A wire of diameter 0.02,m contains 10^{28} fr | vedclass

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(C) The formula for drift velocity is given by $v_d = \frac{I}{neA}$.Here,$I = 100\,A$,$n = 10^{28}\,m^{-3}$,and $e = 1.6 	imes 10^{-19}\,C$.The area

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$2 	imes 10^{-4}\,m/s$

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(C) The formula for drift velocity is given by $v_d = \frac{I}{neA}$.Here,$I = 100\,A$,$n = 10^{28}\,m^{-3}$,and $e = 1.6 	imes 10^{-19}\,C$.The area of cross-section $A = \pi r^2 = \pi (d/2)^2 = \pi (0.01)^2 = \pi 	imes 10^{-4}\,m^2$.Substituting these values into the formula:$v_d = \frac{100}{10^{28} 	imes 1.6 	imes 10^{-19} 	imes \pi 	imes 10^{-4}}$$v_d = \frac{100}{1.6 	imes \pi 	imes 10^5} \approx \frac{100}{5.026 	imes 10^5} \approx 1.989 	imes 10^{-4}\,m/s$.Thus,the drift velocity is nearly $2 	imes 10^{-4}\,m/s$.

$A$ wire of diameter $0.02\,m$ contains $10^{28}$ free electrons per cubic meter. For an electrical current of $100\,A$,the drift velocity of the free electrons in the wire is nearly:

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