Question

Suppose wire A and wire B are made of different metals, and are subjected to the same electric field in two different circuits. Wire B has 4 times the cross-sectional area, 1.7 times as many mobile electrons per cubic centimeter, and 1 times the mobility of wire A. In the steady state, 3e18 electrons enter wire A every second.
How many electrons enter wire B every second?
_________ electrons/second

Answer 1

Answer:

20.4e18 electrons/second ≈ 2e19 electrons/second

Explanation:

Hi!

To solve this problem we are going to use Omh's Law:

V = RI

And the relation ship between the resistance R and conductivity:

R = L/(σA)

*here we are considering a omhic  material*

The conductance σ is related to electron mobility and electron density by:

σ = nμ

Replacing all these relations in the omhs law, we get:

V = (LI)/(σA)

We know that both wire are subject to the same electric field, therefore V is the same for both, moreover, since no additional info for the length of the wires is given we are going to consider that L is the same for both. Therefore

$\frac{V_A}{L_A}=\frac{V_B}{L_B}$

This means that:

$\frac{I_A}{\sigma_A A_A} = \frac{I_B}{\sigma_B A_B}$

From the relation of the conductance and electron mobility and density, and the data given to us, we know that:

$\sigma_B = 1.7 \sigma_A$

Also

$A_B = 4 A_A$

Therefore:

$\frac{I_A}{\sigma_A A_A} = \frac{I_B}{1.7\sigma_A\times4A_A}$

That is:

$I_B = (4\times1.7)I_A = 6.8 I_A$

Since I_A = 3*10^18 e/s

I_B = 20.4*10^18 e/s