If n, e, τ and m respectively represent the 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
$(a) \frac{ml}{ne^{2}\tau A} \quad (b) \frac{m \tau^{2} A}{ne^{2} l} \quad (c) \frac{ne^{2} \tau A}{2ml} \quad (d) \frac{ne^{2} A}{2m \tau l}$
Text Solution
Verified by ExpertsA
$R = \rho \frac{l}{A}$ Since we know resistivity is given by $\rho = \frac{m}{ne^{2}r}$ Resistance $R = \frac{ml}{ne^{2}\tau A}$
where,
m = Mass of electron
$l=$ length of conductor
n = charge density
e= charge of electron
\tau = relaxation time
$A =$ cross-sectional area
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems