Current Electricity

The current $I=80 \text{A}$, and the length of the conductor $L=10 \text{m}$.Expression for Current:Current can be written as $I=n e A V_d$, where $n$ is the number of electrons per unit volume, $e$ is the charge of an electron, $A$ is the cross-sectional area, and $V_d$ is the drift velocity of electrons.Total Number of Electrons:Let $N$ be the total number of free electrons in the length $L$ of the conductor. $N=n A L$.Total Momentum:The total momentum of all free electrons is $P_{\text{total}} = N (m_e V_d)$, where $m_e$ is the mass of an electron.Substitute $N$ :So, $P_{\text{total}} = (n A L) (m_e V_d)$.Rewrite with Current Expression:Since $I=n e A V_d$, we can replace $n A V_d$ with $\frac{I}{e}$.Therefore, $P_{\text{total}} = \frac{I}{e} L m_e$.Simplified Formula:So, $P_{\text{total}} = I \frac{L m_e}{e}$.Plug in the Values:$P_{\text{total}} = 80 \times \frac{10 \times 9.1 \times 10^{-31}}{1.6 \times 10^{-19}}$First, multiply $80 \times 10 = 800$.Now $800 \times 9.1 = 7280$.So, numerator is $7280 \times 10^{-31}$ and denominator is $1.6 \times 10^{-19}$.Divide 7280 by 1.6 to get 4550 .Power of ten: $\frac{10^{-31}}{10^{-19}} = 10^{-12}$Final Answer:$P_{\text{total}} = 4550 \times 10^{-12} \text{N-s} = 4.55 \times 10^{-9} \text{N-s} = 455 \times 10^{-11} \text{N-s}$