Physics · Electromagnetic Induction

JEE Main 2026 — 23 January, Evening Shift — Question 38

Suppose a long solenoid of 100 cm length, radius 2 cm having 500 turns per unit length, carries a current I=10sin⁡(ωt)AI=10 \sin (\omega \mathrm{t}) \mathrm{A}, where ω=1000rad./s\omega=1000 \mathrm{rad} . / \mathrm{s}. A circular conducting loop (B) of radius 1 cm coaxially slided through the solenoid at a speed v=1 cm/s\mathrm{v}=1 \mathrm{~cm} / \mathrm{s}. The r.m.s. current through the loop when the coil B is inserted 10 cm inside the solenoid is α/2μ A\alpha / \sqrt{2} \mu \mathrm{~A}. The value of α\alpha is ____\_\_\_\_ . [Resistance of the loop =10Ω=10 \Omega ]

  1. Option A:

    197

    Correct
  2. Option B:

    80

  3. Option C:

    280

  4. Option D:

    100

Answer: A

Step-by-step solution

EMF induced ε=AdBdt=Aμ0ndidt\varepsilon=\mathrm{A} \frac{\mathrm{dB}}{\mathrm{dt}}=\mathrm{A} \mu_{0} \mathrm{n} \frac{\mathrm{di}}{\mathrm{dt}} ε=Aμ0ni0ωcos⁡ωt\varepsilon=\mathrm{A} \mu_{0} \mathrm{n} \mathrm{i}_{0} \omega \cos \omega \mathrm{t} current induced i=εR=πr2μ0ni0ωRcos⁡ωt\mathrm{i}=\frac{\varepsilon}{\mathrm{R}}=\frac{\pi \mathrm{r}^{2} \mu_{0} \mathrm{ni}_{0} \omega}{\mathrm{R}} \cos \omega \mathrm{t} So i=πr2μ0ni0ω2Ri=\frac{\pi r^{2} \mu_{0} n i_{0} \omega}{\sqrt{2} R} =π×10−4×4π×10−7×500×10×1032×10=\frac{\pi \times 10^{-4} \times 4 \pi \times 10^{-7} \times 500 \times 10 \times 10^{3}}{\sqrt{2} \times 10} =20π22×10−6=\frac{20 \pi^{2}}{\sqrt{2}} \times 10^{-6} ≃1972μ A\simeq \frac{197}{\sqrt{2}} \mu \mathrm{~A}

Answer key and solution verified before publishing.

Practise Electromagnetic Induction

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Exam
JEE Main 2026
Subject
Physics
Chapter
Electromagnetic Induction
Topic
Self-Inductance and Mutual Inductance and Energy Density
Suppose a long solenoid of 100 cm length, radius 2 cm having 500… | JEE Main 2026 PYQ with Solution · DhiX AI