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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

At temperatures near absolute zero, \(B_{\mathrm{c}}\) approaches \(0.142 \mathrm{~T}\) for vanadium, a type-I superconductor. The normal phase of vanadium has a magnetic susceptibility close to zero. Consider a long, thin vanadium cylinder with its axis parallel to an external magnetic field \(\overrightarrow{\boldsymbol{B}}_{0}\) in the \(+x\) -direction. At points far from the ends of the cylinder, by symmetry, all the magnetic vectors are parallel to the \(x\) -axis. At temperatures near absolute zero, what are the resultant magnetic field \(\vec{B}\) and the magnetization \(\vec{M}\) inside and outside the cylinder (far from the ends) for (a) \(\overrightarrow{\boldsymbol{B}}_{0}=(0.130 \mathrm{~T}) \hat{\boldsymbol{\imath}}\) and (b) \(\overrightarrow{\boldsymbol{B}}_{0}=(0.260 \mathrm{~T}) \hat{\imath} ?\)

The free-electron density in a copper wire is \(8.5 \times 10^{28}\) electrons \(/ \mathrm{m}^{3} .\) The electric field in the wire is \(0.0600 \mathrm{~N} / \mathrm{C}\) and the temperature of the wire is \(20.0^{\circ} \mathrm{C}\). (a) What is the drift speed \(v_{\mathrm{d}}\) of the electrons in the wire? (b) What is the potential difference between two points in the wire that are separated by \(20.0 \mathrm{~cm} ?\)

When a resistor with resistance \(R\) is connected to a \(1.50 \mathrm{~V}\) flashlight battery, the resistor consumes \(0.0625 \mathrm{~W}\) of electrical power. (Throughout, assume that each battery has negligible internal resistance.) (a) What power does the resistor consume if it is connected to a \(12.6 \mathrm{~V}\) car battery? Assume that \(R\) remains constant when the power consumption changes. (b) The resistor is connected to a battery and consumes \(5.00 \mathrm{~W}\). What is the voltage of this battery?

An external resistor with resistance \(R\) is connected to a battery that has emf \(\mathcal{E}\) and internal resistance \(r\). Let \(P\) be the electrical power output of the source. By conservation of energy, \(P\) is equal to the power consumed by \(R\). What is the value of \(P\) in the limit that \(R\) is (a) very small; (b) very large? (c) Show that the power output of the battery is a maximum when \(R=r .\) What is this maximum \(P\) in terms of \(\mathcal{E}\) and \(r ?\) (d) A battery has \(\mathcal{E}=64.0 \mathrm{~V}\) and \(r=4.00 \Omega .\) What is the power output of this battery when it is connected to a resistor \(R,\) for \(R=2.00 \Omega, R=4.00 \Omega,\) and \(R=6.00 \Omega ?\) Are your results consistent with the general result that you derived in part (b)?

A magnetic field of \(0.080 \mathrm{~T}\) is in the \(y\) -direction. The velocity of wire segment \(S\) has a magnitude of \(78 \mathrm{~m} / \mathrm{s}\) and components of \(18 \mathrm{~m} / \mathrm{s}\) in the \(x\) -direction, \(24 \mathrm{~m} / \mathrm{s}\) in the \(y\) -direction, and \(72 \mathrm{~m} / \mathrm{s}\) in the \(z\) -direction. The segment has length \(0.50 \mathrm{~m}\) and is parallel to the \(z\) -axis as it moves. (a) Find the motional emf induced between the ends of the segment. (b) What would the motional emf be if the wire segment was parallel to the \(y\) -axis?

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