Question

Drift speed of electrons, when 1.5 A of current flows in a copper wire of cross section 5 mm², is v. If the electron density in copper is 9 × 10²⁸ /m³
the value of v in mm/s is close to (Take charge of electron to be = 1.6 × 10⁻¹⁹ C)
(A) 0.02 (B) 3
(C) 2 (D) 0.2
[JEE Main 2019]

Answer 1

Answer:

A. will be correct answers

Answer 2

The drift speed of electrons = 0.02 $\frac{mm}{s}$

Option (A) is correct.

Explanation:

Given :

The current in wire $(I) =$ 1.5 A

Cross section area $(A) = 5 \times 10^{-6} m^{2}$

Electron density $(n) = 9 \times 10^{28} m^{-3}$

Charge of electron $e = 1.6 \times 10^{-19}$

The current density in wire is given by,

     $J = \frac{I}{A} = nev_{d}$

Where $J =$ current density, $v_{d} =$ drift speed.

    $v_{d} = \frac{1.5}{5 \times 10^{-6} \times 9 \times 10^{28} \times 1.6 \times 10^{-19} }$

    $v_{d} = 0.0208 \times 10^{-3}$

    $v_{d}$ ≅ $0.02 \frac{mm}{s}$