$A$ copper wire of diameter $1.02\, mm$ carries a current of $1.7\, A$. Find the drift velocity $(v_d)$ of electrons in the wire. Given $n$,the number density of electrons in copper $= 8.5 \times 10^{28} /m^3$. (Express the answer in $mm/s$)

The drift velocity of electrons in a silver wire with a cross-sectional area of $3.14 \times 10^{-6} \, m^2$ carrying a current of $20 \, A$ is. Given the atomic weight of $Ag = 108$ and the density of silver $= 10.5 \times 10^3 \, kg/m^3$,the drift velocity is $.......... \times 10^{-4} \, m/s$.

The drift velocity of electrons in a conducting wire connected to a cell is $V_{d}$. If the length of the wire is doubled and the area of cross-section is halved,then the drift velocity of electrons becomes:

$(a)$ Consider the circuit in the figure. How much energy is absorbed by electrons from the initial state of no current (ignore thermal motion) to the state of drift velocity?
$(b)$ Electrons give up energy at the rate of $R I^2$ per second to thermal energy. What time scale would one associate with the energy in problem $(a)$? Given: $n = 10^{29} \, m^{-3}$,length of circuit $l = 10 \, cm$,cross-section $A = (1 \, mm)^2$.

$A$ metal has $9 \times 10^{28}$ conduction electrons per $m^3$ and its resistivity is $1 \times 10^{-8} \Omega \cdot m$. If the drift speed of an electron in the metal is $1.6 \times 10^6 \ m/s$, then its mean free path is (mass of electron $= 9 \times 10^{-31} \ kg$ and charge of electron $= 1.6 \times 10^{-19} \ C$). (in $nm$)

$A$ conductor of length $100 \, cm$ and area of cross-section $1 \, mm^2$ carries a current of $5 \, A$. If the resistivity of the material of the conductor is $3.0 \times 10^{-8} \, \Omega \cdot m$, then the electric field across the conductor is (in $ \, V/m$)