Search any question & find its solution
Question:
Answered & Verified by Expert
A $100 \mathrm{mH}$ coil carries a current of $1 \mathrm{~A}$. Energy stored in the form of magnetic field is
Options:
Solution:
2074 Upvotes
Verified Answer
The correct answer is:
$0.050 \mathrm{~J}$
Energy stored in an inductor $\mathrm{E}=\frac{1}{2} \mathrm{LI}^2$
$\begin{aligned}
& =\frac{1}{2} \times\left(100 \times 10^{-3}\right) \times 1 \\
& =0.05 \mathrm{~J}
\end{aligned}$
$\begin{aligned}
& =\frac{1}{2} \times\left(100 \times 10^{-3}\right) \times 1 \\
& =0.05 \mathrm{~J}
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.