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

Explanation

`M = (mu_(0))/(2b) pia^(2)`

|emf induced in larger coil|`= [((di)/(dt))` in smaller coil]
`e = (mu_(0))/(2b) pia^(2) (4t) = (2 mu_(0)pi a^(2)t)/(b)`
Applying `KVL`,
`+ e - (q)/(C) - iR = 0 rArr (2mu_(0) pia^(2)t)/(b) - (q)/(C) - iR = 0`
Differentiating w.r.t. time, we get
`(2mu_(0)pia^(2))/(b) - (i)/(C) - (di)/(dt) R = 0 = 0`
On solving, `i = (2mu_(0)pia^(2)C)/(b) [1 - e^(-t//RC)]`