Correct Answer: B
Explanation:
Fractional power loss equals ρ(I/A)L/V, where ρ is the resistivity of the cable.
There is an upper limit to the current density I/A that can be conducted; otherwise the wire would get too hot. Hence, the fractional loss of power ΔP/P is proportional to L/V. The field strength is highest at the surface of the cable.