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JEE Main 2017 · Physics · Electromagnetic Induction and Alternating Current · medium

A long solenoid has 1000 turns, length 0.5 m and cross-sectional area 10 cm^2. A small coil of 20 turns and cross-sectional area 2 cm^2 is placed…

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A long solenoid has 10001000 turns, length 0.5 m0.5\ \mathrm{m} and cross-sectional area 10 cm210\ \mathrm{cm^2}. A small coil of 2020 turns and cross-sectional area 2 cm22\ \mathrm{cm^2} is placed coaxially inside it, near its middle. The current in the solenoid is reduced uniformly from 5 A5\ \mathrm{A} to zero in 0.1 s0.1\ \mathrm{s}. The emf induced in the small coil is nearly (μ0=4π×10−7 H m−1\mu_0 = 4\pi\times10^{-7}\ \mathrm{H\,m^{-1}})
  1. A. 0.50 mV0.50\ \mathrm{mV}
  2. B. 2.5 mV2.5\ \mathrm{mV}
  3. C. 0.025 mV0.025\ \mathrm{mV}
  4. D. 5.0 mV5.0\ \mathrm{mV}
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Answer: (A) 0.50 mV0.50\ \mathrm{mV}

Inside a long solenoid B=μ0N1ℓIB = \mu_0\dfrac{N_1}{\ell}I. Only the coil's own area links with this field, so the mutual inductance is M=N2BA2I=μ0N1N2A2ℓ=4π×10−7×1000×20×2×10−40.5≈1.0×10−5 HM = \dfrac{N_2BA_2}{I} = \dfrac{\mu_0N_1N_2A_2}{\ell} = \dfrac{4\pi\times10^{-7}\times1000\times20\times2\times10^{-4}}{0.5} \approx 1.0\times10^{-5}\ \mathrm{H}. The rate of change of current is 5/0.1=50 A s−15/0.1 = 50\ \mathrm{A\,s^{-1}}, so ∣ε∣=M∣dIdt∣=1.0×10−5×50≈5.0×10−4 V=0.50 mV|\varepsilon| = M\left|\dfrac{dI}{dt}\right| = 1.0\times10^{-5}\times50 \approx 5.0\times10^{-4}\ \mathrm{V} = 0.50\ \mathrm{mV}. Using the solenoid area 10 cm210\ \mathrm{cm^2} would wrongly give 2.5 mV2.5\ \mathrm{mV}.
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Original question written in the JEE Main 2017 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake