Question 5.26: A wire having a specific resistance of 1.6 × 10^−8 Ωm, lengt...

First we must find the resistance of the wire and then we can find the voltage drop.

To find the resistance we use:

\begin{aligned}R &=\frac{\rho l}{A}=\frac{1.6 \times 10^{-8} \times 20}{1 \times 10^{-6}} \\ &=32 \times 10^{-2}=0.32 \ \Omega\end{aligned}

To find the voltage drop we can apply Ohm’s Law:

V = I × R = 5 A × 0.32 Ω = 1.6 V

Hence a potential of 1.6 V will be dropped between the ends of the wire.