$A$ current of $10 \,A$ is maintained in a conductor of cross-section $1 \,cm^2$. If the number density of free electrons is $9 \times 10^{28} \,m^{-3}$,the drift velocity of free electrons is .......... $m/s$.

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 $1.6 \times 10^{-19} \, C$).

Assertion: The current density $\vec J$ at any point in an ohmic resistor is in the direction of the electric field $\vec E$ at that point.
Reason: $A$ point charge when released from rest in a region having only an electrostatic field always moves along electric lines of force.

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

$A$ copper wire of diameter $1.02\, mm$ carries a current of $1.7\, A$. Find the drift velocity $(v_d)$ of electrons in the wire. Given $n$,the number density of electrons in copper $= 8.5 \times 10^{28} /m^3$. (Express the answer in $mm/s$)

An ionization chamber with parallel conducting plates as anode and cathode has $5 \times 10^7$ electrons and the same number of singly-charged positive ions per $cm^3$. The electrons are moving at $0.4\, m/s$. The current density from anode to cathode is $4\,\mu A/m^2$. The velocity of positive ions moving towards cathode is ............. $m/s$. [$AIPMT$ $1992$]