Suppose a long solenoid of 100 cm length, radius 2 cm having 500… | JEE Main 2026 (23 Jan, Shift 2)

Q16JEE Main 2026 · 23 JAN · SHIFT 2Difficult

Mutual Inductance

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

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Answer: (d)

Inside a long solenoid $B = \mu_0 n I$ is uniform, so sliding the loop does not change flux; only $I(t)$ does.
$\phi = \mu_0 n I_0 \sin\omega t\,\pi r^2$ , so $e_0 = \mu_0 n I_0 \omega \pi r^2$
$e_0 = 4\pi\times10^{-7}\times500\times10\times1000\times\pi\times10^{-4} = 2\pi^2\times10^{-4} \approx 1.974\times10^{-3}\text{ V}$
$i_0 = e_0/R = 1.974\times10^{-4}\text{ A} \approx 197\,\mu\text{A}$ , so $i_{rms} = \frac{197}{\sqrt{2}}\,\mu\text{A}$
So $\alpha = 197$ , answer (d).

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