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JEE Main 2026 · 8 Apr · Shift 2, Physics question 40

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\times10^{-5} V. The value α\alpha is ________.

Choose one option
Answer and worked solution

Official answer: B

Inductance.
L=μ0n2Al=4π×10−7×(103)2×5×10−4×0.3=1.885×10−4 H.L=\mu_0n^2Al=4\pi\times10^{-7}\times(10^3)^2\times5\times10^{-4}\times0.3=1.885\times10^{-4}\text{ H}.

EMF. dIdt=23.14\dfrac{dI}{dt}=\dfrac2{3.14}, so ε=1.2×10−4\varepsilon=1.2\times10^{-4} V =12×10−5=12\times10^{-5} V.

(B)

Topics:Self inductance
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