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:
[tex]I = \frac{q}{t}[/tex]
[tex]I = \frac{Ne}{t}[/tex]
[tex]I = \frac{6*10^{18}(1.6*10^{-18})C}{1 \ s}[/tex]
[tex]I = 0.96 \ A[/tex]
The magnitude of the electric field is:
E = [tex]\frac{I}{ \mu n eA}[/tex]
E = [tex]\frac{I}{ \mu n e(\pi r^2)}[/tex]
E = [tex]\frac{0.96}{(0.0006 m/s N/C ) (4*10^{28})(1.6*10^{-19}C)(0.55*10^{-3}m)^2}[/tex]
E = 0.2631 N/C
