The resistivity ofgod is 244 x 108 m at room temperature A g

The resistivity ofgod is 2.44 x 10-8 -m at room temperature. A gold wire that is 0.9 mm in diameter and 14 em long carries a current of 940 mA. What is the electric field in the wire? 2

Solution

rho of gold = 2.44*10^8 ohm

Resistance R is given by

R = rho*L/A = rho*L/(pi*r^2)

L = 0.14 m = 14 cm

R = 2.44*10^-8*0.14/(3.14*0.45*10^-3)^2 = 1.71*10^-3 ohm

i = 940*10^-3 Amp.

ohm\'s Law

V = i*R

V = 940*10^-3*1.71*10^-3

V = 1.61*10^-3 V

E = V/d

E = 1.61*10^-3/0.14 = 0.0115 V/m