There is a current of 40 A in a wire of 10-6 m 2 area of cross-section. If the number of free electrons per cubic metre is 1029, then the drift velocity is

Question

There is a current of 40 A in a wire of 10-6 m 2 area of cross-section. If the number of free electrons per cubic metre is 1029, then the drift velocity is (A) 250 × 10 -3 ms-1 (B) 25.0 × 10 -3 ms-1 (C) 2.50 × 10 -3 ms-1 (D) 1.25 × 10 -3 ms-1

Tardigrade Admin · Accepted Answer

Drift velocity vd=(i/n e A) =(40/1029 × 1.6 × 10-19 × 10-6) =2.5 × 10-3 ms -1

Q. There is a current of $40 A$ in a wire of $1 0^{- 6} m^{2}$ area of cross-section. If the number of free electrons per cubic metre is $1 0^{29}$ , then the drift velocity is

$250 \times 1 0^{- 3} m s^{- 1}$

$25.0 \times 1 0^{- 3} m s^{- 1}$

$2.50 \times 1 0^{- 3} m s^{- 1}$

$1.25 \times 1 0^{- 3} m s^{- 1}$

Drift velocity
$v_{d} = \frac{i}{n e A}$
$= \frac{40}{1 0 ^{29} \times 1.6 \times 1 0 ^{- 19} \times 1 0 ^{- 6}}$
$= 2.5 \times 1 0^{- 3} m s^{- 1}$

Q. There is a current of 40A in a wire of 10−6m2 area of cross-section. If the number of free electrons per cubic metre is 1029, then the drift velocity is

250×10−3ms−1

25.0×10−3ms−1

2.50×10−3ms−1

1.25×10−3ms−1