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Physics Question on Electromagnetic induction

There is a current of 40 A40\, A in a wire of 10−6 m210^{-6} \,m ^{2} area of cross-section. If the number of free electrons per cubic metre is 102910^{29}, then the drift velocity is

A

250×10−3 ms−1250 \times 10 ^{-3} \, ms^{-1}

B

25.0×10−3 ms−125.0 \times 10 ^{-3} \, ms^{-1}

C

2.50×10−3 ms−12.50 \times 10 ^{-3} \, ms^{-1}

D

1.25×10−3 ms−11.25 \times 10 ^{-3} \, ms^{-1}

Answer

2.50×10−3 ms−12.50 \times 10 ^{-3} \, ms^{-1}

Explanation

Solution

Drift velocity
vd=ineAv_{d}=\frac{i}{n e A}
=401029×1.6×10−19×10−6=\frac{40}{10^{29} \times 1.6 \times 10^{-19} \times 10^{-6}}
=2.5×10−3 ms−1=2.5 \times 10^{-3} \,ms ^{-1}