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If the current in the inner loop changes...

If the current in the inner loop changes according to `i=2t^(2)` then, find the current in the capacitor as a function of time.

Text Solution

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`M=(mu_(0))/(2b)pia^(2)`
|emf induced in larger coil |`=m[((di)/(dt))" in smaller coil"]implies e=(mu_(0))/(2b)pia^(2)(4t)=(2mu_(0)pia^(2)t)/(b)`
Applying KVL:-
`+e-(q)/( c)-iR=0`
`(2mu_(0)pia^(2)t)/(b)-(q)/( c)-iR=0`
differentiate wrt time:-
`(2mu_(0)pia^(2))/(b)-(i)/(c )-(di)/(dt)R=0`
so solving it
`i=(2mu_(0)pia^(2)C)/(b)[1-e^(-t//RC)]`
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