A cylindrical conductor of length 2 m and ar | vedclass

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(A) Given: Length $l = 2 \ m$, Area $A = 0.2 \ mm^{2} = 0.2 	imes 10^{-6} \ m^{2}$, Current $I = 1.6 \ A$, Voltage $V = 2 \ V$, Concentration $n = 5

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(A) Given: Length $l = 2 \ m$, Area $A = 0.2 \ mm^{2} = 0.2 	imes 10^{-6} \ m^{2}$, Current $I = 1.6 \ A$, Voltage $V = 2 \ V$, Concentration $n = 5 	imes 10^{28} \ m^{-3}$, Charge $e = 1.6 	imes 10^{-19} \ C$.The drift velocity $V_{d}$ is given by $V_{d} = \frac{I}{neA}$.The mobility $\mu$ is defined as $\mu = \frac{V_{d}}{E}$, where $E = \frac{V}{l}$ is the electric field.Substituting $V_{d}$ and $E$, we get $\mu = \frac{I}{neA} 	imes \frac{l}{V}$.Substituting the values: $\mu = \frac{1.6 	imes 2}{5 	imes 10^{28} 	imes 1.6 	imes 10^{-19} 	imes 0.2 	imes 10^{-6} 	imes 2}$.$\mu = \frac{3.2}{5 	imes 10^{28} 	imes 1.6 	imes 10^{-19} 	imes 0.2 	imes 10^{-6} 	imes 2} = \frac{3.2}{3.2 	imes 10^{3}} = 1 	imes 10^{-3} \ m^{2}/V \cdot s$.Comparing with $\alpha 	imes 10^{-3} \ m^{2}/V \cdot s$, we get $\alpha = 1$.

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