The resistivity of gold is $2.44 \times 10 ^ { - 8 } \Omega \cdot \mathrm { m }$ at a temperature of $20 ^ { \circ } \mathrm { C }$. A gold wire that is $0.50 \mathrm {~mm}$ in diameter and $44 \mathrm {~cm}$ long carries a current of $380 \mathrm {~mA}$. How many electrons each second pass a given cross section of the wire at $20 ^ { \circ } \mathrm { C } ? \left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } \right)$

Question

Quizplus · Accepted Answer

The number of electrons passing a given cross section per second is equal to the current divided by the charge of one electron. Given a current of $380 \mathrm {~mA} = 380 \times 10^{-3} \mathrm {~A}$ and the charge of an electron $e = 1.60 \times 10^{-19} \mathrm {~C}$, the calculation is $ \frac{380 \times 10^{-3}}{1.60 \times 10^{-19}} = 2.375 \times 10^{18}$ electrons per second, which rounds to $2.4 \times 10^{18}$.

The Resistivity of Gold Is 2.44×10−8Ω⋅m2.44 \times 10 ^ { - 8 } \Omega \cdot \mathrm { m }2.44×10−8Ω⋅m

The Resistivity of Gold Is $2.44 \times 10 ^ { - 8 } \Omega \cdot \mathrm { m }$