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(B) The total energy in an $LC$ circuit is constant and is given by $U = \frac{q_0^2}{2C}$.At any time $t$,the charge on the capacitor is $q = q_0 \co

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$\frac{\pi}{4} \sqrt{LC}$

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(B) The total energy in an $LC$ circuit is constant and is given by $U = \frac{q_0^2}{2C}$.At any time $t$,the charge on the capacitor is $q = q_0 \cos(\omega t)$,where $\omega = \frac{1}{\sqrt{LC}}$.The energy stored in the electric field is $U_E = \frac{q^2}{2C} = \frac{q_0^2 \cos^2(\omega t)}{2C}$.The energy stored in the magnetic field is $U_B = U - U_E = \frac{q_0^2}{2C} - \frac{q_0^2 \cos^2(\omega t)}{2C} = \frac{q_0^2}{2C} \sin^2(\omega t)$.We are given that the energy is stored equally,so $U_E = U_B$.$\frac{q_0^2}{2C} \cos^2(\omega t) = \frac{q_0^2}{2C} \sin^2(\omega t) \implies \cos^2(\omega t) = \sin^2(\omega t) \implies 	an^2(\omega t) = 1$.Thus,$\omega t = \frac{\pi}{4}$.Substituting $\omega = \frac{1}{\sqrt{LC}}$,we get $t = \frac{\pi}{4} \sqrt{LC}$.

$A$ fully charged capacitor $C$ with initial charge $q_0$ is connected to a coil of self-inductance $L$ at $t = 0$. The time at which the energy is stored equally between the electric and the magnetic fields is:

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