In the given hollow cylindrical conductor,the current density is $J = \frac{J_0}{r^2}$,where $J_0$ is a constant and $r$ is the distance from the axis of the cylinder. If the radius of the inner surface is $a$ and the radius of the outer surface is $2a$,find the current passed through the conductor.

At room temperature,copper has a free electron density of $8.4 \times 10^{28} \, m^{-3}$. The copper conductor has a cross-section of $10^{-6} \, m^2$ and carries a current of $5.4 \, A$. The electron drift velocity in copper is:

Electric current is due to the drift of electrons in

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,$h = 6.63 \times 10^{-34} \, Js$)

$A$ constant potential difference is applied between the ends of a wire. If the length of the wire is elongated to $4$ times its original length,then the drift velocity of electrons will be:

$A$ 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$)