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Chapter 26: Problem 11

When \(230 \mathrm{~V}\) is applied across a wire that is \(14.1 \mathrm{~m}\) long and has a \(0.30 \mathrm{~mm}\) radius, the magnitude of the current density is \(1.98 \times 10^{8} \mathrm{~A} / \mathrm{m}^{2}\). Find the resistivity of the wire.

Short Answer

Expert verified
The resistivity of the wire is \( 8.24 \times 10^{-8} \, \text{Ohm} \cdot \text{m} \).

Step by step solution

01

Understanding 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. In this problem, we know \( J = 1.98 \times 10^8 \, \text{A/m}^2 \) and we need to find the area \( A \) of the wire.
02

Calculate Cross-sectional Area

The cross-sectional area \( A \) of the wire is calculated using the formula for the area of a circle \( A = \pi r^2 \). The radius \( r \) is \( 0.30 \, \text{mm} = 0.0003 \, \text{m} \). Substituting the value, \[ A = \pi (0.0003)^2 = 2.827 \times 10^{-7} \, \text{m}^2. \]
03

Calculate Current through Wire

Using the current density \( J \) formula \( J = \frac{I}{A} \), we can rearrange to find the current \( I = J \times A \). Substituting the known values, \[ I = (1.98 \times 10^8) \times (2.827 \times 10^{-7}) = 55.95 \, \text{A}. \]
04

Understanding Ohm's Law

Ohm's law relates voltage \( V \), current \( I \), and resistance \( R \) by \( V = I \times R \). Rearranging, the resistance is \( R = \frac{V}{I} \). We know \( V = 230 \, \text{V} \) and \( I = 55.95 \, \text{A} \).
05

Calculate Resistance of Wire

Using the Ohm's law rearranged formula \( R = \frac{V}{I} \), we find the resistance \( R \) by substituting \[ R = \frac{230}{55.95} = 4.11 \, \text{Ohms}. \]
06

Relate Resistance to Resistivity

The resistance \( R \) of a uniform wire is also given by \( R = \rho \frac{L}{A} \) where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area. Solve for resistivity: \( \rho = R \frac{A}{L} \).
07

Calculate Resistivity

Substitute the known values into the formula for resistivity: \[ \rho = 4.11 \frac{2.827 \times 10^{-7}}{14.1} = 8.24 \times 10^{-8} \, \text{Ohm} \cdot \text{m}. \]

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Key Concepts

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

Current Density
Current density is a measure of how much electric current is flowing per unit area through a given point in a material, often a wire. It is represented by the letter \( J \), and its unit is amperes per square meter (\( ext{A/m}^2 \)). In practical terms, current density helps us understand how densely packed the electric current flow is within the wire.
To calculate current density, we use the formula:
  • \( J = \frac{I}{A} \)
where \( I \) is the electric current in amperes and \( A \) is the cross-sectional area of the wire in square meters.
Given that we usually know the values of current density (\( J \)) and need to calculate the area (or even the current), it鈥檚 essential to get familiar with manipulating this formula. A higher current density means more current is flowing through a smaller area, which can influence the material's behavior, possibly causing heating effects or requiring better insulation.
Ohm's Law
Ohm鈥檚 Law serves as a foundational principle for understanding how voltage, current, and resistance are interrelated in an electrical circuit. It states that the voltage \( V \) across a conductor is directly proportional to the current \( I \) flowing through it, with proportionality being defined by the resistance \( R \). The mathematical expression for Ohm's Law is:
  • \( V = I \times R \)
This expression can be rearranged in different ways to solve for any one of the three variables if the other two are known. For example, you can find the resistance by rearranging to \( R = \frac{V}{I} \).
  • Example: If a wire carries a current of 55.95 A and the applied voltage is 230 V, the resistance of the wire can be calculated using the rearranged version of Ohm's Law. Here, \( R = \frac{230}{55.95} \). This kind of calculation helps identify how resistant a material is to conducting electric current.
Resistance Calculation
Resistance is a measure of how difficult it is for current to flow through a conductor. Several factors affect the resistance of a wire, including its length, cross-sectional area, and the material it's made of. Resistivity (\( \rho \)) is a property of a material that affects resistance and is measured in ohm-meters (\( \text{Ohm} \cdot ext{m} \)).
When calculating the resistance \( R \) of a wire, we use the formula:
  • \( R = \rho \frac{L}{A} \)
where \( L \) is the wire鈥檚 length, \( A \) is its cross-sectional area, and \( \rho \) is the resistivity.
In practice, once you have calculated the resistance, you can rearrange this formula to find the resistivity if it's unknown:
  • \( \rho = R \frac{A}{L} \)
This relationship illustrates why longer wires with smaller cross-sectional areas exhibit higher resistance, impacting the wire's ability to carry current without overheating or degrading performance.

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Most popular questions from this chapter

Figure \(26-18\) shows wire section 1 of diameter \(D_{1}=4.00 R\) and wire section 2 of diameter \(D_{2}=1.75 R\), connected by a tapered section. The wire is copper and carries a current. Assume that the current is uniformly distributed across any cross-sectional area through the wire's width. The electric potential change \(V\) along the length \(L=2.00 \mathrm{~m}\) shown in section 2 is \(10.0 \mu \mathrm{V}\). The number of charge carriers per unit volume is \(8.49 \times 10^{28} \mathrm{~m}^{-3}\). What is the drift speed of the conduction electrons in section \(1 ?\)

Two conductors are made of the same material and have the same length. Conductor \(A\) is a solid wire of radius \(1.0 \mathrm{~mm}\). Conductor \(B\) is a hollow tube of outside radius \(2.2 \mathrm{~mm}\) and inside radius \(1.0 \mathrm{~mm}\). What is the resistance ratio \(R_{A} / R_{B}\), measured between their ends?

What is the current in a wire of radius \(R=2.67 \mathrm{~mm}\) if the magnitude of the current density is given by (a) \(J_{a}=J_{0} r / R\) and (b) \(J_{b}=J_{0}(1-r / R)\), in which \(r\) is the radial distance and \(J_{0}=5.50 \times 10^{4} \mathrm{~A} / \mathrm{m}^{2} ?\) (c) Which function maximizes the current density near the wire's surface?

A wire initially has length \(L_{0}\) and resistance \(5.00 \Omega\). The resistance is to be increased to \(45.0 \Omega\) by stretching the wire. Assuming that the resistivity and density of the material are unaffected by the stretching, find the ratio of the new length to \(L_{0}\).

A charged belt, \(50 \mathrm{~cm}\) wide, travels at \(30 \mathrm{~m} / \mathrm{s}\) between a source of charge (electrons) and a sphere. The belt carries charge into the sphere at a rate corresponding to \(76 \mu \mathrm{A}\). (a) Compute the surface charge density on the belt. (b) What is the number density (number per unit area) of the electrons on the belt?
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