The magnetic energy' stored in an inductor of inductance $5 \mu \mathrm{H}$ carrying a current of 2 A is
The magnetic energy' stored in an inductor of inductance $5 \mu \mathrm{H}$ carrying a current of 2 A is
- 10 mJ
- 5 mJ
- $10 \mu \mathrm{~J}$
- $5 \mu \mathrm{~J}$
Solution
$\mathrm{U}=\frac{1}{2} \mathrm{LI}^2=\frac{1}{2} \times\left(5 \times 10^{-6}\right) \times 2^2=10 \mu \mathrm{~J}$
.
Asked in: MHT CET 2024 (02 May Shift 1)
Practice more Electromagnetic Induction questions on Aicharya