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NEET 2013 · Self-Inductance & Energy

Self-Inductance & Energy question from NEET 2013

Q79NEET 2013Easy

Self-Inductance & Energy

A current of 2.5 A flows through a coil of inductance 5 H. The magnetic flux linked with the coil is

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Answer: (b)

  1. Flux linked: ϕ=LI\phi = LI
  2. ϕ=5×2.5=12.5 Wb\phi = 5 \times 2.5 = 12.5\text{ Wb}
  3. So the answer is (b) 12.5 Wb.
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