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Electromagnetic Induction question

2024 · 5 Apr · Shift 2 · Q88
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  5. /2024 · 5 Apr · Shift 2 · Q88

Electromagnetic Induction question

2024 · 5 Apr · Shift 2 · Q88

JEE MainPhysicsElectromagnetic InductionNumerical+4 / −1
The current in an inductor is given by I=(3t+8)\mathrm{I}=(3 \mathrm{t}+8)I=(3t+8) where t\mathrm{t}t is in second. The magnitude of induced emf produced in the inductor is 12 mV12 \mathrm{~mV}12 mV. The self-inductance of the inductor ‾mH\underline{\hspace{2cm}}\mathrm{mH}​mH.
Numerical answer
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Correct answer: 4

  1. For an inductor, the magnitude of induced emf is given by ∣e∣=L∣dIdt∣|e| = L\left|\frac{dI}{dt}\right|∣e∣=L​dtdI​​

  2. The current is I=3t+8I = 3t + 8I=3t+8 Therefore, dIdt=3 A/s\frac{dI}{dt} = 3\ \text{A/s}dtdI​=3 A/s

  3. Given induced emf: ∣e∣=12 mV=12×10−3 V|e| = 12\ \text{mV} = 12 \times 10^{-3}\ \text{V}∣e∣=12 mV=12×10−3 V

  4. Using L=∣e∣∣dI/dt∣L = \frac{|e|}{|dI/dt|}L=∣dI/dt∣∣e∣​ we get L=12×10−33=4×10−3 HL = \frac{12\times 10^{-3}}{3} = 4\times 10^{-3}\ \text{H}L=312×10−3​=4×10−3 H

  5. Convert into millihenry: 4×10−3 H=4 mH4\times 10^{-3}\ \text{H} = 4\ \text{mH}4×10−3 H=4 mH

Therefore, the self-inductance of the inductor is 4\boxed{4}4​

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