A 50 mH coil carries a current of 2 A. The energy stored in this coil is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.1 J
The energy stored in the magnetic field of an inductor of inductance L carrying a current I is \( U = \frac{1}{2}LI^{2} \).
\( L = 50\,\mathrm{mH} = 50\times10^{-3}\,\mathrm{H} \)
\( U = \frac{1}{2}\times(50\times10^{-3}\,\mathrm{H})\times(2\,\mathrm{A})^{2} = 0.1\,\mathrm{J} \)