Problem 56 - Fundamentals of Physics

Question

A potential difference of \(1.20 \mathrm{~V}\) will be applied to a \(33.0 \mathrm{~m}\) length of 18 -gauge copper wire (diameter \(=0.0400\) in.). Calculate (a) the current, (b) the magnitude of the current density, (c) the magnitude of the electric field within the wire, and (d) the rate at which thermal energy will appear in the wire.

Step-by-Step Solution

Answer

(a) 1.76 A, (b) \(2.17 \times 10^6 \text{ A/m}^2\), (c) 0.0364 V/m, (d) 2.11 W.

Step 1: Calculate Resistance

To find the current, we need to calculate the resistance using the formula:\[R = \frac{\rho \cdot L}{A}\]where \(\rho = 1.68 \times 10^{-8} \Omega \cdot \text{m}\) is the resistivity of copper, \(L = 33.0\, \text{m}\) is the length of the wire, and \(A\) is the cross-sectional area of the wire.First, convert the diameter from inches to meters: \(0.0400 \text{ in} \times 0.0254 = 0.001016 \text{ m}\).Calculate the area:\[A = \pi \left(\frac{0.001016}{2}\right)^2 = 8.107 \times 10^{-7} \text{ m}^2\]Now, substitute into the resistance formula:\[R = \frac{1.68 \times 10^{-8} \cdot 33.0}{8.107 \times 10^{-7}} = 0.683 \Omega\]

Step 2: Calculate Current

Use Ohm's Law to calculate the current, where \(V = 1.20 \text{ V}\) is the potential difference:\[I = \frac{V}{R} = \frac{1.20}{0.683} = 1.76 \text{ A}\]

Step 3: Calculate Current Density

Current density \(J\) is calculated using the formula:\[J = \frac{I}{A}\]where \(I = 1.76 \text{ A}\) and \(A = 8.107 \times 10^{-7} \text{ m}^2\).Substitute the values:\[J = \frac{1.76}{8.107 \times 10^{-7}} = 2.17 \times 10^6 \text{ A/m}^2\]

Step 4: Calculate Electric Field

The magnitude of the electric field \(E\) inside the wire can be calculated using:\[E = \frac{V}{L}\]where \(V = 1.20 \text{ V}\) and \(L = 33.0 \text{ m}\).Substitute the values:\[E = \frac{1.20}{33.0} = 0.0364 \text{ V/m}\]

Step 5: Calculate Power Dissipated as Thermal Energy

The rate at which thermal energy is generated (power) in the wire is calculated using:\[P = I^2 R\]where \(I = 1.76 \text{ A}\) and \(R = 0.683 \Omega\).Substitute the values:\[P = (1.76)^2 \times 0.683 = 2.11 \text{ W}\]

Key Concepts

Resistivity

Resistivity is a fundamental property of materials that quantifies how strongly a material opposes the flow of electric current. It is denoted by the Greek letter \( \rho \) and is measured in ohm-meters (\( \Omega \cdot \text{m} \)). The resistivity of a wire determines how much resistance it will have overall, influencing the wire's ability to conduct electricity effectively.

To calculate resistance, we use the formula \( R = \frac{\rho \cdot L}{A} \), where \( R \) is resistance, \( L \) is the length of the wire, and \( A \) is its cross-sectional area. The resistivity is material-specific; for instance, copper has a low resistivity of \( 1.68 \times 10^{-8} \Omega \cdot \text{m} \), making it an excellent conductor.

This concept helps in designing electrical circuits because selecting materials with appropriate resistivity can efficiently manage the current flow and energy losses.

Current Density

Current density, represented by \( J \), is a measure of the electric current (\( I \)) flowing per unit area \( A \) across a surface. It is expressed in amperes per square meter (\( \text{A/m}^2\)). Understanding current density is essential in assessing how current is distributed across the wire's cross-section.

We calculate current density using the formula \( J = \frac{I}{A} \). Where \( I \) is the total current flowing through the conductor and \( A \) is its cross-sectional area. In our exercise, the calculated current density was \( 2.17 \times 10^6 \text{ A/m}^2 \), showing that a significant amount of current flows through a small area.

High current density can lead to heating and damage in conductive materials if not managed correctly, making it a critical factor in circuit design.

Electric Field

The electric field \( E \) within a conductor is a vector quantity that describes the force per unit charge exerted on charged particles. It is expressed in volts per meter (\( \text{V/m} \)). The electric field is responsible for driving the flow of electrons, and thus the current, through a conductor.

In our exercise, the electric field is calculated using the formula \( E = \frac{V}{L} \), where \( V \) is the potential difference applied across the conductor, and \( L \) is the length of the conductor. The resulting electric field in this example is \( 0.0364 \text{ V/m} \), indicating how strongly the electrons are being pushed through the wire.

An understanding of electric fields is crucial for analyzing how voltage influences current flow in different materials and geometries.

Thermal Energy

Thermal energy in electrical circuits refers to the energy dissipated as heat due to the resistance that the electrons experience as they move through a conductor. This is commonly known as Joule heating and is quantified by the power formula \( P = I^2 R \), where \( P \) is the power (rate of energy dissipation), \( I \) is the current, and \( R \) is the resistance.

In our problem, this calculation reveals that the wire dissipates energy at a rate of \( 2.11 \text{ W} \). This energy appears as heat in the wire, which could lead to a temperature increase,
highlighting the importance of managing thermal effects in practical applications, such as preventing overheating in electrical components.

Understanding how thermal energy is generated helps in designing cooling systems and selecting suitable materials to enhance the safety and longevity of electrical systems.