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JEE Main 20238 AprilEvening Shift
Exam
JEE Main
Year
2023
Date
8 April
Shift
Evening Shift
Subject
Physics
Chapter
Current Electricity

JEE Main 2023 · 8 April · Evening Shift · Physics · Current Electricity

The number density of free electrons in copper is nearly 8 × 10 ^...

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}.

Correct answer

0

Step-by-step solution

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{cases}3 . 2 & = & 8 \times {10}^{28} \times 2 \times {10}^{- 6} \times 1 . 6 \times {10}^{- 19} \times {v}_{d} \\ & ⇒ & {v}_{d} = \dfrac{3 . 2}{16 \times 1 . 6 \times 1000}m{s}^{- 1} \\ & = & 125 \times {10}^{- 6}m{s}^{- 1}\end{cases}

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