LC circuit

The charge q on the capacitor is
q(t)=A\cos(\omega t+\varphi)
where \omega =\sqrt{ \dfrac{1}{LC}}.
So the current will be given by
i(t)=-A\omega \sin(\omega t+\varphi)
Let at t=t₀, the charge q=q₀=10 C, and i=i₀=20 mA. So that we have
i_0=-A\omega \sin(\omega t_0+\varphi)
and
q_0=A\cos(\omega t_0+\varphi)
Squaring and adding, we get
A=\sqrt{q_0^2+\dfrac{i_0^2}{\omega^2}}
Accordingly the maximum current in the circuit is
i_\textrm{max}=\omega\sqrt{q_0^2+\dfrac{i_0^2}{\omega^2}}=\sqrt{\omega^2q_0^2+i_0^2}=\sqrt{\dfrac{q_0^2}{LC}+i_0^2}
Now plug in the values.

wow!! really good explaination..

wht's the ans..i think there is sum mistake in the values given in the Q