Copper contains 8-4- 1028 free electrons -m3- A copper wire of cross-sectional area 7-4- 10-7m2 carries a current of 1A- The electron drift speed is approximately-

The correct option is D 10^ 3 m/s v=Ine=17.4× 10^ 7× 8.4× 10^28 =10^ 28.4× 1.6× 7.4=10^ 3 . ...

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Byju's Answer

Standard XII

Physics

Drift Velocity

Copper contai...

Question

Copper contains $8.4 \times 10^{28}$ free electrons $/ m^{3}$ . A copper wire of cross-sectional area $7.4 \times 10^{- 7} m^{2}$ carries a current of $1$ A. The electron drift speed is approximately.

10^{- 8}

m/s

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10^{3}

m/s

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1

m/s

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10^{- 3}

m/s

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Solution

The correct option is D $10^{- 3}$ m/s
$v = \frac{I}{n e} = \frac{1}{7.4 \times 10^{- 7}} \times 8.4 \times 10^{28}$
$= \frac{10^{- 2}}{8.4 \times 1.6 \times 7.4} = 10^{-} 3$ .

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Copper contains 8.4×1028 free electrons/m3. A copper wire of cross-sectional area 7.4×10−7m2 carries a current of 1A. The electron drift speed is approximately.

The correct option is D 10−3m/sv=Ine=17.4×10−7×8.4×1028=10−28.4×1.6×7.4=10−3.

Copper contains $8.4 \times 10^{28}$ free electrons $/ m^{3}$ . A copper wire of cross-sectional area $7.4 \times 10^{- 7} m^{2}$ carries a current of $1$ A. The electron drift speed is approximately.

The correct option is D $10^{- 3}$ m/s
$v = \frac{I}{n e} = \frac{1}{7.4 \times 10^{- 7}} \times 8.4 \times 10^{28}$
$= \frac{10^{- 2}}{8.4 \times 1.6 \times 7.4} = 10^{-} 3$ .