The resistivity of a potentiometer wire is `40xx10^(-8)` ohm-m and its area of cross-section is `8 xx 10^(-6) m^(2)`. If a current of 0.4 A is flowing through the wire, then the potential gradient will be

Options

A. `10^(-2) V//m`
B. `10^(-1) V//m`
C. `2xx10^(-2) V//m`
D. `1 V//m`

Answer

C

Explanation

`R= (rho L)/(A)= (40xx10^(-8)xxL)/(8xx10^(-6))=5xx10^(-2)L`
Potential gradient `=(V)/(L)=(IR)/(L)=(0.4xx5xx10^(-2)L)/(L)`
`=2xx10^(-2)V//m`