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

B. 0.05 J

Given,

self-inductance, $L=100mH$

current, $I=1A$

The energy stored in magnetic field, $K=\frac{1}{2}L{I}^2$

$K=\frac{1}{2}\times 100\times {10}^{-3}\times 1^2$

= $0.05J$