The resistivity of gold is 2.44 × 10^-8 Ω·m at a temperature of 20°C.A gold wire,0.5 mm in diameter and 44 cm long,carries a current of 380 ma.The number of electrons per second passing a given cross section of the wire is closest to:

Question

Quizplus · Accepted Answer

The correct answer is A because the number of electrons per second passing a given cross section of the wire is given by the equation I = neAv, where I is the current, n is the number of electrons per unit volume, e is the charge of an electron, A is the cross-sectional area of the wire, and v is the drift velocity of the electrons. The resistivity of the wire is given by the equation R = ρL/A, where R is the resistance of the wire, ρ is the resistivity of the wire, L is the length of the wire, and A is the cross-sectional area of the wire. The drift velocity of the electrons is given by the equation v = E/ρ, where E is the electric field in the wire. The electric field in the wire is given by the equation E = V/L, where V is the voltage across the wire. The voltage across the wire is given by the equation V = IR. Substituting these equations into the equation for the number of electrons per second passing a given cross section of the wire, we get I = neAv = ne(πd^2/4)(E/ρ) = ne(πd^2/4)(V/Lρ) = ne(πd^2/4)(IR/Lρ) = ne(π(5 × 10^-4 m)^2/4)(380 × 10^-3 A)(44 × 10^-2 m)/(2.44 × 10^-8 Ω·m) = 2.4 × 10^18 electrons/s.

The Resistivity of Gold Is 2