Current Electricity

JEE Main level, numerical answer, medium

12 of 380

Question 12 of 380

The number density of free electrons in copper is nearly 8×1028m−38\times {10}^{28}{m}^{-3}. A copper wire has its area of cross-section =2×10−6m2=2\times {10}^{-6}{m}^{2} and is carrying a current of 3.2A3.2A. The drift speed of the electrons is _____×10−6ms−1\times {10}^{-6}m{s}^{-1}.

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Answer: 125

The formula to calculate the current through any conductor is given by
I=nAevd...(1)I=nAe{v}_{d}...\left(1\right)
Substitute the values of the given parameters into equation (1) and solve to calculate the required drift velocity of the electrons.
3.2=8×1028×2×10−6×1.6×10−19×vd⇒vd=3.216×1.6×1000ms−1=125×10−6ms−1\begin{matrix}3.2 & = & 8\times {10}^{28}\times 2\times {10}^{-6}\times 1.6\times {10}^{-19}\times {v}_{d} \\ & \Rightarrow & {v}_{d}=\dfrac{3.2}{16\times 1.6\times 1000}m{s}^{-1} \\ & = & 125\times {10}^{-6}m{s}^{-1}\end{matrix}

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