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JEE Main 2019 (9 Apr, Shift 2) · Mutual Inductance

Mutual Inductance question from JEE Main 2019 (9 Apr, Shift 2)

Q55JEE Main 2019 · 9 APR · SHIFT 2Easy

Mutual Inductance

Two coils 'P' and 'Q' are separated by some distance. When a current of 3 A flows through coil 'P', a magnetic flux of 10–3 Wb passes through 'Q'. No current is passed through 'Q'. When no current passes through 'P' and a current of 2 A passes through 'Q', the flux through 'P' is :-

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Answer: (c)

  1. M=ΦQIP=10−33 HM = \dfrac{\Phi_Q}{I_P} = \dfrac{10^{-3}}{3}\text{ H}
  2. Since MPQ=MQPM_{PQ} = M_{QP}: ΦP=MIQ=10−33×2\Phi_P = MI_Q = \dfrac{10^{-3}}{3} \times 2
  3. ΦP=6.67×10−4 Wb\Phi_P = 6.67 \times 10^{-4}\text{ Wb}
  4. So the answer is (c) 6.67×10−4 Wb6.67 \times 10^{-4}\text{ Wb}.
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