Question

energy stored in a coil of self inductance 40mH carrying a steady current of 2A is

ans:80J

ans should be 0.08J or 80 J

Answer 1

Energy stored in conductor
$$\begin{gathered} =\frac{1}{2}L{I}^2 \\ =\frac{1}{2}\times 40\times {10}^{-3}\times 2^2 \\ =0.08joule \end{gathered}$$