A current of 5 A is passing through a metallic wire of cross-sectional area 4× 10-6m2 . If the density of charge carriers of the wire is 5× 1026 m-3, the drift velocity of the electrons will be

Question

A current of 5 A is passing through a metallic wire of cross-sectional area 4× 10-6m2 . If the density of charge carriers of the wire is 5× 1026 m-3, the drift velocity of the electrons will be (A)  1× 102ms-1  (B)  1.56× 10-2ms-1  (C)  1.56× 10-3ms-1  (D)  1× 10-2ms-1

Tardigrade Admin · Accepted Answer

Drift velocity, vd=(I/neA) ⇒ vd=(5/(5× 1026)× (1.6× 10-19)× (4× 10-6)) =1.56× 10-2ms-1

Q. A current of $5 A$ is passing through a metallic wire of cross-sectional area $4 \times 10^{- 6} m^{2}$ . If the density of charge carriers of the wire is $5 \times 10^{26}$ $m^{- 3},$ the drift velocity of the electrons will be

$1 \times 10^{2} m s^{- 1}$

$1.56 \times 10^{- 2} m s^{- 1}$

$1.56 \times 10^{- 3} m s^{- 1}$

$1 \times 10^{- 2} m s^{- 1}$

Drift velocity, $v_{d} = \frac{I}{n e A}$
$\Rightarrow$ $v_{d} = \frac{5}{( 5 \times 10 ^{26} ) \times ( 1.6 \times 10 ^{- 19} ) \times ( 4 \times 10 ^{- 6} )}$
$= 1.56 \times 10^{- 2} m s^{- 1}$

Q. A current of 5A is passing through a metallic wire of cross-sectional area 4×10−6m2 . If the density of charge carriers of the wire is 5×1026 m−3, the drift velocity of the electrons will be

1×102ms−1

1.56×10−2ms−1

1.56×10−3ms−1

1×10−2ms−1