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Electromagnetic Induction question

2024 · 6 Apr · Shift 2 · Q67
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  5. /2024 · 6 Apr · Shift 2 · Q67

Electromagnetic Induction question

2024 · 6 Apr · Shift 2 · Q67

JEE MainPhysicsElectromagnetic InductionMCQ+4 / −1
In a coil, the current changes from −2 A-2 \mathrm{~A}−2 A to +2 A+2 \mathrm{~A}+2 A in 0.2 s0.2 \mathrm{~s}0.2 s and induces an emf of 0.1 V0.1 \mathrm{~V}0.1 V. The self inductance of the coil is :
  1. A
    4 mH
  2. B
    2.5 mH
  3. C
    1 mH
  4. D
    5 mH
View written solutionFree

Correct answer: D

  1. Use the formula for self-induced emf

For a coil, ∣e∣=L∣dIdt∣|e| = L\left|\frac{dI}{dt}\right|∣e∣=L​dtdI​​ So, L=∣e∣∣dIdt∣L = \frac{|e|}{\left|\frac{dI}{dt}\right|}L=∣dtdI​∣∣e∣​

  1. Find the change in current

The current changes from −2 A-2\,\text{A}−2A to +2 A+2\,\text{A}+2A. Thus, ΔI=2−(−2)=4 A\Delta I = 2 - (-2) = 4\,\text{A}ΔI=2−(−2)=4A

  1. Find the rate of change of current

Given time interval, Δt=0.2 s\Delta t = 0.2\,\text{s}Δt=0.2s Hence, ΔIΔt=40.2=20 A/s\frac{\Delta I}{\Delta t} = \frac{4}{0.2} = 20\,\text{A/s}ΔtΔI​=0.24​=20A/s

  1. Substitute the induced emf

Given, e=0.1 Ve = 0.1\,\text{V}e=0.1V Therefore, L=0.120=0.005 HL = \frac{0.1}{20} = 0.005\,\text{H}L=200.1​=0.005H

  1. Convert into millihenry

0.005 H=5×10−3 H=5 mH0.005\,\text{H} = 5 \times 10^{-3}\,\text{H} = 5\,\text{mH}0.005H=5×10−3H=5mH

  1. Match with the options

So the correct option is: D: 5 mH\boxed{\text{D: } 5\,\text{mH}}D: 5mH​

  1. Comparison with stored answer

Stored correct answer: D

My derived answer is also D. They match.

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