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A very small circular loop of area $5 \times 10^{- 4} m^{2},$ resistance $2 o h m$ and negligible inductance is initially coplanar and concentric with a much larger fixed circular loop of radius $0.1 m$ . A constant current of 1 ampere is passed in the bigger loop and the smaller loop is rotated with angular velocity $ω r a d / s$ about a diameter. Then induced current in the smaller loop as a function of time is $\frac{1}{2} π ω \times 10^{- y} \sin ω t .$ Value of $y$ will be.

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answer is 9.

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Detailed Solution

$B = \frac{μ_{0} I}{2 r}, \Rightarrow ϕ = B π R^{2} \cos ω t = \frac{μ_{0} I}{2 r} π R^{2} \cos ω t$

$ε = \frac{- d ϕ}{d t} = \frac{μ_{0} I π R^{2} ω \sin ω t}{2 r}$

Induced current, $I = \frac{ε}{R e s i s \tan c e} = \frac{\frac{μ_{0} I π R^{2} ω \sin ω t}{2 r}}{2}$

$\Rightarrow I = \frac{μ_{0} I π R^{2} ω \sin ω t}{4 r} = \frac{4 π \times 10^{- 7} \times 1 \times (5 \times 10^{- 4}) \times ω \times sinωt}{4 \times 0.1}$

$\Rightarrow I = \frac{1}{2} π \times ωsinωt \times 10^{- 9}$

hence, y=9.

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