The radius of the thinner wire is 0.22 mm and the radius of the thick wire is 0.55 mm. There are 4.0 x 1028 mobile electrons per cubic meter of this material and the electron mobility is 6.0 x 10-4 (m/s)/(N/C). If 6.0 x 1018 electrons pass location D each second, what is the magnitude of the electric field at location B

0.2631 N/C

Explanation:

Given that:

The radius of the wire r = 0.22 mm = 0.22 × 10⁻³ m

The radius of the thick wire  r' = 0.55 mm = 0.55 × 10⁻³ m

The numbers of electrons passing through B, N = 6.0  × 10¹⁸ electrons

Electron mobility  μ =  6.0 x 10-4 (m/s)/(N/C)

= 0.0006

The number of electron flow per second is calculated as follows:

The magnitude of the electric field is:

E =

E =

E =

E = 0.2631 N/C

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.