Question

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

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

Answer 1

Answer: (B)

0.05 J

Answer 2

Here L = 100 mH $= 100 \times 10^{3}H$

$I = 1A$

As $U = \frac{1}{2}LI^{2} = \frac{1}{2} \times (100 \times 10^{- 3}) \times 1^{2} = 0.05J$