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MHT CET 2024 (2 May, Shift 1) · Self-Inductance & Energy

Self-Inductance & Energy question from MHT CET 2024 (2 May, Shift 1)

Q43MHT CET 2024 · 2 MAY · SHIFT 1Easy

Self-Inductance & Energy

The magnetic energy stored in an inductor of inductance 5 μH5\ \mu\text{H} carrying a current of 2 A is

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Answer: (c)

  1. U=12LI2=12×5×10−6×22U = \tfrac12 L I^2 = \tfrac12\times5\times10^{-6}\times2^2
  2. U=10×10−6 J=10 μJU = 10\times10^{-6}\text{ J} = 10\ \mu\text{J}
  3. So the answer is (c) 10 μ10\ \muJ.
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