Physics: Principles with Applications
Physics: Principles with Applications
6th Edition
ISBN: 9780130606204
Author: Douglas C. Giancoli
Publisher: Prentice Hall
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Chapter 18, Problem 18Q
To determine

To explain:

If the number of free electrons per atom in one wire is double of that in the other, whereas both are geometrically identical with the same atomic density, compare their drift velocities and their electric currents.

Expert Solution & Answer
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Answer to Problem 18Q

Solution:

The drift velocities are the same but the current in one is double of the other.

Formula used:

  •The average drift velocity depends on the structural properties of the material and the external electric field (E), as

                 vd=eτmE

    Where, e is the charge over one electron and vd is drift velocity of the free electrons in the wire at that particular temperature, τ is the average relaxation time between two collisions of free electrons and m is the mass of one electron.

  •The current in a wire of cross-sectional area A, having n number of free electrons per unit volume, is defined as,

                 I=Anevd

    Where, e is the charge over one electron and vd is drift velocity of the free electrons in the wire at that particular temperature.

Explanation of Solution

As it is given that the number of free electrons per atom in the material of one wire is twice of that in the other and also because the density of atoms in both the wires are similar, the number of free electrons per unit volume in one wire will also be twice of that in the other, i.e.,

                 n1=2n2

But because the density of atoms in both the wires is similar, the average relaxation time between two collisions of free electrons will be the same, i.e.

                 τ1=τ2

Therefore, from the formula vd=eτmE, we can conclude that the average drift velocities in the two wires would be same, i.e.,

                 vd1=vd2

But as n1=2n2, therefore from the formula I=Anevd, we can conclude that the current in one wire would be twice of that in the other, i.e.,

                 I1=2I2

Chapter 18 Solutions

Physics: Principles with Applications

Ch. 18 - Prob. 11QCh. 18 - Prob. 12QCh. 18 - When electric lights are operated on low-frequency...Ch. 18 - Prob. 14QCh. 18 - Prob. 15QCh. 18 - Prob. 16QCh. 18 - Prob. 17QCh. 18 - Prob. 18QCh. 18 - Prob. 19QCh. 18 - Prob. 1PCh. 18 - Prob. 2PCh. 18 - Prob. 3PCh. 18 - Prob. 4PCh. 18 - Prob. 5PCh. 18 - Prob. 6PCh. 18 - Prob. 7PCh. 18 - Prob. 8PCh. 18 - Prob. 9PCh. 18 - Prob. 10PCh. 18 - Prob. 11PCh. 18 - Prob. 12PCh. 18 - Prob. 13PCh. 18 - Prob. 14PCh. 18 - Prob. 15PCh. 18 - Prob. 16PCh. 18 - Prob. 17PCh. 18 - Prob. 18PCh. 18 - Prob. 19PCh. 18 - Prob. 20PCh. 18 - Prob. 21PCh. 18 - Prob. 22PCh. 18 - Prob. 23PCh. 18 - Prob. 24PCh. 18 - Prob. 25PCh. 18 - Prob. 26PCh. 18 - Prob. 27PCh. 18 - Prob. 28PCh. 18 - Prob. 29PCh. 18 - Prob. 30PCh. 18 - Prob. 31PCh. 18 - Prob. 32PCh. 18 - Prob. 33PCh. 18 - Prob. 34PCh. 18 - Prob. 35PCh. 18 - Prob. 36PCh. 18 - Prob. 37PCh. 18 - Prob. 38PCh. 18 - Prob. 39PCh. 18 - Prob. 40PCh. 18 - Prob. 41PCh. 18 - Prob. 42PCh. 18 - Prob. 43PCh. 18 - Prob. 44PCh. 18 - Prob. 45PCh. 18 - Prob. 46PCh. 18 - Prob. 47PCh. 18 - Prob. 48PCh. 18 - Prob. 49PCh. 18 - Prob. 50PCh. 18 - Prob. 51PCh. 18 - Prob. 52PCh. 18 - Prob. 53PCh. 18 - Prob. 54PCh. 18 - Prob. 55PCh. 18 - Prob. 56GPCh. 18 - Prob. 57GPCh. 18 - Prob. 58GPCh. 18 - Prob. 59GPCh. 18 - Prob. 60GPCh. 18 - Prob. 61GPCh. 18 - Prob. 62GPCh. 18 - Prob. 63GPCh. 18 - Prob. 64GPCh. 18 - Prob. 65GPCh. 18 - Prob. 66GPCh. 18 - Prob. 67GPCh. 18 - Prob. 68GPCh. 18 - Prob. 69GPCh. 18 - Prob. 70GPCh. 18 - Prob. 71GPCh. 18 - Prob. 72GPCh. 18 - Prob. 73GPCh. 18 - Prob. 74GPCh. 18 - Prob. 75GPCh. 18 - Prob. 76GPCh. 18 - Prob. 77GPCh. 18 - Prob. 78GPCh. 18 - Prob. 79GPCh. 18 - Prob. 80GPCh. 18 - Prob. 81GPCh. 18 - Prob. 82GPCh. 18 - Prob. 83GP
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