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Current Electricity question

2023 · 8 Apr · Shift 1 · Q68
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Current Electricity question

2023 · 8 Apr · Shift 1 · Q68

JEE MainPhysicsCurrent ElectricityNumerical+4 / −1
A current of 2 A2 \mathrm{~A}2 A flows through a wire of cross-sectional area 25.0 mm225.0 \mathrm{~mm}^{2}25.0 mm2. The number of free electrons in a cubic meter are 2.0×10282.0 \times 10^{28}2.0×1028. The drift velocity of the electrons is ‾×10−6 ms−1\underline{\hspace{2cm}}\times 10^{-6} \mathrm{~ms}^{-1}​×10−6 ms−1(given, charge on electron =1.6×10−19C=1.6 \times 10^{-19} \mathrm{C}=1.6×10−19C ).
Numerical answer
View written solutionFree

Correct answer: 25

  1. Use the drift current relation

For a conductor,

I=neAvdI = n e A v_dI=neAvd​

where:

  • I=2 AI = 2\,\mathrm{A}I=2A
  • n=2.0×1028 m−3n = 2.0 \times 10^{28}\,\mathrm{m^{-3}}n=2.0×1028m−3
  • e=1.6×10−19 Ce = 1.6 \times 10^{-19}\,\mathrm{C}e=1.6×10−19C
  • A=25.0 mm2A = 25.0\,\mathrm{mm^2}A=25.0mm2
  • vd=v_d =vd​= drift velocity
  1. Convert area into SI unit

25.0 mm2=25.0×10−6 m225.0\,\mathrm{mm^2} = 25.0 \times 10^{-6}\,\mathrm{m^2}25.0mm2=25.0×10−6m2

  1. Rearrange the formula for drift velocity

vd=IneAv_d = \frac{I}{n e A}vd​=neAI​

Substitute the values:

vd=2(2.0×1028)(1.6×10−19)(25×10−6)v_d = \frac{2}{(2.0 \times 10^{28})(1.6 \times 10^{-19})(25 \times 10^{-6})}vd​=(2.0×1028)(1.6×10−19)(25×10−6)2​

  1. Calculate the denominator

First,

(2.0)(1.6)(25)=80(2.0)(1.6)(25) = 80(2.0)(1.6)(25)=80

and powers of 10:

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

So denominator becomes:

80×103=8.0×10480 \times 10^3 = 8.0 \times 10^480×103=8.0×104

Thus,

vd=28.0×104=2.5×10−5 m s−1v_d = \frac{2}{8.0 \times 10^4} = 2.5 \times 10^{-5}\,\mathrm{m\,s^{-1}}vd​=8.0×1042​=2.5×10−5ms−1

  1. Write in the required form

2.5×10−5=25×10−62.5 \times 10^{-5} = 25 \times 10^{-6}2.5×10−5=25×10−6

So the blank is:

25\boxed{25}25​

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