Mastering Physics: Exercise 25.5

Mastering Physics: Exercise 25.5

Copper has 8.5×10^(28) free electrons per cubic meter. A 71.0 cm length of 12-gauge copper wire, that is 2.05 mm in diameter, carries 4.80 A of current.

Part A

Question: How much time does it take for an electron to travel the length of the wire?

Answer: 

I = n × |q| × v_d × A
v_d = I / (n × |q| × A)
v_d = 4.80 / (8.5×10^(28) 3.57×10^(−6) × 1.602×10^(−19) × π × ((3.57×10^(−3) / 2)^2) )
v_d = 1.06798 ×10^(−4)

t = L / v_d
t = 0.71 / (1.06798 ×10^(−4))

t =  6650  s  

Part B

Question: Repeat part A for 6-gauge copper wire (diameter 4.12 mm ) of the same length that carries the same current.

Answer: 

I = n × |q| × v_d × A

v_d = I / (n × |q| × A)

v_d = 4.80 / (8.5×10^(28) 3.57×10^(−6) × 1.602×10^(−19) × π × ((4.12×10^(−3) / 2)^2) )

v_d = 2.644 ×10^(−5)

t = L / v_d
t = 0.71 / (2.644 ×10^(−5))

t =  2.69×10^4  s

Part C

Question: Generally speaking, how does changing the diameter of a wire that carries a given amount of current affect the drift velocity of the electrons in the wire?

Answer: The drift velocity depends on the diameter of the wire as an inverse square relationship.