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JEE Main 2026 (8 Apr, Shift 2) · Self-Inductance & Energy

Self-Inductance & Energy question from JEE Main 2026 (8 Apr, Shift 2)

Q16JEE Main 2026 · 8 APR · SHIFT 2Medium

Self-Inductance & Energy

A 30 cm long solenoid has 10 turns per cm and area of 5 cm25\text{ cm}^{2}. The current through the solenoid coil varies from 2 A to 4 A in 3.14 s. The e.m.f. induced in the coil is α×10−5 V\alpha \times 10^{-5}\text{ V}. The value α\alpha is ______.

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Answer: (b)

  1. L=μ0n2AlL = \mu_0 n^2 A l with n=1000n = 1000 m−1^{-1}, A=5×10−4A = 5\times 10^{-4} m2^2, l=0.3l = 0.3 m.
  2. L=4π×10−7×106×5×10−4×0.3=6π×10−5L = 4\pi \times 10^{-7} \times 10^6 \times 5\times 10^{-4} \times 0.3 = 6\pi \times 10^{-5} H
  3. dIdt=23.14=2π\dfrac{dI}{dt} = \dfrac{2}{3.14} = \dfrac{2}{\pi} A/s
  4. e=LdIdt=6π×10−5×2π=12×10−5e = L\dfrac{dI}{dt} = 6\pi \times 10^{-5} \times \dfrac{2}{\pi} = 12 \times 10^{-5} V
  5. So α=12\alpha = 12, answer (b).
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