Q. The resistivity of a potentiometer wire is $40 \times 10^{- 8}ohm - m$ and its area of cross-section is $8 \times 10^{- 6}m^{2}.$ If $0.2\ amp$ current is flowing through the wire, the potential gradient will be
A) $10^{- 2}volt/m$
B) $10^{- 1}volt/m$
C) $3.2 \times 10^{- 2}volt/m$
D) $1\ volt/m$
Explanation
(a) Potential gradient $= \frac{V}{L} = \frac{iR}{L} = \frac{i\rho L}{AL} = \frac{i\rho}{A}$
$= \frac{0.2 \times 40 \times 10^{- 8}}{8 \times 10^{- 6}} = 10^{- 2}V/m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.