A copper wire with a resistivity $1.63\times {10}^{-8}\Omega ·m$ has a cross-section area $16.3\times {10}^{-4}c{m}^2$. Calculate the length of the wire required to make a $20\Omega$ coil.

Step 1: Given data

$$\begin{gathered} \rho =1.63\times {10}^{-8}\Omega m \\ A=16.3\times {10}^{-4}c{m}^2=16.3\times {10}^{-8}{m}^2 \\ R=20\Omega \end{gathered}$$

Step 2: To find the length of wire:

$$\begin{gathered} R=\frac{\rho L}{A} \\ L=\frac{RA}{\rho } \\ L=20\times \frac{16.3\times {10}^{-8}}{\left(1.63\times {10}^{-8}\right)}m \\ L=20\times 10m \\ L=200m \end{gathered}$$
Thus, the length of the wire is $L=200m$.