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Current Electricity question

2021 · 24 Feb · Shift 2 · Q63
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  5. /2021 · 24 Feb · Shift 2 · Q63

Current Electricity question

2021 · 24 Feb · Shift 2 · Q63

JEE MainPhysicsCurrent ElectricityNumerical+4 / −1
A cylindrical wire of radius 0.5 mm and conductivity 5 ×\times× 107 S/m is subjected to an electric field of 10 mV/m. The expected value of current in the wire will be x3 π\piπ mA. The value of x is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 5

  1. Use microscopic form of Ohm's law

The current density is J=σEJ = \sigma EJ=σE where

  • σ=5×107 S/m\sigma = 5\times 10^7\ \text{S/m}σ=5×107 S/m
  • E=10 mV/m=10×10−3 V/m=10−2 V/mE = 10\ \text{mV/m} = 10\times 10^{-3}\ \text{V/m} = 10^{-2}\ \text{V/m}E=10 mV/m=10×10−3 V/m=10−2 V/m

So, J=(5×107)(10−2)=5×105 A/m2J = (5\times 10^7)(10^{-2}) = 5\times 10^5\ \text{A/m}^2J=(5×107)(10−2)=5×105 A/m2

  1. Find cross-sectional area of the wire

Radius: r=0.5 mm=0.5×10−3 m=5×10−4 mr = 0.5\ \text{mm} = 0.5\times 10^{-3}\ \text{m} = 5\times 10^{-4}\ \text{m}r=0.5 mm=0.5×10−3 m=5×10−4 m

Area: A=πr2=π(5×10−4)2=π×25×10−8=2.5×10−7π m2A = \pi r^2 = \pi (5\times 10^{-4})^2 = \pi \times 25\times 10^{-8} = 2.5\times 10^{-7}\pi\ \text{m}^2A=πr2=π(5×10−4)2=π×25×10−8=2.5×10−7π m2

  1. Find current

Current is I=JAI = JAI=JA

Thus, I=(5×105)(2.5×10−7π)I = (5\times 10^5)(2.5\times 10^{-7}\pi)I=(5×105)(2.5×10−7π) I=12.5×10−2π AI = 12.5\times 10^{-2}\pi\ \text{A}I=12.5×10−2π A I=0.125π AI = 0.125\pi\ \text{A}I=0.125π A

Convert to mA: I=125π mAI = 125\pi\ \text{mA}I=125π mA

  1. Compare with given form

Given, I=x3π mAI = x^3\pi\ \text{mA}I=x3π mA

So, x3=125x^3 = 125x3=125 x=5x = 5x=5

Final Answer

5\boxed{5}5​

  1. Comparison with stored correct answer

Stored correct answer = 555.

Our derived answer also is 555, so they agree.

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