Q25 / 111 · NTA Abhyas (NEET) 2020 · Physics · Electromagnetic Induction

A current I=20sin100πt A is passed in the first coil, which induces a maximum emf of 10π V in the second coil. The mutual inductance for the pair of coils is

  • A
    10 mH
  • B
    15 mH
  • C
    25 mH
  • Correct answer
    D
    5 mH

Solution

Let I=I0sinωt,

Where I0=20 A, ω=100π rad s-1

Then, ε=MdIdt=MdI0sinωtdt=MI0ωcosωt

⇒εmax=MI0ω=10π

M=10π20×100π

M=0.005 H=5 mH

Hence, the correct answer is option: 'D'.

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