Question

A potential difference of 3.00 V will be applied to a 31.00 m length of 18-gauge platinum wire (diameter = 0.0400 in). Calculate the current (in amps). Calculate the magnitude of the current density. Calculate the magnitude of the electric field within the wire. Calculate the rate at which thermal energy will appear in the wire. Please show units!

Answer 1

diameter is d = 0.04 in = 0.001016 m

radius is r = d/2 = 0.001016/2 = 0.000508 m

Area of cross section is A = pi*r^2 = 3.142*0.000508^2 = 8.1*10^-7 m^2

current density is j = i/A

i = V/R

Resistance is R = rho*l/A

rho is the resistivity of platinum = 10.6*10^-8 ohm-m

R = rho*l/A = 10.6*10^-8*31/(8.1*10^-7) = 4.05 ohm

i = V/R = 3/4.05 = 0.74 A is the current

current density is j = i/A = 0.74/(8.1*10^-7) = 9.135*10^5 A/m^2

Electric field is E = J*rho = 9.135*10^5*10.6*10^-8 = 0.0968 N/C

Power dissipated is P = i^2*R = 0.74^2*4.05
P = 2.22 W