Comparing Current in two different wires

[aba3]

Problem:
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 5 times the cross-sectional area, 1.1 times as many mobile electrons per cubic centimeter, and 2 times the mobility of wire A. In the steady state, 2E18 electrons enter wire A every second. How many electrons enter wire B every second?

Equation:
i = nAuE
n = Electrons/cubic meter
A = cross-sectional area
u = mobility of electrons
E = electric fieldSolution:
i_b=i_a * 2*5*1.1E6
it is 1.1E6 because the mobility factor needs to be in electrons/ cubic meter

i_b = 2.2E25

Webassign says that's wrong. I can't come up with anything else

 

[Delphi51]

ib=ia * 2*5*1.1E6

Omit the E6. The whole story is
i = nAuE = 2*n1*5*A1*1.1*u1*E
= 2*5*1.1*n1*A1*u1*E
= 2*5*1.1*i1
= 2*5*1.1*2E18
If 1.1 times as many electrons per cm^3, then there will be 1.1 times as many electrons per m^3 too.

 

[aba3]

there are a million cubic centimeters in a meter cube. if each cubic cm has 1.1 more electron then the meter cube has 1.1E6 extra electrions
(Also, I already tried that and webassign said I was wrong).

 

[Delphi51]

No. It doesn't give a count of the extra electrons. It says the other wire has 10% more per cm^3. That means 10% more per m^3. Say the first one has 100 per cm^3. Then the second will have 110.
Per m^3 the first one will have 100*1E6 and the second will have
110*1E6. The ratio will be 110E6/100E6 = 1.1Kind of like saying what if every student in the class overeats and comes back to class on Monday 10% heavier. Then the whole class together will be 10% heavier, not 10*the number of students heavier in pounds, kg or tons.

 

[aba3]

You were right. Thanks.