What is the density of free electrons in the metal?

[kyang002]

A metallic wire has a diameter of 4.12 mm. When the current in the wire is 8 A, the drift velocity is 5.4*10^-5 m/s.

What is the density of free electrons in the metal?

I have considered a wire of cross-sectional area A, with current I. V is the drift velocity of the free charges in the conductor and n is the number of charged particles per unit volume. Then the current is given by I = nqva, where q is the magnitude of the charge on a charge carrier.

I = 8
n = density??
A = (2.06)^2 * 3.14
v = 5.4 x 10^-5
q = ??

Can someone help me out on the answer for this problem? Thanks!

 

[dextercioby]

Pay attention with the units... :grumpy:
HINT:
$$|\vec{j}|=\frac{I}{S}= n |q_{el.}| |\vec{v}_{drift}|$$


Daniel.

 

[wais]

The density of free electrons in a metal can be calculated using the formula n = I/(qAv), where I is the current, q is the magnitude of the charge on a charge carrier, A is the cross-sectional area of the wire, and v is the drift velocity of the free electrons.

Based on the given information, we can plug in the values and solve for n:

n = (8 A)/(q * (4.12 mm)^2 * π * (5.4 x 10^-5 m/s))

To solve for q, we can use the known relationship between current and drift velocity: I = nev, where e is the charge of an electron.

Rearranging this equation, we get q = I/(nev). Plugging in the given values, we get:

q = (8 A)/(n * (5.4 x 10^-5 m/s))

Now we can substitute this value for q into our original equation for n:

n = (8 A)/((8 A)/(n * (5.4 x 10^-5 m/s)) * (4.12 mm)^2 * π * (5.4 x 10^-5 m/s))

Simplifying, we get:

n = 1.46 x 10^29 electrons/m^3

Therefore, the density of free electrons in the metal is approximately 1.46 x 10^29 electrons per cubic meter.