The number of years does one electron takes to travel the full length of the cable.

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Answer 1

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Chapter 26, Problem 1P

To determine

The number of years does one electron takes to travel the full length of the cable.

Expert Solution & Answer

Answer to Problem 1P

The number of years that takes one electron to travel the full length of the cable is

$27.0 yr$ .

Explanation of Solution

Given information: Length of the high voltage transmission line is $200 km$ , diameter of the transmission line is $2.00 cm$ , steady current is $1000 A$ , the number of electrons per unit volume is $8.50×1028 electrons/m3$ .

Explanation:

Write the expression for the area of cross section of a conductor.

$A=π4d2$

Here,

$A$ is the area of cross section of a conductor.

$d$ is the diameter of the transmission line.

Substitute $2.00 cm$ for $d$ in equation (1) to find $A$ ,

$A=π4(2.00 cm×1 m100 cm)2=3.14×10−4 m2$

Thus, the area of cross section of a conductor is $3.14×10−4 m2$ .

Formula to calculate the drift velocity of a conductor.

$I=neAVdVd=IneA$ (2)

Here,

$Vd$ is the drift velocity of a conductor.

$n$ is the number of electrons per unit volume.

$I$ is the steady current.

$e$ is the charge of one electron.

The value of charge of one electron is $1.6×10−19 C$ .

Substitute $1000 A$ for $I$ , $8.50×1028 electrons/m3$ for $n$ , $1.6×10−19 C$ for $e$ , $3.14×10−4 m2$ for $A$ in equation (2) to find $Vd$ ,

$Vd=1000 A(8.50×1028 electrons/m3)×(1.6×10−19 C)×(3.14×10−4 m2)=2.34×10−4 m/s$

Thus, the drift velocity of a conductor is $2.34×10−4 m/s$ .

Formula to calculate the number of years that takes one electron to travel the full length of the cable.

$T=DVd$ (3)

Here,

$T$ is the number of years that takes one electron to travel the full length of the cable.

$D$ is the length of the high voltage transmission line.

Substitute $200 km$ for $D$ , $2.34×10−4 m/s$ for $Vd$ in equation (3) to find $T$ ,

$T=200 km×1000 m1 km2.34×10−4 m/s=8.54×108 s×1 yr(365×24×60×60) s=27.0 yr$

Thus, the number of years that takes one electron to travel the full length of the cable is $27.0 yr$ .

Conclusion:

Therefore, the number of years that takes one electron to travel the full length of the cable is $27.0 yr$ .

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Answer 2

Question

100%

Chapter 26, Problem 1P

To determine

The number of years does one electron takes to travel the full length of the cable.

Expert Solution & Answer

Answer to Problem 1P

The number of years that takes one electron to travel the full length of the cable is

$27.0 yr$ .

Explanation of Solution

Given information: Length of the high voltage transmission line is $200 km$ , diameter of the transmission line is $2.00 cm$ , steady current is $1000 A$ , the number of electrons per unit volume is $8.50×1028 electrons/m3$ .

Explanation:

Write the expression for the area of cross section of a conductor.

$A=π4d2$

Here,

$A$ is the area of cross section of a conductor.

$d$ is the diameter of the transmission line.

Substitute $2.00 cm$ for $d$ in equation (1) to find $A$ ,

$A=π4(2.00 cm×1 m100 cm)2=3.14×10−4 m2$

Thus, the area of cross section of a conductor is $3.14×10−4 m2$ .

Formula to calculate the drift velocity of a conductor.

$I=neAVdVd=IneA$ (2)

Here,

$Vd$ is the drift velocity of a conductor.

$n$ is the number of electrons per unit volume.

$I$ is the steady current.

$e$ is the charge of one electron.

The value of charge of one electron is $1.6×10−19 C$ .

Substitute $1000 A$ for $I$ , $8.50×1028 electrons/m3$ for $n$ , $1.6×10−19 C$ for $e$ , $3.14×10−4 m2$ for $A$ in equation (2) to find $Vd$ ,

$Vd=1000 A(8.50×1028 electrons/m3)×(1.6×10−19 C)×(3.14×10−4 m2)=2.34×10−4 m/s$

Thus, the drift velocity of a conductor is $2.34×10−4 m/s$ .

Formula to calculate the number of years that takes one electron to travel the full length of the cable.

$T=DVd$ (3)

Here,

$T$ is the number of years that takes one electron to travel the full length of the cable.

$D$ is the length of the high voltage transmission line.

Substitute $200 km$ for $D$ , $2.34×10−4 m/s$ for $Vd$ in equation (3) to find $T$ ,

$T=200 km×1000 m1 km2.34×10−4 m/s=8.54×108 s×1 yr(365×24×60×60) s=27.0 yr$

Thus, the number of years that takes one electron to travel the full length of the cable is $27.0 yr$ .

Conclusion:

Therefore, the number of years that takes one electron to travel the full length of the cable is $27.0 yr$ .

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Answer 3

Question

100%

Chapter 26, Problem 1P

To determine

The number of years does one electron takes to travel the full length of the cable.

Expert Solution & Answer

Answer to Problem 1P

The number of years that takes one electron to travel the full length of the cable is

$27.0 yr$ .

Explanation of Solution

Given information: Length of the high voltage transmission line is $200 km$ , diameter of the transmission line is $2.00 cm$ , steady current is $1000 A$ , the number of electrons per unit volume is $8.50×1028 electrons/m3$ .

Explanation:

Write the expression for the area of cross section of a conductor.

$A=π4d2$

Here,

$A$ is the area of cross section of a conductor.

$d$ is the diameter of the transmission line.

Substitute $2.00 cm$ for $d$ in equation (1) to find $A$ ,

$A=π4(2.00 cm×1 m100 cm)2=3.14×10−4 m2$

Thus, the area of cross section of a conductor is $3.14×10−4 m2$ .

Formula to calculate the drift velocity of a conductor.

$I=neAVdVd=IneA$ (2)

Here,

$Vd$ is the drift velocity of a conductor.

$n$ is the number of electrons per unit volume.

$I$ is the steady current.

$e$ is the charge of one electron.

The value of charge of one electron is $1.6×10−19 C$ .

Substitute $1000 A$ for $I$ , $8.50×1028 electrons/m3$ for $n$ , $1.6×10−19 C$ for $e$ , $3.14×10−4 m2$ for $A$ in equation (2) to find $Vd$ ,

$Vd=1000 A(8.50×1028 electrons/m3)×(1.6×10−19 C)×(3.14×10−4 m2)=2.34×10−4 m/s$

Thus, the drift velocity of a conductor is $2.34×10−4 m/s$ .

Formula to calculate the number of years that takes one electron to travel the full length of the cable.

$T=DVd$ (3)

Here,

$T$ is the number of years that takes one electron to travel the full length of the cable.

$D$ is the length of the high voltage transmission line.

Substitute $200 km$ for $D$ , $2.34×10−4 m/s$ for $Vd$ in equation (3) to find $T$ ,

$T=200 km×1000 m1 km2.34×10−4 m/s=8.54×108 s×1 yr(365×24×60×60) s=27.0 yr$

Thus, the number of years that takes one electron to travel the full length of the cable is $27.0 yr$ .

Conclusion:

Therefore, the number of years that takes one electron to travel the full length of the cable is $27.0 yr$ .

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Answer 4

Question

100%

Chapter 26, Problem 1P

To determine

The number of years does one electron takes to travel the full length of the cable.

Expert Solution & Answer

Answer to Problem 1P

The number of years that takes one electron to travel the full length of the cable is

$27.0 yr$ .

Explanation of Solution

Given information: Length of the high voltage transmission line is $200 km$ , diameter of the transmission line is $2.00 cm$ , steady current is $1000 A$ , the number of electrons per unit volume is $8.50×1028 electrons/m3$ .

Explanation:

Write the expression for the area of cross section of a conductor.

$A=π4d2$

Here,

$A$ is the area of cross section of a conductor.

$d$ is the diameter of the transmission line.

Substitute $2.00 cm$ for $d$ in equation (1) to find $A$ ,

$A=π4(2.00 cm×1 m100 cm)2=3.14×10−4 m2$

Thus, the area of cross section of a conductor is $3.14×10−4 m2$ .

Formula to calculate the drift velocity of a conductor.

$I=neAVdVd=IneA$ (2)

Here,

$Vd$ is the drift velocity of a conductor.

$n$ is the number of electrons per unit volume.

$I$ is the steady current.

$e$ is the charge of one electron.

The value of charge of one electron is $1.6×10−19 C$ .

Substitute $1000 A$ for $I$ , $8.50×1028 electrons/m3$ for $n$ , $1.6×10−19 C$ for $e$ , $3.14×10−4 m2$ for $A$ in equation (2) to find $Vd$ ,

$Vd=1000 A(8.50×1028 electrons/m3)×(1.6×10−19 C)×(3.14×10−4 m2)=2.34×10−4 m/s$

Thus, the drift velocity of a conductor is $2.34×10−4 m/s$ .

Formula to calculate the number of years that takes one electron to travel the full length of the cable.

$T=DVd$ (3)

Here,

$T$ is the number of years that takes one electron to travel the full length of the cable.

$D$ is the length of the high voltage transmission line.

Substitute $200 km$ for $D$ , $2.34×10−4 m/s$ for $Vd$ in equation (3) to find $T$ ,

$T=200 km×1000 m1 km2.34×10−4 m/s=8.54×108 s×1 yr(365×24×60×60) s=27.0 yr$

Thus, the number of years that takes one electron to travel the full length of the cable is $27.0 yr$ .

Conclusion:

Therefore, the number of years that takes one electron to travel the full length of the cable is $27.0 yr$ .

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Answer 5

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Answer 6

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Answer 7

Chapter 26, Problem 1P

To determine

The number of years does one electron takes to travel the full length of the cable.

Expert Solution & Answer

Answer to Problem 1P

The number of years that takes one electron to travel the full length of the cable is

$27.0 yr$ .

Explanation of Solution

Given information: Length of the high voltage transmission line is $200 km$ , diameter of the transmission line is $2.00 cm$ , steady current is $1000 A$ , the number of electrons per unit volume is $8.50×1028 electrons/m3$ .

Explanation:

Write the expression for the area of cross section of a conductor.

$A=π4d2$

Here,

$A$ is the area of cross section of a conductor.

$d$ is the diameter of the transmission line.

Substitute $2.00 cm$ for $d$ in equation (1) to find $A$ ,

$A=π4(2.00 cm×1 m100 cm)2=3.14×10−4 m2$

Thus, the area of cross section of a conductor is $3.14×10−4 m2$ .

Formula to calculate the drift velocity of a conductor.

$I=neAVdVd=IneA$ (2)

Here,

$Vd$ is the drift velocity of a conductor.

$n$ is the number of electrons per unit volume.

$I$ is the steady current.

$e$ is the charge of one electron.

The value of charge of one electron is $1.6×10−19 C$ .

Substitute $1000 A$ for $I$ , $8.50×1028 electrons/m3$ for $n$ , $1.6×10−19 C$ for $e$ , $3.14×10−4 m2$ for $A$ in equation (2) to find $Vd$ ,

$Vd=1000 A(8.50×1028 electrons/m3)×(1.6×10−19 C)×(3.14×10−4 m2)=2.34×10−4 m/s$

Thus, the drift velocity of a conductor is $2.34×10−4 m/s$ .

Formula to calculate the number of years that takes one electron to travel the full length of the cable.

$T=DVd$ (3)

Here,

$T$ is the number of years that takes one electron to travel the full length of the cable.

$D$ is the length of the high voltage transmission line.

Substitute $200 km$ for $D$ , $2.34×10−4 m/s$ for $Vd$ in equation (3) to find $T$ ,

$T=200 km×1000 m1 km2.34×10−4 m/s=8.54×108 s×1 yr(365×24×60×60) s=27.0 yr$

Thus, the number of years that takes one electron to travel the full length of the cable is $27.0 yr$ .

Conclusion:

Therefore, the number of years that takes one electron to travel the full length of the cable is $27.0 yr$ .

Answer 8

Chapter 26, Problem 1P

To determine

The number of years does one electron takes to travel the full length of the cable.

Expert Solution & Answer

Answer to Problem 1P

The number of years that takes one electron to travel the full length of the cable is

$27.0 yr$ .

Explanation of Solution

Given information: Length of the high voltage transmission line is $200 km$ , diameter of the transmission line is $2.00 cm$ , steady current is $1000 A$ , the number of electrons per unit volume is $8.50×1028 electrons/m3$ .

Explanation:

Write the expression for the area of cross section of a conductor.

$A=π4d2$

Here,

$A$ is the area of cross section of a conductor.

$d$ is the diameter of the transmission line.

Substitute $2.00 cm$ for $d$ in equation (1) to find $A$ ,

$A=π4(2.00 cm×1 m100 cm)2=3.14×10−4 m2$

Thus, the area of cross section of a conductor is $3.14×10−4 m2$ .

Formula to calculate the drift velocity of a conductor.

$I=neAVdVd=IneA$ (2)

Here,

$Vd$ is the drift velocity of a conductor.

$n$ is the number of electrons per unit volume.

$I$ is the steady current.

$e$ is the charge of one electron.

The value of charge of one electron is $1.6×10−19 C$ .

Substitute $1000 A$ for $I$ , $8.50×1028 electrons/m3$ for $n$ , $1.6×10−19 C$ for $e$ , $3.14×10−4 m2$ for $A$ in equation (2) to find $Vd$ ,

$Vd=1000 A(8.50×1028 electrons/m3)×(1.6×10−19 C)×(3.14×10−4 m2)=2.34×10−4 m/s$

Thus, the drift velocity of a conductor is $2.34×10−4 m/s$ .

Formula to calculate the number of years that takes one electron to travel the full length of the cable.

$T=DVd$ (3)

Here,

$T$ is the number of years that takes one electron to travel the full length of the cable.

$D$ is the length of the high voltage transmission line.

Substitute $200 km$ for $D$ , $2.34×10−4 m/s$ for $Vd$ in equation (3) to find $T$ ,

$T=200 km×1000 m1 km2.34×10−4 m/s=8.54×108 s×1 yr(365×24×60×60) s=27.0 yr$

Thus, the number of years that takes one electron to travel the full length of the cable is $27.0 yr$ .

Conclusion:

Therefore, the number of years that takes one electron to travel the full length of the cable is $27.0 yr$ .

Answer 9

Chapter 26, Problem 1P

To determine

The number of years does one electron takes to travel the full length of the cable.

Answer 10

Expert Solution & Answer

Answer to Problem 1P

The number of years that takes one electron to travel the full length of the cable is

$27.0 yr$ .

Answer 11

Expert Solution & Answer

Answer 12

Expert Solution & Answer

Answer 13

Explanation of Solution

Given information: Length of the high voltage transmission line is $200 km$ , diameter of the transmission line is $2.00 cm$ , steady current is $1000 A$ , the number of electrons per unit volume is $8.50×1028 electrons/m3$ .

Explanation:

Write the expression for the area of cross section of a conductor.

$A=π4d2$

Here,

$A$ is the area of cross section of a conductor.

$d$ is the diameter of the transmission line.

Substitute $2.00 cm$ for $d$ in equation (1) to find $A$ ,

$A=π4(2.00 cm×1 m100 cm)2=3.14×10−4 m2$

Thus, the area of cross section of a conductor is $3.14×10−4 m2$ .

Formula to calculate the drift velocity of a conductor.

$I=neAVdVd=IneA$ (2)

Here,

$Vd$ is the drift velocity of a conductor.

$n$ is the number of electrons per unit volume.

$I$ is the steady current.

$e$ is the charge of one electron.

The value of charge of one electron is $1.6×10−19 C$ .

Substitute $1000 A$ for $I$ , $8.50×1028 electrons/m3$ for $n$ , $1.6×10−19 C$ for $e$ , $3.14×10−4 m2$ for $A$ in equation (2) to find $Vd$ ,

$Vd=1000 A(8.50×1028 electrons/m3)×(1.6×10−19 C)×(3.14×10−4 m2)=2.34×10−4 m/s$

Thus, the drift velocity of a conductor is $2.34×10−4 m/s$ .

Formula to calculate the number of years that takes one electron to travel the full length of the cable.

$T=DVd$ (3)

Here,

$T$ is the number of years that takes one electron to travel the full length of the cable.

$D$ is the length of the high voltage transmission line.

Substitute $200 km$ for $D$ , $2.34×10−4 m/s$ for $Vd$ in equation (3) to find $T$ ,

$T=200 km×1000 m1 km2.34×10−4 m/s=8.54×108 s×1 yr(365×24×60×60) s=27.0 yr$

Thus, the number of years that takes one electron to travel the full length of the cable is $27.0 yr$ .

Conclusion:

Therefore, the number of years that takes one electron to travel the full length of the cable is $27.0 yr$ .

Answer 14

Ch. 26.1 - Consider positive and negative charges of equal...Ch. 26.2 - Prob. 26.2QQ Ch. 26.2 - Prob. 26.3QQ Ch. 26.4 - When does an incandescent lightbulb carry more...Ch. 26 - Prob. 1P Ch. 26 - A small sphere that carries a charge q is whirled...Ch. 26 - In the Bohr model of the hydrogen atom (which will...Ch. 26 - Prob. 4P Ch. 26 - Prob. 5P Ch. 26 - Figure P26.6 represents a section of a conductor...

The number of years does one electron takes to travel the full length of the cable.

Solution Summary: The author calculates the number of years that one electron takes to travel the full length of the cable.

Concept explainers

The number of years does one electron takes to travel the full length of the cable.

Answer to Problem 1P

Explanation of Solution

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Chapter 26 Solutions