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Chapter 26: Problem 72

A steel trolley-car rail has a cross-sectional area of \(56.0 \mathrm{~cm}^{2}\). What is the resistance of \(10.0 \mathrm{~km}\) of rail? The resistivity of the steel is \(3.00 \times 10^{-7} \Omega \cdot \mathrm{m}\)

Short Answer

Expert verified
The resistance of the steel rail is approximately 0.54 惟.

Step by step solution

01

Understand the Formula

The resistance (R) of a material is calculated using the formula: \[ R = \frac{\rho \cdot L}{A} \]where \( \rho \) is the resistivity of the material, \( L \) is the length of the material, and \( A \) is the cross-sectional area.
02

Convert Units

Before substituting into the formula, ensure that all units are consistent. The length provided is in kilometers, so we need to convert it to meters (1 km = 1000 m):\[ 10.0 \, \text{km} = 10,000 \, \text{m} \]The cross-sectional area is given in \( \text{cm}^2 \), convert it to \( \text{m}^2 \):\[ 56.0 \, \text{cm}^2 = 0.0056 \, \text{m}^2 \]
03

Substitute Values into the Resistance Formula

Now that all measurements are in consistent units, substitute the known values into the resistance formula:\[ R = \frac{3.00 \times 10^{-7} \, \Omega \cdot \text{m} \times 10,000 \, \text{m}}{0.0056 \, \text{m}^2} \]
04

Perform the Calculation

Calculate the resistance using the formula:\[ R = \frac{3.00 \times 10^{-7} \, \Omega \cdot \text{m} \times 10,000 \, \text{m}}{0.0056 \, \text{m}^2} \]\[ R = \frac{3.00 \times 10^{-3} \, \Omega \cdot \text{m}}{0.0056 \, \text{m}^2} \]\[ R = 0.5357 \, \Omega \]

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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 describes how strongly they resist electric current. It is represented by the symbol \( \rho \) and measured in ohm-meters (\( \Omega \cdot \text{m} \)).

Each material has a unique resistivity which affects how easily current can flow through it.
  • Materials with low resistivity, like copper or silver, allow current to pass through easily and are called conductors.
  • Materials with high resistivity, like rubber or glass, do not allow current to pass easily and act as insulators.

In practical applications, understanding resistivity helps in choosing the right material for electrical wiring and components. In our exercise, we are dealing with steel which has a resistivity of \(3.00 \times 10^{-7} \Omega \cdot \text{m} \). This value is essential for calculating the resistance of the steel rail.
Cross-Sectional Area
Cross-sectional area refers to the area of the slice of an object and is crucial in analyzing how easily current can pass through a material. In our exercise, the cross-sectional area is given as \( 56.0 \, \text{cm}^2 \).

The formula for resistance \( R = \frac{\rho \cdot L}{A} \) reveals that the larger the cross-sectional area, the lower the resistance, since more electrons have the space to flow freely.
  • To calculate resistance accurately, the cross-sectional area must be in the same units as the other measurements, typically in square meters (\( \text{m}^2 \)).
  • Converting \( 56.0 \, \text{cm}^2 \) to \( \text{m}^2 \) is straightforward: \( 56.0 \, \text{cm}^2 = 0.0056 \, \text{m}^2 \).

The reduction in units makes it compatible with the resistivity units, ensuring accurate resistance calculations.
Unit Conversion
Unit conversion is pivotal in ensuring all components of a formula are in harmony. In calculations involving physical dimensions, using consistent units is crucial.

For the exercise, the length of the steel rail is initially given in kilometers. Since resistivity is measured in meters, we need to convert this length.
  • 1 kilometer is equal to 1,000 meters. Therefore, \( 10.0 \, \text{km} \) is converted to \( 10,000 \, \text{m} \) for uniformity.
  • The cross-sectional area goes from \( \text{cm}^2 \) to \( \text{m}^2 \) to match the international SI units used in the resistivity formula.

Proper unit conversion simplifies the process of plugging values into formulas and obtaining precise results, leading to accurate predictions and analyses in physics and engineering.

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Most popular questions from this chapter

A block in the shape of a rectangular solid has a cross-sectional area of \(3.50 \mathrm{~cm}^{2}\) across its width, a front-to-rear length of \(15.8 \mathrm{~cm},\) and a resistance of \(935 \Omega .\) The block's material contains \(5.33 \times 10^{22}\) conduction electrons \(/ \mathrm{m}^{3}\). A potential difference of \(35.8 \mathrm{~V}\) is maintained between its front and rear faces. (a) What is the current in the block? (b) If the current density is uniform, what is its magnitude? What are (c) the drift velocity of the conduction electrons and (d) the magnitude of the electric field in the block?

A cylindrical resistor of radius \(5.0 \mathrm{~mm}\) and length \(2.0 \mathrm{~cm}\) is made of material that has a resistivity of \(3.5 \times 10^{-5} \Omega \cdot \mathrm{m} .\) What are (a) the magnitude of the current density and (b) the potential difference when the energy dissipation rate in the resistor is \(1.0 \mathrm{~W} ?\)

The magnitude \(J\) of the current density in a certain lab wire with a circular cross section of radius \(R=2.00 \mathrm{~mm}\) is given by \(J=\left(3.00 \times 10^{8}\right) r^{2},\) with \(J\) in amperes per square meter and radial distance \(r\) in meters. What is the current through the outer section bounded by \(r=0.900 R\) and \(r=R ?\)

The legend that Benjamin Franklin flew a kite as a storm approached is only a legend-he was neither stupid nor suicidal. Suppose a kite string of radius \(2.00 \mathrm{~mm}\) extends directly upward by \(0.800 \mathrm{~km}\) and is coated with a \(0.500 \mathrm{~mm}\) layer of water having resistivity \(150 \Omega \cdot \mathrm{m} .\) If the potential difference between the two ends of the string is \(160 \mathrm{MV},\) what is the current through the water layer? The danger is not this current but the chance that the string draws a lightning strike, which can have a current as large as 500000 A (way beyond just being lethal).

A human being can be electrocuted if a current as small as 50 mA passes near the heart. An electrician working with sweaty hands makes good contact with the two conductors he is holding, one in each hand. If his resistance is \(2000 \Omega,\) what might the fatal voltage be?
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