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MHT CET 2025 (26 Apr, Shift 2) · Mutual Inductance

Mutual Inductance question from MHT CET 2025 (26 Apr, Shift 2)

Q17MHT CET 2025 · 26 APR · SHIFT 2Easy

Mutual Inductance

Two coils P and Q are kept near each other. When no current flows through coil P and current increases in coil Q at the rate 10 A/s10\text{ A/s}, the emf in coil P is 12 mV. When coil Q carries no current and current of 1.5 A flows through coil P, the magnetic flux linked with the coil Q in mWb is

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Answer: (d)

  1. M=ePdIQ/dt=12×10−310=1.2×10−3M = \dfrac{e_P}{dI_Q/dt} = \dfrac{12\times 10^{-3}}{10} = 1.2\times 10^{-3} H
  2. Same MM both ways: ΦQ=MIP=1.2×10−3×1.5=1.8×10−3\Phi_Q = M I_P = 1.2\times 10^{-3}\times 1.5 = 1.8\times 10^{-3} Wb
  3. So the answer is (d) 1.8 mWb.
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