A 1,m long copper wire carries a current of 1 | vedclass

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(B) Given: Length $l = 1\,m$,Current $i = 1\,A$,Area $A = 2.0\,mm^{2} = 2.0 	imes 10^{-6}\,m^{2}$,Resistivity $ho = 1.7 	imes 10^{-8}\,\Omega\,m$,

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$136$

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(B) Given: Length $l = 1\,m$,Current $i = 1\,A$,Area $A = 2.0\,mm^{2} = 2.0 	imes 10^{-6}\,m^{2}$,Resistivity $ho = 1.7 	imes 10^{-8}\,\Omega\,m$,Charge $e = 1.6 	imes 10^{-19}\,C$.First,calculate the resistance $R$ of the wire:$R = \frac{ho l}{A} = \frac{1.7 	imes 10^{-8} 	imes 1}{2.0 	imes 10^{-6}} = 0.85 	imes 10^{-2}\,\Omega$.The potential difference $V$ across the wire is given by Ohm's Law:$V = iR = 1 	imes 0.85 	imes 10^{-2} = 0.85 	imes 10^{-2}\,V$.The electric field $E$ in the wire is:$E = \frac{V}{l} = \frac{0.85 	imes 10^{-2}}{1} = 0.85 	imes 10^{-2}\,V/m$.The force $F$ experienced by an electron is:$F = eE = 1.6 	imes 10^{-19} 	imes 0.85 	imes 10^{-2} = 1.36 	imes 10^{-21}\,N$.Expressing this in terms of $10^{-23}$:$F = 136 	imes 10^{-23}\,N$.Thus,$x = 136$.

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