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Current Electricity question

2021 · 17 Mar · Shift 1 · Q58
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  5. /2021 · 17 Mar · Shift 1 · Q58

Current Electricity question

2021 · 17 Mar · Shift 1 · Q58

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
A current of 10A exists in a wire of cross-sectional area of 5 mm2 with a drift velocity of 2 ×\times× 10 −-− 3 ms −-− 1. The number of free electrons in each cubic meter of the wire is ‾\underline{\hspace{2cm}}​.
  1. A
    625 ×\times× 1025
  2. B
    1 ×\times× 1023
  3. C
    2 ×\times× 1025
  4. D
    2 ×\times× 106
View written solutionFree

Correct answer: A

  1. Use the drift current relation

    The current in a conductor is I=neAvdI = n e A v_dI=neAvd​ where:

    • I=10 AI = 10\,\text{A}I=10A
    • n=n =n= number of free electrons per unit volume
    • 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=2×10−3 m s−1v_d = 2 \times 10^{-3}\,\text{m s}^{-1}vd​=2×10−3m s−1

  2. Rearrange for nnn

    n=IeAvdn = \frac{I}{eAv_d}n=eAvd​I​

  3. Substitute the values

    n=10(1.6×10−19)(5×10−6)(2×10−3)n = \frac{10}{(1.6 \times 10^{-19})(5 \times 10^{-6})(2 \times 10^{-3})}n=(1.6×10−19)(5×10−6)(2×10−3)10​

  4. Simplify the denominator

    First multiply the numerical factors: 1.6×5×2=161.6 \times 5 \times 2 = 161.6×5×2=16

    Add powers of 10: 10−19×10−6×10−3=10−2810^{-19} \times 10^{-6} \times 10^{-3} = 10^{-28}10−19×10−6×10−3=10−28

    So, eAvd=16×10−28=1.6×10−27eAv_d = 16 \times 10^{-28} = 1.6 \times 10^{-27}eAvd​=16×10−28=1.6×10−27

  5. Compute nnn

    n=101.6×10−27n = \frac{10}{1.6 \times 10^{-27}}n=1.6×10−2710​

    n=101.6×1027=6.25×1027 m−3n = \frac{10}{1.6} \times 10^{27} = 6.25 \times 10^{27}\,\text{m}^{-3}n=1.610​×1027=6.25×1027m−3

  6. Match with options

    Option A is written as: 625×1025=6.25×1027625 \times 10^{25} = 6.25 \times 10^{27}625×1025=6.25×1027

    Hence, the correct option is A.

Final Answer

n=6.25×1027 m−3n = 6.25 \times 10^{27}\,\text{m}^{-3}n=6.25×1027m−3 So the correct choice is A.

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