Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A 5.00-A current runs through a 12-gauge copper wire (diameter 2.05 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) \(3.125 \times 10^{19}\) electrons/s, (b) Current density remains the same, (c) Drift speed decreases, (d) Current density and drift speed decrease with larger diameter.

Step by step solution

01

Calculate the charge of electrons passing through the bulb

The current is given as 5.00 A. Current \( I \) is the rate of charge flow, \( I = \frac{Q}{t} \), where \( Q \) is the total charge and \( t \) is the time. Each electron carries a charge of \( 1.6 \times 10^{-19} \) coulombs. Thus, the number of electrons \( n \) passing through in one second is given by \( n = \frac{I}{e} = \frac{5.00}{1.6 \times 10^{-19}} \approx 3.125 \times 10^{19} \) electrons.
02

Calculate the current density in the wire

Current density \( J \) is defined as the current \( I \) per unit cross-sectional area \( A \). The cross-sectional area of the wire, using the diameter \( d = 2.05 \ mm \) or \( 2.05 \times 10^{-3} \ m \), is \( A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{2.05 \times 10^{-3}}{2}\right)^2 \). The current density \( J = \frac{I}{A} \).
03

Compute the speed of electrons

The drift velocity \( v_d \) can be found using \( I = nAqv_d \), where \( n \) is the number density of electrons, \( A \) is the cross-sectional area, \( q \) is the charge of an electron, and \( v_d \) is the drift velocity. Rearranging gives \( v_d = \frac{I}{nAq} \), where \( n = 8.5 \times 10^{28} \) m⁻³. Substitute the values to find \( v_d \).
04

Analyze effect of resizing the wire

If the wire's diameter is doubled, the cross-sectional area increases by a factor of four since \( A \propto d^2 \). This means the current density would decrease by a factor of four. The number of electrons per second (part a) will stay the same because the current is unchanged, but the drift speed will decrease by a factor of four.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Drift Velocity
Drift velocity is the average velocity that a charge carrier, such as an electron, attains due to an electric field. In a copper wire carrying current, these charge carriers are predominantly free electrons. The flow of electricity in the wire can be thought of as these electrons drifting through the metallic structure of the wire. Although individual electrons move at tremendous speeds randomly, their average net motion is slow and their drift velocity can be calculated using the formula:\[v_d = \frac{I}{nAq}\]where:
  • \(I\) is the current through the wire, which is given as 5.00 A.
  • \(n\) represents the number density of electrons, for copper it is given as \(8.5 \times 10^{28}\) electrons per cubic meter.
  • \(A\) is the cross-sectional area of the wire.
  • \(q\) is the charge of an electron, which is \(1.6 \times 10^{-19}\) coulombs.
Substituting these values calculates how fast electrons drift through the wire, highlighting that the motion responsible for electric current is far slower than one might assume.
Copper Wire
Copper wire is a common conductor in electrical circuits, revered for its excellent conductivity. Conductors allow electrons to move freely through their structure, making them ideal for carrying electric current. Copper’s ability to conduct electricity is tied to its atomic structure, where free electrons can move without significant resistance.For this exercise, the copper wire is specified as a 12-gauge wire with a diameter of 2.05 mm. The wire's cross-sectional area, crucial for calculating current density and drift velocity, can be found using \[A = \pi \left(\frac{d}{2}\right)^2\]where \(d\) is the diameter. This structural property, an integral aspect of its design, ensures the wire can efficiently carry the defined current. If the diameter doubles, the area increases by a factor of four, affecting how current flows through the wire.
Electron Flow
Electron flow refers to the movement of electrons through a conductor, like a copper wire, under the influence of an electric field. Each electron acts as a tiny charge carrier, contributing to the overall electric current. In this setting, electron flow is continuous and quantified as how many electrons pass a point per second.The formula to calculate the number of electrons passing through the light bulb every second is:\[n = \frac{I}{e}\]where:
  • \(I\) represents the electric current (5.00 A),
  • \(e\) is the elementary charge of an electron \(1.6 \times 10^{-19}\) C.
Substituting these values gives the electron flow for any cross-section of the wire, immobilizing the abstract concept of current into an understandable count of electrons.
Current Calculation
Current calculation in a wire involves understanding how current, charge, and time are interconnected. Current, denoted as \(I\), measures how much charge passes a given point in the wire per second.Rowing through the relationship \(I = \frac{Q}{t}\), where \(Q\) is total charge and \(t\) is time, we can measure the flow of electrons in a tangible manner. In the context of this exercise, understanding that current is essentially a count of electrons passing a wire's cross section per second, reifies its definition.The concept of current density, defined as the current per unit area \(J = \frac{I}{A}\), builds on this foundation. For a given wire's cross-sectional area calculated using its diameter, this value indicates how densely packed the current flow is within the cable's structure, informing both theoretical calculations and practical applications, especially in circuit design.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If the conductivity of the thread results from the aqueous coating only, how does the cross-sectional area \(A\) of the coating compare when the thread is 13 mm long versus the starting length of 5 mm? Assume that the resistivity of the coating remains constant and the coating is uniform along the thread. \(A_{13}\) mm is about (a) \\(\frac{1}{10}\\) \(A_5\) mm; (b) \\(\frac{1}{_4}\\) \(A_5\) mm; (c) \\(\frac{2}{_5}\\) \(A_5\) mm; (d) the same as \(A_5\) mm.

A wire 6.50 m long with diameter of 2.05 mm has a resistance of 0.0290 \(\Omega\). What material is the wire most likely made of?

You apply a potential difference of 4.50 V between the ends of a wire that is 2.50 m in length and 0.654 mm in radius. The resulting current through the wire is 17.6 A. What is the resistivity of the wire?

A "540-W" electric heater is designed to operate from 120-V lines. (a) What is its operating resistance? (b) What current does it draw? (c) If the line voltage drops to 110 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 battery-powered global positioning system (GPS) receiver operating on 9.0 V draws a current of 0.13 A. How much electrical energy does it consume during 30 minutes?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free