Question

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

Answer 1

Answer:

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:

$I = \frac{q}{t}$

$I = \frac{Ne}{t}$

$I = \frac{6*10^{18}(1.6*10^{-18})C}{1 \ s}$

$I = 0.96 \ A$

The magnitude of the electric field is:

E = $\frac{I}{ \mu n eA}$

E = $\frac{I}{ \mu n e(\pi r^2)}$

E = $\frac{0.96}{(0.0006 m/s N/C ) (4*10^{28})(1.6*10^{-19}C)(0.55*10^{-3}m)^2}$

E = 0.2631 N/C