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

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Correct Answer

(A) 197

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Detailed explanation

EMF induced \(\varepsilon= A \frac{ dB }{ dt }= A \mu_0 n \frac{ di }{ dt }\)
\(\varepsilon= A \mu_0 n i _0 \omega \cos \omega t\)
current induced \(i =\frac{\varepsilon}{ R }=\frac{\pi r ^2 \mu_0 ni _0 \omega}{ R } \cos \omega t\)
So \(i =\frac{\pi r ^2 \mu_0 ni _0 \omega}{\sqrt{2} R }\)
\(=\frac{\pi \times 10^{-4} \times 4 \pi \times 10^{-7} \times 500 \times 10 \times 10^3}{\sqrt{2} \times 10}\)
\(=\frac{20 \pi^2}{\sqrt{2}} \times 10^{-6}\)
\(\simeq \frac{197}{\sqrt{2}} \mu A\)

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