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 × 10-19C)

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 × 10-19C) (A) 0.2 (B) 3 (C) 2 (D) 0.02

Tardigrade Admin · Accepted Answer

I = neAvd ⇒ vd = (I/neA) = (1.5/9 ×1028 ×1.6 × 10-19 × 5× 10-6) = 0.02 m/s

Q. Drift speed of electrons, when $1.5 A$ of current flows in a copper wire of cross section $5 m m^{2}$ , 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 = $1.6 \times 1 0^{- 19} C$ )

0.2

3

2

0.02

$I = n e A v_{d}$
$\Rightarrow v_{d} = \frac{I}{n e A} = \frac{1.5}{9 \times 1 0 ^{28} \times 1.6 \times 1 0 ^{- 19} \times 5 \times 1 0 ^{- 6}}$
= $0.02 m / s$

Q. Drift speed of electrons, when 1.5A of current flows in a copper wire of cross section 5mm2, 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 = 1.6×10−19C)

0.2

3

2

0.02

I=neAvd​ ⇒vd​=neAI​=9×1028×1.6×10−19×5×10−61.5​ = 0.02m/s

Q. Drift speed of electrons, when $1.5 A$ of current flows in a copper wire of cross section $5 m m^{2}$ , 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 = $1.6 \times 1 0^{- 19} C$ )

$I = n e A v_{d}$
$\Rightarrow v_{d} = \frac{I}{n e A} = \frac{1.5}{9 \times 1 0 ^{28} \times 1.6 \times 1 0 ^{- 19} \times 5 \times 1 0 ^{- 6}}$
= $0.02 m / s$