A cylindrical copper cable 1.50Km long is connected across a 220.0V potential difference.
(a) What should be its diameter so that it produces heat at a rate of 50.0W?
(b) What is the electric field inside the cable under these conditions?
First, the value of the length of the cable will be converted to meters.
Therefore, the length of the cable is L = 1.5*10^3 m = 1500 m.
We know, from enunciation, the value of potential difference:V = 220 volts
The power is given and it's value is of P = 50 Watts
P = V^2/R => R = V^2/P
Since the area of the section of the cable is circular, w'ell recall the formula for the area of the circle:
A = pi*r^2 (1)
A = p*L/R (2), where p = 1.72/10^8 ohm/m
We'll equate (1) and (2) and we'll get:
pi*r^2 = p*L*P/V^2
r^2 = p*L*P/V^2*pi
r = sqrt(p*L*P/V^2*pi)
Diameter is d = 2*r = 2*sqrt(p*L*P/V^2*pi)
d = 2*sqrt1.72*10^-10*15*5*10^3/484
d = 2*sqrt 1.72*10^-7*75/22
d = 1.0325/100
d = 0.010325 meters
The diameter of the cable is of d = 0.010325 meters, such as it is producing heat at a rate of 50W.