A 50 mH coil carries a current of 2 A. The energy stored in this coil is

0.1 J

The correct answer is Option (2) - 0.1 J

The energy stored in an inductor is given by the formula - 

$E=\frac{1}{2}LI^2$

and,

E, Energy level

L, Inductance of coil = $50mH = 50 ×10^{-3}H$ [given]

I, Current = 2A [given]

$∴E=\frac{1}{2}×50×10^{-3}×(2)^2$

$=0.1J$