Q62 / 75 · JEE Main 2026 (23 January Shift 2) · Physics
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 $=10 \Omega$ ]
[Resistance of the loop $=10 \Omega$ ]
- Correct answerA197
- B80
- C100
- D280
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 $i = \dfrac{\pi r^2 \mu_0 n i_0 \omega}{\sqrt{2} R}$
$= \dfrac{\pi \times 10^{-4} \times 4\pi \times 10^{-7} \times 500 \times 10 \times 10^3}{\sqrt{2} \times 10}$
$= \dfrac{20\pi^2}{\sqrt{2}} \times 10^{-6}$
$\simeq \dfrac{197}{\sqrt{2}} \, \mu A$
$\varepsilon = A\mu_0 n \, i_0 \omega \cos\omega t$
current induced $i = \dfrac{\varepsilon}{R} = \dfrac{\pi r^2 \mu_0 n i_0 \omega}{R} \cos\omega t$
So $i = \dfrac{\pi r^2 \mu_0 n i_0 \omega}{\sqrt{2} R}$
$= \dfrac{\pi \times 10^{-4} \times 4\pi \times 10^{-7} \times 500 \times 10 \times 10^3}{\sqrt{2} \times 10}$
$= \dfrac{20\pi^2}{\sqrt{2}} \times 10^{-6}$
$\simeq \dfrac{197}{\sqrt{2}} \, \mu A$
Hence, the correct answer is option: 'A'.