Question

A 100 millihenry coil carries a current of 1 A. Energy stored in its magnetic field is

• A. 0.5 J
• B. 1 A
• C. 0.05 J
• D. 0.1 J

Answer 1

Answer: (C)

$E=\frac{1}{2}L{i}^2=\frac{1}{2}\times (100\times 10^{-3})\times 1^2=0.05J$