# Electric field of a thin/thick wire?

#### ZergLurker

The circuit shown above consists of a single battery, whose emf is 1.4 V, and three wires made of the same material, but having different cross-sectional areas. Each thick wire has cross-sectional area 1.4e-6 m2, and is 21 cm long. The thin wire has cross-sectional area 6.4e-8 m2, and is 7.5 cm long. In this metal, the electron mobility is 6e-4 (m/s)/(V/m), and there are 8e+28 mobile electrons/m3.

Use the appropriate equation(s), plus the equation relating electron current to electric field, to solve for the factor that goes in the blank below:
EF = ______* ED
Use the appropriate equation(s) to calculate the magnitude of ED
ED = ______ V/m
Use the appropriate equation(s) to calculate the electron current at location D in the steady state:
iD = ______

E = V/L doesn't work here. How do I do this?

#### ZergLurker

Nevermind. I figured it out. I'll post the solution here just in case anyone else has a similar problem.

i = naME
i = electron current
n = electron density
a = area
M = electron mobility
E = electric field
For the thin and thick wire, the electron current for both of them is constant. So you can put the two together.
nA(thick)ME(thick) = nA(thin)ME(thin)
A(thick)E(thick) = A(thin)E(thin), n and M are constants, too.
E(thick) = [A(thin)/A(thick)]E(thin)
A(thin) = 6.4E-8 m^2
A(thick) = 1.4E-6 m^2
E(thick) = 0.046E(thin)

You have the equation:
0 = 1.4 - EF*0.21 - ED*0.075 - EF*0.21 (found by using the Loop Rule).
E(thick) = 0.046E(thin).
So, EF = E(thick) and ED = E(thin) from the diagram. Plug it in.
0 = 1.4 - 2(0.21)(0.046)E(thin)
E(thin) = 14.8 V/M

You have i=naME to find i. So plug that in.
M = 6E-4
n = 8E28
A(thin) = 6.4E-8
E = 14.8
i = (6E-4)(8E28)(6.4E-8)(14.8) = 4.5E19 electrons / second

Last edited:
arbolis