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Current Electricity question

2015 · Shift 0 · Q55
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Current Electricity question

2015 · Shift 0 · Q55

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
When 5V5V5V potential difference is applied across a wire of length 0.1m,0.1m,0.1m, the drift speed of electrons is 2.5×10−4  ms−1.2.5 \times {10^{ - 4}}\,\,m{s^{ - 1}}.2.5×10−4ms−1. If the electron density in the wire is 8×1028  m−3,8 \times {10^{28}}\,\,{m^{ - 3}},8×1028m−3, the resistivity of the material is close to :
  1. A
    1.6×10−6Ωm1.6 \times {10^{ - 6}}\Omega m1.6×10−6Ωm
  2. B
    1.6×10−5Ωm1.6 \times {10^{ - 5}}\Omega m1.6×10−5Ωm
  3. C
    1.6×10−8Ωm1.6 \times {10^{ - 8}}\Omega m1.6×10−8Ωm
  4. D
    1.6×10−7Ωm1.6 \times {10^{ - 7}}\Omega m1.6×10−7Ωm
View written solutionFree

Correct answer: B

  1. Given data
  • Potential difference: V=5 VV = 5\,\text{V}V=5V
  • Length of wire: L=0.1 mL = 0.1\,\text{m}L=0.1m
  • Drift speed: vd=2.5×10−4 m s−1v_d = 2.5 \times 10^{-4}\,\text{m s}^{-1}vd​=2.5×10−4m s−1
  • Electron density: n=8×1028 m−3n = 8 \times 10^{28}\,\text{m}^{-3}n=8×1028m−3
  • Charge of electron: e=1.6×10−19 Ce = 1.6 \times 10^{-19}\,\text{C}e=1.6×10−19C

We need resistivity ρ\rhoρ.

  1. Find electric field in the wire

Using

E=VLE = \frac{V}{L}E=LV​

so,

E=50.1=50 V m−1E = \frac{5}{0.1} = 50\,\text{V m}^{-1}E=0.15​=50V m−1
  1. Use drift velocity relation

For a conductor,

vd=eEτmv_d = \frac{eE\tau}{m}vd​=meEτ​

But a more direct relation is

J=nevdJ = ne v_dJ=nevd​

and also

J=σE=EρJ = \sigma E = \frac{E}{\rho}J=σE=ρE​

Therefore,

ρ=EJ=Enevd\rho = \frac{E}{J} = \frac{E}{ne v_d}ρ=JE​=nevd​E​
  1. Calculate current density
J=nevdJ = ne v_dJ=nevd​

Substitute values:

J=(8×1028)(1.6×10−19)(2.5×10−4)J = (8 \times 10^{28})(1.6 \times 10^{-19})(2.5 \times 10^{-4})J=(8×1028)(1.6×10−19)(2.5×10−4)

First multiply numerical factors:

8×1.6=12.8,8 \times 1.6 = 12.8,8×1.6=12.8, 12.8×2.5=3212.8 \times 2.5 = 3212.8×2.5=32

Now powers of 10:

1028×10−19×10−4=10510^{28} \times 10^{-19} \times 10^{-4} = 10^51028×10−19×10−4=105

Hence,

J=32×105=3.2×106 A m−2J = 32 \times 10^5 = 3.2 \times 10^6\,\text{A m}^{-2}J=32×105=3.2×106A m−2
  1. Calculate resistivity
ρ=EJ=503.2×106\rho = \frac{E}{J} = \frac{50}{3.2 \times 10^6}ρ=JE​=3.2×10650​ ρ=15.625×10−6\rho = 15.625 \times 10^{-6}ρ=15.625×10−6 ρ≈1.56×10−5 Ω m\rho \approx 1.56 \times 10^{-5}\,\Omega\,\text{m}ρ≈1.56×10−5Ωm

So the closest value is

1.6×10−5 Ω m\boxed{1.6 \times 10^{-5}\,\Omega\,\text{m}}1.6×10−5Ωm​
  1. Option check
  • A: 1.6×10−6 Ωm1.6 \times 10^{-6}\,\Omega m1.6×10−6Ωm ❌
  • B: 1.6×10−5 Ωm1.6 \times 10^{-5}\,\Omega m1.6×10−5Ωm ✅
  • C: 1.6×10−8 Ωm1.6 \times 10^{-8}\,\Omega m1.6×10−8Ωm ❌
  • D: 1.6×10−7 Ωm1.6 \times 10^{-7}\,\Omega m1.6×10−7Ωm ❌

Therefore, the correct option is B.

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