The number density of free electrons in coppe | vedclass

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(A) Given:Number density of electrons,$n = 8 	imes 10^{28} \ m^{-3}$Area of cross-section,$A = 2 	imes 10^{-6} \ m^2$Current,$I = 3.2 \ A$Charge of

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$125$

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(A) Given:Number density of electrons,$n = 8 	imes 10^{28} \ m^{-3}$Area of cross-section,$A = 2 	imes 10^{-6} \ m^2$Current,$I = 3.2 \ A$Charge of an electron,$e = 1.6 	imes 10^{-19} \ C$The formula for drift velocity $(v_d)$ is given by:$I = n e A v_d$Rearranging for $v_d$:$v_d = \frac{I}{n e A}$Substituting the values:$v_d = \frac{3.2}{(8 	imes 10^{28}) 	imes (1.6 	imes 10^{-19}) 	imes (2 	imes 10^{-6})}$$v_d = \frac{3.2}{25.6 	imes 10^3}$$v_d = 0.125 	imes 10^{-3} \ m/s$$v_d = 125 	imes 10^{-6} \ m/s$Thus,the drift speed is $125 	imes 10^{-6} \ m/s$.

The number density of free electrons in copper is nearly $8 \times 10^{28} \ m^{-3}$. $A$ copper wire has an area of cross-section $= 2 \times 10^{-6} \ m^2$ and is carrying a current of $3.2 \ A$. The drift speed of the electrons is $..... \times 10^{-6} \ m \ s^{-1}$.

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