Target Exam

CUET

Subject

Physics

Chapter

Current Electricity

Question:

The average drift speed of conduction electrons in a copper wire of cross-sectional area $1.0 \times 10^{-7} m^2$ carrying a current of 1.5 A. (Assume the density of conduction electrons to be $\left.9 \times 10^{28} / m^3\right)$

Options:

$2.08 \times 10^{-4} m / s$

$1.04 \times 10^{-3} m / s$

$6.96 \times 10^{-3} m / s$

$2.9 \times 10^{-4} m / s$

Correct Answer:

$1.04 \times 10^{-3} m / s$

Explanation:

The correct answer is Option (2) → $1.04 \times 10^{-3} m / s$

To calculate the average drift speed ($v_d$) of conduction electrons -

$I=neAv_d$

where,

I (current) = 1.5 A

$n$ (number of density per nucleons) = $9×10^{28}m^{-3}$

$v_d=\frac{I}{nAe}=\frac{1.5}{(9×10^{28})(1×10^{-7})(1.6×10^{-19})}$

$≃1.04×10^{-3}m/s$