# Potentiometer wire is of cross sectional area $8 \times 10^{-6}m^2$ and has resistivity $40 \times 10^{-8}$. If $0.2$ amp current flows through it its potential gradient is

$(a)\;10^{-2} v/m \\(b)\;2 \times 10^{-2}\;v/m \\(c)\;10^{-3} \;v/m \\(d)\;3 \times 10^{-3}\;v/m$

Potential gradient is $\large\frac{V}{L}$ where V is potential across wire of length L.
$R= \large\frac{\rho L}{A}$
$\qquad= \large\frac{40 \times 10^{-8} \times L}{8 \times 10^{-6}}$
$\qquad= 5 \times 10^{-2}\;L$
$V= I \times R$
$\qquad= 5 \times 10^{-2} \times L \times 0.2$
$\qquad = 1 \times 10^{-2} \times L$
Potential gradient $=\large\frac{V}{L}$
$\qquad= 10^{-2}\;V/m$
Hence a is the correct answer.