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Current Electricity question

2023 · 8 Apr · Shift 2 · Q58
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Current Electricity question

2023 · 8 Apr · Shift 2 · Q58

JEE MainPhysicsCurrent ElectricityNumerical+4 / −1
The number density of free electrons in copper is nearly 8×1028 m−38 \times 10^{28} \mathrm{~m}^{-3}8×1028 m−3. A copper wire has its area of cross section =2×10−6 m2=2 \times 10^{-6} \mathrm{~m}^{2}=2×10−6 m2 and is carrying a current of 3.2 A3.2 \mathrm{~A}3.2 A. The drift speed of the electrons is ‾×10−6ms−1\underline{\hspace{2cm}}\times 10^{-6} \mathrm{ms}^{-1}​×10−6ms−1
Numerical answer
View written solutionFree

Correct answer: 125

  1. Use the relation between current and drift speed

For a conductor,

I=neAvdI = n e A v_dI=neAvd​

where:

  • I=3.2 AI = 3.2\,\text{A}I=3.2A
  • n=8×1028 m−3n = 8 \times 10^{28}\,\text{m}^{-3}n=8×1028m−3
  • e=1.6×10−19 Ce = 1.6 \times 10^{-19}\,\text{C}e=1.6×10−19C
  • A=2×10−6 m2A = 2 \times 10^{-6}\,\text{m}^2A=2×10−6m2
  • vdv_dvd​ = drift speed

So,

vd=IneAv_d = \frac{I}{n e A}vd​=neAI​
  1. Substitute the values
vd=3.2(8×1028)(1.6×10−19)(2×10−6)v_d = \frac{3.2}{(8 \times 10^{28})(1.6 \times 10^{-19})(2 \times 10^{-6})}vd​=(8×1028)(1.6×10−19)(2×10−6)3.2​
  1. Simplify the denominator

First multiply the numerical factors:

8×1.6×2=25.68 \times 1.6 \times 2 = 25.68×1.6×2=25.6

Now powers of 10:

1028×10−19×10−6=10310^{28} \times 10^{-19} \times 10^{-6} = 10^31028×10−19×10−6=103

Hence denominator is:

25.6×10325.6 \times 10^325.6×103

Therefore,

vd=3.225.6×103v_d = \frac{3.2}{25.6 \times 10^3}vd​=25.6×1033.2​ vd=3.225600=1.25×10−4 m s−1v_d = \frac{3.2}{25600} = 1.25 \times 10^{-4}\,\text{m s}^{-1}vd​=256003.2​=1.25×10−4m s−1
  1. Convert into the asked form

We need

vd=‾×10−6 m s−1v_d = \underline{\hspace{1cm}} \times 10^{-6}\,\text{m s}^{-1}vd​=​×10−6m s−1

Now,

1.25×10−4=125×10−61.25 \times 10^{-4} = 125 \times 10^{-6}1.25×10−4=125×10−6

So the required integer is:

125\boxed{125}125​
  1. Comparison with stored answer

Stored correct answer = 125125125

Our derived answer also = 125125125. So they match.

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