Question

D Question 4 20 pts At room temperature, what is the strength of the electric field in a 12-gauge copper wire (diameter 2.05 mm) that is needed to cause a 4.50-A current to flow? (Hint: See Table 25.1 in textbook for values of 5.211 x 103 V/m O 1.173 x 102 V/m O 2.986 x 102 V/m 9.380 x102 V/m 2.345 x102 V/m

Answer 1

Strength of Electric field is given by:

E = rho*J

rho = resistivity of copper = 1.72*10^-8 ohm-m

J = current density = i/A

E = rho*i/A

A = pi*d^2/4

Using given values:

E = 1.72*10^-8*4.50/(pi*(2.05*10^-3)^2/4)

E = 0.02345 V/m

E = 2.345*10^-2 V/m

Correct option is E.

Please Upvote.