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Chapter 26: Problem 12

A metal rod is \(2 \mathrm{~m}\) long and \(8 \mathrm{~mm}\) in diameter. Compute its resistance if the resistivity of the metal is \(1.76 \times 10^{-8} \Omega \cdot \mathrm{m}\). $$ R=\rho \frac{L}{A}=\left(1.76 \times 10^{-8} \Omega \cdot \mathrm{m}\right) \frac{2 \mathrm{~m}}{\pi\left(4 \times 10^{-3} \mathrm{~m}\right)^{2}}=7 \times 10^{-4} \Omega $$

Short Answer

Expert verified
The resistance is \( 7 \times 10^{-4} \, \Omega \).

Step by step solution

01

Identify the given values

The length of the rod, \( L \), is \( 2 \, \text{m} \). The diameter of the rod is \( 8 \, \text{mm} \) which needs to be converted to meters. The resistivity, \( \rho \), is \( 1.76 \times 10^{-8} \, \Omega \cdot \text{m} \).
02

Calculate the cross-sectional area

First, convert the diameter from millimeters to meters: \( 8 \, \text{mm} = 8 \times 10^{-3} \, \text{m} \). The radius, \( r \), is half of the diameter: \( 4 \times 10^{-3} \, \text{m} \). The cross-sectional area \( A \) of the rod is computed using the formula for the area of a circle: \( A = \pi r^2 = \pi \times (4 \times 10^{-3})^2 \).
03

Compute the resistance using the formula

Substitute the given values into the resistance formula: \( R = \rho \frac{L}{A} = \left(1.76 \times 10^{-8} \, \Omega \cdot \text{m}\right) \frac{2 \, \text{m}}{\pi (4 \times 10^{-3} \, \text{m})^2} \).
04

Simplify the expression to find the resistance

Calculate the value of the denominator: \( \pi (4 \times 10^{-3})^2 = \pi \times 16 \times 10^{-6} = 50.24 \times 10^{-6} \, \text{m}^2 \). Substitute back into the resistance formula: \[ R = \frac{1.76 \times 10^{-8} \times 2}{50.24 \times 10^{-6}} \approx 7 \times 10^{-4} \, \Omega \].
05

Present the final answer

After computing all the expression's calculations, the final resistance of the metal rod is \( 7 \times 10^{-4} \, \Omega \).

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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 represented by the Greek letter \( \rho \) (rho). Metal materials generally have low resistivity because they allow electrons to flow freely, making them good conductors of electricity.

Key points about resistivity include:
  • Units: The resistivity of a material is typically measured in ohm meters \( (\Omega \cdot \text{m}) \).
  • Dependence on temperature: Resistivity can change with temperature; metals often have higher resistivity at increased temperatures.
  • Material's property: Unlike resistance, resistivity is a property of the material itself and does not change with the material's size or shape.
In this exercise, the provided resistivity of the metal rod is \( 1.76 \times 10^{-8} \, \Omega \cdot \text{m} \), which is typical for conducting materials.
Cross-Sectional Area
The cross-sectional area of a rod refers to the area of the face you see if you cut the rod perpendicular to its length. For a cylindrical object like a metal rod, this area is circular. Calculating the cross-sectional area is crucial because it directly affects the rod's electrical resistance.

To calculate the area of a circle, use:
  • Formula: \( A = \pi r^2 \), where \( r \) is the radius.
In the original problem:
  • Convert diameter to radius: Given diameter \( 8 \text{ mm} = 8 \times 10^{-3} \text{ m} \), so \( r = 4 \times 10^{-3} \text{ m} \).
  • Substitute into formula: \( A = \pi (4 \times 10^{-3})^2 \).
This calculation of cross-sectional area is integral to finding the electrical resistance of a rod, as it shows the pathway available for current flow.
Metal Rod
A metal rod is simply a solid cylindrical object made of metal. In electrical contexts, metal rods are often used because metals are good conductors. The configuration of the rod, including its length and diameter, is key to determining its resistance.

Features of a typical metal rod include:
  • Cylindrical shape: Provides a straightforward geometry for calculations involving volume and area.
  • Uniform material: Assumption of uniform resistivity and consistent material properties along the length of the rod.
  • Application: Used frequently in construction, manufacturing, and electrical applications because of their strength and conductivity.
For this exercise, the metal rod has a specific length of \(2 \text{ m}\) and diameter of \(8 \text{ mm}\), allowing us to calculate its resistance using known parameters.
Resistance Formula
The resistance formula is a mathematical way to compute the electrical resistance of an object. Resistance, expressed as \( R \), depends on three key factors: resistivity \( \rho \), the object's length \( L \), and its cross-sectional area \( A \).

The formula is given by:
  • \( R = \rho \frac{L}{A} \)
Breakdown for this exercise:
  • \( \rho \): Represents resistivity \(1.76 \times 10^{-8} \, \Omega \cdot \text{m}\)
  • \( L \): Length of the rod \( 2 \, \text{m}\)
  • \( A \): Cross-sectional area (calculated previously)
Using these, the formula allows us to calculate the resistance of the metal rod. Resistance, found to be approximately \( 7 \times 10^{-4} \, \Omega \), shows how difficult it is for electric current to pass through the rod. Lower resistance indicates better conductivity.

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Most popular questions from this chapter

What is the current through an \(8.0-\Omega\) toaster when it is operating on \(120 \mathrm{~V}\) ?

In the Bohr model, the electron of a hydrogen atom moves in a circular orbit of radius \(5.3 \times 10^{-11} \mathrm{~m}\) with a speed of \(2.2 \times 10^{6}\) \(\mathrm{m} / \mathrm{s}\). Determine its frequency \(f\) and the current \(I\) in the orbit. $$ f=\frac{v}{2 \pi r}=\frac{2.2 \times 10^{6} \mathrm{~m} / \mathrm{s}}{2 \pi\left(5.3 \times 10^{-11} \mathrm{~m}\right)}=6.6 \times 10^{15} \mathrm{rev} / \mathrm{s} $$ Each time the electron goes around the orbit, it carries a charge \(e\) around the loop. The charge passing a point on the loop each second is \(I=e f=\left(1.6 \times 10^{-19} \mathrm{C}\right)\left(6.6 \times 10^{15} \mathrm{~s}^{-1}\right)=1.06 \times 10^{-3} \mathrm{~A}=1.1 \mathrm{~mA}\)

How many electrons flow through a light bulb each second if the current through the light bulb is \(0.75 \mathrm{~A}\) ? From \(I=q / t\), the charge flowing through the bulb in \(1.0 \mathrm{~s}\) is $$ q=I t=(0.75 \mathrm{~A})(1.0 \mathrm{~s})=0.75 \mathrm{C} $$ But the magnitude of the charge on each electron is \(e=1.6 \times 10^{-19}\) C. Therefore, $$ \text { Number }=\frac{\text { Charge }}{\text { Charge } / \text { electron }}=\frac{0.75 \mathrm{C}}{1.6 \times 10^{-19} \mathrm{C}}=4.7 \times 10^{18} $$

Compute the internal resistance of an electric generator that has an emf of \(120 \mathrm{~V}\) and a terminal voltage of \(110 \mathrm{~V}\) when supplying 20 A.

A direct-current generator has an emf of \(120 \mathrm{~V}\); that is, its terminal voltage is \(120 \mathrm{~V}\) when no current is flowing from it. At an output of \(20 \mathrm{~A}\), the terminal potential is \(115 \mathrm{~V}\). (a) What is the internal resistance \(r\) of the generator? \((b)\) What will be the terminal voltage at an output of \(40 \mathrm{~A}\) ? The situation is much like that shown in Fig. \(26-3 .\) Now, however, \(\varepsilon=120 \mathrm{~V}\) and \(I\) is no longer \(25 \mathrm{~A}\). (a) In this case, \(I=20 \mathrm{~A}\) and the p.d. from \(A\) to \(B\) is \(115 \mathrm{~V}\). Therefore, $$ 115 \mathrm{~V}=+120 \mathrm{~V}-(20 \mathrm{~A}) r $$ from which \(r=0.25 \Omega\). (b) Now \(I=40\) A. So Terminal p.d. \(=\varepsilon-I r=120 \mathrm{~V}-(40 \mathrm{~A})(0.25 \Omega)=110 \mathrm{~V}\)
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