Drift speed $(v_d)$ varies with the intensity of the electric field $(E)$ as per the relation:

The drift velocity of electrons in a conducting wire connected to a cell is $V_{d}$. If the length of the wire is doubled and the area of cross-section is halved,then the drift velocity of electrons becomes:

$A$ $1\,m$ long copper wire carries a current of $1\,A$. If the cross-section of the wire is $2.0\,mm^{2}$ and the resistivity of copper is $1.7 \times 10^{-8}\,\Omega\,m$,the force experienced by a moving electron in the wire is $x \times 10^{-23}\,N$. Find the value of $x$. (Charge on electron $= 1.6 \times 10^{-19}\,C$)

If $n, e, \tau$ and $m$ respectively represent the electron density,charge,relaxation time,and mass of the electron,then the resistance of a wire of length $l$ and area of cross-section $A$ will be

$A$ 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 $1.5 \times 10^{16}\,m^{-3}$.

Current density in a cylindrical wire of radius $R$ is given as $J = \begin{cases} J_0 \left( \frac{x}{R} - 1 \right) & 0 \leqslant x < \frac{R}{2} \\ J_0 \frac{x}{R} & \frac{R}{2} \leqslant x \leqslant R \end{cases}$. The current flowing in the wire is: