Electromagnetic Induction

JEE Main level, single correct, medium

117 of 153

Question 117 of 153

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)(B) of radius 1 cm coaxially slided through the solenoid at a speed v=1 cm/sv=1 \mathrm{~cm} / \mathrm{s}. The r.m.s. current through the loop when the coil BB 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 ]

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Answer: (A) 197

EMF induced ε=AdBdt=Aμ0ndidt\varepsilon = A \dfrac{dB}{dt} = A\mu_0 n \dfrac{di}{dt}
ε=Aμ0n i0ωcos⁡ωt\varepsilon = A\mu_0 n \, i_0 \omega \cos\omega t
current induced i=εR=πr2μ0ni0ωRcos⁡ωti = \dfrac{\varepsilon}{R} = \dfrac{\pi r^2 \mu_0 n i_0 \omega}{R} \cos\omega t
So i=πr2μ0ni0ω2Ri = \dfrac{\pi r^2 \mu_0 n i_0 \omega}{\sqrt{2} R}
=π×10−4×4π×10−7×500×10×1032×10= \dfrac{\pi \times 10^{-4} \times 4\pi \times 10^{-7} \times 500 \times 10 \times 10^3}{\sqrt{2} \times 10}
=20π22×10−6= \dfrac{20\pi^2}{\sqrt{2}} \times 10^{-6}
≃1972 μA\simeq \dfrac{197}{\sqrt{2}} \, \mu A

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