LC circuit with AC-supply - current through the coil
$$\begin{align}
Z_R&amp;=10 \;\Omega\\
Z_L&amp;=j\omega L=j \cdot300\times10^{-3}=0.03j\\
Z_C&amp;=\frac{1}{j\omega c}=-0.009j\\
\hline\\
Z_{tot}&amp;=Z_R||Z_L||Z_C=\left({1\over10}+{1\over0.03j}-{1\over0.009j}\right)^{-1}\approx 8.7378 \angle 89.95^{\circ}\;\Omega \longrightarrow \phi=89.95^{\circ}\\
\hline\\
&amp;\text{Using current divider theorem*, } I_L=\frac{Z_{{R||}{C}}}{Z_L||Z_{R||C}}\times I_{tot}=\\
&amp;=\frac{{\left({1\over10}-{1\over0.009j}\right)}^{-1}}{{\left({1\over10}-{1\over0.009j}\right)}^{-1}+0.03j}\times 2\angle\phi=(2\times0.428571..)\angle(\phi-179.926..)\approx 0.857\angle {-89.98}^{\circ} \text{A} \equiv 0.857\angle{-1.57}\;\text{A}\\
&amp;\\
&amp;\text{Hence, } I_L(t)=0.857\sin(\omega t-1.57) \;\text{A} \Longrightarrow \text{Amplitude}=0.857 \;; \text{Phase}=-1.57\; \text{radians}
\end{align}$$ 
*:current divider theorem