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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. |
Answer» Solution :`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)]` |
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