Explanation
$$V_{C}=\frac{1}{2}CV_{0}^{2}\rightarrow max. energy , V_{0}\rightarrow peak voltage$$
$$V_{0}= i_{0}X_{c}=\dfrac{i_{0}}{C\omega }=\dfrac{i_{0}\sqrt{LC}}{C}$$
$$V_{C}=\dfrac{1}{2}C\times\dfrac{\pi ^{2}B^{2}a^{4}}{4R^{2}C^{2}}= \dfrac{\pi ^{2}B^{2}a^{4}}{8R^{2}C} $$
$$P_{ext}=P_{dissipated}=\varepsilon _{0}i_{0}=\dfrac{\pi Ba^{2} }{2\sqrt{LC}}\times \dfrac{\pi Ba^{2} }{2R\sqrt{LC}}$$
$$P_{ext}=\dfrac{\pi ^{2}B^{2}a^{4}}{4LCR}$$
$$<P>= \displaystyle \dfrac{1}{T} \int_0^T i^2 Rdt = \dfrac{(B \pi a^2 \omega)^2}{2R}$$
$$\displaystyle |E_{in}| = |e| = \frac {d \phi} {dt} \Rightarrow |e| = nat^{n\,-\,1} $$
If 0 < n < 1 $$\Rightarrow $$ (n - 1) negative $$ \Rightarrow \displaystyle |e| = \frac {na} {t^{1\,-\,1}} $$ hence |e| decay with time.
If n = 1 $$\Rightarrow$$ = na = constant; If n > 1 $$ \Rightarrow $$ |e| increases with time.
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