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Chapter 26: Problem 64

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 \(1.0 \mathrm{~W} ?\)

Short Answer

Expert verified
(a) Current density is approximately 127.3 A/m虏; (b) Potential difference is about 1.78 V.

Step by step solution

01

Calculate the Resistance

Use the formula for the resistance of a cylindrical resistor \( R = \rho \frac{L}{A} \), where \( \rho = 3.5 \times 10^{-5} \Omega \cdot \mathrm{m} \) is the resistivity, \( L = 0.02 \mathrm{~m} \) is the length, and \( A \) is the cross-sectional area. The cross-sectional area \( A \) can be calculated as \( A = \pi r^2 = \pi (0.005 \mathrm{~m})^2 \). Work out \( A \) first, and then determine the resistance \( R \).
02

Calculate the Power Dissipation Formula

The power dissipated in the resistor can be expressed as \( P = \frac{V^2}{R} \). Here, \( P = 1.0 \mathrm{~W} \). Substitute \( R \) from Step 1 to solve for the potential difference \( V \).
03

Rearrange to Find Current

Use the power formula again to find the current using \( P = I^2 \cdot R \). Rearrange for \( I \) to get \( I = \sqrt{\frac{P}{R}} \). Substitute \( P = 1.0 \mathrm{~W} \) and \( R \) from Step 1 to find \( I \).
04

Calculate the Current Density

Use the formula for current density \( J = \frac{I}{A} \), where \( I \) is the current obtained from Step 3 and \( A \) is the cross-sectional area found in Step 1. Calculate \( J \).

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Key Concepts

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

Understanding Current Density
Current density is a way to express how much electric current flows through a specific area. It's measured in amperes per square meter (A/m虏). Imagine current as water flowing through a pipe; current density would be the flow of water per unit area of the pipe. To calculate the current density, use the formula:
  • \( J = \frac{I}{A} \)
where \( J \) is the current density, \( I \) is the current, and \( A \) is the cross-sectional area of the resistor.
In our example, the cylinder's cross-sectional area is found using the formula \( A = \pi r^2 \), which indicates the area the current passes through. By dividing the current by this area, we determine how concentrated the flow of electricity is.
Decoding Potential Difference
The potential difference, or voltage, is essential in electrical circuits. It's the energy difference that pushes electron flow, making current move through the resistor. Think of potential difference as the pressure difference that drives water through a hose.
For a resistor, you can calculate the potential difference using the power dissipation formula:
  • \( P = \frac{V^2}{R} \)
Solving for \( V \), we have:
  • \( V = \sqrt{P \cdot R} \)
Here, \( P \) represents the power, and \( R \) is the resistance. When you plug the numbers into the equation, \( V \) will show you the voltage required to maintain the given energy dissipation rate.
Exploring Energy Dissipation
Energy dissipation in a circuit, particularly in a resistor, is the process of converting electrical energy into thermal energy (heat). This occurs when electric current passes through a resistor, as it resists the flow of electrons.
The rate of energy dissipation can be determined by the power formula:
  • \( P = I^2 \cdot R \)
This shows that the power (or energy rate) changes with the square of the current. So when calculating energy dissipation, knowing the resistance and the current is key. The dissipated energy can be felt as heat, which is why resistors can get warm during operation.
Cylindrical Resistor Fundamentals
A cylindrical resistor is a specific form of resistor that has a circular cross-section. Understanding its shape is helpful when calculating its resistance and other properties.
When dealing with cylindrical resistors, two factors are key: its length and its cross-sectional area. The formula for resistance in such a resistor is:
  • \( R = \rho \frac{L}{A} \)
where \( \rho \) symbolizes the resistivity of the material, \( L \) is the length, and \( A \) is the area. This formula reflects how both the material and dimensions influence resistance. Cylindrical resistors are common in circuits due to their ease of manufacturing and installation.

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Most popular questions from this chapter

A charged belt, \(50 \mathrm{~cm}\) wide, travels at \(30 \mathrm{~m} / \mathrm{s}\) between a source of charge and a sphere. The belt carries charge into the sphere at a rate corresponding to \(100 \mu \mathrm{A}\). Compute the surface charge density on the belt.

An electric immersion heater normally takes 100 min to bring cold water in a well-insulated container to a certain temperature, after which a thermostat switches the heater off. One day the line voltage is reduced by \(6.00 \%\) because of a laboratory overload. How long does heating the water now take? Assume that the resistance of the heating element does not change.

A wire of Nichrome (a nickel-chromium-iron alloy commonly used in heating elements) is \(1.0 \mathrm{~m}\) long and \(1.0 \mathrm{~mm}^{2}\) in cross- sectional area. It carries a current of \(4.0 \mathrm{~A}\) when a \(2.0 \mathrm{~V}\) potential difference is applied between its ends. Calculate the conductivity \(\sigma\) of Nichrome.

What is the current in a wire of radius \(R=3.40 \mathrm{~mm}\) 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?

Exploding shoes. The rain-soaked shoes of a person may explode if ground current from nearby lightning vaporizes the water. The sudden conversion of water to water vapor causes a dramatic expansion that can rip apart shoes. Water has density \(1000 \mathrm{~kg} / \mathrm{m}^{3}\) and requires \(2256 \mathrm{~kJ} / \mathrm{kg}\) to be vaporized. If horizontal current lasts \(2.00 \mathrm{~ms}\) and encounters water with resistivity \(150 \Omega \cdot \mathrm{m}\), length \(12.0 \mathrm{~cm}\), and vertical cross-sectional area \(15 \times 10^{-5} \mathrm{~m}^{2}\), what average current is required to vaporize the water?
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