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A 5.00 A current runs through a 12 gauge copper wire (diameter \(2.05 \mathrm{~mm}\) ) and through a light bulb. Copper has \(8.5 \times 10^{28}\) free electrons per cubic meter. (a) How many electrons pass through the light bulb each second? (b) What is the current density in the wire? (c) At what speed does a typical electron pass by any given point in the wire? (d) If you were to use wire of twice the diameter, which of the above answers would change? Would they increase or decrease?

Short Answer

Expert verified
a) The number of electrons passing through the light bulb each second is approximately \(3.1 \times 10^{19}\) electrons. b) The current density in the wire is approximately \(1.88 \times 10^6\) Amps/meter². c) A typical electron speeds past any given point in the wire at approximately \(7.4 \times 10^{-5}\) meters/second. d) If the wire's diameter is doubled, both the current density and the speed of electrons would decrease, while the total number of electrons (the current) passing through the wire per second would remain the same.

Step by step solution

01

Calculation of the number of electrons passing through per second

Using the relation \(I = qnAv\), where \(I\) is the current, \(q\) is the charge of an electron (\(1.6 \times 10^{-19}\) coulombs), \(n\) is the number density of electrons, \(A\) is the area of the conductor and \(v\) is the drift velocity. We need to find the number of electrons passing through per second, which is essentially the current divided by the charge of one electron, which we can write as \(\dfrac{I}{q}\). Given that the current \(I = 5.00 A\), we can calculate the number of electrons per second as \(\dfrac{5.00 A}{1.6 \times 10^{-19}}\) coulombs.
02

Calculation of current density

Current density, \(J\), can be given by the current divided by the cross-sectional area of the conductor, \(\dfrac{I}{A}\). We can calculate the area of the copper wire using the formula for the area of a circle \(A= \pi r^2\), where \(r\) is the radius of the wire. Given the diameter is 2.05mm, the radius of the wire would be \(1.025 \times 10^{-3} m\). We then calculate the current density using these values.
03

Calculation of electron drift speed

We can rearrange the equation \(I = qnAv\) to solve for \(v\), the drift speed: \(v= \dfrac{I}{qnA}\). We are given \(n\), the number density of electrons in copper, as \(8.5 \times 10^{28}\) electrons per cubic meter. We substitute these values in to find \(v\).
04

Understanding the effect of increased wire diameter

Doubling the diameter of the wire will quadruple its cross-sectional area, as the area of a circle is proportional to the square of its radius. Therefore, the total number of electrons passing through wire per second (the current) would not change as it is independent of the wire's size. However, the current density would decrease since it is the current per unit area and the area has increased. Similarly, the drift speed of the electrons would also decrease, since the speed is inversely proportional to the cross-sectional area of the wire.

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Key Concepts

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

Current Density
Imagine electrons flowing through a wire like water through a pipe. Just as we can measure the flow rate of water in liters per second, we can gauge the flow of electric charge in a conductor using a concept called current density. Current density is a measure of how much electric current flows through a specific area of a conductor. It's denoted by the symbol J and is calculated by the formula J = I/A, where I is the current in amperes and A is the cross-sectional area in square meters.

In our exercise example, we would calculate the area of the copper wire first. Since it's a circular wire, we use the area formula for a circle, \(A = \pi r^2\), with r being the radius. After finding the area, we can determine the current density by dividing the current by this area.

If we were to increase the diameter of the wire, the area would increase, causing the current density to decrease because the same amount of current would be spread over a larger area. This effect is similar to how water flowing through a wider pipe at the same rate would have a lower flow speed. It's crucial to manage current density in electrical systems to prevent overheating and ensure efficiency.
Electrical Conductivity in Metals
Metals like copper are often used in wiring because they have high electrical conductivity. This refers to the ability of a material to conduct electric current. In metals, conductivity is largely due to the presence of free electrons—electrons that are not bound to any particular atom and can move freely through the metallic structure.

In our discussion, copper has a high number density of free electrons, which means there are a lot of charge carriers that can move through the material. The number density (denoted as n) in our exercise is given as \(8.5 \times 10^{28}\) free electrons per cubic meter of copper. A higher number density generally leads to better conductivity because more electrons are available to carry the electric charge.

However, the conductivity can also be influenced by other factors such as the wire's temperature, the presence of impurities, and the physical structure of the material. In the case of the exercise, if the wire's diameter is doubled, the electrical conductivity itself doesn't change, but the overall resistance of the wire would decrease due to the larger cross-sectional area, allowing the current to flow more easily.
Charge of an Electron
The charge of an electron is a fundamental constant in physics and plays a vital role in our understanding of electric current and circuitry. This tiny charge, although seemingly insignificant when taken alone, cumulatively creates the electric currents that power our world.

To put it into perspective, a single electron carries a charge of approximately \(-1.6 \times 10^{-19}\) coulombs. When we calculate the flow of electrons in a current, as we did in the exercise by finding the number of electrons passing through a point in one second, we divide the total current by this charge. Despite the minuscule size of an electron's charge, billions upon billions of electrons move through conductors to create the macroscopic currents we use every day.

It's fascinating to think that the soft light of a bulb or the rapid processing of a computer all start at this micro level with something as infinitesimal as the electric charge of electrons. As shown in our exercise solution, it's the cumulative effect of these charged particles in motion that forms the basis of electrical phenomena.

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Most popular questions from this chapter

A slender rod, \(0.240 \mathrm{~m}\) long, rotates with an angular speed of \(8.80 \mathrm{rad} / \mathrm{s}\) about an axis through one end and perpendicular to the rod. The plane of rotation of the rod is perpendicular to a uniform magnetic field with a magnitude of \(0.650 \mathrm{~T}\). (a) What is the induced emf in the rod? (b) What is the potential difference between its ends? (c) Suppose instead the rod rotates at \(8.80 \mathrm{rad} / \mathrm{s}\) about an axis through its center and perpendicular to the rod. In this case, what is the potential difference between the ends of the rod? Between the center of the rod and one end?

A coil \(4.00 \mathrm{~cm}\) in radius, containing 500 turns, is placed in a uniform magnetic field that varies with time according to \(B=(0.0120 \mathrm{~T} / \mathrm{s}) t+\left(3.00 \times 10^{-5} \mathrm{~T} / \mathrm{s}^{4}\right) t^{4} .\) The coil is connected to a \(600 \Omega\) resistor, and its plane is perpendicular to the magnetic field. You can ignore the resistance of the coil. (a) Find the magnitude of the induced emf in the coil as a function of time. (b) What is the current in the resistor at time \(t=5.00 \mathrm{~s} ?\)

The armature of a small generator consists of a flat, square coil with 120 turns and sides with a length of \(1.60 \mathrm{~cm} .\) The coil rotates in a magnetic field of \(0.0750 \mathrm{~T}\). What is the angular speed of the coil if the maximum emf produced is \(24.0 \mathrm{mV} ?\)

\(\mathrm{A} \cdot 540 \mathrm{~W} "\) electric heater is designed to operate from \(120 \mathrm{~V}\) lines. (a) What is its operating resistance? (b) What current does it draw? (c) If the line voltage drops to \(110 \mathrm{~V}\), what power does the heater take? (Assume that the resistance is constant. Actually, it will change because of the change in temperature.) (d) The heater coils are metallic, so that the resistance of the heater decreases with decreasing temperature. If the change of resistance with temperature is taken into account, will the electrical power consumed by the heater be larger or smaller than what you calculated in part (c)? Explain.

A long, thin solenoid has 400 turns per meter and radius \(1.10 \mathrm{~cm} \). The current in the solenoid is increasing at a uniform rate \(d i / d t\). The induced electric field at a point near the center of the solenoid and \(3.50 \mathrm{~cm}\) from its axis is \(8.00 \times 10^{-6} \mathrm{~V} / \mathrm{m} .\) Calculate \(d i / d t\)

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