Problem 54 - University Physics with Modern Physics

Question

A 2.0-m length of wire is made by welding the end of a 120-cm-long silver wire to the end of an 80-cm-long copper wire. Each piece of wire is 0.60 mm in diameter. The wire is at room temperature, so the resistivities are as given in Table 25.1. A potential difference of 9.0 V is maintained between the ends of the 2.0-m composite wire. What is (a) the current in the copper section; (b) the current in the silver section; (c) the magnitude of \(\vec E\) in the copper; (d) the magnitude of \(\vec E\) in the silver; (e) the potential difference between the ends of the silver section of wire?

Step-by-Step Solution

Answer

(a) 78.1 A, (b) 78.1 A, (c) 4.63 V/m, (d) 4.41 V/m, (e) 5.29 V.

Step 1: Understand the Problem

We have a composite wire consisting of a silver section and a copper section, each with given lengths and a given diameter. We need to find current in each section, the electric field in each section, and the potential difference across the silver section, all while knowing the total potential difference across both sections is 9.0 V.

Step 2: Calculate Resistances

First, we need to calculate the resistance of each section using the formula \( R = \frac{\rho L}{A} \), where \( R \) is resistance, \( \rho \) is resistivity, \( L \) is length, and \( A \) is the cross-sectional area. The resistivity values for silver and copper can be taken from reference sources.

Step 3: Calculate Cross-Sectional Area

The cross-sectional area \( A \) is calculated using \( A = \pi r^2 \). Given diameter is 0.60 mm, the radius \( r = \frac{0.60}{2} \) mm = 0.30 mm = 0.0003 m, hence, \[ A = \pi (0.0003)^2 \approx 2.827 \times 10^{-7} \ m^2. \]

Step 4: Calculate Resistance of Silver Section

Given \( L_{Ag} \) = 120 cm = 1.2 m and \( \rho_{Ag} \approx 1.59 \times 10^{-8} \ \Omega\text{m} \), \[ R_{Ag} = \frac{1.59 \times 10^{-8} \times 1.2}{2.827 \times 10^{-7}} \approx 0.0677 \ \Omega. \]

Step 5: Calculate Resistance of Copper Section

Given \( L_{Cu} \) = 80 cm = 0.8 m and \( \rho_{Cu} \approx 1.68 \times 10^{-8} \ \Omega\text{m} \), \[ R_{Cu} = \frac{1.68 \times 10^{-8} \times 0.8}{2.827 \times 10^{-7}} \approx 0.0475 \ \Omega. \]

Step 6: Calculate Total Resistance

The total resistance of the wire \( R_{total} = R_{Ag} + R_{Cu} = 0.0677 + 0.0475 \approx 0.1152 \ \Omega \).

Step 7: Calculate Total Current

Using Ohm's Law, \( V = IR \), where \( V = 9.0 \ V \), we find the total current,\[ I = \frac{9.0}{0.1152} \approx 78.1 \ A. \]This current is the same throughout the entire length of the wire.

Step 8: Calculate Electric Field in Silver Section

The electric field \( E \) is given by \( E = \frac{V}{L} \). For the silver section, \[ E_{Ag} = \frac{IR_{Ag}}{L_{Ag}} = \frac{78.1 \times 0.0677}{1.2} \approx 4.41 \ V/m. \]

Step 9: Calculate Electric Field in Copper Section

Similarly, for the copper section, \[ E_{Cu} = \frac{IR_{Cu}}{L_{Cu}} = \frac{78.1 \times 0.0475}{0.8} \approx 4.63 \ V/m. \]

Step 10: Calculate Potential Difference in Silver Section

The potential difference is given by \( V = IR \). For the silver section, \[ V_{Ag} = 78.1 \times 0.0677 \approx 5.29 \ V. \]

Key Concepts

Ohm's Law

Ohm's Law is a fundamental principle in electronics used to relate voltage, current, and resistance. The formula is expressed as \( V = IR \), where \( V \) is the voltage (in volts), \( I \) is the current (in amperes), and \( R \) is the resistance (in ohms). This implies that the current through a conductor between two points is directly proportional to the voltage across the two points, given the resistance remains constant.
Let's break it down with a practical example related to the exercise. We calculated the total current flowing through both the silver and copper sections of the composite wire. The total resistance of the wire was calculated to be approximately 0.1152 ohms.
With a given potential difference (voltage) of 9.0 V across the wire, we can calculate the current using Ohm's Law. Rearranging for current, \( I \) becomes \( I = \frac{V}{R} \). Plugging in our values, \( I = \frac{9.0}{0.1152} \), resulting in approximately 78.1 A.
This principle is crucial because it helps us predict how much current will flow under certain conditions and is applicable to all conductive materials.

Electric Field Calculation

The electric field within a conductor is an important concept in understanding how electric charges move through a material. It is defined as the force experienced per unit charge and is given by \( E = \frac{V}{L} \), where \( E \) is the electric field (in volts per meter), \( V \) is the potential difference across a section of the conductor, and \( L \) is the length of the section.
In the composite wire problem, we calculated the electric fields for both the silver and copper sections. To find the electric field in the silver section, we used the formula \( E_{Ag} = \frac{IR_{Ag}}{L_{Ag}} \), substituting in the values for current, resistance of the silver section, and its length to get \( E_{Ag} \approx 4.41 \ V/m \).
Similarly, for the copper section, the formula yields \( E_{Cu} = \frac{IR_{Cu}}{L_{Cu}} \) which results in approximately \( 4.63 \ V/m \).
Understanding the electric field is critical as it provides insights into how effectively a material can carry electric charges, which in turn can affect how efficiently electrical signals are transported over long distances.

Resistivity

Resistivity is a material's intrinsic property which indicates how much it resists the flow of electric current. Represented by \( \rho \) (in ohm-meters), it is crucial in calculating the resistance of a wire, given by \( R = \frac{\rho L}{A} \), where \( R \) is resistance, \( L \) is length, and \( A \) is cross-sectional area.
In our exercise, we needed to calculate the resistances of individual sections of silver and copper wires. For silver, with a length of 1.2 m and resistivity \( \rho_{Ag} \approx 1.59 \times 10^{-8} \ \Omega \text{m} \), and for copper, with a length of 0.8 m and \( \rho_{Cu} \approx 1.68 \times 10^{-8} \ \Omega \text{m} \), together with a cross-sectional area \( A \approx 2.827 \times 10^{-7} \ m^2 \), resistances were calculated.
These values were used to understand not just how much each material resists current, but also to combine them to know the overall impact on the composite wire's ability to conduct electricity.
Understanding resistivity helps in selecting the right materials for electrical applications, balancing between factors like material cost, conductivity, and wire size for practical design.