Problem 46 - Fundamentals of Physics

Question

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 \(30 \mathrm{~min} ?\)

Step-by-Step Solution

Answer

(a) 1.68 × 10^{-2} V/m; (b) 241.92 J.

Step 1: Identify the relationship between current, resistivity, and electric field

The magnitude of the electric field \( E \) in a wire can be related to the current \( I \), the resistivity \( \rho \), the cross-sectional area \( A \), and the length \( L \) using the formula \( E = \rho \cdot \frac{I}{A} \). These variables describe how an electric field is developed along a conductive wire.

Step 2: Calculate the resistivity of copper

The resistivity \( \rho \) of copper is commonly given as \( 1.68 \times 10^{-8} \ \Omega\,\mathrm{m} \). This is a characteristic constant for copper materials, used in calculations involving copper conductors.

Step 3: Calculate the electric field along the wire

Use the previously identified formula to find the electric field: \[E = \frac{1.68 \times 10^{-8} \ \Omega\,\mathrm{m} \times 2.00 \ \mathrm{A}}{2.00 \times 10^{-6} \ \mathrm{m}^{2}} = 1.68 \times 10^{-2} \ \mathrm{V/m}.\]Therefore, the electric field along the wire is \( 1.68 \times 10^{-2} \ \mathrm{V/m} \).

Step 4: Calculate the potential difference across the wire

The potential difference \( V \) across the wire can be calculated using the formula \( V = E \cdot L \), where \( L \) is the length of the wire. Thus, \[V = 1.68 \times 10^{-2} \ \mathrm{V/m} \times 4.00 \ \mathrm{m} = 6.72 \times 10^{-2} \ \mathrm{V}.\]

Step 5: Calculate the power used by the wire

The power \( P \) used by the wire can be calculated using \( P = I \cdot V \). Substituting the potential difference and the current, \[P = 2.00 \ \mathrm{A} \times 6.72 \times 10^{-2} \ \mathrm{V} = 1.344 \times 10^{-1} \ \mathrm{W}.\]

Step 6: Calculate the energy converted to thermal energy

The energy \( E_{therm} \) converted to thermal energy in time \( t \) is calculated with \( E_{therm} = P \cdot t \), where \( P \) is the power calculated before, and \( t = 30 \ \mathrm{min} = 1800 \ \mathrm{s} \). Therefore,\[E_{therm} = 1.344 \times 10^{-1} \ \mathrm{W} \times 1800 \ \mathrm{s} = 241.92 \ \mathrm{J}.\]

Key Concepts

Current

Current, denoted as \( I \), refers to the flow of electric charge through a conductor, such as a wire. It is measured in amperes (A), which is the rate at which charge flows past a point in the circuit. For example, in this exercise, a current of 2.00 A is flowing through a copper wire.

Understanding current is central to working with electric circuits. It can be thought of as the number of electrons passing a specific point in the conductor per unit of time.

• A consistent current flow, like the one in the exercise, signifies a stable electric system.
• The direction of current flow is conventionally from positive to negative, although electrons move from negative to positive.

In practical applications, adjusting the amount of current can affect how much power an electrical device uses. Hence, monitoring and controlling electric current is essential for efficiency and safety in many technologies.

Resistivity

Resistivity is a fundamental property that quantifies how strongly a material opposes the flow of electric current. It is symbolized by \( \rho \) and has units of \( \Omega\,\mathrm{m} \).

Resistivity can be influenced by several factors, including temperature and the nature of the material itself.

• In copper, a commonly used conductive material, resistivity is typically about \( 1.68 \times 10^{-8} \ \Omega\,\mathrm{m} \). This low resistivity makes it an excellent conductor.
• Higher resistivity will hinder the flow of current and result in more energy being lost as heat.

The resistivity of different materials makes them suitable for different applications; low resistivity materials are used in wiring, while high resistivity materials might be used for insulators.

In the context of the exercise, resistivity is crucial for understanding how the electric field is distributed across the wire. The formula \( E = \rho \cdot \frac{I}{A} \) connects resistivity to the electric field and current, revealing how well the wire can conduct electricity.

Thermal Energy Conversion

Electrical energy can be converted into thermal energy, which is a form of energy that manifests as heat. This transformation is key in understanding many applications, from electric heaters to cooking appliances.

When electric current flows through a conductor, a portion of the electrical energy is converted into thermal energy due to the resistivity of the material.

• This energy conversion in resistive components is described by the formula \( E_{therm} = P \cdot t \), which computes the amount of heat generated over time.
• The power \( P \) in the formula is given by \( P = I \cdot V \), linking it directly to the current and the potential difference across the conductor.
• The time factor \( t \) determines how long the energy conversion process continues, indicating total energy produced.

In our example, the calculation revealed that in 30 minutes, the copper wire converted around 241.92 joules of electrical energy into thermal energy. This conversion illustrates how energy efficiency is impacted by resistance and the duration of current flow.