An electric current of 16A exists in a metal wire of cross-section 10^-6m² and length 1m. Assuming one free electron per atom, the drift speed of the free electrons in the wire will be (Density of the metal $= 5 \times 10^{3} kg / m^{3}$ Atomic Weight )__________

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$5 \times 10^{- 3} m / s$

$2 \times 10^{- 3} m / s$

$4 \times 10^{- 3} m / s$

$7 .5 \times 10^{- 3} m / s$

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{N}{N_{A}} = \frac{m}{M} \Rightarrow N = \frac{m}{M} N_{A} and n = \frac{N}{V} = \frac{m}{M} . \frac{N_{A}}{V} \Rightarrow n = \frac{N_{A}}{M} . d = \frac{6 \times 10^{23} \times 5 \times 10^{3}}{60 \times 10^{- 3}} = 5 \times 10^{28} / m^{3} \\ I = neAV \Rightarrow V = \frac{I}{neA} = \frac{16}{5 \times 10^{28} \times 1 .5 \times 10^{- 19} \times 10^{- 6}} \\ ∴ V = 2 \times 10^{- 3} m / s \end{array}$