A capacitor of capacitance 500 is charged completely using a de...

Energy stored in capacitor

$E = \dfrac{1}{2}C{V}^{2} = \dfrac{1}{2}500 \times {10}^{- 6} \times {10}^{4}$

$⇒ E = \dfrac{5}{2}J$

Current will be maximum when whole energy of capacitor becomes energy of inductor.

$⇒ \dfrac{1}{2}L{I}^{2} = \dfrac{5}{2}$

$⇒ I = \sqrt{\dfrac{5}{L}} = \sqrt{\dfrac{5}{50 \times {10}^{- 3}}} = 10A$