Two coils have a mutual inductance of 0.005 | vedclass

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(B) Given: Mutual inductance $ M = 0.005 \ H $,current $ i = i_{m} \sin \omega t $,peak current $ i_{m} = 10 \ A $,and angular frequency $ \omega = 10

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$ 5 \pi $

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(B) Given: Mutual inductance $ M = 0.005 \ H $,current $ i = i_{m} \sin \omega t $,peak current $ i_{m} = 10 \ A $,and angular frequency $ \omega = 100 \pi \ rad \ s^{-1} $.The induced emf $ \varepsilon $ in the second coil is given by $ \varepsilon = M \frac{di}{dt} $.Substituting the expression for current: $ \varepsilon = M \frac{d}{dt} (i_{m} \sin \omega t) = M i_{m} \omega \cos \omega t $.The maximum value of the induced emf $ \varepsilon_{\max} $ occurs when $ \cos \omega t = 1 $.Therefore,$ \varepsilon_{\max} = M \omega i_{m} $.Substituting the values: $ \varepsilon_{\max} = 0.005 	imes 100 \pi 	imes 10 = 5 \pi \ V $.Thus,the maximum value of the emf induced in the second coil is $ 5 \pi \ V $.

Two coils have a mutual inductance of $ 0.005 \ H $. The current changes in the first coil according to the equation $ i = i_{m} \sin \omega t $,where $ i_{m} = 10 \ A $ and $ \omega = 100 \pi \ rad \ s^{-1} $. The maximum value of the emf induced in the second coil is:

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