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If specific resistance of a potentiometer wire is $10^{-7}\ \Omega$ m and current flowing through it is 0.1 A, cross-sectional area of wire is $10^{-6}$ m$^2$ then potential gradient will be : -
A
$10^{-2}$ V/m
B
$10^{-4}$ V/m
C
$10^{-6}$ V/m
D
$10^{-8}$ V/m
Detailed Solution
Potential gradient $x = \dfrac{V}{l} = \dfrac{IR}{l} = \dfrac{I\rho l}{Al} = \dfrac{I\rho}{A}$
$x = \dfrac{0.1 \times 10^{-7}}{10^{-6}} = 10^{-2}$ V/m
$x = \dfrac{0.1 \times 10^{-7}}{10^{-6}} = 10^{-2}$ V/m