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Standard XII

Physics

Electric Field Due to Charge Distributions - Approach

A uniform wir...

Question

A uniform wire of length $2.0 m$ and cross-sectional area $10^{- 7} m^{2}$ carries a current of $1.6 A$ . If there are $10^{28}$ free electrons per $m^{3}$ in copper, the drift speed of electrons in copper is

2 m m / s

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5 m m / s

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10 m m / s

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1 m / s

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Solution

The correct option is B $10 m m / s$
$V_{d} = \frac{I}{n e A} = \frac{1.6}{10^{28} \times 10^{- 7} \times 1.6 \times 10^{- 19}} = \frac{10^{19}}{10^{28} \times 10^{- 7}} = \frac{1}{100} m / s e c = 10 m m / s e c$

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A uniform wire of length 2.0m and cross-sectional area 10−7m2 carries a current of 1.6A. If there are 1028 free electrons per m3 in copper, the drift speed of electrons in copper is

The correct option is B 10mm/sVd=IneA=1.61028×10−7×1.6×10−19=10191028×10−7=1100m/sec=10mm/sec

A uniform wire of length $2.0 m$ and cross-sectional area $10^{- 7} m^{2}$ carries a current of $1.6 A$ . If there are $10^{28}$ free electrons per $m^{3}$ in copper, the drift speed of electrons in copper is

The correct option is B $10 m m / s$
$V_{d} = \frac{I}{n e A} = \frac{1.6}{10^{28} \times 10^{- 7} \times 1.6 \times 10^{- 19}} = \frac{10^{19}}{10^{28} \times 10^{- 7}} = \frac{1}{100} m / s e c = 10 m m / s e c$