The resistivity of potentiometer wire is `40xx10^(-8)` ohm-metre and its area of cross-section is `8xx10^(-6) m^(2)`. If 0.2 ampere current is flowing through the wire, the potential gradient of the wire is

Options

A. `10^(-1) V//m`
B. `10^(-2) V//m`
C. `10^(-3) V//m`
D. `10^(-4) V//m`

Answer

B

Explanation

`R=(rho L)/(A)`
`therefore (R )/(L)=(rho)/(A)=(40xx10^(-8))/(8xx10^(-6))=5xx10^(-2)`
and Potential gradient `=(V)/(L)=(IR)/(L)=I . (rho)/(A)`
`= 0.2xx5xx10^(-2)`
`= 10^(-2)V//m`