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

20.4e18 electrons/second ≈ 2e19 electrons/second

Step-by-step 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

`(V_A)/(L_A)=(V_B)/(L_B)`

This means that:

`(I_A)/(\sigma_A A_A) = (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:

`(I_A)/(\sigma_A A_A) = (I_B)/(1.7\sigma_A*4A_A)`

That is:

`I_B = (4*1.7)I_A = 6.8 I_A`

Since I_A = 3*10^18 e/s

I_B = 20.4*10^18 e/s