The resistivity of a potentiometer wire is $40 \times {10^{ - 8}}\,ohm - m$ and its area of cross-section is $8 \times {10^{ - 6}}\,{m^2}$. If $0.2\, amp$ current is flowing through the wire, the potential gradient will be
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Potential gradient $ = \frac{V}{L} = \frac{{iR}}{L} = \frac{{i\rho L}}{{AL}} = \frac{{i\rho }}{A}$
$ = \frac{{0.2 \times 40 \times {{10}^{ - 8}}}}{{8 \times {{10}^{ - 6}}}} = {10^{ - 2}}\,V/m$
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