$A$ potential difference of $V$ is applied at the ends of a copper wire of length $l$ and diameter $d$. On doubling only $d$,drift velocity

Drift speed of electrons,when $1.5 \, A$ of current flows in a copper wire of cross-section $5 \, mm^2$,is $v$. If the electron density in copper is $9 \times 10^{28} \, m^{-3}$,the value of $v$ in $mm/s$ is close to (Take charge of electron to be $1.6 \times 10^{-19} \, C$).

$A$ copper wire of length $1 \ m$ and radius $1 \ mm$ is connected in series with an iron wire of length $2 \ m$ and radius $3 \ mm$. If a current flows through both wires,the ratio of the current density in the copper wire to that in the iron wire will be:

$A$ conductor of length $\ell$ with a circular cross-section is shown in the figure,carrying a current $i$. The radius of the cross-section varies linearly from $a$ to $b$. Assuming $(b - a) << \ell$,calculate the current density at a distance $x$ from the left end.

For a cylinder of radius $R$,the current density is given by $J = J_0 \frac{r}{R}$,where $J_0$ is a constant and $r$ is the distance from the axis. Calculate the total current.

$A$ metal has $9 \times 10^{28}$ conduction electrons per $m^3$ and its resistivity is $1 \times 10^{-8} \Omega \cdot m$. If the drift speed of an electron in the metal is $1.6 \times 10^6 \ m/s$, then its mean free path is (mass of electron $= 9 \times 10^{-31} \ kg$ and charge of electron $= 1.6 \times 10^{-19} \ C$). (in $nm$)