Physics

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?

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  1. 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.

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