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Current Electricity question

2019 · 9 Jan · Shift 1 · Q47
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Current Electricity question

2019 · 9 Jan · Shift 1 · Q47

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
Drift speed of electrons, when 1.5 A of current flows in a copper wire of cross section 5 mm2, is υ\upsilonυ. If the electron density in copper is 9 ×\times× 1028/m3 the value of υ\upsilonυ. in mm/s is close to (Take charge of electron to be = 1.6 ×\times× 10 −-− 19C)
  1. A
    0.02
  2. B
    3
  3. C
    2
  4. D
    0.2
View written solutionFree

Correct answer: A

  1. Use the drift current relation

For a conductor,

I=neAvdI = n e A v_dI=neAvd​

where:

  • I=1.5 AI = 1.5\,\text{A}I=1.5A
  • n=9×1028 m−3n = 9 \times 10^{28}\,\text{m}^{-3}n=9×1028m−3
  • e=1.6×10−19 Ce = 1.6 \times 10^{-19}\,\text{C}e=1.6×10−19C
  • A=5 mm2=5×10−6 m2A = 5\,\text{mm}^2 = 5 \times 10^{-6}\,\text{m}^2A=5mm2=5×10−6m2
  • vd=?v_d = ?vd​=?

So,

vd=IneAv_d = \frac{I}{n e A}vd​=neAI​
  1. Substitute the values
vd=1.5(9×1028)(1.6×10−19)(5×10−6)v_d = \frac{1.5}{(9 \times 10^{28})(1.6 \times 10^{-19})(5 \times 10^{-6})}vd​=(9×1028)(1.6×10−19)(5×10−6)1.5​
  1. Simplify the denominator

First the numerical part:

9×1.6×5=729 \times 1.6 \times 5 = 729×1.6×5=72

Now the 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

72×103=7.2×10472 \times 10^3 = 7.2 \times 10^472×103=7.2×104

Therefore,

vd=1.57.2×104v_d = \frac{1.5}{7.2 \times 10^4}vd​=7.2×1041.5​ vd=2.08×10−5 m/sv_d = 2.08 \times 10^{-5}\,\text{m/s}vd​=2.08×10−5m/s
  1. Convert to mm/s

Since 1 m=1000 mm1\,\text{m} = 1000\,\text{mm}1m=1000mm,

vd=2.08×10−5×103v_d = 2.08 \times 10^{-5} \times 10^3vd​=2.08×10−5×103 vd=2.08×10−2 mm/sv_d = 2.08 \times 10^{-2}\,\text{mm/s}vd​=2.08×10−2mm/s vd≈0.02 mm/sv_d \approx 0.02\,\text{mm/s}vd​≈0.02mm/s
  1. Check options
  • A: 0.020.020.02 ✅
  • B: 333 ❌
  • C: 222 ❌
  • D: 0.20.20.2 ❌

So the correct option is A.

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