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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?

Short Answer

Expert verified
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 by step solution

01

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.
02

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.
03

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 \( \vec{E} = \frac{V}{L} \).
04

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 \( \vec{E} = \frac{V}{L} \).
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A typical cost for electrical power is $0.120 per kilowatthour. (a) Some people leave their porch light on all the time. What is the yearly cost to keep a 75-W bulb burning day and night? (b) Suppose your refrigerator uses 400 W of power when it's running, and it runs 8 hours a day. What is the yearly cost of operating your refrigerator?

An overhead transmission cable for electrical power is 2000 m long and consists of two parallel copper wires, each encased in insulating material. A short circuit has developed somewhere along the length of the cable where the insulation has worn thin and the two wires are in contact. As a power-company employee, you must locate the short so that repair crews can be sent to that location. Both ends of the cable have been disconnected from the power grid. At one end of the cable (point \(A\)), you connect the ends of the two wires to a 9.00-V battery that has negligible internal resistance and measure that 2.86 A of current flows through the battery. At the other end of the cable (point \(B\)), you attach those two wires to the battery and measure that 1.65 A of current flows through the battery. How far is the short from point A?

A ductile metal wire has resistance \(R\). What will be the resistance of this wire in terms of \(R\) if it is stretched to three times its original length, assuming that the density and resistivity of the material do not change when the wire is stretched? (\(Hint:\) The amount of metal does not change, so stretching out the wire will affect its cross-sectional area.)

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?

The potential difference across the terminals of a battery is 8.40 V when there is a current of 1.50 A in the battery from the negative to the positive terminal. When the current is 3.50 A in the reverse direction, the potential difference becomes 10.20 V. (a) What is the internal resistance of the battery? (b) What is the emf of the battery?

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