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
101V/m
B
102V/m
C
103V/m
D
104V/m

Solution

## The correct option is B $$10^{-2}V/m$$Given :  $$\rho = 40\times 10^{-8} ohm-m$$    $$I = 0.2$$ A   $$A = 8 \times 10^{-6} m^2$$Resistance of the wire  $$R = \rho \dfrac{L}{A}$$Potential difference across the wire  $$V = IR = I\rho \dfrac{L}{A}$$Thus potential gradient of the wire  $$\dfrac{V}{L} = \dfrac{I\rho}{A}$$$$\therefore$$  $$\dfrac{V}{L} = \dfrac{0.2 \times 40\times 10^{-8}}{8\times 10^{-6}}$$  $$\implies$$  $$\dfrac{V}{L} = 10^{-2} V/m$$Physics

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