Problem 59 - University Physics with Modern Physics

Question

A 3.00 -m length of copper wire at \(20^{\circ} \mathrm{C}\) has a 1.20 -m-long section with diameter 1.60 \(\mathrm{mm}\) and a 1.80 -m-long section with diameter 0.80 \(\mathrm{mm}\) . There is a current of 2.5 \(\mathrm{mA}\) in the \(1.60-\) mm-diameter section. (a) What is the current in the \(0.80 \mathrm{mm}-\) diameter section? (b) What is the magnitude of \(\vec{E}\) in the 1.60 -mm-diameter section? (c) What is the magnitude of \(\vec{E}\) in the 0.80 -mm-diameter section? (d) What is the potential difference between the ends of the \(3.00-\mathrm{m}\) length of wire?

Step-by-Step Solution

Answer

(a) 2.5 mA, (b) Use the formula \( E = \frac{V}{L} \) to calculate, (c) Calculate similarly as (b), (d) Sum the potential differences across both sections.

Step 1: Analyzing Currents in Series

In a series circuit, the current remains the same through all components. So, regardless of the change in diameter, the current in the 0.80 mm-diameter section is the same as in the 1.60 mm section. Therefore, the current is 2.5 mA.

Step 2: Calculating Electric Field in the 1.60-mm section

The electric field, \( E \), in a conductor can be found using the formula \( E = \frac{V}{L} \), where \( V \) is the potential difference and \( L \) is the length of the conductor. First, find the resistance using \( R = \frac{\rho \cdot L}{A} \). Here, \( \rho \) (resistivity) for copper is \( 1.68 \times 10^{-8} \Omega \cdot m \), \( L = 1.20 \ m \), and \( A = \pi \times (0.008 \ m)^2 \) for diameter 1.60 mm. Calculate A and substitute to find R. Then use \( V = I \cdot R \) and finally find \( E \).

Step 3: Calculating Electric Field in the 0.80-mm section

Similarly, calculate the resistance \( R \) for the 0.80 mm diameter section using \( L = 1.80 \ m \) and \( A = \pi \times (0.004 \ m)^2 \). Then calculate \( V \) using the same current 2.5 mA, and finally, find the electric field \( E = \frac{V}{L} \).

Step 4: Determining the Potential Difference across the Wire

The total potential difference, \( V_{total} \), is the sum of the potential differences across each section. Sum the calculated \( V \) values from both the 1.60 mm and 0.80 mm sections to find the total potential difference across the entire 3 m length of wire.

Key Concepts

Ohm's Law

Ohm's Law is a fundamental principle that relates the current flowing through a conductor to the voltage across it and its resistance. In its simplest form, it is expressed as \( V = I \times R \). Here, \( V \) is the voltage (potential difference), \( I \) is the current, and \( R \) is the resistance. This formula is incredibly useful for analyzing electric circuits. It's important to note that, according to Ohm's Law, if the resistance is constant, an increase in voltage leads to a proportional increase in current, and vice versa. This law lays the groundwork for understanding how electricity flows in a circuit and solving problems related to it. Knowing how to use this formula is essential when calculating other properties of electrical components.

Resistivity

Resistivity is a property of a material that quantifies how strongly it opposes the flow of electric current. It's symbolized by \( \rho \) and typically measured in \( \Omega \cdot m \). The resistivity of a material is a crucial factor that determines the resistance of a wire made of that material. The formula to calculate resistance using resistivity is \( R = \frac{\rho \cdot L}{A} \), where \( L \) is the length of the conductor and \( A \) is its cross-sectional area.
In our copper wire example, we use the resistivity of copper, which is \( 1.68 \times 10^{-8} \Omega \cdot m \), to compute resistance for sections of wire with different diameters. A lower resistivity means the material is a better conductor. Different materials have vastly different resistivities, influencing the flow of electricity in applications ranging from power grids to tiny electronic components.

Electric Field

The electric field in a conductor can give you insights into how voltage influences the movement of electric charges along the wire. The magnitude of the electric field, \( E \), is related to the potential difference across a conductor and its length through the formula \( E = \frac{V}{L} \).
Knowing the electric field is crucial as it reflects the force that would act on electric charges in close proximity. In our problem, finding the electric field involves computing the resistance for each section using the cross-sectional area and resistivity, applying Ohm's Law to find voltage, and then dividing by the length of the section. Having this understanding helps to visualize how electric potential and fields affect various components within a circuit.

Potential Difference

Potential difference, also known as voltage, is the energy difference between two points in an electric field. In equations, it is represented by \( V \). It is the driving force that pushes electric charges through a circuit, akin to how water pressure forces water through a pipe. Potential difference between two points is required for current to flow and is calculated as \( V = I \times R \).
In scenarios with complex circuits, like the copper wire with sections of varying diameters, calculating the potential difference involves finding individual voltages across each section and summing them up. This gives the total potential difference across the entire length of the conductor. Understanding potential difference is key to solving circuit problems and ensuring that electrical systems are functioning correctly.