Drift speed of electrons, when 1.5 A of current flows in a copper wire of cross section $5 \text{ mm}^2$, is v. If the electron density in copper is $9 \times 10^{28}/\text{m}^3$ the value of v in mm/s is close to (Take charge of electron to be $= 1.6 \times 10^{-19}$ C)

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Current Electricity

Drift speed of electrons, when 1.5 A of current flows in a copper wire of cross section

5 mm^{2}

, is v. If the electron density in copper is

9 \times 1 0^{28} / m^{3}

the value of v in mm/s is close to (Take charge of electron to be

= 1.6 \times 1 0^{- 19}

Answer: (A)

Drift speed follows from

i = n e A v_{d}

, so

v_{d} = \frac{i}{n e A}

. Put in

i = 1.5

n = 9 \times 1 0^{28} m^{- 3}

e = 1.6 \times 1 0^{- 19}

C and

A = 5 \times 1 0^{- 6} m^{2}

: the denominator is

9 \times 1 0^{28} \times 1.6 \times 1 0^{- 19} \times 5 \times 1 0^{- 6} = 7.2 \times 1 0^{4}

, so

v_{d} = \frac{1.5}{7.2 \times 1 0 ^{4}} \approx 2.1 \times 1 0^{- 5}

m/s. In the units asked that is about

0.02

mm/s.

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Single correct option