A current of 5 A is passing through a metallic wire of cross-sectional area 4 × 1 0 − 6 m 2 . If the density of charge carriers of the wire is 5 × 1 0 26 m − 3 , then the drift velocity of the electrons will be

Drift velocity, v d = n e A 1 ⇒ v d = ( 5 × 1 0 26 ) × ( 1.6 × 1 0 − 19 ) × ( 4 × 1 0 − 6 ) 5 = 64 1 = 1.56 × 1 0 − 2 ms − 1

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PhysicsCurrent ElectricityMicroscopicMedium2 minPYQ_2011

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A current of 5 A is passing through a metallic wire of cross-sectional area $4 \times 1 0^{- 6} m^{2}$ . If the density of charge carriers of the wire is $5 \times 1 0^{26} m^{- 3}$ , then the drift velocity of the electrons will be

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Solution:

Drift velocity, $v_{d} = \frac{1}{n e A}$ $\Rightarrow v_{d} = \frac{5}{( 5 \times 1 0 ^{26} ) \times ( 1.6 \times 1 0 ^{- 19} ) \times ( 4 \times 1 0 ^{- 6} )}$ $= \frac{1}{64} = 1.56 \times 1 0^{- 2} ms^{- 1}$

Stream:JIPMERSubject:PhysicsTopic:Current ElectricitySubtopic:Microscopic

⏱ 2mℹ️ Source: PYQ_2011

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