$A$ potential difference $V$ is applied across the ends of a metallic wire of length $l$. The drift velocity will be doubled if:

$A$ 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$)

Two cylindrical rods of uniform cross-sectional area $A$ and $2A$,having free electrons per unit volume $2n$ and $n$ respectively,are joined in series. $A$ current $I$ flows through them in a steady state. Then the ratio of the drift velocity of free electrons in the left rod to the drift velocity of electrons in the right rod is $\left( \frac{v_L}{v_R} \right)$.

Consider a metallic cube of edge length $L$. Its resistance,$R$,measured across its opposite faces is $R = \frac{m_e v}{n e^2 L^2}$,where $n$ is the number density and $v$ is the drift speed of electrons in the cube,and $e$ and $m_e$ are the charge and mass of an electron respectively. Assuming the de-Broglie wavelength of the electron to be $L$,the maximum resistance of the sample is closest to ............. $\Omega$ ($e = 1.60 \times 10^{-19} \, C$; $m_e = 9.11 \times 10^{-31} \, kg$; Planck's constant,$h = 6.63 \times 10^{-34} \, Js$)

What is current density? Derive Ohm's law in the form of current density.

$A$ charged particle having a drift velocity of $7.5 \times 10^{-4} \, m s^{-1}$ in an electric field of $3 \times 10^{-10} \, V m^{-1}$ has a mobility in $m^{2} V^{-1} s^{-1}$ of: