A 1   m long copper wire carries a current of 1   A . If the cross section of the wire is 2 . 0…

Given here, length $l=1 \mathrm{m}$, current $i=1 \mathrm{A}$, area $A=2\times {10}^{-6} \mathrm{m}$ and resistivity $\rho =1.7\times {10}^{-8} \mathrm{\Omega } \mathrm{m}$.

Resistance of wire is $R=\frac{\rho l}{A}=\frac{1.7\times {10}^{-8}\times 1}{2\times {10}^{-5}}=\frac{1.7}{2}\times {10}^{-2} \mathrm{\Omega }$

Now, electric field in the wire is $E=\rho J$, where, $J=\frac{i}{A}$ is current density.

Thus, $E=\frac{i\rho }{A}=iR=1\times \frac{1.7}{2}\times {10}^{-2} \mathrm{V} {\mathrm{m}}^{-1}$

Electric force in terms of electric field is given by $F=qE$

Therefore, force experienced by moving electron is  $F=1.6\times {10}^{-19}\times \frac{1.7}{2}\times {10}^{-2}$

$=1.36\times {10}^{-21}$$=136\times {10}^{-23} \mathrm{N}$