Answer:
30.22 hours
Explanation:
Given data:
A= l² = (2 x [tex]10^{-3}[/tex])² = 4 x [tex]10^{-6}[/tex] m²
Length 'L' = 5m
current '[tex]I[/tex]' = 2 A
density of free electrons 'n'= 8.5 x [tex]10^{28}[/tex] /m³
Current Density 'J' = [tex]I[/tex]/ A
J= 2/4 x [tex]10^{-6}[/tex]
J= 5 x [tex]10^{5}[/tex] A/m²
We can determine the time required for an electron to travel the length of the wire by
T= L/ Vd
Where,
L is length and Vd is drift velocity.
Vd can be defined by J/ n|q|
where,
n is the charge-carrier number density
|q| is is the charge carried by each charge carrier =>1.6 x [tex]10^{-19}[/tex]C
T= L/ Vd
Therefore,
T= L . n|q| / J
T= (4 x 8.5 x [tex]10^{28}[/tex] x |1.6 x [tex]10^{-19}[/tex]|)/5 x [tex]10^{5}[/tex]
T= 108800 seconds =>1813.33 minutes
Converting minute into hours:
T= 30.22 hours
Thus, time that is required for an electron to travel the length of the wire is 30.22 hours
