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

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

This means that:

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

Also

Therefore:

That is:

Since I_A = 3*10^18 e/s

I_B = 20.4*10^18 e/s

d they tend to lose electrons and become cations.

explanation:

small = small current large mag large current