The relationship between the electric field $E$ and the current density $J$ is given by:

The drift velocity of an electron is $v_d$ in a conductor of area of cross-section $A$ and carries a current $I$. Now, the area of cross-section and current flowing through the conductor are doubled, then the new drift velocity of the electron is . . . . . . .

$A$ silicon block at $300\,K$ has a length of $10\,cm$ and a cross-sectional area of $1 \times 10^{-4}\,m^2$. If a $2\,V$ battery is connected across its ends parallel to the length,find the current flowing through it. The electron mobility is $0.14\,m^2V^{-1}s^{-1}$ and the electron number density is $1.5 \times 10^{16}\,m^{-3}$.

$A$ constant electric current $I$ is passed through a straight conductor of length $l$. If $S$ is the specific charge of an electron,then the total momentum of the electrons is:

$A$ current $I$ is passing through a wire having two sections $P$ and $Q$ of uniform diameters $d$ and $d/2$ respectively. If the mean drift velocity of electrons in sections $P$ and $Q$ is denoted by $v_P$ and $v_Q$ respectively,then

The drift velocity of free electrons in a conductor is $v$ when a current $i$ is flowing in it. If both the radius and current are doubled,then the drift velocity will be