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

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

$A$ potential difference is applied across the ends of a metallic wire. If the potential difference is doubled,then the drift velocity

Which of the following is a vector quantity?

$A$ copper wire of cross-sectional area $1.0 \ mm^2$ carries a current of $1.34 \ A$. Assuming that each copper atom contributes one free electron,calculate the drift velocity of the free electrons in the wire in $mm/s$. (Given: density of copper = $8990 \ kg/m^3$,atomic mass = $63.50 \ g/mol$)

$A$ silver wire of length $3 \,m$ and of cross-sectional area $6.14 \times 10^{-6} \,m^2$ carries a current of $6 \,A$. The atomic weight and density of silver are $108 \,g/mol$ and $10500 \,kg/m^3$,respectively. $A$ silver atom contributes one free electron for conduction. The Avogadro number is $6.023 \times 10^{23} /mol$. The drift velocity of electrons in silver is close to: