The resistivity of a potentiometer wire is 40×10−8Ω−m and its area of cross section is8×10−6m2. If 0.2A current is flowing through the wire, the potential gradient will be

dvdx=iρA=0.2×40×10−88×10−6=10−2V/m

dvdx=iρA=0.2×40×10−88×10−6=10−2V/m

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The resistivity of a potentiometer wire is $40 \times 10^{- 8} Ω - m$ and its area of cross section is $8 \times 10^{- 6} m^{2}$ . If 0.2A current is flowing through the wire, the potential gradient will be

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$10^{- 2} v / m$

$10^{- 1} v / m$

$3.2 \times 10^{- 2} v / m$

1v m

answer is A.

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$\frac{dv}{dx} = \frac{iρ}{A} = \frac{0 .2 \times 40 \times 10^{- 8}}{8 \times 10^{- 6}} = 10^{- 2} V / m$

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