Question

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

• A. $10^{-1} V /m $
• B. $10^{-2} V /m $
• C. $10^{-3} V /m $
• D. $10^{-4} V /m $

Answer 1

Answer: (B)

Given : $\rho=40 \times 10^{-8}$ ohm $- m\,\,\, I=0.2 A \,\,\,A=8 \times 10^{-6} m ^{2}$
Resistance of the wire $R=\rho \frac{L}{A}$
Potential difference across the wire $V = IR = I \rho \frac{ L }{ A }$
Thus potential gradient of the wire $\frac{ V }{ L }=\frac{ I \rho}{ A }$
$\therefore \frac{ V }{ L }=\frac{0.2 \times 40 \times 10^{-8}}{8 \times 10^{-6}}$
$\Rightarrow \frac{ V }{ L }=10^{-2} V / m$