$A$ 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$)

$A$ copper wire has a square cross-section,$2.0 \, mm$ on a side. It carries a current of $8 \, A$ and the density of free electrons is $8 \times 10^{28} \, m^{-3}$. The drift speed of electrons is equal to

When the current $i$ is flowing through a conductor,the drift velocity is $v$. If $2i$ current is flowed through the same metal but having double the area of cross-section,then the drift velocity will be

$A$ cylindrical conductor of length $2 \ m$ and area of cross-section $0.2 \ mm^{2}$ carries an electric current of $1.6 \ A$ when its ends are connected to a $2 \ V$ battery. Mobility of electrons in the conductor is $\alpha \times 10^{-3} \ m^{2}/V \cdot s$. The value of $\alpha$ is: (electron concentration $n = 5 \times 10^{28} \ m^{-3}$ and electron charge $e = 1.6 \times 10^{-19} \ C$)

Charge passing through a conductor of cross-section $0.3 \, m^2$ is given by $q = (3t^2 + 5t + 2) \, C$ where '$t$' is in seconds. The drift velocity at $t = 2 \, s$ is (Concentration of electrons in the conductor $= 2 \times 10^{25} \, m^{-3}$)

There is a current of $20 \, A$ in a copper wire of $10^{-6} \, m^2$ area of cross-section. If the number of free electrons per cubic meter is $10^{29}$,then the drift velocity is: