Current Electricity

JEE Main level, numerical answer, medium

13 of 380

Question 13 of 380

A current of 2A2A flows through a wire of cross-sectional area 25.0mm225.0{mm}^{2}. The number of free electrons in a cubic meter are 2.0×10282.0\times {10}^{28}. The drift velocity of the electrons is _____ ×10−6ms−1\times {10}^{-6}{ms}^{-1}
(given, charge on electron =1.6×10−19C=1.6\times {10}^{-19}C).

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Answer: 25

The drift velocity is given by
vd=IneA{v}_{d}=\dfrac{I}{neA}
⇒I=nAevd...(i)\Rightarrow I=nAe{v}_{d}...(i)
The given data is
I=2AA=25×10−6m2n=2×1028e=1.6×10−19CI=2A A=25\times {10}^{-6}{m}^{2} n=2\times {10}^{28} e=1.6\times {10}^{-19}C
Substituting the given values in equation (i)
2=2×1028×25×10−6×1.6×10−19×vd⇒vd=22×1028×25×10−6×1.6×10−192=2\times {10}^{28}\times 25\times {10}^{-6}\times 1.6\times {10}^{-19}\times {v}_{d} \Rightarrow {v}_{d}=\dfrac{2}{2\times {10}^{28}\times 25\times {10}^{-6}\times 1.6\times {10}^{-19}}
⇒vd=125×1.6×1000ms−1=25×10−6ms−1\Rightarrow {v}_{d}=\dfrac{1}{25\times 1.6\times 1000}m{s}^{-1}=25\times {10}^{-6}m{s}^{-1}

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