The drift velocity of electrons in silver wire with cross-sectional area 3.14 × 1 0 − 6 m 2 carrying a current of 20 A is. Given atomic weight of Ag = 108 , density of silver = 10.5 × 1 0 3 kg / m 3

Number of electrons per kg of silver = 108 6.023 × 1 0 26 Number of electrons per unit volume of silver n = 108 6.023 × 1 0 26 × 10.5 × 1 0 3 v d = n e A I = 6.023 × 1 0 26 × 10.5 × 1 0 3 × 1.6 × 1 0 − 19 × 3.14 × 1 0 − 6 20 × 108 = 6.798 × 1 0 − 4 m / sec .

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PhysicsCurrent ElectricityMicroscopicMedium2 minPYQ_2016

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The drift velocity of electrons in silver wire with cross-sectional area $3.14 \times 1 0^{- 6} m^{2}$ carrying a current of $20 A$ is. Given atomic weight of $Ag =$ $108,$ density of silver $= 10.5 \times 1 0^{3} kg / m^{3}$

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Answer:

Solution:

Number of electrons per kg of silver $= \frac{6.023 \times 1 0 ^{26}}{108}$ Number of electrons per unit volume of silver $n = \frac{6.023 \times 1 0 ^{26}}{108} \times 10.5 \times 1 0^{3} v_{d} = \frac{I}{n e A} = \frac{20}{6.023 \times 1 0 ^{26} \times 10.5 \times 1 0 ^{3} \times 1.6 \times 1 0 ^{- 19} \times 3.14 \times 1 0 ^{- 6}} \times 108 = 6.798 \times 1 0^{- 4} m / sec .$

Stream:BITSATSubject:PhysicsTopic:Current ElectricitySubtopic:Microscopic

⏱ 2mℹ️ Source: PYQ_2016

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