Density of electrons in a metal (electricity)

[Talz1994]

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 :@

 

[dikmikkel]

You already found the number of free electrons pr unit volume, which i think is the density.

 

[Talz1994]

seriously i have? how?

 

[darkxponent]

What is the dimensions of 'n' in the formula you used. Can you evalute its dimensions using dimensional analysis?

 

[asteriatic]

As a scientist, it is important to understand the basic principles and equations related to the topic at hand. In this case, we are dealing with the density of free electrons in a metal, which is related to electricity. The given equation, I=nqv(drift)a, is known as the current density equation, which relates the current (I) to the concentration of charged particles (n), their charge (q), and their drift velocity (v(drift)) through a given cross-sectional area (a).

To solve this problem, we need to rearrange the equation to solve for the concentration of charged particles (n), which is the density of free electrons in this case. This is the first step you have already correctly done. However, the next step is to understand what each variable in the equation represents.

I represents the current, which is given as 8.00 A in the problem.
a represents the cross-sectional area, which can be calculated using the diameter of the wire (4.12 mm) and the formula for the area of a circle (a=πr^2).
v(drift) represents the drift velocity, which is given as 5.40*10^-5 m/s in the problem.
q represents the magnitude of the charge on a charge carrier, which in this case is the charge of an electron, given as 1.60*10^-19 C.

Substituting these values into the rearranged equation, we get:
n = I/av(drift)q
= (8.00 A)/[(π(4.12*10^-3 m/2)^2)*(5.40*10^-5 m/s)*(1.60*10^-19 C)]
= 6.95*10^28 electrons/m^3

Therefore, the density of free electrons in the metal is 6.95*10^28 electrons/m^3. This can also be written as 6.95*10^28 m^-3, as requested in the question.

It is important to note that this is an approximation, as the actual number of free electrons in a metal can vary depending on factors such as temperature and impurities. However, this calculation gives us an estimate of the density of free electrons in the metal.