Current Electricity

To find the resistivity of the potentiometer wire, we are provided with the following information:Area of cross-section (A): $4 \text{cm}^2 = 4 \times 10^{-4} \text{m}^2$Current (i): 1 APotential Gradient $(\;\text{V/l}) : 7.5 \text{V/m}$We need to calculate the resistivity ( $\rho$ ) of the wire.Calculation:From Ohm's law, we know the following relationship:$V = i \cdot R$The resistance $R$ for a wire can be expressed in terms of its resistivity:$R = \rho \; \frac{l}{A}$Thus, substituting $R$ in the equation for $V$ :$V = i \cdot \; \frac{\rho l}{A}$Rearranging this equation to solve for resistivity ( $\rho$ ):$\rho = \; \frac{V}{l} \cdot \; \frac{A}{i}$Substitute the known values:$\rho = \left(7.5 \text{V/m}\right) \cdot \left(\; \frac{4 \times 10^{-4} \text{m}^2}{1 \text{A}}\right)$Calculate $\rho$ :$\rho = \; \frac{7.5 \times 4 \times 10^{-4}}{1}$ $\; \rho = 30 \times 10^{-4}$$\; \rho = 3 \times 10^{-3} \Omega\text{-m}$Therefore, the resistivity of the potentiometer wire is $3 \times 10^{-3} \Omega\text{-m}$.