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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
1.35 x 10^5 A/m虏; 0.094 V

Step by step solution

01

Identify Given Values

We need to identify the given values from the problem. The radius of the cylinder is given as \( r = 5.0 \text{ mm} = 0.005 \text{ m} \). The length of the cylinder (which is the same as the height for a resistor) is \( L = 2.0 \text{ cm} = 0.02 \text{ m} \). The resistivity of the material is \( \rho = 3.5 \times 10^{-5} \Omega \cdot \text{m} \). The power dissipation rate is \( P = 1.0 \text{ W} \).
02

Calculate Cross-Sectional Area

The cross-sectional area \( A \) of the cylindrical resistor is the area of a circle, which is calculated using the formula \( A = \pi r^2 \). Substituting the radius, we have \( A = \pi (0.005)^2 \approx 7.85 \times 10^{-5} \text{ m}^2 \).
03

Apply the Formula for Resistance

The resistance \( R \) of the cylindrical resistor is calculated using the formula \( R = \frac{\rho L}{A} \). Substituting the given values, we have \( R = \frac{3.5 \times 10^{-5} \cdot 0.02}{7.85 \times 10^{-5}} \approx 0.0089 \Omega \).
04

Use Power Formula to Find Current

We know that power \( P \) is given by \( P = I^2 R \) where \( I \) is the current. Rearranging to solve for \( I \), we have \( I = \sqrt{\frac{P}{R}} \). Substituting the values, we find \( I = \sqrt{\frac{1.0}{0.0089}} \approx 10.6 \text{ A} \).
05

Calculate Current Density

The current density \( J \) is the current per unit area, given by \( J = \frac{I}{A} \). Substituting the known values, we have \( J = \frac{10.6}{7.85 \times 10^{-5}} \approx 1.35 \times 10^5 \text{ A/m}^2 \).
06

Calculate Potential Difference

Using Ohm's Law, the potential difference \( V \) across the resistor is given by \( V = IR \). Substituting, we obtain \( V = 10.6 \times 0.0089 \approx 0.094 \text{ V} \).

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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 that quantifies how strongly a given material opposes the flow of electric current. It is often denoted by the symbol \( \rho \) and is measured in ohm-meters (\( \Omega \cdot \text{m} \)). Materials with high resistivity make poor conductors of electricity; conversely, low resistivity indicates good conductivity. Some important points to remember about resistivity include:
  • It is intrinsic to the material type: different materials have characteristic resistivities.
  • Resistivity is affected by temperature. For most conductors, resistivity increases with temperature.
  • In the context of a resistor, resistivity helps determine the overall resistance using the formula \( R = \frac{\rho L}{A} \), where \( R \) is resistance, \( L \) is length, and \( A \) is the cross-sectional area.
The knowledge of resistivity can help us design circuits with the desired resistance by selecting the appropriate material, size, and shape.
Current Density
Current density \( J \) is a measure of how much electric current (\( I \)) flows through a given cross-sectional area (\( A \)) of a conductor. It is expressed in the unit of amperes per square meter (\( \text{A/m}^2 \)). The formula for calculating current density is:
  • \( J = \frac{I}{A} \)
This concept is crucial because it provides insight into how the current is distributed over a surface. Current density can vary greatly depending on both the geometry of the conductor and the total current passing through it. Here are key pointers on current density:
  • High current density can lead to overheating and potentially damage the conductor.
  • Engineers often manage current density to ensure efficient power delivery and avoid safety risks.
Understanding current density is vital for designing electrical components that are both safe and efficient.
Potential Difference
Potential difference, often called voltage (\( V \)), is the electrical pressure that drives current through a circuit. It is measured in volts (\( V \)) and correlates to the energy per unit charge supplied by a source like a battery. A few essential aspects of potential difference include:
  • It is the difference in electric potential between two points in a circuit.
  • Potential difference is what causes current to flow. Without it, electrons remain stationary.
  • It can be calculated using Ohm's Law in the context of a resistor: \( V = IR \), where \( I \) is current and \( R \) is resistance.
In practical terms, potential difference helps us understand how components within a circuit interact; ensuring a circuit functions as intended requires appropriate voltage levels for each component.
Ohm's Law
Ohm's Law is a fundamental principle in the study of electricity. It relates the current flow in a conductor to the potential difference (voltage) across it and the resistance it offers. It is represented by the equation:
  • \( V = IR \)
where \( V \) is voltage in volts, \( I \) is current in amperes, and \( R \) is resistance in ohms. Important insights derived from Ohm's Law include:
  • The current through a conductor between two points is directly proportional to the voltage across the two points.
  • For a constant resistance, increasing voltage will increase the current.
  • It forms the basis for many practical calculations in electronic circuits, helping to design systems with specific current and voltage requirements.
Utilizing Ohm's Law is essential for understanding and predicting electrical behavior in both simple and complex circuits.

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Most popular questions from this chapter

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 fuse in an electric circuit is a wire that is designed to melt, and thereby open the circuit, if the current exceeds a predetermined value. Suppose that the material to be used in a fuse melts when the current density rises to \(440 \mathrm{~A} / \mathrm{cm}^{2} .\) What diameter of cylindrical wire should be used to make a fuse that will limit the current to \(0.50 \mathrm{~A}\) ?

A certain wire has a resistance \(R\). What is the resistance of a second wire, made of the same material, that is half as long and has half the diameter?

A copper wire of cross-sectional area \(2.00 \times 10^{-6} \mathrm{~m}^{2}\) and length \(4.00 \mathrm{~m}\) has a current of \(2.00\) A uniformly distributed across that area. (a) What is the magnitude of the electric field along the wire? (b) How much electrical energy is transferred to thermal energy in 30 min?

A small but measurable current of \(1.2 \times 10^{-10} \mathrm{~A}\) exists in a copper wire whose diameter is \(2.5 \mathrm{~mm}\). The number of charge carriers per unit volume is \(8.49 \times 10^{28} \mathrm{~m}^{-3}\). Assuming the current is uniform, calculate the (a) current density and (b) electron drift speed.
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