A 1m long copper wire carries a current of 1A . If the cross...

Given here, length $l = 1m$, current $i = 1A$, area $A = 2 \times {10}^{- 6}m$ and resistivity $ρ = 1 . 7 \times {10}^{- 8}Ωm$.

Resistance of wire is $R = \dfrac{ρl}{A} = \dfrac{1 . 7 \times {10}^{- 8} \times 1}{2 \times {10}^{- 5}} = \dfrac{1 . 7}{2} \times {10}^{- 2}Ω$

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

Thus, $E = \dfrac{iρ}{A} = iR = 1 \times \dfrac{1 . 7}{2} \times {10}^{- 2}V{m}^{- 1}$

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

Therefore, force experienced by moving electron is $F = 1 . 6 \times {10}^{- 19} \times \dfrac{1 . 7}{2} \times {10}^{- 2}$

$= 1 . 36 \times {10}^{- 21}$$= 136 \times {10}^{- 23}N$