Chapter 26

# Chapter 26 - CHAPTER 26 Electric Current and Direct-Current...

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Unformatted text preview: CHAPTER 26 Electric Current and Direct-Current Circuits 1* ∙ In our study of electrostatics, we concluded that there is no electric field within a conductor in electrostatic equilibrium. How is it that we can now discuss electric fields inside a conductor? When a current flows, the charges are not in equilibrium. In that case, the electric field provides the force needed for the charge flow. 2 ∙ A physics professor has assembled his class at the baggage-claim carousel of the local airport to demonstrate an analog of electrical current. “Think of each suitcase on the conveyor belt as a package of electrons carrying one coulomb of charge,” he says. Counting and timing the suitcases reveals that the conveyor belt represents a wire carrying a constant 2-A current (constant as long as annoyed travellers could be kept away from their baggage by some of the huskier students). ( a ) How many suitcases will go by a given point in 5.0 min? ( b ) How many electrons does that represent? ( a ) ∆ Q = I ∆ t ( b ) n = ∆ Q / e ∆ Q = 600 C = 600 suitcases n = 600/1.6 × 10 –19 = 3.75 × 10 21 electrons 3 ∙ A 10–gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. I / A = nev d ; n = 8.47 × 10 28 m –3 (Example 26-1) A = 5.26 × 10-6 m 2 (Table 26-2) v d = 20/(8.47 × 10 28 × 5.26 × 10 –6 × 1.6 × 10 –19 ) m/s = 0.281 mm/s 4 ∙ In a fluorescent tube of diameter 3.0 cm, 2.0 × 10 18 electrons and 0.5 × 10 18 positive ions (with a charge of + e ) flow through a cross-sectional area each second. What is the current in the tube? The positive and negative charges flow in opposite directions. 1. Find I electron 2. Finbd I ion ; I = I electron + I ion I electron = 2 × 10 18 × 1.6 × 10 –19 A = 0.32 A I ion = 0.08 A; I = 0.40 A 5* ∙ In a certain electron beam, there are 5.0 × 10 6 electrons per cubic centimeter. Suppose the kinetic energy of each electron is 10.0 keV, and the beam is cylindrical, with a diameter of 1.00 mm. ( a ) What is the velocity of an electron in the beam? ( b ) Find the beam current. ( a ) m K/ 2 = v e ; K = 10 4 × 1.6 × 10 –19 J ( b ) I = nevA ; A = π D 2 /4 v = 5.93 × 10 7 m/s I = 37.2 µ A Chapter 26 Electric Current and Direct-Current Circuits 6 ∙∙ A charge + q moves in a circle of radius r with speed v . ( a ) Express the frequency f with which the charge passes a particular point in terms of r and v . ( b ) Show that the average current is qf and express it in terms of v and r . ( a ) T = 2 π r / v = 1/ f ; f = v /2 π r . ( b ) I = ∆ Q / ∆ t = q / T = qf = qv /2 π r . 7 ∙∙ A ring of radius a with a linear charge density λ rotates about its axis with angular velocity ω . Find an expression for the current....
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## This homework help was uploaded on 04/09/2008 for the course PHYS 161,260,27 taught by Professor Staff during the Spring '08 term at Maryland.

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Chapter 26 - CHAPTER 26 Electric Current and Direct-Current...

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