Question

A copper wire of resistivity $1.6\times {10}^{-8}$ ohms metre has a cross-sectional area of $10\times {10}^{-4}c{m}^2$. Calculate the length of this wire required to make a $10$ ohm coil.

Answer 1

Given,

$R=10\Omega$

$A=10\times {10}^{-8}{m}^2$

$\rho =1.6\times {10}^{-8}\Omega /m$

We know that,

$R=\rho \frac{l}{A}$

$l=\frac{RA}{\rho }=\frac{10\times 10\times {10}^{-8}}{1.6\times {10}^{-8}}$

$l=62.5m$

The length of wire copper wire is $62.5m$