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JEE MAIN 2026 · Electromagnetic Induction and Alternating Currents

JEE Main 2026 Session 2 Question Paper: 8 Apr, Shift 2 (B.E./B.Tech), question 40

Updated
Single correct+4 / -1Answer matches NTA’s final key

A 30 cm long solenoid has 10 turns per cm and area of 5 cm². 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\alpha \times 10^{-5} V. The value α\alpha is ______.

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Solution

Solution

Self-inductance: L=μ0n2AlL = \mu_0 n^2 A l with n=1000n = 1000 m−1^{-1}, A=5×10−4A = 5 \times 10^{-4} m², l=0.3l = 0.3 m:

L=4π×10−7×106×5×10−4×0.3=6π×10−5 H.L = 4\pi \times 10^{-7} \times 10^6 \times 5 \times 10^{-4} \times 0.3 = 6\pi \times 10^{-5} \text{ H}.

ε=LΔIΔt=6π×10−5×23.14=12×10−5 V.\varepsilon = L\frac{\Delta I}{\Delta t} = 6\pi \times 10^{-5} \times \frac{2}{3.14} = 12 \times 10^{-5} \text{ V}.

Answer: α=12\alpha = 12.

Answer checked against the official answer key. The solution is our own working.

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