Suppose a long solenoid of 100… - JEE Main 2026 Physics PYQ

JEE Main 2026 — Physics Electromagnetism

2026mcqhard

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

[Resistance of the loop $=10 \Omega$ ]

Official previous-year question

Held on 23 Jan 2026 · Verified 6 Jul 2026.

Options

A
197
B
80
C
100
D
280

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Solution

EMF induced $\varepsilon = A \dfrac{dB}{dt} = A\mu_0 n \dfrac{di}{dt}$

$\varepsilon = A\mu_0 n \, i_0 \omega \cos\omega t$

current induced $i = \dfrac{\varepsilon}{R} = \dfrac{\pi r^2 \mu_0 n i_0 \omega}{R} \cos\omega t$

So $i = \dfrac{\pi r^2 \mu_0 n i_0 \omega}{\sqrt{2} R}$

$= \dfrac{\pi \times 10^{-4} \times 4\pi \times 10^{-7} \times 500 \times 10 \times 10^3}{\sqrt{2} \times 10}$

$= \dfrac{20\pi^2}{\sqrt{2}} \times 10^{-6}$

$\simeq \dfrac{197}{\sqrt{2}} \, \mu A$

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Solution

EMF induced $\varepsilon = A \dfrac{dB}{dt} = A\mu_0 n \dfrac{di}{dt}$

$\varepsilon = A\mu_0 n \, i_0 \omega \cos\omega t$

current induced $i = \dfrac{\varepsilon}{R} = \dfrac{\pi r^2 \mu_0 n i_0 \omega}{R} \cos\omega t$

So $i = \dfrac{\pi r^2 \mu_0 n i_0 \omega}{\sqrt{2} R}$

$= \dfrac{\pi \times 10^{-4} \times 4\pi \times 10^{-7} \times 500 \times 10 \times 10^3}{\sqrt{2} \times 10}$

$= \dfrac{20\pi^2}{\sqrt{2}} \times 10^{-6}$

$\simeq \dfrac{197}{\sqrt{2}} \, \mu A$

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Solution

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Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

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