Question

A copper wire has a square cross section 2.0 mm on a side. The wire is 5.0 m long and carries a current of 2.0 A. The density of free electrons is 8.5*1028/m3. How much time is required for an electron to travel the length of the wire?

Answer 1

Answer:

30.22 hours

Explanation:

Given data:

A= l² = (2 x $10^{-3}$)² = 4 x $10^{-6}$ m²

Length 'L' = 5m

current '$I$' = 2 A

density of free electrons 'n'= 8.5 x $10^{28}$ /m³

Current Density 'J' = $I$/ A

J= 2/4 x $10^{-6}$

J= 5 x $10^{5}$ 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 $10^{-19}$C

T= L/ Vd

Therefore,

T= L . n|q| / J

T= (4 x 8.5 x $10^{28}$ x |1.6 x $10^{-19}$|)/5 x $10^{5}$

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