Mastering Physics: Exercise 25.23

Mastering Physics: Exercise 25.23

A current-carrying gold wire has a diameter of 0.86 mm . The electric field in the wire is 0.51 V/m .

Part A

Question: What is the current carried by the wire?

Answer: 

p (resistivity) of Au = 2.4×10^(−8)

A = π × r^2

A = π × ((0.86×10^(−3) / 2)^2)

A = 5.8088 ×10^(−7)

p = E / J = E / (I / A) = EA / I

=> I = E×A / I

I = 0.51 × 5.8088 ×10^(−7) / (2.4×10^(−8))

I =  12  A

Part B

Question: What is the potential difference between two points in the wire 6.6 m apart?

Answer: 

V = E × L

V = 0.51 × 6.6

V =  3.4  V  

Part C

Question: What is the resistance of a 6.6-m length of this wire?

Answer: 

R = V / J

R = 3.366 / 12.3477

R =  0.28  Ω