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$\frac{ml}{ne^2 \tau A}$
- B$\frac{m \tau^2 A}{ne^2 l}$
- C$\frac{ne^2 \tau A}{2ml}$
- D$\frac{ne^2 A}{2m \tau l}$
Explore More
Similar Questions
The relaxation time in conductors
Medium
View SolutionWhich of the following is a vector quantity?
Easy
View SolutionFor 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.
Medium
View SolutionThe area of cross-section of a copper wire is $4 \times 10^{-7} \,m^2$ and the number of free electrons per cubic metre in copper is $8 \times 10^{28}$. If the wire carries a current of $6.4 \,A$,then the drift velocity of the electrons (in $10^{-3} \,m \,s^{-1}$) is:
$A$ straight conductor of uniform area of cross-section carries a current $I$. If $s$ is the specific charge (charge-to-mass ratio) of the electron,what is the total momentum of all the electrons per unit length of the conductor due to their drift velocity?
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo