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MHT CET 2024 (9 May, Shift 1) · Mutual Inductance

Mutual Inductance question from MHT CET 2024 (9 May, Shift 1)

Q26MHT CET 2024 · 9 MAY · SHIFT 1Easy

Mutual Inductance

Two coils have a mutual inductance 5×10−3 H5 \times 10^{-3}\text{ H}. The current changes in the first coil according to the equation I1=I0sin⁡ωtI_1 = I_0 \sin \omega t where I0=10 AI_0 = 10\text{ A} and ω=100π rad/s\omega = 100\pi\text{ rad/s}. What is the value of the maximum e.m.f. in the coil?

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Answer: (d)

  1. e=−MdI1dt=−MI0ωcos⁡ωte = -M\dfrac{dI_1}{dt} = -MI_0\omega\cos\omega t, so emax=MI0ωe_{max} = MI_0\omega
  2. emax=5×10−3×10×100π=5π Ve_{max} = 5 \times 10^{-3} \times 10 \times 100\pi = 5\pi\text{ V}
  3. So the answer is (d) 5π V5\pi\text{ V}.
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