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

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Current Electricity

The number density of free electrons in copper is nearly

8 \times 1 0^{28} m^{- 3}

. A copper wire has its area of cross section

= 2 \times 1 0^{- 6} m^{2}

and is carrying a current of 3.2 A . The drift speed of the electrons is ____

\times 1 0^{- 6} ms^{- 1}

Answer: 125

The current carried by the drifting electrons is

I = n e A v_{d}

, where

n

is the free-electron density,

e

the electronic charge and

A

the cross-section. Solving for the drift speed,

v_{d} = \frac{I}{n e A} = \frac{3.2}{( 8 \times 1 0 ^{28} ) ( 1.6 \times 1 0 ^{- 19} ) ( 2 \times 1 0 ^{- 6} )}

. The denominator equals

2.56 \times 1 0^{4}

, so

v_{d} = 1.25 \times 1 0^{- 4} m s^{- 1} = 125 \times 1 0^{- 6} m s^{- 1}

. The drift speed is

125 \times 1 0^{- 6} m s^{- 1}

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Integer answer