The resistivity of a potentiometer wire is `40xx10^(-8)Omega-m` and its area of cross section is `8xx10^(-6)m^(2)`. If 0.2 A current is flowing through the wire the potential gradient will be

Options

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

Answer

B

Explanation

Given , `rho = 40 xx 10^(-8) Omegam` .
`A = 98 xx 10^(-6) m^(2), I = 0.2 A`
As, resistance is given as, `R = (pl)/(A)`
`:. R/ I = rho/A = (40 xx 10^(-8))/(8 xx 10^(-6)) = 5 xx 10^(-2)`
The potential gradient of the were is
`V/ I = (IR)/(I) = 0.2 xx 5 xx 10^(-2) = 10^(-2) V//m`