Physics · Electromagnetic Induction

JEE Main 2026 — 8 April, Evening Shift — Question 12

A $30\mathrm{cm}$ long solenoid has $10\mathrm{turns}$ per cm and area of $5\mathrm{cm}^2$ . 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 $\alpha \times 10^{-5}\mathrm{V}$ . The value $\alpha$ is

Option A:
60
Option B:
12
Correct
Option C:
120
Option D:
34

Answer: B

Step-by-step solution

Self inductance $L = \mu_0 n^2 A l = 4\pi\times10^{-7} \times (1000)^2 \times 5\times10^{-4} \times 0.3 = 4\pi\times10^{-7}\times10^6\times1.5\times10^{-4} = 6\pi\times10^{-5}$ H. $\frac{di}{dt} = \frac{2}{3.14} = \frac{2}{\pi}$ . Induced emf $\varepsilon = L\frac{di}{dt} = 6\pi\times10^{-5} \times \frac{2}{\pi} = 12\times10^{-5}$ V, so $\alpha = 12$ .

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Exam: JEE Main 2026
Paper: 8 April, Evening Shift
Subject: Physics
Chapter: Electromagnetic Induction
Topic: Self-Inductance and Mutual Inductance and Energy Density