Finding electron drift speed

[jheld]

Homework Statement

1.0 * 10^20 electrons flow through a cross section of a 2.0-mm-diameter iron wire in 5.0 s. What is the electron drift speed?

Homework Equations

v_d = e*tao/m * E
i_e = n_e * e * tao * A/m * E

The Attempt at a Solution

I can't seem to get very far. I think that I need to find E, but I'm having no luck actually deriving it.
Any suggestions?

 

[jheld]

Well I figured it out!

relevant equation:
N_e = n_e * A * v_d * delta(t)
N_e = total electrons in cross-section = 10^20 electrons
n_e = number density of electrons in iron = 8.5*10^28
A = pi*r^2 = pi * (10^-3)^2
delta(t) = 5
so solve for v_d...I get 7.5*10^-5 m/s

 

[nlightner]

As a scientist, we can approach this problem by using the equation for electron drift speed:

v_d = i_e / (n_e * e * A)

Where:
v_d = electron drift speed
i_e = electric current
n_e = electron density
e = elementary charge
A = cross-sectional area of the wire

We are given the value for electric current, i_e = 1.0 * 10^20 electrons in 5.0 s. We can calculate the electron density using the formula:

n_e = i_e / (e * tao * A)

Where:
tao = time (5.0 s)
A = cross-sectional area (2.0 mm diameter = 1.0 mm radius = 0.001 m radius)
e = elementary charge (1.6 * 10^-19 C)

Substituting these values into the equation, we get:

n_e = (1.0 * 10^20) / (1.6 * 10^-19 * 5.0 * 0.001 * 0.001 * pi) = 3.98 * 10^27 electrons/m^3

Now, we can substitute this value for n_e into the equation for electron drift speed and solve for v_d:

v_d = (1.0 * 10^20) / (3.98 * 10^27 * 1.6 * 10^-19 * 0.001 * 0.001 * pi) = 7.9 * 10^-3 m/s

Therefore, the electron drift speed in the iron wire is approximately 7.9 * 10^-3 m/s.