Q 12-06-030JEE MainJEE Main 2026 (8 Apr, Shift 2)Medium
A $30$ cm long solenoid has $10$ turns per cm and area of $5\ \text{cm}^2$. 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 $\alpha \times 10^{-5}$ V. The value $\alpha$ is ______.
Answer: (B) $12$
$n = 10$ turns/cm $= 1000$ turns/m, $l = 0.3$ m, $A = 5 \times 10^{-4}\ \text{m}^2$.
Self-inductance: $L = \mu_0n^2Al = 4\pi \times 10^{-7} \times 10^6 \times 5 \times 10^{-4} \times 0.3 = 6\pi \times 10^{-5}$ H.
$e = L\dfrac{\Delta I}{\Delta t} = 6\pi \times 10^{-5} \times \dfrac{2}{3.14} = 12 \times 10^{-5}$ V, so $\alpha = 12$.
Solution by Sreeraj P, M.Sc Physics