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A copper wire of cross-sectional area \(2.40 \times 10^{-6} \mathrm{~m}^{2}\) and length \(4.00 \mathrm{~m}\) has a current of \(2.00 \mathrm{~A}\) 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 \(30 \mathrm{~min} ?\)

Short Answer

Expert verified
(a) The electric field magnitude is approximately 0.014 V/m. (b) 201.6 J of energy is transferred to thermal energy in 30 minutes.

Step by step solution

01

Calculate the Resistance of the Wire

The resistance of the wire, denoted as \( R \), can be calculated using the formula for resistivity: \( R = \frac{\rho \cdot L}{A} \), where \( \rho \) is the resistivity of copper (approximately \( 1.68 \times 10^{-8} \, \Omega \cdot \text{m} \)), \( L \) is the length of the wire (\( 4.00 \text{ m} \)), and \( A \) is the cross-sectional area (\( 2.40 \times 10^{-6} \, \text{m}^2 \)). Substituting these values gives \( R = \frac{1.68 \times 10^{-8} \times 4.00}{2.40 \times 10^{-6}} \approx 0.028 \Omega \).
02

Calculate the Electric Field Along the Wire

To find the electric field, use the relation \( V = E \cdot L \) and Ohm's Law \( V = I \cdot R \), combining them we have \( E = \frac{V}{L} = \frac{I \cdot R}{L} \). Substituting \( I = 2.00 \text{ A} \), \( R \approx 0.028 \Omega \), and \( L = 4.00 \text{ m} \), gives \( E = \frac{2.00 \times 0.028}{4.00} \approx 0.014 \, \text{V/m} \).
03

Calculate Energy Transferred in 30 Minutes

The electrical power \( P \) is given by \( P = I^2 \cdot R \). With \( I = 2.00 \text{ A} \) and \( R \approx 0.028 \Omega \), we find \( P = 2.00^2 \times 0.028 \approx 0.112 \, \text{W} \). The energy transferred over \( 30 \) minutes (or \( 1800 \) seconds) is \( E = P \times t = 0.112 \times 1800 \approx 201.6 \, \text{J} \).

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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 measures their ability to resist the flow of electric current. Essentially, it tells us how strongly a material opposes the movement of electric charges within it. The resistivity of a conductor depends on its material, length, and cross-sectional area, governed by the formula:
  • \[ R = \frac{\rho \cdot L}{A} \]
  • Where \( R \) is the resistance, \( \rho \) is the resistivity, \( L \) is the length of the conductor, and \( A \) is its cross-sectional area.
For example, copper is widely used in electrical wiring because it has a low resistivity value of approximately \( 1.68 \times 10^{-8} \, \Omega \cdot \text{m} \). This means it allows electric current to pass through it easily with minimal resistance.
The resistivity formula helps calculate the resistance of a wire. This is significant as it directly impacts how well electrical devices can perform. The lower the resistance, the more efficiently electricity can be conducted, minimizing energy losses.
Ohm's Law
Ohm's Law is a key principle in physics that relates the electric current flowing through a conductor to the voltage across it and its resistance. The law is succinctly expressed by the equation:
  • \[ V = I \cdot R \]
  • Where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance.
This relationship indicates that the voltage across a conductor is directly proportional to the current flowing through it, assuming the resistance remains constant.
In practical terms, Ohm's Law can be useful for determining any one of these three variables if the other two are known. In the context of the exercise, it allowed us to find the electric field along a copper wire. By combining Ohm's Law with the formula \( V = E \cdot L \) where \( E \) is the electric field, we can determine:
  • \[ E = \frac{I \cdot R}{L} \]
  • This showcases the interconnectedness of electric fields, current, voltage, and resistance in a conduction scenario.
Electrical Energy Transformation
Electrical energy transformation is the process by which electrical energy is converted into another form, typically thermal energy, when it passes through a conductor with resistance. This energy transformation is a critical concept for understanding power dissipation in electrical circuits.
The rate at which electrical energy is converted into heat energy in a resistor is given by the power formula:
  • \[ P = I^2 \cdot R \]
  • Where \( P \) is the power, \( I \) is the current, and \( R \) is the resistance.
This equation shows that the power, and thus the amount of energy converted to heat, is proportional to the square of the current flowing through the resistor.
In real-world applications, this energy conversion explains why electrical appliances heat up during operation. For instance, the exercise demonstrates how much electrical energy is transformed into thermal energy over \( 30 \) minutes by using the wire's resistance and the constant current passed through it.
  • The total energy transformed was calculated using: \[ E = P \times t \]
  • Where \( E \) is the energy in joules, \( P \) is power in watts, and \( t \) is the time in seconds.

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Most popular questions from this chapter

Kiting during a storm. 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 \(1.80 \mathrm{~km}\) 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 \(213 \mathrm{MV}\), 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 \(500000 \mathrm{~A}\) (way beyond just being lethal).

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