ohm-meter (\(\Omega \cdot \text{m}\)). A material with low resistivity allows easy flow of electric charges, while a high resistivity means the opposite. For practical purposes, resistivity helps determine how well an electrical conductor performs.

For instance, in the exercise, the resistivity calculated was approximately \(3.65 \times 10^{-8} \, \Omega \cdot \text{m}\), indicating that the material is a good conductor of electricity. You can compute resistivity using the formula:
  • \(\rho = \frac{R \cdot A}{L}\)
where \(R\) is resistance, \(A\) is cross-sectional area, and \(L\) is the length of the conductor. This formula shows how resistivity relates to other properties like resistance, area, and length.
Ohm's Law
Ohm’s Law is a fundamental principle that describes the relationship between voltage, current, and resistance in an electrical circuit. It states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points, assuming the temperature remains constant.

The mathematical expression for Ohm's Law is:
  • \(V = I \cdot R\)
where \(V\) is voltage, \(I\) is current, and \(R\) is resistance. This relationship shows how changing the voltage or resistance affects the current.

In the exercise, we used Ohm's Law to find the current flowing through a conductor with known resistance and electric field strength. Rearranging the formula to solve for current gives:
  • \(I = \frac{E}{R}\)
where \(E\) represents the electric field magnitude. This simplicity makes Ohm's Law incredibly useful in designing and analyzing electrical circuits.
Drift Speed
Drift speed refers to the average speed that free charge carriers, like electrons, move through a conductor under the influence of an electric field. Although individual electrons move rapidly within a conductor, their average velocity (drift speed) is much slower due to frequent collisions with atoms.

The formula for calculating the drift speed is:
  • \(v_d = \frac{I}{n \cdot A \cdot e}\)
where:
  • \(I\) is the current flowing through the conductor,
  • \(n\) is the number of charge carriers per unit volume,
  • \(A\) is the cross-sectional area of the conductor, and
  • \(e\) is the charge of an electron (approximately \(1.6 \times 10^{-19} \text{C}\)).


Drift speed is an important concept as it helps us understand how fast electrons travel within a conductor, which is critical for the functioning of electronic components and systems. In the exercise, we calculated the drift speed to be approximately \(1.87 \times 10^{-4} \, \text{m/s}\), revealing the microscopic level interaction in conductors.
Electrical Conductors
Electrical conductors are materials that allow electric charges to flow through them easily. These materials contain atoms that have loosely bound electrons, which are free to move through the material, facilitating electrical conduction. Common conductors include metals like copper, aluminum, and gold.

Factors affecting a material's conductivity include resistivity, temperature, and the presence of impurities. In practice, conductors are used in electrical wiring and components because they efficiently transmit electricity with minimal resistance.

In the exercise, the conductor in question supports significant electric current flow, indicated by its low resistivity. Additionally, the knowledge of the number of free electrons per cubic meter assists in accurately determining the material's electricity conducting ability, leading to precise calculations of quantities like drift speed.

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Most popular questions from this chapter

Current passes through a solution of sodium chloride. In \(1.00 \mathrm{s}, 2.68 \times 10^{16} \mathrm{Na}^{+}\) ions arrive at the negative electrode and \(3.92 \times 10^{16} \mathrm{Cl}^{-}\) ions arrive at the positive electrode. (a) What is the current passing between the electrodes? (b) What is the direction of the current?

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

A 5.00 -A current runs through a 12 -gauge copper wire (diameter 2.05 \(\mathrm{mm}\) ) and through a light bulb. Copper has \(8.5 \times\) \(10^{28}\) free electrons per cubic meter, (a) How many electrons pass through the light bulb each second? (b) What is the current density in the wire? (c) At what speed does a typical electron pass by any given point in the wire? (d) If you were to use wire of twice the diameter, which of the above answers would change? Would they increase or decrease?

The following measurements of current and potential difference were made on a resistor constructed of Nichrome wire: $$\begin{array}{llll}{\boldsymbol{I}(\mathbf{A})} & {0.50} & {1.00} & {2.00} & {4.00} \\ {\boldsymbol{V}_{a b}(\mathbf{V})} & {1.94} & {3.88} & {7.76} & {15.52}\end{array}$$ (a) Graph \(V_{a b}\) as a function of \(I\) (b) Does Nichrome obey Ohm's law? How can you tell? (c) What is the resistance of the resistor in ohms?

A typical cost for electric power is \(\$ 0.120\) per kilowatt- hour. (a) Some people leave their porch light on all the time. What is the yearly cost to keep a \(75-\) W bulb buming 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?

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