1) Consider an electric power transmission line that carries a constant electric current of i = 500 A. The cylindrical copper cable used to transmit this current has a diameter o = 2.00 cm and a length L = 150 km. If there are 8.43x10^28 free electrons per cubic meter (m^3 ) in the cable, calculate how long it would take for an electron to cross the entire length of the transmitter line.
answer= 10 x 10^6 years
Answers
Answer:
please make me as brain list
Given info : an electric power transmission line that carries a constant electric current of i = 500 A. The cylindrical copper cable used to transmit this current has a diameter, d = 2 cm and a length,l = 150 km. if there are 8.43 × 10^28 free electrons/cm³ in the cable.
To find : the time taken by an electron to cross the entire length of the transmitter line is. ..
solution : drift velocity, v = i/neA
where A = cross sectional area of wire = πr² = 3.14 × (1 × 10¯² m)² = 3.14 × 10¯⁴ m²
i = current = 500A
l = length of transmitter line = 150 km = 150 × 10³ m
n = 8.43 × 10^28 /cm³
e = 1.6 × 10^-19 C
now, v = 500/(8.43 × 10^28 × 1.6 × 10^-19 × 3.14 × 10^-4)
= 1.18 × 10¯⁴ m/s
now time taken by an electron to cross the transmitter line = length of line/drift velocity of electron
= 150 × 10³/(1.18 × 10¯⁴) s
= 127.118 × 10^7 sec
= 1.27118 × 10^9 sec
= 40.309 year
Therefore the time taken by an electron to cross the transmitter line is 40.309 years.