There is a current of 40 ampere in a wire of 10-6m2 area of cross-section. If the number of free electron per m3 is 1029 , then the drift velocity will be

Question

There is a current of 40 ampere in a wire of 10-6m2 area of cross-section. If the number of free electron per m3 is 1029 , then the drift velocity will be (A)  1.25× 103 m/s  (B)  2.50× 10-3 m/s  (C)  25.0× 10-3 m/s (D)  250× 10-3 m/s

Tardigrade Admin · Accepted Answer

The relation between drift velocity vd and current i are related as given by i=A n e vd, So, vd=(i/A n e) vd=(40/10-6 × 1.6 × 10-19 × 1029) =(40/16) × 10-3 m / s =2.5 × 10-3 m / s

Q. There is a current of $40$ ampere in a wire of $1 0^{- 6} m^{2}$ area of cross-section. If the number of free electron per $m^{3}$ is $1 0^{29}$ , then the drift velocity will be

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

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

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

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

The relation between drift velocity $v_{d}$ and current $i$ are related as given by
$i = A n e v_{d},$ So, $v_{d} = \frac{i}{A n e}$
$v_{d} = \frac{40}{1 0 ^{- 6} \times 1.6 \times 1 0 ^{- 19} \times 1 0 ^{29}}$
$= \frac{40}{16} \times 1 0^{- 3} m / s$
$= 2.5 \times 1 0^{- 3} m / s$

Q. There is a current of 40 ampere in a wire of 10−6m2 area of cross-section. If the number of free electron per m3 is 1029 , then the drift velocity will be

1.25×103m/s

2.50×10−3m/s

25.0×10−3m/s

250×10−3m/s