Electromagnetic Induction

JEE Main level, numerical answer, medium

126 of 153

Question 126 of 153

Two coils have mutual inductance 0.002H0.002H. The current changes in the first coil according to the relation i=i0sinωti={i}_{0}sin\omega t, where i0=5A{i}_{0}=5A and ω=50π\omega =50\pi rads−1rad{s}^{-1}. The maximum value of emfemf in the second coil is παV\dfrac{\pi }{\alpha }V. The value of α\alpha is

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Answer: 2

The amount of flux induced in the second coil due to the first one is given by
ϕ2=Mi=Mi0sinωt...(1)\begin{matrix}{\phi }_{2} & = & Mi \\ & = & M{i}_{0}sin\omega t...\left(1\right)\end{matrix}
where, MM is the mutual inductance of the coils.
The formula to calculate the emf induced can be written as
ϵ=−dϕ2dt...(2)\epsilon =-\dfrac{d{\phi }_{2}}{dt}...\left(2\right)
From equation (2), it follows that,
ϵmax⁡=Mi0ω=0.002×5×50πV=π2V\begin{matrix}{\epsilon }_{\max } & = & M{i}_{0}\omega \\ & = & 0.002\times 5\times 50\pi V \\ & = & \dfrac{\pi }{2}V\end{matrix}
Hence, α=2\alpha =2.

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