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MHT CET 2024 (3 May, Shift 2) · Mutual Inductance

Mutual Inductance question from MHT CET 2024 (3 May, Shift 2)

Q27MHT CET 2024 · 3 MAY · SHIFT 2Easy

Mutual Inductance

Two coils are kept near each other. When no current passes through first coil and current in the 22nd coil increases at the rate 10 A/s10\text{ A/s}, the e.m.f. in the 11st coil is 20 mV. When no current passes through 22nd coil and 3.6 A current passes through 11st coil the flux linkage in coil 2 is

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Answer: (d)

  1. M=e1dI2/dt=20×10−310=2×10−3 HM = \dfrac{e_1}{dI_2/dt} = \dfrac{20 \times 10^{-3}}{10} = 2 \times 10^{-3}\text{ H}
  2. M12=M21=MM_{12} = M_{21} = M, so flux linkage in coil 2: Φ2=MI1=2×10−3×3.6\Phi_2 = MI_1 = 2 \times 10^{-3} \times 3.6
  3. Φ2=7.2×10−3 Wb\Phi_2 = 7.2 \times 10^{-3}\text{ Wb}
  4. So the answer is (d) 7.2×10−3 Wb7.2 \times 10^{-3}\text{ Wb}.
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