- #1

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resisivity, p of copper = 1.68 * 10^-8 ohms meter

The answer in the book apparently is 2 * 10^-5 m/s.

Please help by showing all your work. Your attempts will be greatly appreciated.

- Thread starter eno
- Start date

- #1

- 7

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resisivity, p of copper = 1.68 * 10^-8 ohms meter

The answer in the book apparently is 2 * 10^-5 m/s.

Please help by showing all your work. Your attempts will be greatly appreciated.

- #2

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please read this before posting.......

https://www.physicsforums.com/showthread.php?t=28 :tongue2:

- #3

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i have tried it many times and it did not reach tat ans.

- #4

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[tex] I=\rho e v A [/tex]

ring the bell?

since you have everything except v, this is one equation with one unknown... solving it is straight forward....

you must be careful on the unit conversion, it is a little bit tricky....

the area is not given directly, but you have radii, the area shold be piece of cake

charge of electron is on the back of your textbook.... if not, look it up on internet

show me your calculation .....

- #5

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total

1.0 A = 6.24*10^18 e = 10^29 * v * (π * (0.001)^2)

v = 2 * 10^-6 m/s

therefore, does not match with the ans in the book.

- #6

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L = v * 1 sec = v

totaledrifted within 1 sec is 1 Ampere.

1.0 A = 6.24*10^18 e = 10^29 * v * (π * (0.001)^2)

v = 2 * 10^-6 m/s

therefore, does not match with the ans in the book.

what the hack are you doing here?

- #7

- 609

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[tex] \rho [/tex] is the free electronss density

[tex] e [/tex] is the charge per electron....

[tex] A [/tex] is the cross section area

[tex] I [/tex] is the current

plug in all number and solve for v....

- #8

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it appears that I have punched in the wrong notation the whole time. ths for ur help.

- #9

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the formula i use is

I = NaQv

so V = I / NAQ

A (cross sectional area) = (0.001)^2 multiplied by Pi

Q = 1.6 * 10^-19

N= 10^29

I = 1

is this right so far?

Regards,

Mo

- #10

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Daniel.

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