Density of electrons in a metal (electricity)

  • Thread starter Talz1994
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  • #1
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Homework Statement


A metallic wire has a diameter of 4.12mm. When the current in the wire is 8.00 A, the drift velocity is 5.40 10^-5 m/s

What is the density of free electrons in the metal?
Express your answer numerically in m^-3 to two significant figures.



Homework Equations


I=nqv(drift)a, where I is current, a is cross sectional area, v(drift) is the drift velocity, n is the number of charged particles per unit volume (i.e., the concentration of charged particles), and q is the magnitude of the charge on a charge carrier.


The Attempt at a Solution


This really confused me, but here's the equation that was given as a hint for the question when asked:
I=nqv(drift)a :where I is current, a is cross sectional area, v(drift) is the drift velocity, n is the number of charged particles per unit volume (i.e., the concentration of charged particles), and q is the magnitude of the charge on a charge carrier.

this is another hint given to question when asked:
Recall the definition of current as the net charge flowing through a given cross-sectional area per unit time. If you express the net charge flowing through the area in terms of the concentration of the charge carriers per unit volume, you will find that the current depends on the concentration and charge of the charged particles, as well as on the magnitude of their drift velocity. Recall that in metals the charge carriers are free electrons.

I really don't understand this question, can somebody please walk me through it? step by step,

i rearranged the equation
I/Avq=n
n=6.945314735*10^28

what do i do next, how am i supposed to find density from number of electrons :@
 

Answers and Replies

  • #2
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You already found the number of free electrons pr unit volume, which i think is the density.
 
  • #3
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seriously i have? how?
 
  • #4
387
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What is the dimensions of 'n' in the formula you used. Can you evalute its dimesions using dimensional analysis?
 

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