The area of cross-section of a copper wire is $4 \times 10^{-7} \,m^2$ and the number of free electrons per cubic metre in copper is $8 \times 10^{28}$. If the wire carries a current of $6.4 \,A$,then the drift velocity of the electrons (in $10^{-3} \,m \,s^{-1}$) is:

When a potential difference is applied across the ends of a linear metallic conductor,

$(a)$ 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 $A = (1 \, mm)^2$.

$A$ potential difference of $V$ is applied at the ends of a copper wire of length $l$ and diameter $d$. On doubling only $d$,drift velocity

The total momentum of electrons in a straight wire of copper of length $1\, m$ carrying a current of $16\, A$ is

Charge passing through a conductor of cross-section $0.3 \, m^2$ is given by $q = (3t^2 + 5t + 2) \, C$ where '$t$' is in seconds. The drift velocity at $t = 2 \, s$ is (Concentration of electrons in the conductor $= 2 \times 10^{25} \, m^{-3}$)