The electric intensity $E$,current density $j$,and specific resistance $\rho$ (often denoted as $k$ in some texts) are related to each other by the relation:

The drift velocity of electrons in a current-carrying wire of radius $r$ is $v$. If we want a drift velocity of $2v$ in a wire of the same material but with half the radius,what current must be passed through it?

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:

$A$ copper wire of length $10 \, m$ and radius $(10^{-2} / \sqrt{\pi}) \, m$ has an electrical resistance of $10 \, \Omega$. The current density in the wire for an electric field strength of $10 \, V/m$ is:

In a copper wire,each atom releases one free electron. If the diameter is $1\, mm$ and the current is $1.1\, A$,find the drift velocity. (Given: Density $\rho = 9 \times 10^{3}\, kg/m^3$,Molar mass $M = 63\, g/mol$,Avogadro number $N_A = 6.023 \times 10^{23}\, mol^{-1}$,Elementary charge $e = 1.6 \times 10^{-19}\, C$)

What is relaxation time $(\tau )$? And what is drift velocity $(v_d)$?