Problem 55 - University Physics with Modern Physics

Question

A 3.00-m length of copper wire at 20\(^\circ\)C has a 1.20-mlong section with diameter 1.60 mm and a 1.80-m-long section with diameter 0.80 mm. There is a current of 2.5 mA in the 1.60- mm-diameter section. (a) What is the current in the 0.80-mmdiameter 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-m length of wire?

Step-by-Step Solution

Answer

The current remains 2.5 mA throughout, \( \vec{E} \) can be calculated using \( V = IR \) and \( \vec{E} = \frac{V}{L} \), and voltage sums for potential difference.

Step 1: Understand Ohm's Law and Resistivity

We need to analyze sections of the wire using Ohm's Law, which states \( V = IR \), where \( V \) is voltage, \( I \) is current, and \( R \) is resistance. The resistance \( R \) of a wire can also be expressed in terms of resistivity \( \rho \), length \( L \), and cross-sectional area \( A \): \( R = \frac{\rho L}{A} \). In steady state, the current \( I \) should be the same throughout the wire if all sections are connected in series.

Step 2: Calculate Current in 0.80 mm Section

Since the current in the 1.60 mm section is given as 2.5 mA, and the wire is continuous, by conservation of charge, current remains the same through the wire. Therefore, the current in the 0.80 mm section is also 2.5 mA.

Step 3: Calculate Electric Field in 1.60-mm Section

The magnitude of the electric field \( \vec{E} \) in a conductor is related to the current and resistivity: \( \vec{E} = \frac{V}{L} \). First, find the resistance of this section. Cross-sectional area \( A = \pi \left( \frac{1.60}{2} \right)^2 \), and \( R = \frac{\rho L}{A} \). Use \( V = IR \) to find \( V \), then \( \vec{E} = \frac{V}{L} \).

Step 4: Calculate Electric Field in 0.80-mm Section

Similarly, calculate the cross-sectional area for the 0.80-mm section: \( A = \pi \left( \frac{0.80}{2} \right)^2 \). Find resistance \( R \), then voltage \( V = IR \) considering the same current of 2.5 mA. Finally, calculate \( \vec{E} = \frac{V}{L} \).

Step 5: Calculate the Potential Difference Across the 3.00-m Length of Wire

The potential difference along the entire wire is the sum of potential differences across each section. Use \( V_{total} = V_1 + V_2 \) where \( V_1 \) and \( V_2 \) are the voltages across the 1.20 m and 1.80 m sections calculated by \( V = IR \) for each.

Key Concepts

Electrical Resistance

Electrical resistance is a crucial concept in understanding how electrical circuits work. It represents how much a material resists the flow of electric current. The resistance of a conductor is determined by several factors:

• Length: The longer the conductor, the greater the resistance, as it provides more material for the electrons to collide with.
• Cross-sectional Area: A wider conductor has a lower resistance because it allows more electrons to flow through it at once.
• Material: Different materials have different inherent resistances, known as resistivity.

When calculating resistance in the context of Ohm's Law, we use the formula \( R = V/I \), where \( V \) is voltage, \( I \) is current, and \( R \) is resistance. In exercises involving wires, it's common to express the resistance \( R \) in terms of the wire's resistivity \( \rho \), its length \( L \), and its cross-sectional area \( A \): \( R = \frac{\rho L}{A} \). Understanding resistance helps us predict how much energy will be lost as electrical current passes through a wire, crucial for designing efficient circuits. Breaking up the text simplifies understanding the direct relationships these factors have with resistance.

Resistivity

Resistivity is a material-specific property that measures how strongly a material opposes the flow of electric current. It's denoted by the Greek letter \( \rho \) and typically measured in ohm-meters \((\Omega \cdot m)\). Unlike resistance, which is dependent on the dimensions and configuration of the conductor, resistivity is an inherent property of the material itself.

A few key points on resistivity:

• Temperature Dependency: Resistivity of materials often changes with temperature. For most conductive materials, resistivity increases as temperature rises.
• Material Variation: Conductors, semiconductors, and insulators all have different typical resistivities, affecting how readily they allow current to pass.

Knowing the resistivity of a material helps in calculating the resistance of the wire via the formula \( R = \frac{\rho L}{A} \), where \( L \) is the length and \( A \) is the cross-sectional area.
When solving problems involving resistivity, it's essential to factor in the type of material and its environmental conditions, as these directly impact the resultant electrical resistance.

Electric Field

An electric field \( \vec{E} \) is a vector field surrounding electric charges that exert forces on other charges within the field. It's a crucial concept when examining how charges move in a circuit. The electric field inside a conductor is directly related to the potential difference across it and the length over which the potential difference exists.

To calculate the electric field within a conductor:

• Voltage Relationship: The electric field can be expressed as \( \vec{E} = \frac{V}{L} \), where \( V \) is the potential difference and \( L \) is the length of the conductor section.
• Current Flow Influence: The electric field within a conductor drives the current flow through it, meaning higher fields result in higher currents, assuming resistance remains constant.

In practice, calculating the electric field is crucial for understanding how circuits operate at a fundamental level, ensuring components function within their design specifications and circuits are powered efficiently and safely. This keeps devices running smoothly and prevents overheating or damage.