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Chapter 26: Problem 23

When \(115 \mathrm{~V}\) is applied across a wire that is \(10 \mathrm{~m}\) long and has a \(0.30 \mathrm{~mm}\) radius, the magnitude of the current density is \(1.4 \times\) \(10^{8} \mathrm{~A} / \mathrm{m}^{2}\). Find the resistivity of the wire.

Short Answer

Expert verified
The resistivity of the wire is \( 2.66 \times 10^{-8} \mathrm{~\Omega \cdot m} \).

Step by step solution

01

Understanding the Current Density Formula

The current density \( J \) is given by the formula \( J = \frac{I}{A} \), where \( I \) is the current and \( A \) is the cross-sectional area of the wire. We need to find the current \( I \) first in order to calculate the resistivity.
02

Calculate the Cross-sectional Area

The cross-sectional area \( A \) of the wire is a circle with radius \( r = 0.30 \times 10^{-3} \) m. Use the formula \( A = \pi r^2 \) to find \( A \). Substitute the values: \( A = \pi (0.30 \times 10^{-3})^2 \).
03

Calculate the Current (I)

Using the current density \( J = 1.4 \times 10^{8} \mathrm{~A/m^2} \) and the cross-sectional area \( A \) from Step 2, calculate the current \( I \) using the formula \( I = J \times A \).
04

Understand Ohm's Law and Resistivity

Ohm's Law relates voltage \( V \), current \( I \), and resistance \( R \) with the formula \( V = IR \). The resistance \( R \) of the wire can also be expressed in terms of resistivity \( \rho \) using the formula \( R = \rho \frac{L}{A} \), where \( L \) is the length of the wire.
05

Calculate Resistance Using Ohm's Law

From Step 1, \( V = 115 \) V and \( I \) is calculated. Use \( R = \frac{V}{I} \) to find the resistance of the wire.
06

Calculate Resistivity with Resistance Formula

Now use the resistance \( R \) from Step 5 and the known length \( L = 10 \) m to find the resistivity \( \rho \) using the formula \( \rho = R \frac{A}{L} \). Use the value of \( A \) from Step 2.

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Key Concepts

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

Ohm's Law
Ohm's Law is a fundamental principle in the study of electricity. It states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points. This relationship can be described mathematically as:
  • \( V = IR \)
Where:
  • \( V \) is the voltage (in volts)
  • \( I \) is the current (in amperes)
  • \( R \) is the resistance (in ohms)
This law allows us to calculate any one of the three variables (voltage, current, or resistance) if the other two are known. In our exercise, we use it to find the resistance of the wire by rearranging the formula to \( R = \frac{V}{I} \). This step is crucial for finding resistivity later on, as resistivity is directly related to the resistance.
Current Density
Current density is a measure of the electric current per unit area of cross-section of a conductor. It is denoted by the symbol \( J \) and is calculated using the formula:
  • \( J = \frac{I}{A} \)
where:
  • \( J \) is the current density (in amperes per square meter \( A/m^2 \))
  • \( I \) is the current (in amperes)
  • \( A \) is the cross-sectional area (in square meters)
In our scenario, the current density is provided as \( 1.4 \times 10^8 \) A/m虏. To find the current flowing through the wire, we multiply the current density by the cross-sectional area of the wire (calculated in a later step). Understanding current density is essential as it helps assess how efficiently a wire can carry electricity without overheating or losing much energy.
Cross-sectional Area
The cross-sectional area of a wire is crucial for calculating both current and resistance. In the case of a cylindrical wire, such as the one in the exercise, the cross-sectional area is a circle. Calculating this area involves the formula:
  • \( A = \pi r^2 \)
where:
  • \( A \) is the cross-sectional area
  • \( \pi \) (pi) is approximately 3.14159
  • \( r \) is the radius of the wire (in meters)
For the given wire with a radius of 0.30 mm (or 0.30 x 10^{-3} m), substituting this value into the formula gives the area that we use to compute both the current and the resistivity. The accurate calculation of the cross-sectional area ensures precise determination of other electrical properties.
Electrical Resistance
Electrical resistance is a key concept in understanding how materials affect electric current flow. Resistance is a measure of how much a material opposes the flow of electric current. Ohm's Law gives us the ability to calculate resistance with:
  • \( R = \frac{V}{I} \)
Once resistance \( R \) is known, we can determine the resistivity \( \rho \), which is a material characteristic, using the formula:
  • \( \rho = R \frac{A}{L} \)
where:
  • \( \rho \) is the resistivity
  • \( R \) is the resistance
  • \( A \) is the cross-sectional area
  • \( L \) is the length of the wire
In the exercise, this understanding allows us to express the resistivity based on the resistance value we calculated from the known voltage and derived current. Resistivity provides insight into the intrinsic properties of the wire material, explaining how well it conducts electricity.

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Most popular questions from this chapter

The magnitude \(J(r)\) of the current density in a certain cylindrical wire is given as a function of radial distance from the center of the wire's cross section as \(J(r)=B r\), where \(r\) is in meters, \(J\) is in amperes per square meter, and \(B=2.00 \times 10^{5} \mathrm{~A} / \mathrm{m}^{3}\). This function applies out to the wire's radius of \(2.00 \mathrm{~mm}\). How much current is contained within the width of a thin ring concentric with the wire if the ring has a radial width of \(10.0 \mu \mathrm{m}\) and is at a radial distance of \(1.20 \mathrm{~mm} ?\)

The current density in a wire is uniform and has magnitude \(2.0 \times 10^{6} \mathrm{~A} / \mathrm{m}^{2}\), the wire's length is \(5.0 \mathrm{~m}\), and the density of conduction electrons is \(8.49 \times 10^{28} \mathrm{~m}^{-3} .\) How long does an electron take (on the average) to travel the length of the wire?

A caterpillar of length \(4.0 \mathrm{~cm}\) crawls in the direction of electron drift along a 5.2-mm-diameter bare copper wire that carries a uniform current of 12 A. (a) What is the potential difference between the two ends of the caterpillar? (b) Is its tail positive or negative relative to its head? (c) How much time does the caterpillar take to crawl \(1.0 \mathrm{~cm}\) if it crawls at the drift speed of the electrons in the wire? (The number of charge carriers per unit volume is \(8.49 \times 10^{28} \mathrm{~m}^{-3}\).)

A potential difference of \(1.20 \mathrm{~V}\) will be applied to a \(33.0 \mathrm{~m}\) length of 18 -gauge copper wire (diameter \(=0.0400\) in.). Calculate (a) the current, (b) the magnitude of the current density, (c) the magnitude of the electric field within the wire, and (d) the rate at which thermal energy will appear in the wire.

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 \(3.00 \times 10^{-7} \Omega \cdot \mathrm{m}\)
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