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MHT CET 2026 (15 May, Shift 2) · Mutual Inductance

Mutual Inductance question from MHT CET 2026 (15 May, Shift 2)

Q1MHT CET 2026 · 15 MAY · SHIFT 2Easy

Mutual Inductance

Two coils PP and QQ are separated by some distance. When a current of 3 A3\text{ A} flows through coil PP, a magnetic flux of 10−310^{-3} Wb passes through QQ. No current is passed through QQ. When no current passes through PP and a current of 2 A2\text{ A} passes through QQ, the flux through PP is x×10−3x \times 10^{-3} Wb. The value of xx is

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Answer: (b)

  1. Mutual inductance is the same both ways: M=ΦQIP=10−33M = \dfrac{\Phi_Q}{I_P} = \dfrac{10^{-3}}{3} H.
  2. ΦP=MIQ=10−33×2=0.667×10−3\Phi_P = M I_Q = \dfrac{10^{-3}}{3}\times 2 = 0.667\times 10^{-3} Wb
  3. So x=0.667x = 0.667, answer (b).
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