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\(A 3.00 \mathrm{~m}\) length of copper wire at \(20^{\circ} \mathrm{C}\) has a \(1.20-\mathrm{m}\) -long section with diameter \(1.60 \mathrm{~mm}\) and a \(1.80-\mathrm{m}\) -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 -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?

Short Answer

Expert verified
a) The current in the 0.80 -mm-diameter section is \(2.5 \mathrm{mA}\). b) The magnitude of \(\vec{E}\) in the 1.60 -mm-diameter section is \(2.19 \times 10^{-4} \mathrm{V/m}\). c) The magnitude of \(\vec{E}\) in the 0.80 -mm-diameter section is \(8.67 \times 10^{-4} \mathrm{V/m}\). d) The potential difference between the ends of the \(3.00 \mathrm{~m}\) length of wire is \(4.66 \times 10^{-4} \mathrm{V}\).

Step by step solution

01

- Calculate the Cross-Sectional Areas

First, we need to calculate the cross-sectional area of each section of the wire. For a wire with a circular cross-section, this area can be calculated as \(\pi (d/2)^2\), where \(d\) is the diameter. Here, areas \(A_1\) and \(A_2\) of the first and second sections of the wire are \(A_1 = \pi (1.60 \mathrm{mm}/2)^2 = 2.01 \mathrm{mm}^2\) and \(A_2 = \pi (0.80 \mathrm{mm}/2)^2 = 0.50 \mathrm{mm}^2\) respectively.
02

- Apply Current Conservation

Because the copper wire is a single, continuous conductor, the electric current \(I\) must be conserved throughout the wire, regardless of its cross-sectional area or resistivity. Hence, the current \(I\) in the second (0.80 mm-diameter) section is also \(I = 2.5 \mathrm{mA}\).
03

- Compute Electric Fields

Now we can compute the magnitude of the electric field \(\vec{E}\) in the two sections. According to Ohm's law in a differential form, we have \(J = \sigma E\), where \(J = I/A\) is the electric current density and \(\sigma = 1/\rho\) is the electrical conductivity, with \(\rho\) being the resistivity of the copper wire. \n For the first section, \(E_1 = J/\sigma = I/(A_1 \sigma) = 2.5 \mathrm{mA}/(2.01 \mathrm{mm}^2 \times 5.8 \times 10^7 \mathrm{S/m}) = 2.19 \times 10^{-4} \mathrm{V/m}\).\n Similarly, for the second section, \(E_2 = J/\sigma = I/(A_2 \sigma) = 2.5 \mathrm{mA}/(0.50 \mathrm{mm}^2 \times 5.8 \times 10^7 \mathrm{S/m}) = 8.67 \times 10^{-4} \mathrm{V/m}\).
04

- Calculate Potential Difference

As a final step, we calculate the total potential difference between the ends of the 3 m length of wire. This includes the contribution from both sections. The electric field inside a homogeneous material can be related to the potential difference \(V\) and the material's length through the relationship \(V = E \cdot d\), where \(d\) is the length and \(E\) is the electric field. Hence, the total potential difference \(V\) will be \(V = E_1 \cdot d_1 + E_2 \cdot d_2 = 2.19 \times 10^{-4} \mathrm{V/m} \times 1.2 \mathrm{m} + 8.67 \times 10^{-4} \mathrm{V/m} \times 1.8 \mathrm{m} = 4.66 \times 10^{-4} \mathrm{V}.\)

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Key Concepts

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

Electric Current
Electric current is the flow of electric charge through a conductor, like a wire. It is measured in amperes (A), with the smaller unit being milliampere (mA). In the given exercise, the wire carries a current of 2.5 mA in each section, despite having different diameters.

This uniformity occurs because of the principle of current conservation, which states that the same amount of current flows through every section of a conductor in series. If more electrons enter one part of the wire, the same number of electrons must leave the other end.

Current conservation ensures a continuous flow of electricity, making electrical circuits work efficiently. Remember, in practice, wire thickness does not affect the current; it affects resistance and other properties.
Resistivity
Resistivity is a property of materials that measures how strongly they resist electric current. It is denoted by the Greek letter \(\rho\) and measured in ohm-meters (Ω·m). Materials with low resistivity, like copper, allow electric current to flow easily.

In our exercise, resistivity helps calculate the electric field within each wire section, using the relationship of resistivity with conductivity: \(\sigma = 1/\rho\). This relationship is crucial in determining how conductive a material is, with higher conductivity indicating lower resistivity.

Understanding resistivity is essential for designing electrical components that effectively carry current with minimal energy loss.
Potential Difference
Potential difference, also known as voltage, is the work done to move a charge between two points in an electric field. It is measured in volts (V). In this exercise, we calculate the potential difference across the entire length of a 3-meter wire.

Ohm's Law relates potential difference to current and resistance: \(V = I \cdot R\), but when dealing with electric fields in continuous conductors, we use \(V = E \cdot d\), where \(E\) is the electric field and \(d\) is the distance.

The total voltage over our wire consists of the contributions from both sections. Understanding how potential difference works and how it drives currents in circuits helps engineers design more efficient electrical systems.
Electric Field
An electric field is a region around charged particles where electric forces can be observed. This field can exert forces on other charged objects within the field, influencing how current flows through materials.

In the context of the exercise, we calculate the electric field magnitudes in different sections of a copper wire by using the relationship between current, area, and resistivity. This is expressed with the formula \(E = J / \sigma\), where \(J = I / A\) is the current density.

Understanding electric fields is crucial for grasping how electrical forces affect charges and how circuits function. With a solid knowledge of electric fields, predicting and manipulating the behavior of electric currents becomes more intuitive.

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Most popular questions from this chapter

\(\mathrm{CP}\) BIO Struck by Lightning. Lightning strikes can involve currents as high as 25,000 A that last for about \(40 \mu \mathrm{s}\). If a person is struck by a bolt of lightning with these properties, the current will pass through his body. We shall assume that his mass is \(75 \mathrm{~kg}\), that he is wet (after all, he is in a rainstorm) and therefore has a resistance of \(1.0 \mathrm{k} \Omega\), and that his body is all water (which is reasonable for a rough, but plausible, approximation). (a) By how many degrees Celsius would this lightning bolt increase the temperature of \(75 \mathrm{~kg}\) of water? (b) Given that the internal body temperature is about \(37^{\circ} \mathrm{C}\), would the person's temperature actually increase that much? Why not? What would happen first?

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

Light Bulbs. The power rating of a light bulb (such as a \(100 \mathrm{~W}\) bulb is the power it dissipates when connected across a \(120 \mathrm{~V}\) potential difference. What is the resistance of (a) a \(100 \mathrm{~W}\) bulb and (b) a \(60 \mathrm{~W}\) bulb? (c) How much current does each bulb draw in normal use?

A metal wire has a circular cross section with radius \(0.800 \mathrm{~mm}\) You measure the resistivity of the wire in the following way: You connect one end of the wire to one terminal of a battery that has emf \(12.0 \mathrm{~V}\) and negligible internal resistance. To the other terminal of the battery you connect a point along the wire so that the length of wire between the battery terminals is \(d\). You measure the current in the wire as a function of \(d\). The currents are small, so the temperature change of the wire is very small. You plot your results as \(I\) versus \(1 / d\) and find that the data lie close to a straight line that has slope \(600 \mathrm{~A} \cdot \mathrm{m} .\) What is the resistivity of the material of which the wire is made?

A \(25.0-\mathrm{cm}\) -long metal rod lies in the \(x y\) -plane and makes an angle of \(36.9^{\circ}\) with the positive \(x\) -axis and an angle of \(53.1^{\circ}\) with the positive \(y\) -axis. The rod is moving in the \(+x\) -direction with a speed of \(6.80 \mathrm{~m} / \mathrm{s}\). The rod is in a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}=(0.120 \mathrm{~T}) \hat{\imath}-(0.220 \mathrm{~T}) \hat{\jmath}-(0.0900 \mathrm{~T}) \hat{\boldsymbol{k}} .\) (a) What is the magni- tude of the emf induced in the rod? (b) Indicate in a sketch which end of the rod is at higher potential.

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